Cfe Higher Human Biology Unit 1: Human Cells

SCHOLAR Study Guide CfE Higher Human Biology Unit 1: Human Cells

Authored by: Eoin McIntyre (Previously Auchmuty High School)

Reviewed by: Sheena Haddow (Perth College)

Previously authored by: Mike Cheung Eileen Humphrey Eoin McIntyre Jim McIntyre

Heriot-Watt University Edinburgh EH14 4AS, United Kingdom. First published 2014 by Heriot-Watt University. This edition published in 2016 by Heriot-Watt University SCHOLAR. Copyright © 2016 SCHOLAR Forum. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no proﬁt accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide.

Distributed by the SCHOLAR Forum. SCHOLAR Study Guide Unit 1: CfE Higher Human Biology 1. CfE Higher Human Biology Course Code: C740 76 ISBN 978-1-909633-16-2

Print Production and Fulfilment in UK by Print Trail www.printtrail.com Acknowledgements Thanks are due to the members of Heriot-Watt University’s SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum.

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Contents

1 Stem cells 1 1.1 Introduction ...... 2 1.2 What are stem cells? ...... 3 1.3 Embryonic stem cells ...... 4 1.4 Tissue (adult) stem cells ...... 7 1.5 Learning points ...... 9 1.6 Extended response question ...... 9 1.7 End of topic test ...... 10

2 Differentiation in cells 13 2.1 Mitosis and meiosis ...... 15 2.2 Differentiation in somatic cells ...... 21 2.3 Human body tissues ...... 22 2.4 Formation of the body organs ...... 29 2.5 Differentiation of germline cells ...... 29 2.6 Mutations in germline and somatic cells ...... 30 2.7 Learning points ...... 32 2.8 Extended response question ...... 33 2.9 End of topic test ...... 33

3 Research: Stem cells and cancer cells 37 3.1 Introduction ...... 39 3.2 Stem cell research ...... 39 3.3 Therapeutic uses of stem cells ...... 44 3.4 Ethical issues and the regulation of stem cell use ...... 47 3.5 The biology of cancer cells ...... 50 3.6 Learning points ...... 53 3.7 Extended response question ...... 54 3.8 End of topic test ...... 55

4 DNA structure and replication 59 4.1 Introduction ...... 60 4.2 DNA structure ...... 61 4.3 Arrangement of DNA in chromosomes ...... 68 4.4 DNA replication ...... 70 4.5 Learning points ...... 74 4.6 Extended response question ...... 75 4.7 End of topic test ...... 76

5 Gene expression in human cells 79 ii CONTENTS

5.1 Introduction ...... 80 5.2 Gene expression through protein synthesis ...... 81 5.3 Structure and functions of RNA ...... 83 5.4 Transcription of DNA into mRNA ...... 86 5.5 Translation of mRNA into a polypeptide ...... 93 5.6 Single gene, several proteins ...... 100 5.7 Learning points ...... 103 5.8 Extended response question ...... 105 5.9 End of topic test ...... 106

6 Genes and proteins in health and disease 111 6.1 Introduction ...... 113 6.2 Proteins ...... 114 6.3 Different levels of protein structure ...... 115 6.4 Functions of proteins ...... 120 6.5 Mutations and genetic disorders ...... 122 6.6 Single gene mutations ...... 124 6.7 The effect of mutations on the structure and function of proteins . . . . 135 6.8 Chromosome structure mutations ...... 137 6.9 Learning points ...... 140 6.10 Extended response question ...... 142 6.11 End of topic test ...... 143

7 Human genomics 145 7.1 Introduction ...... 146 7.2 Sequencing DNA ...... 146 7.3 Bioinformatics ...... 148 7.4 Systematics ...... 150 7.5 Personalised medicine ...... 151 7.6 Ampliﬁcation and detection of DNA sequences ...... 153 7.7 DNA probes ...... 157 7.8 Medical and forensic applications ...... 159 7.9 Learning points ...... 163 7.10 Extended response question ...... 165 7.11 End of topic test ...... 166

8 Cell metabolism 169 8.1 Introduction ...... 171 8.2 Types of metabolic pathway ...... 172 8.3 Control of metabolic pathways - the action of enzymes ...... 177 8.4 The role of the active site ...... 180 8.5 Control of metabolic pathways through enzyme inhibition ...... 185 8.6 Learning points ...... 189 8.7 Extended response question ...... 191 8.8 End of topic test ...... 191

9 Cellular respiration I: Glycolysis 195 9.1 Introduction ...... 196 9.2 Summary of glucose breakdown ...... 197 9.3 Roles of ATP in the cell ...... 199

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9.4 Glycolysis ...... 202 9.5 Regulation of the respiratory pathway ...... 208 9.6 Learning points ...... 210 9.7 Extended response question ...... 212 9.8 End of topic test ...... 212

10 Cellular Respiration II: Citric acid cycle 215 10.1 Introduction ...... 216 10.2 The mitochondrion ...... 217 10.3 The citric acid cycle ...... 219 10.4 Alternative substrates for respiration ...... 224 10.5 Learning points ...... 226 10.6 Extended response question ...... 228 10.7 End of topic test ...... 228

11 Cellular Respiration III: Electron transport system 231 11.1 Introduction ...... 232 11.2 The electron transport chain ...... 233 11.3 ATP synthesis ...... 236 11.4 Learning points ...... 238 11.5 Extended response question ...... 239 11.6 End of topic test ...... 239

12 Energy systems in muscle cells 241 12.1 Introduction ...... 242 12.2 Creatine phosphate ...... 244 12.3 Lactic acid metabolism ...... 246 12.4 Types of skeletal muscle ﬁbres ...... 248 12.5 Learning points ...... 252 12.6 Extended response question ...... 254 12.7 End of topic test ...... 254

13 End of unit test 257

Glossary 265

Answers to questions and activities 270 1 Stem cells ...... 270 2 Differentiation in cells ...... 274 3 Research: Stem cells and cancer cells ...... 279 4 DNA structure and replication ...... 284 5 Gene expression in human cells ...... 289 6 Genes and proteins in health and disease ...... 295 7 Human genomics ...... 301 8 Cell metabolism ...... 306 9 Cellular respiration I: Glycolysis ...... 311 10 Cellular Respiration II: Citric acid cycle ...... 316 11 Cellular Respiration III: Electron transport system ...... 320 12 Energy systems in muscle cells ...... 322 13 End of unit test ...... 326

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Topic 1

Stem cells

Contents

1.1 Introduction ...... 2 1.2 What are stem cells? ...... 3 1.3 Embryonic stem cells ...... 4 1.4 Tissue (adult) stem cells ...... 7 1.5 Learning points ...... 9 1.6 Extended response question ...... 9 1.7 End of topic test ...... 10

Learning objectives By the end of this topic, you should be able to:

• explain the term ’cellular differentiation’;

• describe the properties of stem cells;

• describe the differences between embryonic and tissue stem cells;

• list the types of cell into which bone marrow cells can develop. 2 TOPIC 1. STEM CELLS

1.1 Introduction Learning objective By the end of this section, you should be able to:

• brieﬂy describe the process of embryonic development after fertilisation. All living things are characterised by levels of organisation that are hierarchical. The cell is the lowest level of organisation that can exist independently. Multicellular organisms, like humans, have cells organised into groups of cells called tissues, the next level of organisation. Tissues are formed from specialised cells that carry out a particular function. The columnar cells in the lining of the intestine, for example, are specialised for absorption (of nutrients), the stomach is made up of mucus membrane tissue, muscle tissue and a layer of tissue lining the abdomen. Tissues can themselves become grouped together to form an organ. Most organs, such as the heart, lung and liver, are also specialised for a certain function. The ﬁnal level of organisation is the organ system, where a group of organs work together at a particular function. The systems in our bodies include integumentary (skin, hair, nails), cardiovascular (circulatory), digestive, muscular and the nervous system.

Organisation: cells to tissues to organs to systems After fertilisation, the zygote undergoes repeated mitotic divisions to form first a solid ball of cells and then a hollow ball with a fluid-filled interior. Early in this process, the cells are unspecialised and are capable of developing into any of the body’s cell types. Later, the cells lose this general ability and become increasingly limited in terms of their potential functions. However, it is important to remember that all the cells of the body carry the same genetic information in the nucleus. Cellular differentiation is the process by which an unspecialised cell develops specific functions. It does this by expressing only those genes which are characteristic of that type of cell. At the same time, genes which would give it other capabilities are switched off.

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Thus, once a cell becomes differentiated, it only expresses the genes that produce the proteins characteristic of that type of cell. In this way, a developing muscle cell activates the genes which control the production of the contractile proteins, but switches off those which control the production of a digestive enzyme such as amylase.

Introduction: Questions Q1: What is meant by the term ’body tissue’? ...... Go online

Q2: "The heart is an organ". Explain this statement......

Q3: What is a human body system? Give two examples......

1.2 What are stem cells? Learning objective By the end of this section, you should be able to:

• state that stem cells are found only in multicellular organisms;

• explain how stem cells are different from other body cells;

• state that stem cells can self-renew and differentiate. In the early embryo all the cells are able to divide, whereas in the adult body only relatively few have this ability. Such cells are called stem cells. As the embryo grows, different groups of cells start to become specialised to carry out particular functions. At the same time, their potential to do other things is lost as the genes associated with these activities are turned off. This is the process of differentiation. Stem cells are different from other body cells because they have the following characteristics:

• undifferentiated (unspecialised cell type), allowing them to divide and maintain a supply of stem cells for the body;

• found in all multicellular organisms;

• self-renewing and can differentiate: in some organs, like the gut, stem cells regularly divide to repair and replace worn out or damaged tissues.

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The two types of stem cells found in humans are:

1. embryonic stem cells; 2. tissue (adult) stem cells.

A stem cell line is a group of constantly dividing cells from a single parent group of stem cells. They can be obtained from human or animal tissues, and can replicate for long periods of time in vitro (within glass; or, commonly, in the lab, in an artiﬁcial environment). These cell cultures may be used for a wide range of research purposes and may even (occasionally) be used to clone entire organisms.

Stem cells: Questions Q4: What do you understand by the term ’differentiation’ in regard to stem cells? Go online ...... Q5: What is the unique property of stem cells which makes them different from specialised cells? ...... Q6: Give examples of differentiated cells and their functions......

1.3 Embryonic stem cells Learning objective By the end of this section, you should be able to:

• explain what a blastocyst is;

• explain why embryonic stem cells are pluripotent;

• list some diseases that might be treated by transplanting cells generated from human embryonic stem cells. The ﬁrst successful isolation and culturing of embryonic stem cells took place in the 1980s. This work involved mouse embryos, but by the late 1990s a method to extract stem cells from human embryos had been successfully developed. Scientists were able to use embryos initially created for reproductive purposes through in vitro fertilisation. Many more eggs are fertilised in this process than are eventually implanted into the mother’s womb. With the informed consent of the donor, the extra embryos are made available for research. At this stage, the embryos are balls of 70-100 undifferentiated cells. However, there are considerable ethical issues concerning the use of human embryos in this way. Most embryonic stem cells are derived from embryos that are ready for implantation. The diagram shows stages in the development of the embryo up to the point of implantation.

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Stages in the early development of the human embryo After fertilisation, the zygote undergoes rapid cell division (stage A) and produces a multicellular hollow ball of cells which contains a fluid-filled cavity, the blastocoel (stage B). Once this ball comprises 70-100 cells, there is an inner cell mass at one end which will eventually develop into the embryo. At this point, the ball of cells is called a blastocyst and it is ready to implant into the endometrium of the uterus. Once implanted, the inner cell mass starts to differentiate and is known as the epiblast (stage C). At the pre-implantation stage, the inner cell mass consists of undifferentiated cells which are pluripotent, i.e. they have the potential to form any cell in the body (apart from the extra-embryonic tissues such as the amnion). It is these cells which are used for stem cell research. Human embryonic stem cells can be formed by transferring cells from a preimplantation- stage embryo into a culture dish that contains a nutrient broth, known as culture medium. As well as having the ability to undergo cell division, embryonic stem cells are able to undergo cell differentiation to generate new functional cells. In more recent research into embryonic stem cells, scientists have reliably directed the differentiation of embryonic stem cells into specific cell types. They are able to use the resulting differentiated cells to treat certain diseases. Diseases that might be treated by transplanting cells generated from human embryonic stem cells include Parkinson’s disease, diabetes, traumatic spinal cord injury, vision and hearing loss, Duchenne’s muscular dystrophy and heart disease. The techniques used for this research work will be studied in a later topic.

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Embryonic stem cells: Questions Q7: Put the following stages into the correct order to explain how to use human Go online embryonic stem cells to form specialised cells: • Formation of mass of undifferentiated stem cells • Stem cell cultured in the laboratory • Embryo stem cell removed • Undifferentiated stem cells cultured in different culture conditions • Formation of specialised cells: nerve cell, muscle cell, gut cells • Early human embryo Blastocyst

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Q8: What is a blastocyst? ......

Q9: Where do we ﬁnd embryonic stem cells? ......

Q10: For what reason are the cells of the inner cell mass useful for stem cell research? ......

Q11: List three diseases that might be treated using transplanted cells originated from embryonic stem cells......

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1.4 Tissue (adult) stem cells Learning objective By the end of this section, you should be able to:

• state that bone marrow contains two types of tissue stem cell;

• explain why tissue stem cells are multipotent;

• explain that tissue stem cells can only make certain types of specialised cells. Tissue stem cells are undifferentiated cells which can be found among differentiated cells in a tissue or organ. The primary roles of tissue stem cells in a living organism are to maintain and repair the tissue in which they are found. Work on tissue stem cells began in the 1950s when researchers discovered that bone marrow contains stem cells which develop into red blood cells, platelets, and the various forms of phagocytes and lymphocytes. Tissue stem cells have limited capabilities compared with embryonic stem cells. Tissue stem cells can only make the type of cells they belong to: they are called multipotent. They can only make certain types of specialised cells, not all of the kinds of cells in the body, so they are limited in their ability to differentiate. For example, bone tissue contains two types of tissue stem cells: haematopoietic stem cells, which form all the types of blood cells in the body; bone marrow stem cells, which generate bone, cartilage, fat, cells that support the formation of blood, and fibrous connective tissue. Tissue stem cells have been identified in many organs and tissues, including brain, bone marrow, peripheral blood, blood vessels, skeletal muscle and skin. Although they are found in so many types of tissues, only a very small number of stem cells actually occur in these tissues. This makes the study and the generation of tissue stem cells outside the body very challenging. Tissue stem cells are most commonly obtained from the outside part of the pelvis called the iliac crest. A needle is inserted into the iliac bone and bone marrow is withdrawn. It is likely that several samples may be obtained from this area. These stem cells are then separated from other cells in the marrow. Under ideal laboratory conditions they are grown, or expanded, in a process that can take between 7 and 21 days. After the formation of these ’new’ stem cells, they are placed in a specific tissue environment such as the bone. The tissue stem cells can become activated and they start to divide. A set of new stem cells and second generation (progenitor) cells is formed. It is these progenitor cells that differentiate into newer cells with the same phenotype as the host tissue. In 2006, it was shown that tissue stem cells isolated from adult mice could be induced to become pluripotent, i.e. capable of dividing and differentiating to form any type of body cell. When, one year later, it was found that human cells could be similarly induced, this opened the door to much wider research using stem cells without the controversy which surrounds the use of embryos.

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Tissue stem cells: Questions Q12: For each term or statement listed below, decide if it refers to embryonic or tissue Go online stem cells:

1. Pluripotent 2. Develop into all cell types 3. Ability to differentiate into some cells in the body 4. Give rise to a limited range of cell types 5. Develop into cell types that are closely related to the tissue in which they are found 6. Found in developing embryo - blastocyst 7. Have the capacity to become all cell types 8. Multipotent 9. Found in body tissues 10. Ability to differentiate into all of the cell types

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Q13: What are the two types of tissue stem cells found in bone marrow? What types of tissues can they generate? ......

Q14: What do you understand when tissue stem cells are described as multipotent? ......

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1.5 Learning points

Summary

• Cellular differentiation is the process by which a cell develops more specialised functions.

• Cellular differentiation occurs by expressing only the genes characteristic of that type of cell.

• Once a cell becomes differentiated, it only expresses the genes that produce the proteins characteristic for that type of cell.

• Stem cells are relatively unspecialised cells.

• Stem cells are able to continue to divide to produce more stem cells or cells which will differentiate into specialised cells of one or more types.

• During embryological development, the unspecialised cells of the early embryo differentiate into cells with specialised functions.

• Embryonic stem cells are found in the inner cell mass (epiblast) of the blastocyst.

• Embryonic stem cells are capable of developing into any of the body’s cell types (pluripotent).

• Tissue (adult) stem cells replenish differentiated cells that need to be replaced.

• Tissue (adult) stem cells are found in very small numbers throughout the body tissues.

• Tissue (adult) stem cells give rise to a limited range of cell types typical of the organ of which they are a part (multipotent).

1.6 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of embryonic and tissue stem cells before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: Embryonic and tissue stem cells Compare the location and functions of embryonic and tissue (adult) stem cells. (7 marks) ...... 15 min

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1.7 End of topic test End of Topic 1 test Q15: Complete the statements by matching the parts on the left with the parts on the Go online right. (12 marks) 5 min Cellular differentiation: adult stem cells.

Muscle cells only express stem cells.

Produced from bone marrow stem inner cell mass of the blastocyst. cells:

the genes characteristic of that type Cells capable of repeated division: of cell.

Stem cells: replenish differentiated cells.

Dividing stem cells produce dividing stem cells.

Specialised cells are produced from embryonic stem cells.

Embryonic stem cells are located in red blood cells, platelets, the phagocytes and lymphocytes.

cell develops more specialised Through the body: functions.

Tissue stem cells a limited range of cell types.

Can develop into any cell type: more stem cells.

Tissue stem cells can produce relatively unspecialised.

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Q16: What is meant by the term ’cellular differentiation’? (1 mark) ......

Q17: Why does a muscle cell only produce the proteins typical of its cell type? (1 mark) ......

Q18: List the types of blood cell into which bone marrow stem cells can differentiate. (1 mark) ......

Q19: List two general features of stem cells. (2 marks) ......

Q20: When they divide, what are the two types of cell that a stem cell can produce? (2 marks) ......

Q21: Where are embryonic stem cells found? (1 mark) ......

Q22: Where are tissue (adult) stem cells found? (1 mark) ......

Q23: State the function of tissue stem cells. (1 mark) ......

Q24: Compare the potential of the cells produced by embryonic and tissue stem cells. (1 mark) ......

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Topic 2

Differentiation in cells

Contents

2.1 Mitosis and meiosis ...... 15 2.1.1 Mitosis ...... 16 2.1.2 Meiosis ...... 17 2.2 Differentiation in somatic cells ...... 21 2.3 Human body tissues ...... 22 2.3.1 Epithelial tissue ...... 22 2.3.2 Connective tissue ...... 23 2.3.3 Muscle tissue ...... 24 2.3.4 Nervous tissue ...... 26 2.4 Formation of the body organs ...... 29 2.5 Differentiation of germline cells ...... 29 2.6 Mutations in germline and somatic cells ...... 30 2.7 Learning points ...... 32 2.8 Extended response question ...... 33 2.9 End of topic test ...... 33

Prerequisite knowledge In the previous topic, the process by which the relatively unspecialised stem cells in the embryo and adult tissues are able to differentiate into all the cell types of the body was outlined. These cells belong to two fundamental types: germline cells, which give rise to gametes by dividing by mitosis and meiosis, and somatic cells, which produce all the other cells of the body by mitosis. In what follows in this topic, we will look at some of the ways in which the various types of cell in the body become specialised to their different functions. Learning objectives By the end of this topic, you should be able to:

• describe the production and differentiation of somatic cells;

• describe the specialisations and functions of the four types of body tissue;

• describe the types of cell division in germline cells which produce more germline cells and gametes; 14 TOPIC 2. DIFFERENTIATION IN CELLS

• explain the signiﬁcance of mutations in somatic and germline cells.

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2.1 Mitosis and meiosis Learning objective By the end of this section, you should be able to:

• outline the process of mitosis and explain its signiﬁcance in the production of diploid somatic and germline cells;

• outline the process of meiosis and explain its signiﬁcance in the production of haploid gametes. In this topic you will be learning about somatic cells and germline cells, and how these cells are involved in the process of differentiation. From previous study, you should be familiar with the two types of cell division: mitosis and meiosis. The details of these processes are not required for this course, but it is helpful to be familiar with them to fully understand their signiﬁcance in relation to somatic and germline cells. Every normal human body cell has 46 chromosomes arranged in 23 pairs. These cells are known as diploid cells (because they contain a double set of chromosomes). Every normal human gamete (sex cell) contains only 23 unpaired chromosomes. Such cells are called haploid (because they contain a single set of chromosomes). After fertilisation, the newly formed zygote now contains two sets of chromosomes. In other words, it is diploid. The zygote continues to divide by mitosis and eventually develops into an adult. The cycle of sexual reproduction can now begin all over again. These events are summarised in the following diagram.

The sexual life cycle in animals Haploid cells are produced by a special type of cell division known as meiosis. Both mitosis and meiosis are types of nuclear division. At the end of each process, however, the cytoplasm divides and daughter cells are formed. It is important to understand that mitosis and meiosis serve fundamentally different purposes.

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Mitosis produces two cells which are genetically identical to the parent cell. The process of copying and allocating the DNA to the daughter cells is virtually foolproof; when it rarely does fail, there are cellular checking systems which correct any errors. This ensures that all the cells of the body contain exactly the same inherited information and so can be relied upon to react to stimuli like hormones in a consistent and predictable way. Meiosis produces gametes (sex cells) which are intended to fuse with another gamete to produce a zygote, and hence a new individual with a complete genetic complement. To achieve this, every gamete must contain an example of each gene and each chromosome. This is brought about in Meiosis I where the diploid number of chromosomes in the gamete mother cells is reduced to the haploid number in the gametes by the pairing and separation of the homologous chromosomes. The crossing over of portions of chromatid in Meiosis I, which is an important source of variation, also causes the two chromatids of each chromosome to no longer be identical. In Meiosis II, a division that is very similar to mitosis, the two dissimilar chromatids of each chromosome are separated into daughter cells, ensuring that every gamete only carries one allele of each gene. The products of mitosis are two identical diploid cells; in contrast, meiosis produces four non-identical haploid cells. In order to help you understand the differences between mitosis and meiosis, we will very quickly review the stages involved in mitosis.

2.1.1 Mitosis Mitosis produces two daughter cells which are genetically identical to the parent cell.

Stages of mitosis The stages can be described as follows:

1. Shows the cell just before it begins mitosis. The chromosomes as single chromatids have replicated so that each chromosome now consists of two chromatids joined at the centromere. 2. Shows the chromosomes have coiled up to become shorter and thicker and can be seen to be double stranded. Each double stranded chromosome consists of a pair of chromatids joined together at the centromere. The nuclear membrane is beginning to break down. 3. Shows the spindle forming and the double-stranded chromosomes migrating to, and line up along, the equator of the cell. Each chromosome becomes attached to a separate spindle ﬁbre.

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4. Shows the chromatids being pulled apart to opposite poles of the cell by the spindle ﬁbres.

5. Shows a new nuclear membrane forming around each set of single-stranded chromosomes. Division of the cytoplasm follows.

Two identical daughter cells have been produced from a diploid parent cell.

Mitosis: Question Q1: Decide on the correct order of the illustrations to show the stages of mitosis: Go online 5 min

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2.1.2 Meiosis All of our body cells are diploid, that is they contain 23 pairs of chromosomes (46 in all). Gametes, however, are haploid, containing 23 unpaired chromosomes. This is important so that a diploid zygote (fertilised egg) is produced at fertilisation. Meiosis is the type of nuclear division which reduces the number of chromosomes, thus producing haploid gametes. Meiosis consists of two separate divisions, the ﬁrst and second meiotic divisions, and therefore results in four haploid daughter cells. We will look at each division in turn.

Stages of the ﬁrst meiotic division during meiosis The stages can be described as follows:

1. Shows the cell just before meiosis begins. It is at this point that the chromosomes are replicated.

2. Shows that the chromosomes have shortened and thickened. They can be seen to consist of pairs of chromatids joined at the centromere.

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3. Shows the chromosomes ﬁnding their homologous partners and moving very close to each other. The chromatids become intertwined at points called chiasmata.At these points the chromatids may break and rejoin so that some genes may cross over from one chromatid to the other. This does not happen in mitosis.

4. Shows the pairs of homologous chromosomes migrating to the equator of the cell as a unit and lining up independently of any other pair. The spindle forms and each homologous pair of chromosomes becomes attached to the same spindle ﬁbre.

5. Shows the homologous chromosomes (still double-stranded) being pulled apart towards opposite ends of the cell by the motor proteins of the spindle ﬁbres.

6. Shows the formation of new nuclear membranes around each set of chromosomes. Then the cell divides in two. Note that each new cell is haploid, that is they possess only a single set of (double-stranded) chromosomes.

Stages of the second meiotic division during meiosis Each cell now undergoes what is essentially a mitotic division to separate the double- stranded chromosomes:

1. Shows the cells just before they begin the second meiotic division. Note that there is no replication of the chromosomes since they are already double-stranded.

2. Shows the nuclear membranes of the cells breaking down, the spindle forming, the chromosomes migrating to the equator, and each chromosome becoming attached to a separate spindle ﬁbre.

3. Shows the chromosomes (now single-stranded) being pulled to opposite poles of the cells.

4. Shows a new nuclear membrane forming around each set of chromosomes. The cytoplasm divides again, producing four haploid gametes.

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Meiosis: Questions Q2: Decide on the correct order of the illustrations to show the stages of the ﬁrst meiotic division during meiosis: Go online 10 min

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Q3: Decide on the correct order of the illustrations to show the stages of the second meiotic division during meiosis:

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Q4: There are 46 chromosomes (23 pairs of chromosomes) in a human cell. How many chromosomes are present in each daughter cell after mitosis? ......

Q5: How many chromosomes are present in each daughter cell after the ﬁrst meiotic division? ......

Q6: How many chromosomes are present in each daughter cell after the second meiotic division? ......

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A cell entering meiosis

Q7: Which label in the illustration indicates a pair of homologous chromosomes? ......

Q8: Which label in the illustration indicates a single chromosome? ......

Q9: Which label in the illustration indicates a chromatid? ......

Q10: Which label in the illustration indicates a centromere? ......

Q11: Which label in the illustration indicates a chiasma? ......

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2.2 Differentiation in somatic cells Learning objective By the end of this section, you should be able to:

• deﬁne a somatic cell;

• give examples of somatic cells. A somatic cell is a non-sex cell; they make up all the cells in the human body except the reproductive cells (gametes). Somatic cells are the differentiated cells that form the different types of human body tissues. Somatic cells make up all the organs, skin, bones and blood. Some examples of somatic cells are smooth muscle cells, red blood cells, B lymphocytes, and epithelial cells. Somatic cells in the body contain the diploid number of chromosomes and some can undergo mitosis, giving rise to daughter cells. These daughter cells then grow, and may themselves divide to form more cells. The result of these two processes is an increase in the size of the body and its organs.

A micrograph of typical stained somatic cells

Differentiation in somatic cells: Questions Q12: How many chromosomes are found in somatic cells? ...... Go online

Q13: What are differentiated cells? ......

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2.3 Human body tissues Learning objective By the end of this section, you should be able to:

• list the four types of body tissues as epithelial, connective, muscle and nervous;

• give examples of each type of tissue;

• state the functions of each type of tissue. The somatic cells of the body of all multicellular animals form four types of tissue (each of which can be subdivided). These tissues are:

1. connective - gives shape to organs and supports them, including blood and bone;

2. epithelial - covers the organ surfaces, including the skin and the lining of the gut;

3. muscle - causes locomotion or movement within organs by the action of contractile cells;

4. nervous - transmits messages within the central nervous system and between the central nervous system and the rest of the body.

2.3.1 Epithelial tissue Epithelial tissues cover the whole surface of the body. They are made up of cells closely packed together and arranged in one or more layers. These tissues are specialised to form the covering or lining of all internal and external body surfaces, e.g. the skin, the lining of the airways, the gut lining and the reproductive tracts. Depending on its location and function, the epithelium may have various structures. Epithelial tissue has two general functions:

1. protection, which it does by having very tight junctions between its cells, e.g. the skin epidermis, the epidermis lining of the mouth, and the surface layer of the endometrium;

2. secretion and absorption, e.g. secretion of digestive enzymes by the intestinal glands in the small intestine and absorption into the villi of the small intestine or absorption into the alveoli of the lung.

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Different types of epithelium

Epithelial tissue: Question Q14: List the main functions of epithelial tissues...... Go online

2.3.2 Connective tissue The main function of connective tissue is to support the human body and to connect tissues together. It is strong enough to give a mechanical framework for the skeleton and so it also plays an important role in body movement. Connective tissue has three main components, all of which are embedded in the body ﬂuids:

1. cells; 2. ﬁbres; 3. extracellular matrix (the extracellular part of the animal tissue that provides structural support).

Cells, called ﬁbroblasts, are responsible for the production of many forms of connective tissue. Examples of the different functions of connective tissues are:

• connecting body organs - blood; • linking connective tissue to muscle tissue - cartilage (in tendons); • protection - skull bones; • structural framework - ribs; • storage of energy - adipose tissue.

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Connective tissue of different types may be grouped together, e.g. in an artery wall.

Structure of an artery Cartilage is a good example of connective tissue. As well as forming the basic structure of all bones, it is found in the joints between bones, the rib cage, the ear, the nose, the elbow, the knee and the ankle. Over 150 disorders have been identiﬁed related to connective tissue, e.g. cellulitis (a result of an infection), scars (injuries to connective tissue) and genetic disorders such as Marfan syndrome.

Connective tissue: Questions Q15: Where do you ﬁnd cartilage connective tissues? Go online ......

Q16: List the main functions of connective tissue......

2.3.3 Muscle tissue The function of muscle cells is to produce force, and, as a result, motion. This is achieved by protein ﬁlaments in the cells sliding past each other, changing both the length and the shape of the cell. Muscle cells can only contract and relax, they cannot push; hence the need for antagonistic pairs of muscles at joints (e.g. the biceps and triceps at the elbow).

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There are three types of muscle tissue:

1. Skeletal (voluntary) muscle is attached by cartilage to different bones at two or more points, which allows it to bring about locomotion and to maintain posture. This muscle appears striped under the microscope because of the arrangement of the contractile proteins. The contraction of this muscle is under voluntary control; although posture is maintained by unconscious reﬂexes, the muscles can also be contracted voluntarily. Examples are the quadriceps of the upper leg, and the biceps and triceps of the upper arm. The ’-ceps’ refers to the number of attachment points (origins) at the inner end of the muscle.

Skeletal muscle structure

2. Smooth (involuntary) muscle is found widely throughout the body in the walls of structures, e.g. arteries (contributing to blood ﬂow in major arteries and controlling access to capillary beds in arterioles), the stomach and all of the alimentary canal (causing peristalsis), and the bladder (causing elimination of urine). The erector pili, which cause ’goose ﬂesh’ when we are cold, are also smooth muscle, as are the muscles which control the diameter of the pupil of the eye by means of the iris. Smooth muscle is not under conscious control.

Smooth muscle structure

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3. Cardiac muscle is a form of involuntary muscle which is unique to the heart. Like voluntary muscle, it has a striped appearance under the microscope which is caused by the arrangement of the contractile proteins within the cells. It has a very high resistance to fatigue.

Cardiac muscle structure

Muscle tissue: Question Q17: Use the word list to complete the gaps in the sentences below. Words may be Go online used more than once.

5 min

The body contains three types of muscle tissue: , and

. is unique in that it is only found in one organ, the .

Two of the muscle types are not under conscious control and are called muscle.

The other type of muscle, , is under conscious control and is called

muscle. The two muscle types which are striped are and .

The muscle type which forces food through the small intestine is .

The muscle type which maintains posture is , but its control is both and . Word list: cardiac, heart, involuntary, skeletal, smooth, voluntary......

2.3.4 Nervous tissue The human nervous system consists of the central nervous system (CNS), comprising the brain and spinal cord, and the peripheral nervous system (PNS), comprising the cells which connect the CNS to the rest of the body. At the most primitive level, the tissue which makes up the nervous system is specialised to react to stimuli and to conduct impulses to various organs in the body, bringing about a response to each stimulus. It comprises two groups of cell: neurons and glial cells.

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Neurons Neurons (often referred to loosely as ’nerve cells’), which conduct impulses, have three types:

1. Sensory - carry information as impulses from sense receptors into the central nervous system (CNS);

2. Motor - carry impulses from the CNS out to muscles and glands (effectors);

3. Interneurons - form connections between neurons in the CNS.

Sensory neurons are adapted to take messages to the brain from sensory receptors in the skin, the specialised sense organs, the nose, tongue and ears.

Note that the cell body sits part way along the axon and there are no dendrites on the cell body.

A motor neuron has many dendrites protruding from the cell body, which is end-on to the axon.

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An interneuron has many dendrites and axon terminals to connect with other neurons.

Glial cells Glial cells (neuroglia) also have several forms and functions:

• to maintain a constant environment for the neurons by supplying nutrients and removing waste;

• make the myelin which ensheaths some neurons and provides insulation;

• provide support and protection in the nervous tissue.

Nervous tissue: Question Q18: Complete the following sentences using terms from the list below. Go online consists of neurons and glial cells.

conduct impulses into the CNS.

conduct impulses out from the CNS.

connect other neurons.

maintain a constant environment for neurons. Word list: Glial cells, Interneurons, Motor neurons, Nervous tissue, Sensory neurons......

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2.4 Formation of the body organs Learning objective By the end of this section, you should be able to:

• explain that body organs can be formed from different body tissues. There are almost 70 organs in a human body which vary according to their sizes, functions or actions. Each organ has its own recognisable structure like the heart, lungs, liver, eyes, ears and stomach, and each performs its own speciﬁc functions. Tissues are grouped together so as to perform a single or several functions in the human body. For example, the gall bladder contains different types of tissues such as the epithelial tissues, forming a lining to protect the organ from the effects of bile, and muscle tissue that contracts and expels bile.

Formation of the body organs: Question Q19: Why is the gall bladder a good example of an organ? ...... Go online

2.5 Differentiation of germline cells Learning objective By the end of this section, you should be able to:

• deﬁne a germline cell;

• state the functions of a germline cell. The germline of an individual is the sequence, or line, of cells that is capable of passing its genetic material on to the next generation. A germ cell is one which can give rise to gametes. The germline cells segregate from the other body cells very early in the formation of embryo. As the body grows and develops, the germline cells release hormones into the blood circulation; in turn, they are controlled by hormones released by the pituitary gland. Thus, the body develops the characteristics of the male or the female and the germline cells are stimulated to start meiosis as well as mitosis so that, in addition to producing more diploid germline cells, haploid gametes are formed.

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The sexual life cycle in animals In contrast to most cells, which can only divide 30-50 times in the growing embryo, germline cells (like stem cells) can divide indeﬁnitely.

2.6 Mutations in germline and somatic cells Mutations are changes to the genetic material, affecting either chromosomes or individual genes. They may have no effect, alter the product of a gene or metabolic process, or prevent a gene from functioning properly. The details of these various mutations are dealt with in a later topic. Mutations have a bad press. This is undeserved; without mutations, and the consequent production of new combinations of DNA within genes and new arrangements of genes on chromosomes, natural selection would have had no material to work on and evolution would not have been possible. Some mutations arise without the action of outside inﬂuences. They may be caused by errors in the DNA replication or repair processes, or by bonds breaking by hydrolysis in a natural process of chemical decay. Alternatively, mutations can arise as a result of exposure to outside agents and the list of such mutagenic agents, or mutagens, is long. It includes:

• various forms of radiation, e.g. X-rays, gamma rays, ultraviolet light, radon;

• a wide range of chemicals, e.g. nitrosamines (tobacco), mustard gas, bromine, benzene;

• several metals and their compounds, e.g. arsenic, cadmium, chromium;

• some bacteria and viruses, e.g. Helicobacter pylori, hepatitis B virus.

The effects and signiﬁcance of mutations which take place in germline and somatic cells are very different.

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Germline cell mutation After the egg and sperm fuse at fertilisation, the zygote divides by mitosis to give rise to all the other cells of the body. If a mutation has occurred in the germline cells that produced one of these gametes, this mutation will be transmitted to every single cell of the resulting organism. Germline mutations can be found in every cell which originates from the zygote to which the gamete containing the mutation contributed. An oft-quoted example of such a mutation involves Queen Victoria. There is no record of haemophilia in her family history, yet her youngest son was a haemophiliac and two of her five daughters were carriers. The mutation must have occurred in the gamete mother cells of Victoria’s father and he then passed the mutant allele on to his only child, thus making her a carrier of the condition. (The expected outcome of such a situation, in which the mother is a carrier and the father is unaffected, would be that, on average, half of her sons would be haemophiliacs and half of her daughters would be carriers. Victoria had four sons and five daughters so one out of four and two out of five is well within the expected range.) Somatic cell mutation In somatic cell mutation, only cells within the body of the person affected will contain the mutation. If the cell containing the mutation has already undergone differentiation, then only it will be affected; if the cell is a stem cell, then all of its progeny will also carry the mutation. Compared to germline mutations, this type of mutation is localised in its presence within the body and is non-inheritable. Cancer tumours are the result of one type of somatic cell mutation. In this case, the mutation affects a gene involved in cell division (a proto-oncogene), causing uncontrolled cell division and leading to the collection of undifferentiated cells that is a tumour. Although the mutation is restricted to this tissue, its effects may of course extend to the whole body.

Mutations in germline and somatic cells: Question Q20: State one similarity and two differences between mutations to germline and somatic cells. Go online ......

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2.7 Learning points

Summary

• All of the cells of the body are somatic cells apart from gametes and the cells which produce them.

• Examples of somatic cells are skin epidermis, B lymphocytes and motor neurons.

• Somatic cells are produced by mitosis.

• Some somatic cells are able to divide by mitosis to form more somatic cells.

• Dividing by mitosis ensures that somatic cells maintain the diploid chromosome number and the genetic complement.

• Somatic cells can differentiate to form different types of tissue.

• The four types of tissue found in the body are epithelial, connective, muscle and nervous.

• The functions of epithelial tissue are protection (e.g. skin epidermis), secretion (e.g. intestinal glands), and absorption (e.g. villi of small intestine).

• The functions of connective tissue are protection (e.g. skull bones), structural framework (e.g. ribs), storage of energy (e.g. adipose tissue), connecting body organs (e.g. blood), and linking connective tissue to muscle tissue (cartilage of tendons).

• The function of muscle tissue is to produce force and motion, e.g. skeletal muscle (movement), smooth muscle (arteriole contraction) and cardiac muscle (heart contractions).

• The function of nervous tissue is to react to stimuli and conduct impulses to various parts of the body, e.g. neurons (carry impulses) and glial cells (maintain a constant environment for neurons).

• Body organs are all formed from a variety of these tissues.

• A germline cell is one which is capable of passing on its genetic information to the next generation by giving rise to gametes.

• Some germline cells divide by mitosis to produce more diploid germline cells.

• Some germline cellsdivide by meiosis to produce haploid gametes.

• Mutations that occur in germline cells will be passed to offspring.

• Mutations in somatic cells will not be passed to offspring.

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2.8 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of the four types of human body tissue before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: Four types of human body tissue Give an account, with examples, of the different body tissue types and their functions. (8 marks) 15 min ......

2.9 End of topic test End of Topic 2 test Q21: Complete the sentences by matching the parts on the left with the parts on the right. (8 marks) Go online 20 min

Somatic cells are found differentiation.

Germline cells are B-lympbocytes and motor neurons.

Examples of somatic cells are more somatic cells.

Dividing somatic cells produce cell or tissue involved.

Mitosis germline.

Somatic cells form different types of throughout the body. tissue by

Cells capable of producing gametes maintains the chromosome number. are

Mutations in somatic cells affect the gamete mother cells.

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......

Q22: Complete the statements by matching the deﬁnitions on the left with the terms and values on the right. (7 marks)

Somatic cells divide: 23

Number of types of tissue in mitosis stomach:

Produces more germline cells: four

Germline cells produce gametes: affect all cells of offspring

Germline chromosome complement: mitosis

Gamete chromosome complement: diploid

Mutations in germline cells: meiosis ......

Q23: Complete the paragraph using the words from the list. (11 marks)

Skin epidermis is an example of tissue, the function of which is .

tissue is used to store and is an example of tissue. muscle has the function of producing and is found in

cells are found in tissue where they make the of some neurons. Word list: adipose, artery walls, connective, energy, epithelial, force and motion, glial, myelin sheath, nervous, protection, smooth......

Q24: Where are somatic cells found? (1 mark) ......

Q25: Name two examples of somatic cells. (1 mark) ......

Q26: When they divide, what do some somatic cells produce? (1 mark) ......

Q27: State the type of cell division found in somatic cells. (1 mark) ......

Q28: Explain the signiﬁcance of somatic cells dividing in this way. (1 mark) ......

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Q29: Name the process by which somatic cells form different types of tissue. (1 mark) ......

Q30: Complete the table by listing the four types of body tissue, a function for each, and an example of each. (12 marks)

Tissue Function Example

......

Q31: Which tissues form an organ such as the stomach? (1 mark) ......

Q32: State the meaning of the term ’germline cell’. (1 mark) ......

Q33: Name the process by which germline cells produce more germline cells. (1 mark) ......

Q34: Name the process by which germline cells produce gametes. (1 mark) ......

Q35: List the terms used to describe the chromosome complement of germline cells and gametes, and explain their meaning. (2 marks) ......

Q36: Compare the signiﬁcance of mutations which take place in the somatic cells and germline cells. (1 mark) ......

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©HERIOT-WATT UNIVERSITY 37

Topic 3

Research: Stem cells and cancer cells

Contents

3.1 Introduction ...... 39 3.2 Stem cell research ...... 39 3.2.1 Cell growth and differentiation ...... 43 3.2.2 Gene regulation ...... 43 3.2.3 Drug testing ...... 43 3.2.4 Development of disease ...... 44 3.3 Therapeutic uses of stem cells ...... 44 3.4 Ethical issues and the regulation of stem cell use ...... 47 3.5 The biology of cancer cells ...... 50 3.6 Learning points ...... 53 3.7 Extended response question ...... 54 3.8 End of topic test ...... 55

Prerequisite knowledge You should remember from Topic 1 that stem cells in multicellular organisms have the ability to develop into different types of body cells during early embryonic development and growth. In certain types of body tissues, tissue stem cells can become an internal repair system; they can divide without limit to replenish other cells. Learning objectives By the end of this topic, you should be able to: Stem cells

• describe the contribution of stem cell research to our understanding of cell processes such as growth, differentiation and gene regulation;

• explain why stem cells have therapeutic uses;

• describe the use of stem cells in the repair of damaged or diseased organs and tissues;

• explain the use of stem cells as models to study disease development; 38 TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS

• describe the use of stem cells in drug testing;

• outline the principal ethical issues associated with stem cell use in research;

• describe the legal regulation of stem cell use;

Cancer cells

• describe the differences between cancer and normal cells;

• explain the formation of a tumour;

• explain the differences between malignant and benign tumours;

• state the meaning of the terms primary tumour and secondary tumour.

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3.1 Introduction In an earlier topic, you learned about embryonic and tissue stem cells. In this topic, you will study the use of stem cell research and therapy. You will also consider some of the moral and ethical issues related to the wider use of stem cells, especially the use of embryonic stem cells. With cancer being one of the major causes of death in the UK, in this topic you will also study the biology of cancer cells and the development of the different types of tumour.

3.2 Stem cell research Learning objective By the end of this section, you should be able to:

• give examples of research work done on stem cells;

• deﬁne stem cell lines;

• describe the use of stem cells in skin grafts;

• brieﬂy describe how stem cells can be used in bone marrow transplants;

• brieﬂy describe how stem cells can be used in cornea transplants;

• describe how induced pluripotent stem cells (iPSC) are produced;

• give examples of the use of stem cells for drug testing. Scientists have been interested in the study of cell biology since the discovery of microscopes in the 1800s. In the early 1900s, researchers discovered that blood cells, i.e. white blood cells, red blood cells and platelets, all originated from a particular stem cell type. In the 1950s, bone marrow transplants involving adult stem cells were first used in patients receiving radiation and chemotherapy. The world’s first human embryonic stem cell line was developed by James Thomson in 1998 at the University of Wisconsin in Madison. He successfully removed cells from spare embryos at fertility clinics and cultured them in the laboratory. Several thousand research papers on embryonic and adult stem cells are published annually in scientific journals. Although clinical trials using embryonic stem cells are still in their infancy, tissue stem cells are already being used in the treatment of many conditions including leukaemia and heart disease.

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Since James Thomson’s work in 1998, there has been large amount of research work done on stem cells. Here are just a few of the signiﬁcant ﬁndings in stem cell research over the last decade:

• 2000s: Several reports of adult stem cell plasticity are published.

• 2005: Researchers at Kingston University in UK discovered a third category of stem cell called cord-blood-derived embryonic-like stem cells (CBEs). These cells came from the blood of the umbilical cord. The researchers claimed that the stem cells were capable of differentiating into more types of tissue than adult stem cells.

• October 2006: Scientists at Newcastle University in UK used umbilical cord blood stem cells to create the ﬁrst ever artiﬁcial liver cells.

• October 2007: Capecchi, Evans and Smithies won the Nobel Prize for Physiology or Medicine. They were rewarded for their work on mouse embryonic stem cells in which they used gene targeting strategies to produce genetically engineered mice for gene research. They called the mice "Knockout mice".

• October 2008: German scientist Conrad and colleagues generated pluripotent stem cells from sperm mother cells in the adult human testis.

• 2011: Israeli scientist Inbar Friedrich Ben-Nun and his team produced the ﬁrst stem cells from endangered species; this could be a breakthrough that could save animals in danger of extinction.

• 2013: A team from the University of Pittsburgh, Pennsylvania, grew human heart stem cells in culture dishes and created conditions in which they formed a very primitive heart.

• 2013: Mark Post of Maastricht University produced meat from muscle stem cells, which was eaten as a burger.

The research work on stem cell differentiation and development can provide us with a lot of information about the human body. If we can understand which genes turn ’on’ and ’off’, and the effect this has on our development, we will have a better understanding of diseases like cancer and other growth abnormalities of the body’s tissues and organs. The reasons why stem cells are desirable for research are that they are pluripotent and can be cultured indeﬁnitely in the laboratory. This means that stem cells with particular properties can be maintained in repositories and appropriate cells can be supplied for particular research projects. Such lines include cells with different genetic abnormalities, including Duchenne muscular dystrophy and thalassaemia. Studying these cells enables recognition of mechanisms at the molecular level that might possibly be blocked and so impede the disease progression. Thus embryonic stem cell lines originating from embryos with genetic and chromosomal abnormalities enhance our understanding of the pathways of genetic defects.

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What are stem cell lines? You should remember that embryonic stem cells are derived from blastocysts. These human blastocysts, used to produce stem cell lines, have been donated for research from in vitro fertilisation clinics. In order to obtain an embryonic stem cell line, scientists have to remove the cells from the inner cell mass region of a blastocyst. This is because these cells are pluripotent. Once these cells are removed, they are placed in culture plates with the correct nutrients and growth factors. An embryonic cell line is established when these cells multiply and divide. Under the right conditions, these cell lines can be maintained indeﬁnitely and can be used in all types of stem cell research. The next stage involves the addition of different growth factors to induce these cells to differentiate into different cell types. The newly formed cells can then be used in therapies to replace damaged cells or organs. Stages of stem cell line production

1. Stages 1 and 2 involve the isolation of the embryonic stem cells.

2. In stages 3 and 4, the cells were tested or characterised. This should provide enough information to let the researchers know if these cells were stem cells and if they have the potential to differentiate.

3. Finally, stage 5 involves scaling up so as to provide enough cells for further clinical studies.

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The production of induced pluripotent stem cells (iPSC) An alternative method of producing pluripotent stems cells is by using a process called induced pluripotent stem cells (iPSC). Normally, the embryos are destroyed if we are to harvest stem cells from donated embryos. This provokes wide discussions about the ethical and moral issues of using embryos in stem cell work. Research work has shown that by using differentiated adult cells, like skin cells, it is possible to ’reprogramme’ these cells by switching off the genes that make another adult differentiated cell. Through the use of a virus as carrier (vector) to transfer genes, the reprogramming will change the adult skin cell into one that has the properties of an embryonic stem cell. The reprogrammed pluripotent stem cells behave like embryo stem cells, but are actually from an adult. Although these cells meet the deﬁning criteria for pluripotent stem cells, it is not known if iPSCs and embryonic stem cells are clinically different. The iPSC process is illustrated in the following diagram:

Induced pluripotent stem cell (iPSC) A major problem with the use of viruses to alter the genome of the somatic cell is that it can trigger the expression of oncogenes, which cause tumours to develop. Recently, it has been found that it is possible to induce somatic cells to become pluripotent by treating them with certain proteins, thus opening the door to many more applications of iPSCs. Researchers are hoping to use iPSC in transplantation medicine, and also for drug development and the modelling of diseases. At present, viruses are used for the reprogramming of embryonic stem cells into adult cells; the process must be carefully controlled and tested before this technique can be used for the treatment of human diseases. The production of iPSC creates pluripotent stem cells which, with the studies of other types of pluripotent stem cells, will help researchers learn how to reprogramme cells to repair damaged tissues in the human body.

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Stem cell research: Questions Q1: What are stem cell lines? ...... Go online Q2: Where do embryonic stems cells come from in the production of a stem cell line? ...... Q3: Where do the cells used for iPSC originate from? ...... Q4: Why do you think people who are against embryonic stem cell research would like to see further studies in iPSC? ......

3.2.1 Cell growth and differentiation In 2013, researchers at the Austrian Academy of Sciences successfully grew complex human brain tissue which formed ’mini-brains’ with a cerebral cortex and retina. Their research began with human embryonic stem cell lines and induced pluripotent stem cells from mouse embryos. By carefully adjusting the nutrient medium in which they were cultured, the scientists caused the cells to divide and differentiate into nervous tissue. These fragments were attached to a 3D scaffold, and, within 3-4 weeks, deﬁned brain regions had formed. These mini-brains were maintained for several months. The value of this work is its potential to give new insights into developmental brain disorders. Already, it has been used to analyse the onset of microcephaly, a genetic condition in which the brain is severely reduced in size.

3.2.2 Gene regulation At several universities, the molecular causes of Rett syndrome are being studied. Rett syndrome is an autism spectrum disorder which is the second most frequent form of intellectual disability in girls after Down’s syndrome. In most cases, this is caused by the mutation of an X-linked gene, the protein product of which functions as a regulator of gene expression. The researchers are using induced pluripotent stem cells from Rett syndrome patients to study the effect of key hormones and their inhibitors on neurone function, in particular the growth of dendrites.

3.2.3 Drug testing One of the important potential outcomes of embryonic stem cell research is that these cells might be used to create very speciﬁc cell cultures for the initial stages of drug testing. For example, by maintaining them in an appropriate nutrient solution, embryonic stem cells could be stimulated to generate brain cells showing the effects of Alzheimer’s disease. Researchers could then screen millions of potential drugs against this more realistic cellular model and identify the most effective compounds before going forward to animal testing and clinical trials. The speed and accuracy of drug screening would be improved by the use of real human cells that mimic diseases.

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3.2.4 Development of disease A potentially exciting use for iPS cells is the development of cell models of Parkinson’s disease. In theory, scientists could use cells from people living with Parkinson’s disease to create iPS cell models of the disease that have the same intrinsic cellular machinery as a Parkinson’s patient. Researchers could use these cell models to evaluate genetic and environmental factors implicated in Parkinson’s disease. Many scientists use animals to model human diseases (mice can be obese or display symptoms of Parkinson’s disease, while rats get Alzheimer’s and diabetes), but animal models are seldom perfect and so scientists are looking at the use of induced pluripotent stem cells (iPS cells). People with Parkinson’s disease don’t have enough dopamine, a chemical that allows messages to be sent to the parts of the brain that control movement. The disease kills dopamine-producing nerve cells, or neurons, in part of the brain called the substantia nigra. Parkinson’s is also linked to formation of clumps of a protein called alpha- synuclein in the brain. Although the underlying cause of Parkinson’s disease is unknown, scientists do know which cells and areas of the brain are involved. Researchers are already using stem cells to grow dopamine-producing nerve cells in the lab so that they can study the disease. Because a single, well-deﬁned type of cell is affected, it may also be possible to treat Parkinson’s by replacing the lost nerve cells with healthy new ones. In the study of motor neuron disease, iPS cells can create motor neurons that grow in a Petri dish and show the same characteristic blobs and tangles in the long ﬁbres of the nerve cells as are found in patients with the condition.

3.3 Therapeutic uses of stem cells Learning objective By the end of this section, you should be able to:

• brieﬂy describe the therapeutic use of stem cells in skin grafts;

• list other therapeutic uses of stem cells - bone marrow and cornea grafts. ’Therapeutic uses’ simply refers to use in medical treatments. The generation of cells and tissues used for cell-based therapies is one of the most important potential applications of human stem cells. The availability of donated organs and tissues is limited compared to the demand for transplant operations. The possibility of using stem cells which have been stimulated to differentiate into speciﬁc cell types can be a valuable source for replacing ailing or damaged cells and tissues. Stem cells are already available for therapeutic use in diabetes, osteoarthritis, Alzheimer’s disease, stroke, burns, and rheumatoid arthritis.

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Here are three examples of the therapeutic use of stem cells: 1. Skin grafts At present, patients with a full thickness burn to the skin, where the epidermis and the delicate tissue of the dermis are damaged, are often treated using skin grafts of their own skin (autologous grafts). During the period of 3 weeks that it takes for the graft to be prepared, the patient is at risk of dehydration and infection. The diagram shows a section through the human skin and the position of the major tissues, the epidermis and the dermis.

A section of human skin To prepare a graft, a section of healthy skin is removed from an undamaged part of the patient’s body. This is then grown in special nutrient solutions, which encourage the epidermal stem cells to divide, and a graft for the burn site is produced from this culture. It takes up to three weeks between the harvesting of the skin and growth of sufﬁcient new epidermis for grafting. An alternative procedure is to use skin donated from deceased donors, but this risks problems of rejection. In order to improve the outcome of the skin grafting treatment, studies have been conducted into the possibility of using embryonic stem cells to generate a ready source of skin cells for temporary grafts while patients are waiting for their autologous grafts. The process involves culturing embryonic stem cells on a framework made of collagen and the ﬁbroblast cells which form the underlying structure of tissues and are involved in healing. The stem cells are induced to develop into epidermal cells by maintaining them in a specialised medium that encourages cell differentiation. After several rounds of subculturing and replication, a layer of tissue develops which can be used to quickly provide a temporary graft. It is hoped that, in the future, these cells can be stimulated to generate a graft containing the dermis layer, where the nerve endings, hair follicles, oil glands and sweat glands are located.

©HERIOT-WATT UNIVERSITY 46 TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS

2. Bone marrow transplants A bone marrow transplant involves taking healthy stem cells from the bone marrow of one person and transferring them to the bone marrow of another person, or back into the donor at a later date. The cells removed are the haematopoietic stem cells from the red bone marrow, most often from the upper edge of the pelvis at the iliac crest. These are multipotent stem cells which can divide and differentiate into all of the various types of blood cell. The patient’s own bone marrow cells must be killed, either by radiation or drugs, then the donor stem cells are infused. If they are compatible with the patient, they will travel to the bone marrow and start to produce new blood cells. Bone marrow transplants are often used to treat conditions that damage bone marrow so that it is no longer able to produce blood cells. In cases where a bone marrow transplant is necessary to treat congenital defects or autoimmune diseases, the stem cells are transplanted from a donor. Examples of conditions treated in this way are:

• severe aplastic anaemia (bone marrow failure);

• certain genetic blood and immune system disorders, such as sickle cell anaemia, thalassaemia and some severe immune system diseases.

In the case of patients with cancer who require a bone marrow transplant, the patient’s own marrow is harvested and reintroduced into their body after radiation or chemotherapy has controlled the disease. Examples of conditions treated in this way are:

• some forms of leukaemia - cancer of the white blood cells;

• some non-Hodgkin’s lymphomas - cancer of the lymphatic system.

3. Corneal repair The cornea is the transparent front part of the eye that covers the iris and pupil. It consists of a layer of epithelial tissue called the corneal epithelium. If a person is to have excellent vision, the corneal epithelium must be transparent and not damaged in any way. The conjunctiva covers the white of the eye and is connected to the corneal epithelium by a group of tissues called the limbus. It is in the limbus that stem cells are found. If the limbus is damaged, then the person is unable to make corneal epithelial stem cells to replace any damaged area of the cornea. The cornea of the eye will be covered with blood vessels and eventually the person can lose his or her sight. Traditionally, the person would have to wait for donated corneal tissues to become available. In this treatment, the limbus of the person remains damaged and unable to produce new corneal epithelium. With the therapeutic use of stem cells, it is possible to isolate stem cells from the patient’s healthy eye and culture them in the laboratory. The new tissue is then transplanted back into the damaged eye where the new cells will be able to repair the damaged area of the corneal epithelium.

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Therapeutic uses of stem cells: Questions Q5: Explain why embryonic stem cells are potentially more useful as sources of tissue for skin grafts than epidermal stem cells. Go online ......

Q6: Why would the production of tissue for temporary skin grafts improve the outcome for burns patients awaiting a graft of their own tissue? ......

Q7: Why is it that blood stem cells are not suitable for other types of bone marrow transplant? ......

Q8: Why is it important that bone marrow used in transplants is of the same genetic tissue type as the patient? ......

Q9: Why is corneal repair using stem cells better than the traditional organ donor scheme for damaged corneas? ......

3.4 Ethical issues and the regulation of stem cell use Learning objective By the end of this section, you should be able to:

• explain that the ethical objections to the use of embryonic stem cells are principally that a zygote is a human being with a right to life, and that human beings in the form of embryos should not be artiﬁcially created;

• state that in the UK, stem cell research is regulated by The Human Tissue Authority and The Human Fertilisation and Embryology Authority. Ethics Tissue stem cells have been used in medicine for many years. It is the use of embryonic stem cells in research and their possible application in medicine that has raised ethical concerns. One of the main issues is the sourcing of embryonic stem cells. Until recently, these came from embryos donated for research because they were the unused embryos from in vitro fertilisation (IVF) programmes. As up to thirty eggs may be fertilised in the procedure, couples using IVF have four choices for the left-over embryos: have them destroyed, give them to other couples, store them for an annual fee, or donate them to research.

©HERIOT-WATT UNIVERSITY 48 TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS

Rather than have the surplus embryos simply destroyed, many parents prefer them to be used for research purposes. Storage allows for future use by the couple, but also puts off taking the decision about the unused embryos. The key ethical objections to the use of embryonic stem cells that have been raised are as follows:

• Human life should not be artiﬁcially created, an objection which also applies to in vitro fertilisation. This is the very strictest stance; others ﬁnd it acceptable to use IVF as long as the gametes used are from the parents to be, but the use of donor sperm would not be permissible.

• Life begins at fertilisation, so a fertilised egg (and, of course, blastocyst) is a human being with the same right to life as anyone else. Others argue that life begins at implantation, and here the debate shifts to the early termination of pregnancy by abortion.

Within the law, it is up to individuals to decide whether the rights of an embryo or foetus outweigh the potential beneﬁts from stem cell research and therapeutic work. The range of types of stem cell available for research today is wide, and, in some situations, tissue stem cells and induced pluripotent stem cells provide a preferable, less problematic, alternative to embryonic stem cells. Legal In the UK, there are very strict laws controlling the use of embryonic stem cells. All stem cell research work must be licensed by The Human Tissue Authority and The Human Fertilisation and Embryology Authority. The embryo must be less than the development stage of 14 days, and researchers must justify the creation of such an embryo for research work. Most recently, the law has changed to allow the creation of human embryos speciﬁcally for research work. These embryos must only be used for research and never for implantation into a womb. All work on therapeutic treatments with stem cells is restricted to tissue stem cells in the UK; the use of embryonic stem cells for therapeutic use remains a subject of great debate. Is using human admixed embryos by means of the nuclear transfer technique a possible solution? This is a relatively new process involving the creation of human admixed embryos. The process involves mixing human tissue stem cells with the embryonic stem cell of an animal. These could provide a source of embryonic stem cells without the need for human eggs.

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Admixed embryo production The following provides a summary of admixed embryo production. Go online

Human admixed embryos ......

Ethical issues and the regulation of stem cell use: Question Q10: List two arguments for the use of embryonic stem cells in research and two arguments against their use. Go online ......

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3.5 The biology of cancer cells Learning objective By the end of this section, you should be able to:

• state that, compared to normal cells, cancer cells divide uncontrollably to produce a mass of abnormal cells called a tumour;

• state that, compared to normal cells, cancer cells do not respond to normal regulatory signals which would instruct them to stop dividing, when necessary;

• explain that a tumour develops when mutations take place in the genes controlling cell division, programmed cell death, and differentiation;

• state that a benign tumour is one which does not invade neighbouring tissues or spread throughout the body;

• explain that a malignant tumour is one in which the cells lose the molecules on their surface that normally hold them in place so they can therefore become detached from their neighbours and spread through the body in the blood or lymphatic circulation;

• state that the original tumour is called the primary tumour and those forming from the detached cells are called secondary tumours. The term ’cancer’ refers to a broad group of over 200 diseases which are characterised by unregulated cell growth. The cancer cells divide excessively to produce a mass of cells called a tumour. They do not respond to regulatory signals and may fail to attach to each other. If these cells invade neighbouring tissues and spread to other parts of the body in the blood or lymphatic systems, they are classed as malignant tumours. The tumours which develop as a result of this spread are known as secondary tumours. If the tumour cells do not invade other tissues or spread, they are classed as benign, e.g. moles. So-called benign tumours, though not actually cancers, can of course cause very serious or fatal problems if they develop in areas where they put pressure on other tissues, nerves or blood vessels, e.g. in the brain. Causes of cancer The causes of cancer are only partially understood because they are very varied and complicated. An increased risk of cancer is associated with:

• ionising radiation (X-rays, gamma-rays, radioactive substances);

• UV radiation (sunlight, sunbeds);

• chemicals (tobacco smoke contains over 50 known carcinogens, asbestos, benzene);

• viruses (human papilloma virus, hepatitis B, hepatitis C);

• poor diet, physical inactivity and obesity;

©HERIOT-WATT UNIVERSITY TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS 51

• inheritance (breast cancer, ovarian cancer); • hormones (oestrogen and progesterone with breast and ovarian cancer, testosterone with prostate cancer).

Development of a tumour Cancers develop when the regulation of tissue growth fails as a result of the mutation of the genes which control cell division and differentiation. The genes involved are:

• Tumour suppressor genes, which produce proteins that slow down the cell division cycle and stimulate programmed cell death; • Proto-oncogenes, which promote cell growth and differentiation, and on mutation can give rise to tumour forming oncogenes.

The diagram shows the steps in the development of a tumour.

Steps in the development of a tumour

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Under the microscope, tumour cells show several differences from normal cells in that they have little cytoplasm, multiple nuclei, large nucleoli, and coarse chromosomal material. The most signiﬁcant differences between them lie in their behaviour:

• the cells divide uncontrollably to produce a mass of abnormal cells;

• they do not respond to normal regulatory signals which would instruct them to stop dividing when necessary;

• in a malignant tumour, they lose the molecules on their surface that normally hold them in place; they can therefore become detached from their neighbours and spread through the body to form secondary tumours.

The biology of cancer cells: Questions Q11: State two characteristics of cancer cells. Go online ......

Q12: Explain the steps by which a malignant tumour produces secondary tumours......

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3.6 Learning points

Summary Stem cells:

• Stem cells are useful for research because they are capable of giving rise to a range of cell types and can be cultured indeﬁnitely in the laboratory.

• Stem cell research has contributed to our understanding of cell processes such as growth, differentiation and gene regulation.

• Stem cells have therapeutic uses because they are pluripotent and can be stimulated in culture to form a wide variety of tissues.

• Stem cells can be used in the repair of damaged or diseased organs and tissues.

• In corneal repair, the use of corneal stem cells from the patient’s own tissue speeds up the process, creates new healthy tissue, and reduces the chance of rejection.

• In skin grafting, the use of epidermal stem cells allows the much more rapid production of a temporary graft, which reduces the risk of dehydration or infection.

• Stem cells cultured from patients suffering from particular conditions can be used as models to study disease development, e.g. Parkinson’s disease, motor neuron disease.

• Stem cells cultured from patients suffering from particular conditions can be used to produce very speciﬁc cultures for use in testing a wide range of drugs, e.g. Alzheimer’s disease.

• The principal ethical issues associated with stem cell use in research apply to embryonic stem cell use.

• The ethical objections to the use of embryonic stem cells are principally that a zygote is a human being with a right to life and that human beings in the form of embryos should not be artiﬁcially created.

• In the UK, stem cell research is regulated by The Human Tissue Authority and The Human Fertilisation and Embryology Authority.

Cancer cells:

• Compared to normal cells, cancer cells:

◦ divide uncontrollably to produce a mass of abnormal cells called a tumour; ◦ do not respond to normal regulatory signals which would instruct them to stop dividing, when necessary.

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Summary continued • A tumour develops when mutations take place in the genes controlling cell division, programmed cell death and differentiation.

• A benign tumour is one which does not invade neighbouring tissues or spread throughout the body.

• A malignant tumour is one in which the cells lose the molecules on their surface that normally hold them in place; they can therefore become detached from their neighbours and spread through the body in the blood or lymphatic circulation.

• The original tumour is called the primary tumour and those forming from the detached cells are called secondary tumours.

3.7 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of stem cell research work before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: Stem cell research work Describe the way in which the use of stem cells has contributed to: 20 min A) corneal transplants; (3 marks)

B) skin grafts. (3 marks)

©HERIOT-WATT UNIVERSITY TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS 55

......

3.8 End of topic test End of Topic 3 test Q13: Complete the sentences by matching the parts on the left with the parts on the right. (10 marks) Go online

One reason stem cells are useful motor neuron disease. for research is that

Cells which can differentiate into all reducing the chance of rejection. cell types are

Stem cell research has contributed reducing the chance of dehydration. to understanding

The use of stem cells has improved Alzheimer’s disease. corneal repair by

The use of stem cells has improved they can be cultured indeﬁnitely. skin graft procedures by

The development of this can be The Human Tissue Authority. studied using stem cells

The use of stem cell cultures to test a wide range of drugs helps embryonic stem cells. treatment of

Ethical objections have been raised a zygote is a human being. against the use of

An ethical objection against the use pluripotent. of stem cells is that

This regulates cell research in the cell growth and differentiation. UK ......

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Q14: Complete the sentences by matching the parts on the left with the parts on the right. (8 marks)

Unlike normal cells, cancer cells mutations.

Cancer cells do not respond to a mass of abnormal cells.

A tumour is divide uncontrollably.

its cells lose the molecules that Tumours develop as a result of normally hold them in place.

The genes involved in tumour normal regulatory signals. development control cause

slower cell division and programmed A tumour is benign if cell death.

cells detach from a primary tumour A tumour becomes malignant if and spread through the body.

it does not invade neighbouring Secondary tumours develop when tissues......

Q15: List two reasons that stem cells are useful for research. (2 marks) ......

Q16: To which cell processes has stem cell research contributed most? (2 marks) ......

Q17: Describe the improvement that the use of stem cells has made to the therapeutic procedures of:

i. corneal repair; ii. skin grafts.

(2 marks) ......

Q18: Name two diseases, the development of which can be studied using stem cell lines. (1 mark) ......

Q19: How can stem cell cultures contribute to the treatment of Alzheimer’s disease? (1 mark) ......

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Q20: The scientiﬁc use of which stem cells has given rise to ethical objections? (1 mark) ......

Q21: State two of the ethical objections to the use of stem cells. (2 marks) ......

Q22: Name the authorities which regulate stem cell research in the UK. (2 marks) ......

Q23: State the two key differences in behaviour between cancer cells and normal cells. (2 marks) ......

Q24: What causes a tumour to develop? (1 mark) ......

Q25: What would cause a tumour to be classed as benign? (1 mark) ......

Q26: How does a tumour become malignant? (1 mark) ......

Q27: How do secondary tumours develop? (2 marks) ......

©HERIOT-WATT UNIVERSITY 58 TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS

©HERIOT-WATT UNIVERSITY 59

Topic 4

DNA structure and replication

Contents

4.1 Introduction ...... 60 4.2 DNA structure ...... 61 4.3 Arrangement of DNA in chromosomes ...... 68 4.4 DNA replication ...... 70 4.5 Learning points ...... 74 4.6 Extended response question ...... 75 4.7 End of topic test ...... 76

Learning objectives By the end of this topic, you should be able to:

• explain the role that DNA plays in the inheritance of characteristics;

• state that DNA directs its own replication;

• describe the structure of a DNA molecule;

• describe the arrangement of DNA molecules in chromosomes;

• explain the process of DNA replication. 60 TOPIC 4. DNA STRUCTURE AND REPLICATION

4.1 Introduction Learning objective By the end of this section, you should be able to:

• state that DNA is the molecule of inheritance and can direct its own replication;

• explain that all cells store their genetic information in the base sequence of DNA;

• state that the genotype is determined by the sequence of DNA bases;

• state that DNA is found on chromosomes. Proteins are essential to the functioning of all biological systems. They perform a vast array of functions. As enzymes and hormones, proteins are involved in controlling cell activities, e.g. catalase and growth hormone. Some are structural proteins, e.g. collagen in cartilage and bone, keratin in hair, feathers and scales, and elastin in artery walls. Yet others are involved in cell signalling and transport. Each protein is composed of amino acids joined together by peptide bonds into a unique sequence. The type and order of the amino acids give the protein its characteristics. They are determined by the deoxyribonucleic acid (DNA) of the gene which codes for that protein. Speciﬁcally, it is the sequence of the bases in the DNA which stores the genetic information, the genotype and the genome, of the organism. DNA, or its close relative RNA (ribonucleic acid), is found in all life-forms on Earth, and wherever it occurs, its structure and function are fundamentally the same. Not only is DNA the molecule of inheritance, but, by the fact that it controls the production of proteins, it also directs its own replication.

Introduction: Questions Q1: Of what are proteins made? Go online ......

Q2: List three examples of proteins and state their general functions......

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4.2 DNA structure Learning objective By the end of this section, you should be able to:

• state that the basic unit of DNA is the nucleotide;

• explain that a DNA nucleotide is composed of deoxyribose sugar, phosphate and an organic nitrogenous base;

• state that the 5 carbon atoms of the deoxyribose molecule are numbered (1 = one prime) according to their position in the molecule;

• state that nucleotides are joined by strong covalent bonds to form a sugar- phosphate backbone;

• state that there are four bases, adenine (A), thymine (T), guanine (G) and cytosine (C);

• state that these bases join to the equivalent base on the opposite strand by weak hydrogen bonds;

• explain that each base can only bond to one other of the four to form a complementary pair;

• state that the complementary pairs are A-T and G-C;

• explain that the complementary base pairs hold the two strands by hydrogen bonds forming a double helix;

• state that DNA is a double-stranded, antiparallel structure, with the strands running in opposite directions;

• explain that the direction of the strand is identiﬁed by the carbon atoms of the deoxyribose which are found at the end of the strand;

• state that one strand goes from 3 to 5 the other from 5 to 3;

• explain that deoxyribose is found at the 3 end of the DNA strand and phosphate is found at the 5 end of the DNA strand. Chromosomes are thread-like structures found in the nucleus of the cell; they contain DNA, which is tightly wound around alkaline proteins called histones. On the following page is the famous ’Photograph 51’, an X-ray crystallography image which was taken in 1952 by Raymond Gosling, under the supervision of Rosalind Franklin. It was she who concluded, from measurements of the structures in the photograph, that the DNA molecule was a helix. Based on this, James Watson and Francis Crick developed their double helix model in 1953, which was further reﬁned by Maurice Wilkins. This work was all conducted in labs in Cambridge, and Crick, Watson and Wilkins were awarded the Nobel Prize in 1962 for it; Franklin died in 1958 and so was not eligible.

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Photograph 51 Watson and Crick’s model of the DNA molecule comprises two strands of repeating units, called nucleotides, hydrogen-bonded together to form a double helical structure. These two strands run in opposite, or antiparallel, directions.

Interactive DNA structure An activity that shows the double helix structure of DNA as an interactive 3D model is Go online available in the online materials at this point. The following illustration gives an idea of what to expect.

The double helix structure of DNA ......

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Nucleotide structure Each nucleotide consists of the following three parts.

• A central 5-carbon sugar; deoxyribose in the case of DNA and ribose in the case of RNA. These carbons are numbered 1 to 5, with four of them being arranged in a ring that is completed by an oxygen atom, and the ﬁfth standing off to the side. In the diagram below, the carbon atoms are represented by the numbers 1 (one prime) to 5, and the other elements by their symbols.

A central 5-carbon sugar If there was an OH attached at carbon 2, rather than one of the hydrogens, then the sugar would be ribose.

• A phosphate group attached to carbon 5.

• An organic nitrogenous base attached to carbon 1, of which there are four: adenine, thymine, guanine and cytosine.

The components of a nucleotide

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Nucleotide bases A strong chemical (covalent) bond forms between the phosphate group of one nucleotide and carbon 3 of the deoxyribose sugar of another. These bonds are not easily broken and join neighbouring nucleotides into a permanent strand with a sugar-phosphate backbone. The DNA molecule comprises two of these strands, joined together by the weak hydrogen bonds that form between their bases. These hydrogen bonds can easily be broken so that the strands can be separated when necessary. Each type of base can only bond with one other base: adenine (A) always bonds with thymine (T), by two hydrogen bonds; guanine (G) always bonds with cytosine (C), by three hydrogen bonds. A-T and G-C are called complementary base pairs, and the bases A and T, and G and C, are complementary to each other.

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Part of a double-stranded DNA molecule The resultant double-stranded molecule is DNA, and its two strands are arranged in a twisted coil called a double helix. It is like a spiral ladder in which the sugar-phosphate molecules form the upright parts and the base pairs form the rungs. If the DNA in a single human diploid cell was completely unravelled, it could extend to about 1.8 metres. The following diagram provides a more detailed view of the chemical structure of DNA. Notice that each set of base pairs have a total of three ’rings’ so are a constant width, but that the numbers of hydrogen bonds between the base pairs are different.

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The chemical structure of DNA Notice also that the two strands are running in opposite, or antiparallel, directions. The ends of the sugar-phosphate backbones are different. On the left, the strand ends with the phosphate group on the 5 carbon of the deoxyribose, whereas on the right, the strand ends with the 3 carbon of the sugar. These are referred to as the 5 (ﬁve-prime) and 3 (three-prime) ends of the strands. Here are two fascinating facts.

1. DNA is not restricted to the nucleus. It is also found in the mitochondrion (and the chloroplast in plants), reﬂecting the ancient incorporation of these organelles as bacteria into the eukaryote cell. The DNA which controls the mitochondria is split between the nucleus and the mitochondria, and it is thought that the genes now found in the nucleus have ’migrated’ there from the mitochondria. We all inherit our mitochondria from our mothers in the cytoplasm of the egg.

2. ATP (adenosine tri-phosphate) is very similar to a nucleotide. This ubiquitous co-enzyme energy carrier comprises a ribose sugar with an adenine base attached to carbon 1, and a chain of three phosphate groups attached to carbon 5.

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DNA structure: Questions Q3: What shape is the DNA molecule? ...... Go online

Q4: What type of bonding holds the two strands together in DNA? ......

Q5: What feature of these bonds allows the two strands of DNA to be easily separated? ......

Q6: Complete the table by matching each DNA base with its complementary base.

DNA Base Complementary base Adenine Cytosine Guanine Thymine ......

Q7: Which base pair is more easily separated and why? ......

Q8: State the components of an RNA nucleotide......

Q9: What term describes the opposite directions of the two strands of DNA? ......

Q10: What are found at the 3 and 5 ends of the DNA molecule? ......

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4.3 Arrangement of DNA in chromosomes Learning objective By the end of this section, you should be able to:

• explain that chromosomes are made up of DNA;

• explain that chromosomes consist of tightly coiled DNA, packaged with associated proteins. Chromosomes tend to be illustrated as condensed and elongated structures during the process of cell division. However, when cells are not undergoing division, chromosomes do not look like this at all. For example, in nerve cells, which do not divide, the chromosomes never assume a condensed shape. The main function of DNA is to carry the genetic information which determines the proteins that make up the human body. Each human cell contains nearly 2 metres of DNA, and this has to be packaged into the nucleus which is about 6 μm(micrometre) in diameter. This is like packaging about 50 km of ﬁne threads into a tennis ball. The only way that this can be achieved is for specialised proteins to bind and fold the DNA, forming a series of loops and coils, but at the same time preventing the DNA from becoming all tangled together. Although DNA is compact in the cell, it still has to allow for cell replication, repair, and for the functioning of genes. The DNA in the human body is split between different chromosomes. The human genome contains about 3.2 × 10 7 nucleotides; these are distributed between 24 different chromosomes. Each chromosome has a single linear DNA molecule with the proteins that fold and pack the strands of DNA into a compact structure. The materials from which chromosomes are made are chromatin, histones, and non-histone proteins. The packaging of DNA into a chromosome can be divided into four stages:

1. Nucleosomes, or ’beads on a string’, are the basic unit of DNA packaging, consisting of a segment of DNA wound around histone proteins. This stage is often compared to a thread wrapped around a spool.

2. Formation of chromatin ﬁbre. The nucleosomes are coiled and stacked.

3. Looped ﬁbres. Chromatin ﬁbres are looped along the protein scaffold.

4. Compacted chromosome. More folds and coiling of coils result in the most compacted chromosome. The most condensed form of DNA is packaged during metaphase in cell division.

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Arrangement of DNA from double helix to chromosome

Arrangement of DNA in chromosomes: Questions Q11: By what means is so much DNA packed into the cell nucleus? ...... Go online

Q12: What forms the scaffold around which DNA is arranged? ......

Q13: A human somatic cell contains 46 chromosomes, but the genome is distributed between 24 different chromosomes. Explain this apparent contradiction......

©HERIOT-WATT UNIVERSITY 70 TOPIC 4. DNA STRUCTURE AND REPLICATION

4.4 DNA replication Learning objective By the end of this section, you should be able to:

• state that DNA is unwound and unzipped by an enzyme (helicase);

• explain that this forms two template strands;

• state that DNA polymerase is the enzyme which adds nucleotides to the new DNA strand;

• explain that DNA polymerase needs a primer to start replication;

• state that DNA polymerase can only add complementary nucleotides to the deoxyribose / 3 end of the DNA strand;

• explain that this results in 3 strand / the leading strand being continuously replicated;

• state that the 5 strand / the lagging strand is replicated in fragments;

• state that these fragments are joined together by the enzyme ligase. The two processes that require new cells to be formed in the body are growth and repair. The replication of DNA allows cells to pass this information to the new cells. In order for an organism to function in a controlled and organised way, the information in the nucleus of all cells must be the same. For this to be possible, the DNA replication process (and the cell division process) must take place in a highly structured manner which virtually eliminates mistakes. Occasional errors do occur in the process of DNA replication, but there are systems in place to check for and correct them. As each daughter molecule contains one of the strands from the original DNA molecule, it is referred to as being ’semi-conservative’.

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General diagram of DNA replication

©HERIOT-WATT UNIVERSITY 72 TOPIC 4. DNA STRUCTURE AND REPLICATION

Summary of DNA replication The stages of DNA replication are summarised below. This account includes rather more detail than is strictly required for the syllabus, but it is given to provide a more complete, and hopefully comprehensible, picture of the process.

1. The double helix of the DNA molecule is unwound by the enzyme topoisomerase.

2. The two strands are separated by the enzyme helicase, and binding proteins stabilise the single strands of DNA.

3. Two replicating forks form, so named because of the forked shape of the unzipped DNA strands. The exposed bases of these strands form the templates for the construction of new complementary DNA strands.

4. Because of the antiparallel arrangement of the strands, one of the separated strands ends with an exposed deoxyribose (3) end; this is designated the leading strand. The other ends with a 5 end; this is designated the lagging strand.

Detailed model of DNA replication

5. As DNA polymerase can only attach complementary bases to the deoxyribose (3) end of a DNA strand, primers must be made. These are RNA (ribonucleic acid) nucleotides, which are attached by the enzyme DNA primase to exposed bases at the end of the leading DNA strand and at various points along the lagging DNA strand. They provide the necessary 3 starting points for the enzyme.

6. DNA polymerase replaces DNA primase, begins to attach nucleotides to the primers, and so synthesises a new strand starting from the 3 end. This is continuous on the leading strand, but discontinuous on the lagging strand as it has to be read in reverse order.

7. The gaps in the sugar phosphate backbone of the new strands are then closed by the enzyme ligase. The result is two identical double strands of DNA.

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DNA replication: Question Q14: Complete the sentences by matching the parts on the left with the parts on the right. Go online

DNA is unwound and the enzyme helicase. unzipped to form

DNA is unzipped by ligase.

At the end of the 3 two template strands. strand is

At the end of the 5 leading 3 strand. strand is

To start attaching nucleotides, DNA deoxyribose. polymerase requires

DNA polymerase attaches nucleotides to lagging 5 strand. the

The strand that is continuously replicated is ATP and free nucleotides. the

The strand replicated in fragments is the 3 end of the DNA strand.

The enzyme which joins a primer. DNA fragment is

In addition to a DNA molecule and enzymes, phosphate. DNA replication needs ......

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4.5 Learning points

Summary DNA

• DNA is the molecule of inheritance and can direct its own replication.

• All cells store their genetic information in the base sequence of DNA.

• The genotype is determined by the sequence of DNA bases.

• DNA is a component of chromosomes.

• Chromosomes consist of tightly coiled DNA, packaged with associated proteins.

Structure of DNA

• The basic unit of DNA is the nucleotide.

• A DNA nucleotide is composed of deoxyribose sugar, phosphate and an organic nitrogenous base.

• The 5 carbon atoms of the deoxyribose molecule are numbered (e.g. 1 = one prime) according to their position in the molecule.

• Nucleotides are joined by strong covalent bonds to form a sugar-phosphate backbone.

• There are four bases: adenine (A), thymine (T), guanine (G) and cytosine (C).

• These bases join to the equivalent base on the opposite strand by weak hydrogen bonds.

• Each base can only bond to one other of the four to form a complementary pair.

• The complementary pairs are A-T and G-C.

• The complementary base pairs hold the two strands of the DNA molecule by hydrogen bonds, forming a double helix.

• DNA is a double-stranded, antiparallel structure with the strands running in opposite directions.

• The direction of the strand is identiﬁed by the carbon atoms of the deoxyribose which are found at the end of the strand.

• One strand goes from 3 to 5 the other from 5 to 3.

• Deoxyribose is found at the 3 end of the DNA strand and phosphate is found at the 5 end of the DNA strand.

DNA replication

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Summary continued • DNA is unwound and unzipped by an enzyme (helicase).

• This forms two template strands.

• DNA polymerase is the enzyme which adds nucleotides to the new DNA strand.

• DNA polymerase needs a primer to start replication.

• DNA polymerase can only add complementary nucleotides to the deoxyribose / 3 end of the DNA strand.

• This results in the 3 strand / the leading strand being continuously replicated.

• The 5 strand / the lagging strand is replicated in fragments.

• These fragments are joined together (by the enzyme ligase).

4.6 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of the location and structure of DNA, and DNA replication, before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: The location and structure of DNA, and DNA replication A) Describe the location and structure of DNA. (8 marks)

B) Give an account of the replication of DNA. (7 marks)

......

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4.7 End of topic test End of Topic 4 test Q15: Complete the sentences by matching the parts on the left with the parts on the Go online right. (10 marks)

DNA is the molecule of inheritance held together by hydrogen bonds. and is

The base sequence of DNA is able to direct its own replication.

DNA is stored on chromosomes and adenine, cytosine, guanine and is thymine.

used by cells to store genetic The basic unit of DNA is information.

identiﬁed by their position in the A nucleotide is molecule.

The carbons in deoxyribose are tightly coiled with associated proteins.

The sugar-phosphate backbone is the nucleotide.

adenine and thymine, guanine and The bases of DNA are cytosine.

composed of deoxyribose, phosphate The base pairings of DNA are and a base.

held together by strong covalent The two strands of DNA are bonds.

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......

Q16: Complete the table using the words and phrases from the list. (8 marks)

The shape of the DNA molecule is a The two strands of DNA are At the 3 and 5 ends of the DNA strands are The enzyme that adds nucleotides to the new DNA strand is Before DNA replication can start, what is required is a Complementary nucleotides are added to the 3 end of the The DNA strand which is continuously replicated is the The enzyme that joins fragments of DNA together is

Word/phrase list: antiparallel, deoxyribose and phosphate, DNA polymerase, double helix, ligase, new DNA strand, primer, the 3 or leading strand......

Q17: Explain the signiﬁcance of DNA. (1 mark) ......

Q18: How do cells store their genetic information? (1 mark) ......

Q19: How is DNA stored in chromosomes? (1 mark) ......

Q20: Name the basic unit of DNA and describe its composition. (2 marks) ......

Q21: How are the carbon atoms on deoxyribose identiﬁed? (1 mark) ......

Q22: How is the sugar-phosphate backbone of DNA held together? (1 mark) ......

Q23: Name the nitrogenous bases found in DNA and list the complementary pairs they form. (2 marks) ......

Q24: How are the two strands of DNA held together? (1 mark) ......

Q25: Why are the two strands of DNA relatively easy to separate? (1 mark) ......

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Q26: What shape is the DNA molecule? (1 mark) ......

Q27: What term refers to the fact that the strands of the DNA molecule run in opposite directions? (1 mark) ......

Q28: What is found at the 3 and 5 ends of the DNA strand? (1 mark) ......

Q29: How is the DNA molecule formed into two template strands? (1 mark) ......

Q30: Name the enzyme which adds nucleotides to the new DNA strand. (1 mark) ......

Q31: What is the function of a primer? (1 mark) ......

Q32: To where on a DNA strand can complementary nucleotides be added? (1 mark) ......

Q33: Which DNA strand is continuously replicated? (1 mark) ......

Q34: How are the fragments of DNA replicated from the other strand joined together? (1 mark) ......

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Topic 5

Gene expression in human cells

Contents

Prerequisite knowledge You will remember from previous studies that DNA codes for proteins in the cell. The phenotype of a person is the characteristics, or traits, of that person, and is determined by their genome, which is the entire genetic structure of the individual. This is contained in the nucleus of each cell. The replication of DNA allows cells to pass this information to the new cells. Learning objectives By the end of this topic, you should be able to:

• explain the link between the genome and the phenotype of an individual;

• describe the structure and the functions of RNA;

• describe the transcription of DNA into messenger RNA (mRNA);

• describe the translation of mRNA into polypeptides and proteins;

• explain the fact that a single gene can give rise to several different proteins. 80 TOPIC 5. GENE EXPRESSION IN HUMAN CELLS

5.1 Introduction The many thousands of proteins that our cells use are synthesised inside the cells by a complex process involving the nucleic acids DNA and RNA, and structures called ribosomes which are mainly found on the rough endoplasmic reticulum. In addition, free nucleotides, amino acids, many enzymes and ATP are required. You will remember from previous studies that a protein is composed of amino acids joined together in a specific sequence. The information that determines this sequence is contained in the DNA in the nucleus. In this topic, we will study how the instructions for making a protein are transferred into the cytoplasm using a form of RNA, and how a polypeptide is first constructed on the ribosomes and then modified to form the final protein. We will also see that the old maxim of ’one gene - one enzyme’ has had to be modified to ’one gene - many enzymes’ as we have discovered more about the mechanism of converting the DNA code into protein. Gene expression involves two major stages. The first process is transcription, in which the DNA is copied to produce an RNA molecule called messenger RNA (mRNA). Although there are some differences, this mRNA carries the same code in its base sequence as the DNA of the original gene. This takes place in the nucleus. The second stage is known as translation because the code of the base sequence of the nucleic acid is used to produce a sequence of amino acids in the form of a polypeptide chain. This takes place on the ribosomes, which are mainly located on the rough endoplasmic reticulum. The polypeptide is then modified in various ways, e.g. folding or adding phosphate, to produce the final protein. An analogy with writing might help you to remember these terms. If you write down a sentence from a computer screen, you have taken information in one form and put the same information into a different form: you have written it across, or trans-scribed it. If you then turn the sentence from English into Gaelic, you have taken the meaning of the information and constructed something new: you have carried the meaning across, or trans-lated it.

DNA makes mRNA makes protein Process: transcription translation ribosomes in Occurs in: nucleus cytoplasm

Protein synthesis summary

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5.2 Gene expression through protein synthesis Learning objective At the end of this section, you should be able to:

• state that genes are expressed by using their DNA code to produce a protein;

• state that phenotype is determined by the proteins produced by gene expression;

• state that in any one cell, only a fraction of the genome is expressed;

• explain how gene expression is inﬂuenced by intracellular factors, e.g. phosphofructokinase activated by high levels of ADP;

• explain how gene expression is inﬂuenced by extracellular factors, e.g. low blood sugar levels activating the alpha cells of the islets of Langerhans in the pancreas to produce glucagon;

• state that gene expression has two major stages, transcription and translation;

• explain that transcription takes place in the nucleus and is the transfer of the genetic code from the DNA of the gene to mRNA;

• state that translation takes place at the ribosome in the cytoplasm and is the formation of a polypeptide from the code carried on the mRNA;

• describe how gene expression may be controlled at any of the many steps of both transcription and translation;

• state that the code of the gene is a series of triplets of bases in DNA;

• explain that there are 64 possible combinations of three bases, hence 64 different triplets. Chromosomes carry the genetic information of the cell in the form of genes. Gene expression is the process by which the information stored in a gene is used to synthesise a product, which is almost always a protein. Each gene has a speciﬁc base sequence which is hundreds or thousands of nucleotides long. This sequence codes for the linear order of amino acids in a polypeptide chain. Proteins are made up of one or more polypeptide chains. The nucleus of every cell carries the entire genome of the individual, but only a proportion of these genes will be expressed in any one cell. All cells will have the genes controlling glycolysis, but only a few will activate the gene to produce amylase. Inactivation at this level takes place during differentiation. Gene expression may also be switched on and off temporarily, by both extra- and intracellular inﬂuences. There are many ways in which this is achieved, including regulating the production or activity of an enzyme. The regulation of blood sugar levels is an example of extracellular regulation. The detection of low blood glucose levels by the alpha cells of the islets of Langerhans, in the pancreas, causes the gene for the production of the peptide hormone glucagon to be

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’switched on’ so that the hormone is manufactured and released into the blood stream. The glucagon is detected when it attaches to glucagon receptors, present on the cell membrane of liver and muscle cells, where it triggers the activation of the enzymes that break down glycogen to glucose. At the intracellular level, the enzyme phosphofructokinase, which controls a key step in the sequence of reactions that is glycolysis, is activated by increasing levels of ADP. As ADP is formed when ATP is broken down, its presence at high levels indicates a low level of available energy within the cell in the form of ATP. Although it is convenient to refer to genes being switched ’on’ or ’off’, as later sections in this topic will show, the process of gene expression is much more complex than this simple picture suggests. Transcription and translation involve several steps, each depending on the action of several enzymes, all of which may be used to regulate the process of turning the triplet code of the DNA into the gene-product as a protein. The genetic code and protein synthesis Inherited characteristics are the result of many biochemical processes controlled by enzymes. In humans, for example, certain enzymes control the biochemical pathways which cause a particular eye colour to develop. Enzymes are one form of protein. Their exact molecular structure, shape and ability to carry out functions all depend on the sequence of their amino acids. This sequence, in turn, is determined by the sequence of bases in the organism’s DNA. In this way, DNA controls the structure of enzymes and, in so doing, determines the organism’s inherited characteristics. Remember that there are about 20 different amino acids. DNA contains only four different nucleotides which are deﬁned by their bases. If each were to specify a single amino acid, only four amino acids could be coded for. A two base code would give 16 (42) possible arrangements - still not enough to code for all 20 amino acids. The shortest unit that can code for all amino acids is a triplet code, which produces 64 (43) possibilities - more than enough!

Gene expression through protein synthesis: Questions Q1: A gene encodes the information for a speciﬁc protein.

Go online a) True b) False

......

Q2: The regulation of blood sugar levels by hormones is an example of which type of regulation of gene expression? ......

Q3: High levels of which chemical in the cell activates phosphofructokinase? ......

Q4: At what stages of gene expression may regulation of gene expression take place? ......

Q5: How many more possible combinations of bases does a triplet code provide than a code based on only two bases?

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......

5.3 Structure and functions of RNA Learning objective At the end of this section, you should be able to:

• state that ribonucleic acid (RNA) is a single strand of nucleic acid;

• describe the structure of RNA as comprising ribose sugar, a phosphate group and one of four bases;

• explain that RNA contains three of the bases in DNA (namely adenine, cytosine, and guanine), plus uracil, which replaces the thymine of DNA;

• state that RNA bases pair A-U and G-C, both with DNA and other RNA molecules;

• state that there are three forms of RNA;

• state that messenger RNA (mRNA) carries the code of the gene from the DNA of the nucleus out to a ribosome in the cytoplasm;

• state that transfer (tRNA) carries speciﬁc amino acids to the mRNA, attached to a ribosome;

• describe how base pairing between different parts of the mRNA strand folds the molecule, exposing three bases at one end (the anticodon) and an attachment point for an amino acid at the other end;

• state that ribosomal RNA (rRNA) reads the mRNA code in the ribosome; Ribonucleic acid (RNA) provides a bridge between DNA and protein synthesis. RNA consists of nucleotides with the generalised structure shown below.

An RNA nucleotide

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Important points to remember about RNA ribonucleotide structure:

• RNA nucleotides contain the sugar ribose as opposed to the deoxyribose of DNA, a small difference that signiﬁcantly alters the shape of the molecule;

• the RNA strand has A (adenine), U (uracil), G (guanine) and C (cytosine) as its bases;

• RNA has the nitrogenous base (U) uracil rather than its close relative (T) thymine (as found in DNA);

• RNA molecules are usually single-stranded.

There are three main types of RNA involved in protein synthesis:

• mRNA (messenger RNA), which transcribes the code from the DNA molecule and carries it out of the nucleus to the ribosomes in the cytoplasm where the proteins are synthesised;

• tRNA (transfer RNA), molecules of which are found in the cytoplasm, becoming attached to speciﬁc amino acids and bringing them to the ribosomes where they are joined to form a polypeptide;

• rRNA (ribosomal RNA), which catalyses the combination of mRNA and tRNA during translation.

For the synthesis of a particular protein, the particular sequence of bases on the DNA is ﬁrst transcribed into the complementary sequence of mRNA. This messenger Messenger RNA (mRNA) RNA can then carry the information for a particular protein through the nuclear envelope to the sites of protein synthesis (the ribosomes). This type of RNA is responsible for the transport and transfer of individual amino acids during protein synthesis. Amino acids are transported by speciﬁc tRNA molecules, which recognise the genetic code presented Transfer RNA (tRNA) by the mRNA. The 3 bases exposed at the bottom form the anticodon. This is the complementary base sequence to the base sequence on mRNA coding for a particular amino acid. This type of RNA is bound to structural proteins to form a Ribosomal RNA (rRNA) ribosome. The ribosome is used in the synthesis of proteins.

The three forms of RNA The triplet codes which deﬁne the proteins are carried in the DNA of the nucleus; this code is then carried in the matching codons of the mRNA to the ribosomes in the cytoplasm. Just as living cells carry their inherited information in the form of nucleic acid

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(usually DNA, occasionally RNA, e.g. polio virus), ribosomes are also found in all living cells. They are complex molecular machines which build proteins, and they comprise two major parts: a small subunit, which reads the mRNA, and a larger subunit, which constructs the amino acid chain of the polypeptide. Each of these units consists of rRNA and a variety of proteins.

Structure and functions of RNA: Questions Q6: Complete the diagram of the RNA nucleotide structure using the labels and structural components shown below. Go online

...... Q7: Complete the table using the properties from the list.

RNA DNA Structure Preferred form Number of types Present in Nitrogenous bases

Properties: >1; 1; adenine, thymine, guanine, cytosine; adenine, uracil, guanine, cytosine; double helix; not a double helix; double-stranded; single-stranded; the nucleus; the cytoplasm and the nucleus...... Q8: Name the three types of RNA found in the cell......

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Q9: What are main functional differences between mRNA and tRNA? ......

Q10: Nucleotides are the building blocks of:

a) DNA only b) RNA only c) both DNA and RNA d) neither DNA nor RNA

......

Q11: Name the two components of a ribosome......

......

5.4 Transcription of DNA into mRNA Learning objective At the end of this section, you should be able to:

• deﬁne transcription as the transfer of the gene information from the triplet code on the DNA template to mRNA;

• state that the DNA double helix is unwound and unzipped by the enzyme RNA polymerase;

• state that RNA nucleotides pair with the appropriate exposed bases of the DNA template;

• state that the base on each RNA nucleotide pairs with its complementary DNA base;

• state that RNA polymerase causes the RNA nucleotides to bind into a single strand;

• state that this strand of mRNA is the primary transcript;

• state that the primary transcript contains both protein-coding regions (exons) and non-coding regions (introns);

• explain that, after transcription, the introns are removed from the primary transcript and the exons are joined together;

• state that the joining of the exons is called RNA splicing;

• state that the ﬁnal form of the mRNA is the mature transcript.

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The nucleus of every cell contains one or more structures called nucleoli. Each nucleolus is composed of RNA, DNA and protein, but is not bounded by a membrane. Nucleoli are the site of the synthesis of RNA and other components of ribosomes. These components pass out of the nucleus through pores and are assembled in the cytoplasm. Polypeptides are made on the ribosomes in the cytoplasm. The instructions (or code) telling the ribosomes what proteins to make are contained in the DNA, which never leaves the nucleus. This code is carried from the nucleus to the ribosomes by a form of RNA called messenger RNA (mRNA). The DNA code is transcribed from one of the DNA strands using free RNA nucleotides present in the cell’s nucleus. Transcription is the ﬁrst step in protein synthesis. During DNA replication, the DNA double helix is unwound by the enzyme helicase as DNA polymerase in unable to do this. In mRNA synthesis, the DNA is unwound to expose the DNA template strand by the RNA polymerase enzyme itself. Free RNA nucleotides in the nucleus bind to complementary DNA nucleotides using the base pair rules shown in the following table.

DNA nucleotide RNA nucleotide A pairs with U T pairs with A G pairs with C C pairs with G

Base pairing rules The enzyme RNA polymerase then joins the nucleotides to each other by attaching the phosphate group of one nucleotide to the ribose of the next, and the strand separates from the DNA template. The RNA molecule produced during transcription of DNA is messenger RNA (mRNA) and is called the primary transcript. It contains both protein coding sections, called exons, and non-coding sections, called introns. As the synthesis of a new strand of mRNA is in the 5 to 3 direction, only the exposed DNA strand which runs in the 3 to 5 direction is used as a template to build the mRNA. This apparent oddity results from the fact that, in order to be recognised when it reaches the ribosome, the mRNA must start with a special cap which only attaches to the 5 end of the molecule (5 cap). This and other changes to the original mRNA strand, such as the removal of introns, are called post-transcriptional regulation, and they produce the mature transcript which leaves the nucleus for the ribosomes. The removal of introns and the joining-up of the exons are achieved by the process of RNA splicing.

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Transcription of DNA into mRNA: Steps The following provides a summary of nucleotide structure and the steps involved in the Go online process of RNA transcription. 10 min

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...... Most cells contain thousands of ribosomes in their cytoplasm. Ribosomes are important as the site of protein synthesis in the cell and contain enzymes essential for protein synthesis. Cells that have a lot of ribosomes synthesise more protein than those that have fewer ribosomes.

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Transcription of DNA into mRNA: Questions Q12: What process takes place on ribosomes? Go online ......

Q13: What are the functions of the enzymes helicase and RNA polymerase? ......

Q14: Which of the following best describes transcription?

a) DNA → DNA b) DNA → RNA c) DNA → protein d) RNA → DNA

......

Q15: The complementary messenger RNA strand that would be synthesized from the DNA base sequence of CTGAC would be:

a) GACTG b) UGACU c) AGTUC d) GACUG

......

Q16: Why is only one strand of DNA transcribed into mRNA? ......

Q17: State two changes which take place to the primary transcript to produce the mature transcript......

Q18: Name the process by which the mature transcript is produced......

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5.5 Translation of mRNA into a polypeptide Learning objective At the end of this section, you should be able to:

• deﬁne translation as the conversion of the gene information in the form of mRNA into a polypeptide at a ribosome;

• state that ribosomes comprise rRNA and proteins;

• state that mRNA carries the genetic code in the form of sets of three bases, called codons;

• state that each tRNA has three exposed bases, called an anticodon;

• state that each amino acid is carried by a different type of tRNA;

• state that, at the ribosome, rRNA reads the incoming mRNA;

• explain that one codon acts as start signal and others act as stop signals, indicating the start and ﬁnish of the codons to be translated;

• explain that, when the start codon is reached, the ribosome starts attaching tRNA bases to the mRNA strand;

• state that the anticodons of the tRNA molecules are matched to complementary codons on the mRNA;

• state that the enzymes in a ribosome join the amino acids together with peptide bonds;

• state that the tRNA then disengages from the amino acid and the ribosome, and goes off to collect another molecule of its speciﬁc amino acid;

• explain that, when the stop codon is reached, the ribosome disassociates and the chain of amino acids is released as a polypeptide. Once the DNA of a gene has been transcribed into mRNA, the mRNA molecules pass through the nuclear pores into the cytoplasm. Here, translation of mRNA into polypeptides takes place on ribosomes. This involves a third type of RNA, transfer RNA (tRNA). Each molecule of tRNA is a single-stranded sequence of between 70 and 90 nucleotides, which is folded into a clover-leaf shape as a result of base pairing, using hydrogen bonds, between different regions of the strand. At the 3 end of the strand is a site to which an amino acid attaches by a covalent bond, and at the opposite end of the molecule are three nucleotides with exposed bases.

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tRNA These exposed bases match up against the codons of the mRNA and are called anticodons. As it is possible for any of the four RNA bases (A, U, G and C) to be present on each of these nucleotides, there are 64 possible combinations of bases in the anticodon.

tRNA The following provides structural information about a tRNA molecule, in this case with Go online the amino acid serine attached to it. 15 min

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...... However, in the translation of the mRNA there must be codons which signal where translation should start (AUG, also methionine) and stop (UAG, UAA and UGA). It is changes to the DNA which produce such ’stop’ combinations that lead to the nonsense mutation mentioned in the next topic. This leaves far more combinations than are minimally necessary to code for the 20 common amino acids. As a result, several mRNA codons signify the same amino acid and several tRNA molecules with the corresponding anticodons will attach to it.

The genetic code table of amino acids and codon letters The process of translation begins when the small subunit of a ribosome binds to the 5 end of the mRNA strand. The small subunit then moves along the mRNA until it encounters the AUG start codon, at which point it is joined by the large subunit of the ribosome. A tRNA molecule with anticodon bases complementary to the ﬁrst codon of the mRNA transcript is then attached to the exposed bases by the large subunit. The subunits of the ribosome then move along the mRNA strand, attaching the appropriate tRNA molecules to the exposed codons. As they do so, enzymes in the large subunit form peptide bonds between the amino acids, creating the polypeptide molecule. The tRNA then releases its amino acid, disengages from the mRNA and goes off to recharge with another amino acid.

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When the stop codon is reached, rather than another tRNA attaching, release factors are produced which cause the ribosome subunits to disassociate, releasing the polypeptide.

Translation: Steps The following provides a summary of the steps involved in the translation of mRNA into Go online a protein. 15 min

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......

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Translation of mRNA into a polypeptide: Questions Q19: Select the correct answers from the options listed after each question. Go online 1. The triplet code in the nucleus is stored in: mRNA, tRNA, DNA or rRNA? 2. The nucleic acid which carries amino acids is: mRNA, tRNA, DNA or rRNA? 3. The cloverleaf shape of tRNA is caused by base-pairing using: covalent bonds, peptide bonds or hydrogen bonds? 4. At the 3 end of a charged tRNA is found the: codon, anticodon or amino acid? 5. The codon AUG indicates: start, stop or nonsense? 6. The mRNA is read by the: ribosome, small subunit or large subunit? 7. The type of nucleic acid found in the small subunit of the ribosome is: mRNA, tRNA, DNA or rRNA? 8. The bonds formed between the amino acids are: covalent bonds, peptide bonds or hydrogen bonds? 9. The stop codon causes: tRNA attachment, tRNA detachment, polypeptide release or ribosome subunits disassociation?

......

Q20: Try using the genetic code table of amino acids and codon letters to complete the sequence of each of the proteins by typing the appropriate amino acid name into each of the answer boxes

The genetic code table of amino acids and codon letters

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......

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5.6 Single gene, several proteins Learning objective At the end of this section, you should be able to:

• state that the human body contains many more types of protein and mRNA than there are genes in the DNA;

• state that one gene must be able to code for several proteins;

• state that different mRNA molecules can be produced from the same primary transcript;

• explain that this is achieved by different mRNA segments being treated as exons and introns;

• state that this is called alternative RNA splicing;

• state that proteins may be modiﬁed by the cutting and combining of polypeptide chains;

• state that proteins may be modiﬁed by adding phosphate or carbohydrate groups;

• state that these are examples of post-translational modiﬁcation of protein structure. Recent studies about the human genome have revealed that there are about 21,000 protein-coding genes. Yet, in the human body, the number of proteins are in excess of 25,000, and earlier estimates were that the human genome would comprise 100,000 genes. In fact, only about 1.5% of the genome represents protein-coding genes, the rest being associated with non-coding RNA (e.g. tRNA), the regulation of gene expression, and introns. There are also many sequences for which no function has yet been found. Therefore, within the human body, one gene must be able to encode many proteins, i.e. a variety of proteins can be expressed by the same gene. This is achieved by two processes: alternative RNA splicing and post-translational modiﬁcation (PTM). Alternative RNA splicing This process allows a variety of different proteins to be expressed from one gene. The human genome consists of exons and introns. These terms might seem a little counterintuitive, but exon is short for ’expressed region’ and intron for ’intragenic region’. The coding sections of a gene (the exons) can be split into several sections by non- coding sections (the introns). The name ’exon’ is short for ’expressed region’; ’intron’ is perhaps best remembered as the ’in-between’ sections. Splicing occurs after transcription when the introns are removed from the primary transcript. For many introns, splicing is done in a series of reactions which are catalysed by a complex of small, nuclear ribonucleoproteins (snRNPs). NB: snRNPs are not in the syllabus, but are included here because of their name, ’snurps’.

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Particular exons can either be included or excluded from the mature transcript. It is also possible that two splice sites are produced at one end of an exon, making it possible to produce multiple transcripts by ’alternative splicing’. The joining of different splice sites allows individual genes to express multiple mRNAs that encode proteins with diverse and even antagonistic functions. The process allows a single gene to express a variety of proteins. The advantage of the splicing process is that a single primary transcript can be spliced into different mRNA by the inclusion of different sets of exons. The process of alternative splicing explains how the 21,000 genes in the human genome are able to encode the 120,000 different translated mRNA molecules reported to exist in the human cells. In humans, it is estimated that alternative splicing occurs in more than 60% of genes.

Alternative RNA splicing In the example in the diagram above, when the introns are removed, exons 1 and 2 may either be spliced with exon 3 or with exon 4. Post-translational modification The chemical modification of a polypeptide after its translation is called post-translational modification (PTM). The process is a late stage of protein synthesis which also allows a gene to produce many different proteins. PTM of the polypeptide alters the function of the protein. This is achieved by either adding other biochemical groups, such as phosphate, lipids and carbohydrates, or cutting the polypeptide chain and reassembling it into a modified version. Also, the amino acid coded by the start codon, methionine, is usually removed. Effectively PTM will change the structural or functional nature of the protein. Two examples of PTM are as follows.

• A protease enzyme cuts the polypeptide chain in two places. The middle section of the protein is then removed and the structure re-formed. An example of this is pro-insulin→insulin formation. The succession of changes leading to the ﬁnal mature version of insulin begins in the membrane of the endoplasmic reticulum, continues in the Golgi apparatus, and is completed in the secretory vesicle, which will excrete the insulin from the cell.

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• Proteins are able to be phosphorylated with the addition of a phosphate group by a kinase enzyme. This is a way to control the activity of many enzymes and receptors. The addition of a phosphate group causes a change in the structure of the protein, allowing enzymes and receptors to be switched ’on’ or ’off’.

Another post-translational modiﬁcation is ’protein folding’. The correct three-dimensional structure is determined by interactions between the amino acids and is essential to the correct functioning of the protein.

Single gene, several proteins: Question

Q21: Complete the paragraph using the words from the list.

Go online One gene can be expressed as many as a result of RNA and post-translational . Different molecules are produced from the

same transcript depending on which RNA segments are treated as exons

and . Protein structure can be altered post-translation by:

cutting and combining chains, e.g. the formation of insulin from ; adding or groups to the protein. Word list: carbohydrate, introns, modiﬁcation, mRNA, phosphate, polypeptide, primary, pro-insulin, proteins, splicing......

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5.7 Learning points

Summary Gene expression

• Genes are expressed by using their DNA code to produce a protein.

• The phenotype is determined by the proteins produced by gene expression.

• In any one cell, only a fraction of the genome is expressed.

• Gene expression is inﬂuenced by intracellular factors, e.g. phosphofructokinase activated by high levels of ADP.

• Gene expression is inﬂuenced by extracellular factors, e.g. low blood sugar levels activating the alpha cells of the islets of Langerhans in the pancreas to produce glucagon.

• Gene expression has two major stages: transcription and translation.

• Transcription takes place in the nucleus and is the transfer of the genetic code from the DNA of the gene to mRNA.

• Translation takes place at the ribosome in the cytoplasm and is the formation of a polypeptide from the code carried on the mRNA.

• Gene expression may be controlled at any of the many steps of both transcription and translation.

• The code of the gene is a series of triplets of bases in DNA.

• There are 64 possible combination of three bases, so 64 different triplets.

Structure and functions of RNA

• Ribonucleic acid (RNA) is a single strand of nucleic acid.

• RNA comprises ribose sugar, a phosphate group and one of four bases.

• RNA contains 3 of the bases in DNA, namely adenine, guanine and cytosine, plus uracil which replaces the thymine of DNA.

• RNA bases pair A-U and G-C, both with DNA and other RNA molecules.

• There are three forms of RNA.

• Messenger RNA (mRNA) carries the code of the gene from the DNA of the nucleus out to a ribosome in the cytoplasm.

• Transfer (tRNA) carries speciﬁc amino acids to the mRNA attached to a ribosome.

• Base pairing between different parts of the mRNA strand fold the molecule, exposing three bases at one end (the anticodon) and an attachment point for an amino acid at the other end.

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Summary continued • Ribosomal RNA (rRNA) reads the mRNA code in the ribosome.

Transcription of DNA into mRNA

• Transcription is the transfer of the gene information from the triplet code on the DNA template to mRNA.

• The DNA double helix is unwound and unzipped by the enzyme RNA polymerase.

• RNA nucleotides pair with the appropriate exposed bases of the DNA template.

• The base on each RNA nucleotide pairs with its complementary DNA base.

• RNA polymerase causes the RNA nucleotides to bind into a single strand.

• This strand of mRNA is the primary transcript.

• The primary transcript contains both protein-coding regions (exons) and non-coding regions (introns).

• After transcription, the introns are removed from the primary transcript and the exons are joined together.

• The joining of the exons is called RNA splicing.

• The ﬁnal form of the mRNA is the mature transcript.

Translation of mRNA into polypeptides and proteins

• Translation is the conversion of the gene information, in the form of mRNA, into a polypeptide at a ribosome.

• Ribosomes comprise rRNA and proteins.

• mRNA carries the genetic code in the form of sets of 3 bases called codons.

• Each tRNA has three exposed bases called an anticodon.

• Each amino acid is carried by a different type of tRNA.

• At the ribosome, rRNA reads the incoming mRNA.

• One codon acts as start signal and others act as stop signals, indicating the start and ﬁnish of the codons to be translated.

• When the start codon is reached, the ribosome starts attaching tRNA bases to the mRNA strand.

• The anticodons of the tRNA molecules are matched to complementary codons on the mRNA.

• Enzymes in ribosome join the amino acids by peptide bonds.

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Summary continued • The tRNA then disengages from the amino acid and the ribosome, and goes off to collect another molecule of its speciﬁc amino acid.

• When the stop codon is reached, the ribosome disassociates and the chain of amino acids is released as a polypeptide.

One gene, many proteins

• The human body contains many more types of protein and mRNA than there are genes in the DNA.

• One gene must be able to code for several proteins.

• Different mRNA molecules can be produced from the same primary transcript.

• This is achieved by different mRNA segments being treated as exons and introns.

• This is called alternative RNA splicing.

• Proteins may be modiﬁed by the cutting and combining of polypeptide chains.

• Proteins may be modiﬁed by adding phosphate or carbohydrate groups.

• These are examples of post-translational modiﬁcation (PTM) of protein structure.

5.8 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of transcription and post-translational modiﬁcation (PTM) before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: Transcription and post-translational modiﬁcation (PTM) Give an account of gene expression under the following headings.

A) Transcription (6 marks) B) Post-translational modiﬁcation (PTM) (4 marks)

......

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5.9 End of topic test End of Topic 5 test Q22: Complete the paragraph using the words from the list. (10 marks)

Go online The individual’s phenotype is determined by the proteins produced by gene .

These proteins are generated from the in the nucleus of each cell, but in any

one cell only a fraction of the is expressed. Intracellular factors can inﬂuence gene expression, e.g. is activated by high levels of ADP. factors

may also have an effect, e.g. low blood blood sugar levels activate the production of by the alpha cells of the pancreas. Gene expression has two stages,

in the nucleus and translation at the , both of which may be used to control the

process. The code of a gene consists of a series of of bases in the DNA. A

triplet of bases, of which there are types, can represent 64 different instructions. Word list: DNA, expression, extracellular, four, genome, phosphofructokinase, glucagon, ribosome, transcription, triplets......

Q23: Complete the table using the words and phrases from the list. (9 marks)

RNA: Phosphate, base, ... Adenine, cytosine, guanine, ... G-C, ... Messenger, transfer, ... mRNA: tRNA: rRNA: Anticodon:

Word/phrase list: 3 exposed bases on tRNA, A-U, amino acid to ribosome, anticodon, nucleus to cytoplasm, reads mRNA code, ribose sugar, ribosomal, single strand of nucleic acid, uracil.

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......

Q24: Complete the table by matching the parts on the left with the parts on the right. (9 marks)

Transcription: protein-coding regions of mRNA.

RNA base uracil: joining up of exons.

RNA polymerase: DNA triplet code to mRNA. mRNA codon: ﬁnal form of mRNA.

DNA unwound and unzipped/RNA Primary transcript: nucleotides join.

Exons: DNA base adenine pairs with it.

Introns: DNA triplet of bases matches with it.

RNA splicing: ﬁrst RNA strand formed from DNA.

Mature transcript: non-coding regions of mRNA......

Q25: Rearrange the following steps to describe the order in which translation takes place. (9 marks)

• Enzymes in ribosome join the amino acids by peptide bonds. • Translation takes place at the ribosome. • When a start codon is reached, the ribosome starts attaching tRNA bases to the mRNA. • tRNA disengages and goes off to ﬁnd another of its amino acids. • Ribosomes consist of rRNA and proteins. • Each tRNA molecule carries a speciﬁc amino acid. • At the ribosome, rRNA reads the incoming mRNA. • When a stop codon is reached, the ribosome diassociates and the polypeptide is released. • mRNA has sets of three bases called codons and tRNA has matching sets called anticodons.

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......

Q26: Complete the sentences by matching the parts on the left with the parts on the right. (6 marks)

The body contains many more exons and introns. proteins than there are

One gene must be able to code for several different mRNA molecules.

One primary transcript can produce post-translational modiﬁcation.

Alternative RNA splicing treats carbohydrate groups. different RNA segments as

Post-translational modiﬁcation may genes. involve adding

Cutting and combining of several proteins. polypeptides is another form of ......

Q27: By what means is phenotype linked to genotype? (1 mark) ......

Q28: Give an example of an intracellular factor inﬂuencing gene expression. (2 marks) ......

Q29: Describe the way in which low blood sugar levels act as an extracellular factor. (2 marks) ......

Q30: Explain why only 4 bases can give rise to 64 different codes for amino acids. (1 mark) ......

Q31: When RNA is being formed using DNA as a template, what are the three base pairings which can occur (letters are acceptable)? (1 mark) ......

Q32: Name the three forms of RNA involved in gene expression (letters are not acceptable except for RNA). (2 marks) ......

Q33: State the meaning of the term ’transcription’. (1 mark) ......

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Q34: How is the mature transcript produced from the primary transcript? (2 marks) ......

Q35: What is RNA splicing? (1 mark) ......

Q36: Describe the structure of tRNA. (2 marks) ......

Q37: Explain the fact that 20 amino acids are coded for by 64 possible combinations of bases in the anticodons of tRNA. (1 mark) ......

Q38: Explain what happens when the end of the mRNA section to be translated is reached? (1 mark) ......

Q39: Explain how different mRNA molecules can be produced from the same primary transcript. (1 mark) ......

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©HERIOT-WATT UNIVERSITY 111

Topic 6

Genes and proteins in health and disease

Contents

6.1 Introduction ...... 113 6.2 Proteins ...... 114 6.3 Different levels of protein structure ...... 115 6.4 Functions of proteins ...... 120 6.5 Mutations and genetic disorders ...... 122 6.6 Single gene mutations ...... 124 6.7 The effect of mutations on the structure and function of proteins ...... 135 6.8 Chromosome structure mutations ...... 137 6.9 Learning points ...... 140 6.10 Extended response question ...... 142 6.11 End of topic test ...... 143

Prerequisite knowledge You will have come across proteins before and covered some of the many roles they undertake in the human body. Proteins are essential in all biological systems because they control most of the work going on in the cell. They are also important components of cellular structures and, as enzymes and hormones, are involved in controlling cell activities. As described in the previous topic, proteins are speciﬁed by the DNA code of a gene; anything which changes that code will almost always cause changes to the protein it deﬁnes, with possibly fatal results. Learning objectives By the end of this topic, you should be able to:

• state that proteins show a large variety of structures and shapes, resulting in a wide range of functions; • state that amino acids combine to form polypeptide chains; • explain that these chains can fold into three-dimensional shapes; • state that genetic disorders are caused by changes to genes or chromosomes; 112 TOPIC 6. GENES AND PROTEINS IN HEALTH AND DISEASE

• state that single gene mutations involve the alteration of a DNA nucleotide sequence;

• explain that the alteration of a DNA nucleotide sequence is a result of the substitution, insertion or deletion of nucleotides;

• describe the altered protein structure and function in genetic disorders;

• understand the link between altered chromosomal structure and genetic disorders.

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6.1 Introduction Proteins are essential in all biological systems. They control most of the work going on in the cell and are an important component of cellular structures.

Protein groups In this topic we will explore how the great diversity in the structure and function of proteins arises. Any changes to the sequence of amino acids in the protein have the potential to cause very signiﬁcant changes to its structure and function. Such changes arise from mutations to DNA code which determines the amino acid sequence. In the second half of the topic, we will examine the range of mutations which can affect individual genes or whole chromosomes, and consider some examples of the medical conditions which result.

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6.2 Proteins Learning objective By the end of this section, you should be able to:

• describe the structure of proteins. Proteins are organic compounds which are made up of the elements carbon, hydrogen, oxygen, nitrogen, and sometimes other elements such as sulfur. They are complex molecules with three dimensional structures which are unique for each protein. Proteins are composed of building blocks called amino acids, of which twenty kinds are commonly found in natural proteins. From these twenty, the cell can synthesise thousands of different proteins. The amino acids are linked together by strong bonds, called peptide bonds, to form polypeptide chains. The polypeptide chains can be any size, several can be linked together, and they may have other groups such as carbohydrates, phosphates and metal ions bound into them.

Proteins: Questions Q1: List the elements found in all proteins. Go online ......

Q2: Name the basic building blocks of proteins and state how many common forms of them exist......

Q3: Name the type of strong bonds which link amino acids into chains......

Q4: What name is given to the chains of amino acids which make up proteins? ......

Q5: Explain the three sources of variation in structure and function of proteins......

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6.3 Different levels of protein structure Learning objective By the end of this section, you should be able to:

• state that function of an enzyme depends on its three-dimensional structure;

• explain that an area of the enzyme molecule is folded to form an ’active site’;

• state that the active site is exposed so that it readily combines with its substrate;

• state that proteins are organised at four different structural levels;

• explain that:

1. primary structure consists of the sequence of amino acids; 2. secondary structure involves the folding of the polypeptide chains, which are then held in place by weak hydrogen bonds; 3. tertiary structure involves further folding held in place by strong (disulphide) bonds between amino acids; 4. quaternary structure is formed when two or more polypeptide chains associate together to form the ﬁnal 3D protein. The function of a protein depends on its three-dimensional (3D) structure. Proteins are organised at four different structural levels.

Different levels of protein structure: Visualisations

Go online

Primary and secondary level protein structure

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Tertiary protein structure

Quaternary protein structure

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Mature proteins fall into two main groups as shown below.

Mature protein types ......

Structures of hexokinase and haemoglobin The tertiary structure for the protein hexokinase is shown. Hexokinase is an enzyme that phosphorylates (adds phosphate to) glucose molecules in the start of its oxidation. Go online You can see the substrate (glucose) at the active site of the upper enzyme. The function of an enzyme depends on its three-dimensional structure. One part of the molecule has been folded to form an ’active site’, to which the substrate molecules temporarily become attached. The active site is exposed so that it readily combines with its substrate(s).

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Structure of hexokinase

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An interactive molecular model of amylase, an enzyme that causes hydrolysis of polysaccharides to glucose molecules (presented in 3D), is available in the online materials at this point. Haemoglobin is an example of a protein with a quaternary structure. It is found in red blood cells and is responsible for carrying oxygen around the body. It is composed of four chains of the protein globin, with two each of alpha- and beta-globin. An interactive molecular model showing the structure of haemoglobin (presented in 3D) is available in the online materials at this point. A diagram of the structure is shown below.

The structure of haemoglobin ......

Different levels of protein structure: Question Q6: Complete the table using the phrases from the list. Go online

Level of structure Description Primary: Secondary: Tertiary: Quaternary:

Phrase list: held by strong (disulphide) bonds between amino acids, held by weak hydrogen bonds between amino acids, the sequence of amino acids, two or more polypeptide chains associated together......

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6.4 Functions of proteins Learning objective By the end of this section, you should be able to:

• list the functions of proteins as enzymes, some hormones, muscular contraction, transport, antibodies and structural support;

• give at least one example of each. Proteins are fundamental to life on earth; they are the other side of the coin to nucleic acids. Consequently, they are to be found involved in all the functions of the body, a few of which are mentioned below. The most ubiquitous of functions, that of enzyme, is covered in other topics. The table shows the variety of proteins found in human cells.

Protein group Example Function form part of cellular structural collagen, elastin structures allow muscle cells to contractile actin and myosin shorten involved in controlling hormones insulin blood sugar levels recognises insulin on the receptors insulin receptor cell membrane allows glucose to enter the transport proteins glucose transporter cell recognise molecules of defence proteins antibodies invading organisms cut, assemble, digest enzymes lipase, amylase molecules

Protein diversity Hormones Hormones are chemical messengers produced by special glands. They are secreted into the bloodstream and are carried to their site of action where they play an important role in regulating growth and metabolism. Many hormones, for example insulin and ADH (anti-diuretic hormone), are proteins. Glucagon, oxytocin and prolactin are also peptide hormones. Muscles Muscles are unique; their fibres are made up from a number of different proteins, for example actin and myosin, arranged in a very specific manner. The appearance of skeletal and cardiac muscle is characterised by the presence of a number of relatively light and dark bands across the filaments.

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Transport Many proteins transport substances around the body. Myoglobin and haemoglobin, for example, are proteins involved in the transport of oxygen. Other proteins found in cell membranes transport chemicals across the membrane. Antibodies Antibodies are proteins produced by certain white blood cells, called lymphocytes, and are very important in preventing and ﬁghting disease. This will be discussed in more detail in a later topic. Structural Bones, tendons, ligaments, skin and hair are all mainly composed of structural proteins such as collagen, elastin and keratin. Collagen is the most common protein in the body, contributing up to a third of the total protein content. It is found in all connective tissue, as well as being a major component of tendons, ligaments and cartilage. As its name implies, elastin forms ﬁbres which will contract to their original length after being stretched. It is found in artery walls, the lungs, the bladder, skin and the elastic cartilage of the nose and ears. Keratin is a key structural protein, making up the much of the outer layers of the skin as well as hair and nails. In the animal kingdom, it also is found in horns, feathers and scales.

Interactive antibody structure An interactivity showing the molecular model of an antibody (presented in 3D) is available in the online materials at this point. Go online ......

Functions of proteins: Questions Q7: Complete the table using the types of protein from the list. Go online 20 min Type of protein Example collagen, elastin actin and myosin in muscle cells insulin insulin receptor in liver cells (forming part of the structure of the plasma membrane) transporter of glucose into the cell immunoglobulins lipase, pepsin, maltase

Type of proteins: contractile, defence proteins, enzymes, hormones, receptors, structural, transport proteins.

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......

6.5 Mutations and genetic disorders Learning objective By the end of this section, you should be able to:

• deﬁne the term mutagenic agent (mutagens);

• list examples of mutagens. The occurrence of mutations A mutation is a change in genetic material. An individual gene expressing a mutation is referred to as a mutant. The tissue in which the mutation occurs is of great importance. A mutation in a somatic cell or a tissue stem cell will only affect that cell or its daughter cells; only if it affects the proto-oncogenes and leads to the development of a tumour is this likely to be serious. However, a mutation that occurs in the germline cells of an organism can be inherited from one generation to the next; if it is, it will be present in all the cells of the offspring. Mutation provides a source of variation in a species. Whereas the shuffling of genes during meiosis provides new combinations of existing alleles, mutation is the only source of new alleles and genes. Without mutation, evolution would not be possible. Given that only a very small percentage of the total DNA in the nucleus is protein-coding, most changes to the order of bases are unlikely to change a gene. However, changes to start and stop codons can have profound effects, as can alterations to the promoters which provide an attachment for RNA polymerase near a gene. Types of mutation There are different types of mutation: they might cause a change in the structure of a chromosome or alter a single base in a gene. The effect of such changes can range from none to being fatal. In humans, a huge range of inherited disorders are the result of single gene mutations. Despite this, the fact is that most mutations have no effect on survival. The majority of mutations in single genes are recessive. A recessive mutation in one allele of a gene cannot normally be observed in an individual because the mutated allele is masked by the original allele that is still present. Only if two individuals who are heterozygous for the mutant allele have offspring will there be a possibility (1 in 4) that an affected individual will be produced. Mutagenic agents A mutagen is an agent that increases the rate of mutation. In any individual, there is a low rate of spontaneous mutation caused, for example, by decay of the bonds in DNA or errors in replication but, as cells have very efficient checking systems for errors in DNA replication, any mistakes are normally very quickly rectified.

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A wide variety of chemicals (e.g. benzene, arsenic) and some types of radiation, such as ultra-violet (UV) light and X-rays, are mutagenic. The major source of human exposure to UV light is the sun. The two-fold increase in the incidence of skin cancer (which is caused by mutations in skin cells) in Scotland over the last twenty years is thought to be due, at least in part, to the rising number of holidays taken in sunny destinations such as Spain. Another signiﬁcant factor is the increased use of sun or tanning beds.

Treatment of bacteria with UV light: Question A freshly grown culture of bacteria was diluted, and samples were spread out onto agar plates to give approximately 300 bacterial cells per plate. Each plate was then Go online exposed for up to 3 minutes to a UV light source. Following UV exposure, the plates 20 min were incubated for two days before counting the bacterial colonies that grew on them.

UV exposure

Q8: Describe the relationship between the number of bacteria that are killed and the duration of exposure to UV light......

Mutations and genetic disorders: Questions Q9: What is a mutagen? ...... Go online

Q10: Give two examples of mutagens......

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6.6 Single gene mutations Learning objective By the end of this section, you should be able to:

• state that gene mutations are due to the alteration of a DNA nucleotide sequence;

• describe the effects of the different types of gene mutation: substitution, deletion and insertion;

• state that a mutation that affects a single nucleotide is called a point mutation;

• explain that the substitution of one nucleotide will alter a maximum of one amino acid (and may alter none);

• explain that insertions and deletions may alter from no amino acids to the whole DNA sequence;

• state that nucleotide insertions or deletions can also result in an expansion of a nucleotide sequence repeat;

• state that Huntingdon’s disease is a condition associated with nucleotide repeat expansion. Single gene mutations are those which take place within the DNA of a single gene or regions of the DNA which affect its expression. Such mutations in the sequence of bases in DNA can be described according to the effect that they have on the base sequence. Substitution - in this type of mutation, one nucleotide is replaced with another.

A substitution mutation

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Single-nucleotide substitutions include:

• missense, in which one nucleotide is changed in the DNA of the gene, and hence the complementary mRNA codon is altered and may code for a different amino acid;

• nonsense, when the change to the DNA triplet code causes an amino acid mRNA codon to be replaced by a stop codon so that during translation no amino acid is attached and the process stops;

• splice-site mutations, which either create or destroy codons for exon-intron splicing.

A substitution mutation within a protein-coding gene may not always lead to a change in the amino acid sequence of the encoded protein. In this case, the mutation is described as silent. As there are 64 possible combinations of bases in a DNA base triplet, many of the 20 amino acids are coded for by more than one combination. Consequently, a change to a single nucleotide may leave the amino acid sequence, and hence the protein, unaltered. Insertion - this describes the addition of one or more nucleotides into the DNA.

An insertion mutation If an insertion mutation occurs in a protein-coding gene, it usually has a dramatic effect on the amino acid sequence of the encoded protein. The diagram below illustrates this point.

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A frameshift mutation In part (a), the DNA is represented as a repeat sequence of the base triplet GAT. The DNA is transcribed into mRNA to produce a sequence of codons (in this case CUA, encoding the amino acid leucine). An insertion mutation - the addition of a C between the ﬁrst and second base triplets in the DNA - is shown in part (b). This mutation disturbs the pattern of the DNA triplets and alters the way in which they are transcribed into mRNA (see part (c)). All of the codons in the mRNA that appear after the mutation have changed, and, as a result, the amino acids in the protein are also different. A type of mutation which has this effect is called a frameshift mutation. Deletion - this refers to the removal of one or more nucleotides from the DNA.

A deletion mutation

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As with an insertion mutation, a deletion mutation alters the pattern of base triplets in the DNA and is also an example of a frameshift mutation.

Effects of gene mutations on amino acid sequences A gene is a region of DNA which consists of a speciﬁc sequence of nucleotide bases arranged in triplets. Every amino acid is coded for by one or more of these triplets. Go online Therefore, the sequence of bases determines the sequence in which amino acids are joined together to form a polypeptide or protein. Thus, a gene codes for a particular protein or polypeptide. A gene mutation is a change in the sequence or type of nucleotide bases in a strand of DNA. This can lead to a change in the sequence of amino acids, and thus to a change in the protein which is synthesised in the ribosomes. Sometimes a mutation causes only a minor change, perhaps affecting only one amino acid. If this is the result of a substitution, then a maximum of one amino acid is altered and the protein produced may be little altered. Other forms of point mutation, i.e. insertions or deletions, can cause a major change affecting the coding for many amino acids. Such a mutation is known as a frameshift mutation and leads to a completely different protein being produced, which cannot carry out the required function. Each of the three types of gene mutation are described in the following diagrams, with particular respect to changes in the amino acid sequences, which show the codons of part of a DNA strand and the amino acids which are coded for by them. Remember that the same amino acid can be coded for by more than one DNA triplet.

Q11:

This is a deletion mutation. The ﬁrst ’A’ nucleotide in the original DNA strand has been removed (or deleted). Answer the following questions: How many amino acids in this sequence have been changed? Is a deletion mutation a frameshift mutation? ......

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Q12:

This is an insertion mutation. The ’T’ nucleotide has been inserted between the ’C’ and ’A’ nucleotides in the original DNA strand. Answer the following questions: How many amino acids in this sequence have been changed? Is a deletion mutation a frameshift mutation? ......

Q13:

This is a substitution mutation. The ’T’ nucleotide in the middle of the original DNA strand has been replaced (or substituted) by a ’G’ nucleotide. Answer the following questions: How many amino acids in this sequence have been changed? Is a substitution mutation a frameshift mutation?

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Point and frameshift mutations Any mutation involving a single nucleotide is a point mutation. Substitutions, and insertions or deletions involving a single nucleotide, are all point mutations. However, their impact on the make-up of the ensuing polypeptide can be very different. If the number of nucleotides inserted or deleted is not a multiple of three, the impact of the mutation is likely to be seen in all the subsequent triplets and hence the amino acids for which they code. The pattern of base triplets in the DNA that is transcribed into mRNA is called the reading frame; an insertion that changes the reading frame is called a frameshift mutation. A frameshift mutation will also alter the ﬁrst triplet that signals a stop codon, potentially producing a much longer polypeptide than intended; equally, if it generates a new stop codon, a much shorter polypeptide will result. Neither will function correctly. Expansion of a nucleotide sequence repeat Repeat sequences of DNA nucleotide triplets (trinucleotides) are common in genes, where they code for a sequence of the same amino acid being repeated several times in the polypeptide. Because there is a repeating set of bases, during DNA replication it is possible for the copying process to ’lose the place’, causing sections of the new DNA strand to loop out and additional new trinucleotides to be added against the original template strand. The result is known as a trinucleotide repeat expansion. The occurrence of deletion and insertion point mutations can have the effect of triggering this slippage in the DNA replication process, increasing the number of repeated amino acids in the ﬁnal polypeptide. The greater the number of repeats added, the greater the effect of the mutation. A small expansion usually has no noticeable effect, but a characteristic of this type of mutation is that the expansion tends to increase with each generation. Eventually, the protein produced fails to function normally and symptoms become apparent. Two conditions resulting from trinucleotide repeat expansions are Huntingdon’s disease and Fragile X syndrome. Huntingdon’s disease is a trinucleotide repeat disorder of the HTT gene on chromosome 4. The repeated base sequence is CAG, which codes for glutamic acid. Individuals with 26 or fewer CAG repeats are not affected, whereas 40 and over show progressive degeneration of the nervous system, affecting muscle coordination and cognitive ability. It typically only becomes noticeable in mid-adult life. Fragile X syndrome is associated with the expansion of a sequence of repeated CGG trinucleotides (coding for arginine) in one of the genes on the X-chromosome. It is the most common single-gene cause of autism and inherited intellectual disability, although there is a wide spectrum of associated effects. As with any condition controlled by genes on the X-chromosome, it affects males at a higher rate than females (although the difference is much less than might be expected).

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Gene mutations: Examples The following provides examples of the three types of gene mutations. Go online 30 min

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Gene mutations: Questions Q14: A substitution mutation involves: Go online a) one nucleotide 20 min b) two nucleotides c) three nucleotides

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Q15: A mutation which results in a new stop codon: a) missense b) nonsense c) splice-site

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Q16: A mutation which affects exons and introns: a) missense b) nonsense c) splice-site

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Q17: A mutation which causes a change to one mRNA codon: a) missense b) nonsense c) splice-site

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Q18: A substitution mutation will lead to a different amino acid: a) always b) sometimes c) never

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Q19: An insertion mutation affects the triplets which come: a) before it b) after it c) at any point

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Q20: A deletion mutation affects the triplets which come: a) before it b) after it c) at any point

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Q21: The minimum number of amino acids affected by a point mutation:

a) 0 b) 1 c) 2

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Q22: A frameshift mutation can alter codons for:

a) amino acids b) start c) stop d) all of these

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Complete the sequences of amino acids after mutation in each of the examples by referring to the genetic code table.

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Q23:

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Q24:

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Q25:

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Q26: Why does a frameshift mutation arise only if the number of nucleotides inserted or deleted is not a multiple of three? ......

Q27: Describe a nucleotide sequence repeat (also known as a trinucleotide repeat sequence)? ......

Q28: What may trigger the expansion of a trinucleotide repeat sequence? ......

Q29: Name a trinucleotide repeat disorder......

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6.7 The effect of mutations on the structure and function of proteins Learning objective By the end of this section, you should be able to:

• give examples of inherited diseases caused by different single gene mutations. As mentioned in the previous section, single gene mutations can have effects ranging from none to being fatal, depending on which proteins are affected and the impact of that effect on their functioning. The following is a list of inherited diseases caused by single gene mutations. The syllabus does not specify any that should be known about in particular, but you should be able to quote details about a couple of examples with different causes.

• Sickle-cell disease (mis-sense, autosomal codominant) - causes the production of abnormal haemoglobin which makes the red cells take on a rigid, sickle shape, causing obstructions in small vessels as well as a wide range of other serious effects. The condition survives because people who are heterozygous for it suffer much reduced symptoms if infected with malaria, compared to people homozygous for the normal haemoglobin gene.

• Phenylketonuria (PKU) (mis-sense, autosomal recessive) - causes the production of an inactive form of the enzyme (phenylalanine hydroxylase, PAH) which breaks down the amino acid phenylalanine, resulting in a range of symptoms including impaired brain function.

• Beta (β) thalassemia (splice-site, autosomal recessive) - causes reduced, or no, production of one of the polypeptide chains (the beta chain) which make up haemoglobin. The symptoms range from none to severe anaemia, depending on where the mutation occurs on the gene.

• Duchenne muscular dystrophy (DMD) (nonsense, X-linked recessive) - caused by mutation to the largest gene on the X chromosome, resulting in failure to produce dystrophin, an important structural protein in muscle. It causes muscle degeneration and early death.

• Tay-Sachs syndrome (frameshift insertion, autosomal recessive) - caused by mutation to the hexoaminidase A, or HEXA, gene on chromosome 15, resulting in progressive destruction of the central nervous system and early death.

• Cystic ﬁbrosis (frameshift deletion, autosomal recessive) - caused by a mutation to the gene CFTR which regulates the movement of chloride and sodium ions across epithelial membranes. It results in the secretory glands of the lungs producing very sticky mucus, poor growth and frequent chest infections. Untreated, it leads to death in infancy.

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The effect of mutations on the structure and function of proteins: Question Q30: Complete the table using the causes from the list. Go online

Inherited disease Cause Beta thalassemia Cystic ﬁbrosis Duchenne muscular dystrophy (DMD) Phenylketonuria (PKU) Sickle-cell disease Tay-Sachs syndrome

Cause list: frameshift deletion, autosomal recessive; frameshift insertion, autosomal recessive; missense, codominant; missense, autosomal recessive; nonsense, X-linked recessive; splice-site, autosomal recessive......

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6.8 Chromosome structure mutations Learning objective By the end of this section, you should be able to:

• state that chromosome structural mutations are due to the addition, removal or reversal of segments of chromosome;

• explain that any mutation which alters gene expression is likely to have a serious effect, e.g. by altering stop/start codons, exons, introns and promoters;

• describe the different forms of chromosome mutation, i.e. deletion, duplication, translocation and inversion;

• give examples of conditions which result from each type of chromosome mutation. Chromosome mutations occur during cell division, both in mitosis or meiosis. They can cause a change in either the number or structure of the chromosomes, leading to numerical or structural abnormalities. Numerical abnormalities arise during meiosis I or II and lead to conditions such as Down’s syndrome (3 of chromosome 21), Edwards syndrome (3 of chromosome 18), Turner syndrome (only one sex chromosome, X), and Klinefelter syndrome (male with an extra X, so XXY). Here we consider only (some of) the mutations which produce structural abnormalities.

• Deletion: a segment of a chromosome, and its associated genes, is lost during crossover in meiosis I. The effect depends on exactly which sections of DNA are lost and varies from no effect to being fatal. An example is Cri-du-chat syndrome, which arises from deletion of 10-20% from the end of the short arm of chromosome 5. It is so named as the cry of affected children sounds like the meowing of a cat because of problems with the nervous system and the larynx. There are many other symptoms, including poor growth, severe speech and cognitive delays, behavioural problems, and heart defects. A second example is Williams syndrome, which is caused by a deletion of about 26 genes from chromosome 7. It results in a neurodevelopmental disorder, generally linked with an unusually cheerful character and ease with strangers, delayed mental development coupled with strong language skills, and cardiovascular problems.

• Duplication: a chromosome gains a segment that has been lost by another chromosome, so extra genetic material is present as genes are repeated on the same chromosome. It may arise in various ways, e.g. during crossing-over in meiosis I or during DNA replication. As the extra genes are non-functional, they can mutate without any deleterious effect on the individual, and thus provide a valuable source of genetic variation for natural selection. Duplication of oncogenes is a cause of many cancers. Being restricted to the somatic cells, such mutations are not inherited, but have potentially fatal effects for the individual concerned. Common cancers caused are breast, cervical, colorectal, cervical, ovarian and gastric.

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• Inversion: a segment of a chromosome becomes reversed because it has broken off, turned round, and become reattached during meiosis I. This may well go unnoticed unless a gene is broken, in which case it may well be fatal. The most common effect is a reduction in fertility.

• Translocation: segments of two (or more) chromosomes are exchanged or an entire chromosome is attached to another. An example is chronic myeloid leukaemia, which results from a gene from chromosome 22 fusing with a gene on chromosome 9. It causes a cancer of stem cells in the bone marrow, which leads to overproduction of certain white blood cells.

These mutations are illustrated in the following diagram, where the colours represent sections of chromosome rather than individual genes.

Chromosome mutations

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Chromosome structure mutations: Question Q31: Complete the sentences by matching the parts on the left with the parts on the right. Go online

Deletion and duplication are examples of deletion.

Chromosome mutations occur during reduced fertility.

Deletion is loss of a segment of chromosome structural abnormality. during

The condition Cri-du-Chat is caused by a myeloid leukaemia.

Extra chromosomal material will be present oncogenes. because of

Duplication of these genes is the cause of many meiosis I. cancers

The most common effect of an inversion translocation. mutation is

A gene from chromosome 22 fusing a gene on duplication. chromosome 9

A cancer of stem cells in the bone marrow cell division......

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6.9 Learning points

Summary Proteins

• Proteins are composed of building blocks called amino acids.

• Amino acids are linked together by strong bonds, called peptide bonds, to form polypeptide chains.

• The diversity of structure and function among proteins is due to:

◦ the arrangement of amino acids in polypeptide chains; ◦ the association, in many cases, of more than one polypeptide chain in the mature protein; ◦ the inclusion of non-protein structures such as phosphates, metal ions and carbohydrates.

• The 3D structure of the enzyme is vital to its function.

• An area of the enzyme is folded so that an ’active site’ is exposed, which readily combines with its substrate.

• Proteins are organised at four different structural levels:

◦ primary structure, consisting of the sequence of amino acids; ◦ secondary structure, involving the folding of the polypeptide chains, which are then held in place by weak hydrogen bonds; ◦ tertiary structure, involving further folding held in place by strong bonds between amino acids; ◦ quaternary structure, formed when two or more polypeptide chains associate together to form the ﬁnal three dimensional protein structure.

• Proteins have many functions, including: enzymes, some hormones, muscular contraction, transport, antibodies and structural support.

• Many hormones, for example insulin and ADH, are proteins.

• Many proteins transport substances around the body.

• Myoglobin and haemoglobin, for example, are involved in the transport of oxygen.

• Bones, tendons, ligaments, skin and hair are all mainly composed of structural proteins such as collagen, elastin and keratin.

Gene mutations

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Summary continued • A gene mutation is a random change in the number or sequence of bases.

• A gene mutation results in no protein, or a faulty protein, being produced.

• Mutations that involve a single nucleotide are called point mutations.

• A substitution mutation occurs when one base is replaced by another.

• Single-nucleotide substitutions include:

◦ missense, when one amino acid codon replaces another, e.g. sickle cell disease; ◦ nonsense, when an amino acid codon is replaced with a stop codon, e.g. Duchenne muscular dystrophy (DMD); ◦ splice-site mutations, resulting in the creation or destruction of the codons for exon-intron splicing, e.g. Beta (β) thalassemia.

• Substitution mutations do not usually have a large effect on the amino acid sequence coded unless they involve stop or start codons, introns or exons.

• A deletion mutation involves the removal of a base.

• An insertion mutation involves the addition of a base.

• Insertion and deletion mutations are known as frameshift mutations because they alter the entire sequence of amino acids coded after them.

• The effect of a frameshift mutation is greater the earlier in the gene that it occurs as more codons, introns, exons and stop codons will be affected.

• Frameshift mutations are likely to result in no protein being produced, and so may often be fatal.

• An example of a condition caused by a frameshift insertion mutation is Tay- Sachs syndrome.

• An example of a condition caused by a frameshift deletion mutation is Cystic Fibrosis;

• Nucleotide insertions or deletions can also result in an expansion of a nucleotide sequence repeat - a condition associated with nucleotide repeat expansion is Huntingdon’s disease. Chromosome mutations • Chromosome structural mutations are due to the addition, removal or reversal of segments of chromosome.

• Any mutation which alters gene expression is likely to have a serious effect, e.g. by altering start and stop codons, exons, introns and promoters.

• Chromosome mutations can take the form of:

◦ a deletion, involving the loss of a segment of a chromosome;

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Summary continued ◦ duplication, when a segment of chromosome is repeated; ◦ translocation, the rearrangement of chromosomal material involving two or more chromosomes; ◦ inversion, the reversal of a segment of a chromosome.

• The substantial changes in chromosome mutations often make them lethal.

• Cri-du-chat syndrome is an example of deletion: a part of the short arm of chromosome 5 is missing.

• Common cancers, e.g. breast, colorectal and gastric, are often associated with duplication.

• Chronic myeloid leukaemia (CML) is an example of reciprocal translocation: a gene from chromosome 22 is fused with a gene on chromosome 9.

• Inversion mutations usually have no noticeable effects, although they may reduce fertility.

6.10 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of proteins and gene mutation before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: Proteins and gene mutation A) Give an account of the structure and function of proteins. (8 marks)

B) Give an account of gene mutation. (9 marks)

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6.11 End of topic test End of Topic 6 test Q32: Complete the paragraph using the words from the list. Some of the words may be

used more than once each. (12 marks) Go online

Proteins are composed of chains of joined together by to form

a . They are held in complex shapes by bonds and interactions between individual . In addition, several chains may

associate together and non-protein groups such as and metal ions may be

bound into the ﬁnal protein structure. Proteins have many functions including acting as

enzymes (e.g. ), (e.g insulin), transport (e.g. ), and

structural components (e.g. ). Word list: amino acids, carbohydrates, collagen, haemoglobin, hormones, hydrogen, peptide bonds, phosphofructokinase, polypeptide, three-dimensional......

Q33: Complete the table about gene mutations using the phrases from the list. (8 marks)

Term Deﬁnition Point mutation Substitution Missense Nonsense Splice-site Deletion Insertion Frameshift

Phrase list: addition of a base, base sequence changed, new exon-intron codons, new stop codon, one base replaced, one codon replaced, removal of a base, single base involved.

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Q34: Complete the table about chromosome mutations by selecting the correct answers from the options listed. (8 marks)

addition / removal / Of what are chromosome structural mutations the result? reversal / all of these Which type of mutation results from the loss of a segment of deletion / duplication / chromosome? translocation exons / introns / stop Which codons are potentially altered by structural mutations? codons / all of these deletion / duplication / Which type of mutation causes Cri-du-Chat? translocation / inversion deletion / duplication / Which type of mutation is associated with breast cancer? translocation / inversion deletion / duplication / Which type of mutation causes chronic myeloid leukaemia? translocation / inversion deletion / duplication / Which type of mutation usually has not noticeable effect? translocation / inversion What is the likely effect of substantial changes to the no effect / moderate chromosome? effect / lethal effect ......

Q35: What is the name of the bonds which hold amino acids together? (1 mark) ......

Q36: What is a polypeptide? (1 mark) ......

Q37: Explain the function of the protein structure of hydrogen bonds. (1 mark) ......

Q38: Name two types of non-protein groups that may be bound into the protein structure. (2 marks) ......

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Topic 7

Human genomics

Contents

7.1 Introduction ...... 146 7.2 Sequencing DNA ...... 146 7.3 Bioinformatics ...... 148 7.4 Systematics ...... 150 7.5 Personalised medicine ...... 151 7.6 Ampliﬁcation and detection of DNA sequences ...... 153 7.7 DNA probes ...... 157 7.8 Medical and forensic applications ...... 159 7.9 Learning points ...... 163 7.10 Extended response question ...... 165 7.11 End of topic test ...... 166

Learning objectives By the end of this topic, you should be able to:

• understand DNA sequencing and how the sequence of bases can be determined for individual genes and entire genomes;

• describe the use of bioinformatics;

• explain the role of systematics;

• describe how the analysis of an individual’s genome may lead to personalised medicine through understanding the genetic component of risk of disease;

• describe the ampliﬁcation and detection of DNA sequences;

• explain the Polymerase Chain Reaction (PCR);

• explain the use of DNA probes;

• give examples of medical and forensic applications through ampliﬁcation and detection of DNA sequences. 146 TOPIC 7. HUMAN GENOMICS

7.1 Introduction DNA can now be sequenced, synthesised and recombined in the laboratory. These important advances in technology have led to the new sciences of bioinformatics and genomics with their implications for understanding evolutionary relationships and developing personalised medicine. The ability to replicate DNA in vitro has implications for fundamental research and for medical and forensic applications, and provides opportunities for informed discussion of the associated social, moral and ethical issues facing society. There are undoubtedly huge beneﬁts to be gained by using DNA technology, but it also raises many moral and ethical issues. Two techniques that are very important to the Human Genome Project are described in detail here, namely the polymerase chain reaction (PCR) and DNA probes.

7.2 Sequencing DNA Learning objective By the end of this section, you should be able to:

• explain that sequencing DNA and the sequence of bases can be used to determine individual genes and entire genomes. The Human Genome Project began in October 1990 and was completed in 2003. The project involved the discovery of all the estimated 20,000 human genes, making them available for further studies. The project also led to the discovery of the complete sequence of the 3 billion DNA sub-units (the bases in the human genome). In April 2003, the completion of the human DNA sequence coincided with the 50th anniversary of Crick and Watson’s description of the DNA structure in 1953. Only about 3 percent of the human genome is actually used as the set of instructions to make proteins. These regions are called coding regions. At present, little is known functionally for most of the remaining 97 percent of the genome. These regions are called non-coding regions, but that should not be interpreted as indicating that they have no function, e.g. some act as introns, which can have a wide variety of functions, some are gene promoters, which initiate DNA transcription, and the function of the rest is the subject of much current research.

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Human proteins categorized by function, given both as number of encoding genes and percentage of all genes Single nucleotide polymorphisms, or SNPs (pronounced "snips"), are DNA sequence variations that occur when a single nucleotide (A, C, G, or T) in the genome sequence is altered. For example, a SNP might change the DNA sequence ACGGCTCA to ATGGCTCA. Such substitution point mutations are the simplest form of mutation, affecting at most a single amino acid in the protein for which a gene codes. Generally they have little or no effect on the phenotype, but they are more frequent in non-coding regions of the genome as these are not subjected to the pressures of natural selection. A variation is considered a SNP when it occurs in at least 1% of the population. SNPs make up about 90% of all human genetic variation and occur every 100 to 300 bases along the 3-billion-base human genome. About 66% of SNPs involve the replacement of cytosine (C) with thymine (T), which can occur in coding and non-coding regions of the genome. The genome of Craig Venter (American biologist and entrepreneur, and founder of the Celera Genomics) was published in 2007. His genome contains 4.1 million variations, 3.2 million of which were SNPs. Today, the emphasis has shifted from sequencing the entire genome to focussing solely on the exons, the so-called exome, as it is in these coding regions that 85% of human inherited disorders are found. Although SNPs have no effect on cell function, SNP maps help to identify the multiple genes associated with complex ailments such as cancer, diabetes, vascular disease, and some forms of mental illness. They are also very important in the process of DNA ﬁngerprinting. Although SNPs do not usually cause diseases, they can help determine the likelihood of someone developing a particular illness, e.g. apolipoprotein E, or ApoE, a gene associated with Alzheimer’s disease. It is also believed that SNPs may help in the discovery and cataloguing of the unique sets of changes involved in different types of cancers, and so offer new methods of combatting such diseases. Scientists are trying to identify all the different SNPs in the human genome. They are sequencing the genomes of a large number of people and then comparing the base sequences to discover SNPs. The sequence data is being stored in computers that can generate a single map of the human genome containing all possible SNPs.

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Sequencing DNA: Questions Q1: In terms of base pairs, how big is the human genome (in billions)? Go online ......

Q2: What are SNPs? ......

Q3: What do we call the sections of the human genome which are used as a set of instructions to make proteins? ......

7.3 Bioinformatics Learning objective By the end of this section, you should be able to:

• explain that the enormous amounts of data produced by DNA and protein sequencing can be managed and analysed using computer technology, an area known as bioinformatics. Bioinformatics is a field of study which draws on the techniques of computer science, statistics, mathematics and engineering to process biological data. It uses software tools to develop and improve methods for storing, retrieving, organising and analysing information. In the context of this topic, computer technology has been used to identify DNA sequences. DNA and protein sequencing generate vast quantities of data which can only be managed by the use of computer technology. This has been used to identify gene sequences by searching for DNA sequences which are similar to known genes, start codons, or sequences lacking stop codons. In the same way, base sequences can be identified which correspond to the amino acid sequence of a particular protein. A good example of bioinformatics in action is the ’1000 Genome Project’ which was launched in 2008. Its aim was to catalogue human genetic variation by sequencing the genomes of at least 1000 anonymous volunteers from a variety of ethnic groups from around the world. By 2011, over 1150 prokaryote and 310 eukaryote genomes had been published. At least two-thirds of genes known to be involved with cancer have also been found in fruit flies, Drosophila melanogaster. Finding out how these genes work in a much simpler organism should provide researchers a better understanding of how they operate in humans. Ultimately, this should give us a better understanding of how to control or prevent the onset of certain types of cancer.

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The table shows a comparison of genome sizes between human and some other organisms.

Estimated size Chromosome Estimated gene Organism (base pairs) number number Human 3 billion 46 ≈20,000 (Homo sapiens) Fruit ﬂy (Drosophila 165 million 8 13,000 melanogaster) Mouse 2.9 billion 40 ≈25,000 (Mus musculus) Gorilla 2.9 billion 48 ≈25,000 Roundworm (Caenorhabditis 97 million 12 19,000 elegans) Bacteria 4.6 million 1 3,200 (Escherichia coli) Yeast (Saccharomyces 12 million 32 6,000 cerevisiae)

Comparison of genome sizes between different organisms In the future, bioinformatics will allow us to predict:

• how genes are regulated;

• how the proteins will be synthesised;

• how they will fold and function in the cell;

• how they will interact with other molecules, be they natural or synthetic.

Bioinformatics: Questions Q4: What techniques does bioinformatics draw on to make sense of biological data? ...... Go online

Q5: What do computer programs use to identify gene sequences? ......

Q6: How do computer programs identify protein sequences? ......

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7.4 Systematics Learning objective By the end of this section, you should be able to:

• explain that systematics, that is the comparison of human genome sequence data (and genomes of other species), provides information on evolutionary relationships and origins. Systematics is the comparison of human genome sequence data. When the Human Genome Project was being proposed, some argued that only the 3% of the genome which encodes proteins should be sequenced. This would involve collecting mRNAs to make cDNA libraries. Since some genes are only expressed at certain times during development and in certain cell types, this approach could mean some genes would be missed. Other sequences important for gene expression could also be missed, which is why it was decided to sequence the entire genome in the Human Genome Project. Another argument for sequencing the entire genome is that it provides the opportunity to study chromosome structure and evolution. In addition to mapping the human genome, the genomes of other species are also being mapped. These include species important to biological research and agriculture, such as the mouse, chicken, pig, cow, rice, wheat, Caenorhabditis elegans (nematode), Drosophila melanogaster (fruit fly), Saccharomyces cerevisiae (yeast), Escherichia coli (E. coli), and other prokaryotes. The genomes of some of these organisms, such as E. coli, yeast, the nematode and the fruit fly, have now been completely mapped and sequenced. These maps can be used to locate homologous genes in the human genome and to help in determining gene function. Comparative genome analysis is being used to find out more about evolution. In the early 1950s, insulin became the first protein to have its amino acid sequence worked out. The amino acid sequence for a protein in one species can be compared with that from another species. The number of differences in an amino acid sequence can be used to calculate the time since two species diverged from a common ancestor. If there are lots of differences between the maps, it can be deduced that the species diverged a longer time ago than if there are only a few differences. This type of information is used alongside other methods for measuring the rate of evolution.

Gene maps can be used to predict gene order. If gene X is found next to gene Y and Z in one species, the likelihood is that it will be found next to the same two genes in another closely related species. Comparative maps will be used to ﬁnd candidate genes for phenotypes mapped in species as diverse as chicken and human.

Systematics: Questions Q7: Comparative genome analysis has shown that some genes have been conserved Go online in evolution. Genes that have evolved from a common ancestor are described as: a) Complementary b) Linked c) Homologous d) Identical

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Q8: Using comparative genome analysis, how is it possible to work out if two species are from the common ancestor? ......

7.5 Personalised medicine Learning objective By the end of this section, you should be able to:

• explain that the analysis of an individual’s genome may lead to personalised medicine as a result of understanding the genetic component of risk of disease. The completion of the Human Genome Project in 2003 has led to the further understanding of genetic contributions to human health; the ﬁeld of genomic medicine was born as a result. Genetics can play an increasingly important role in the diagnosis, monitoring, and treatment of diseases. It has given us a better understanding of the importance of distinguishing between neutral mutations and harmful mutations, and also the complex nature of many diseases. Pharmacogenetics is the use of genome information in the choice of effective drugs. Pharmaceutical companies maintain that pharmacogenetics has several advantages in the treatment of human diseases:

• it allows the production of more ’powerful’ medicines through the discovery of drugs based on the enzymes, proteins and RNA molecules associated with genes and diseases;

• it allows the production of better and safer drugs more quickly than the standard trial and error method;

• it gives more accurate methods of determining drug dosages;

• it allows for advanced screening for a particular disease;

• knowledge of their genetic code will allow individuals to make adequate lifestyle and environmental changes at an early age, thus avoiding or lessening the severity of a genetic disorder;

• pharmaceutical companies will be able to discover potential therapies more easily using genome targets.

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However, there are challenges for pharmacogenetics:

• it is a complex process to ﬁnd gene variations that affect drug response because millions of SNPs must be identiﬁed and analysed to determine their involvement (if any) in drug response;

• there are limited drug alternatives because only one or two approved drugs are available for treatment of a particular condition;

• drug companies are reluctant to make multiple pharmacogenetics products;

• the introduction of multiple pharmacogenetic products to treat the same condition will complicate the process of prescribing and dispensing drugs, and it will not be cost effective to educate healthcare providers.

Personalised medicine: Question Q9: List four advantages of using pharmacogenetics in the treatment of human Go online diseases. (4 marks) ......

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7.6 Ampliﬁcation and detection of DNA sequences Learning objective By the end of this section, you should be able to:

• state that the polymerase chain reaction (PCR) is a technique for amplifying (increasing in quantity) a sample of DNA;

• explain that PCR is useful in any area requiring larger samples of DNA, e.g. forensics and archaeology;

• state that PCR is a four step process, involving:

1. Mixing together the components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase). 2. Heating (90 - 95◦C) separates the DNA strands. 3. Annealing (54◦C) is the binding of the primers, which mark the start and end of the sequence to be ampliﬁed as well as preventing reattachment of the separated DNA strands. 4. Extension (72◦C) of the primers to complete the complementary strand is carried out by heat stable DNA polymerase.

• describe how repeated cycles of heating and cooling amplify this region of DNA;

• explain that PCR uses the same mechanism for copying the DNA as our cells use for DNA replication;

• state that PCR uses the enzyme DNA polymerase to replicate, or ’amplify’, the DNA strand. For a range of purposes, it is essential to copy samples of DNA many times to provide sufﬁcient material for analysis. The main areas involved are:

• medical applications, e.g. genetic testing, tissue typing; • the study of infectious disease, e.g. human immumodeﬁciency virus (HIV), tuberculosis (TB); • forensic applications, e.g. genetic ﬁngerprinting; • research applications, e.g. analysis of DNA from extinct species, study of gene expression.

The technique most frequently used is the Polymerase Chain Reaction (PCR), although there are now a number of alternative approaches. PCR can be used to amplify a desired DNA sequence of any origin (virus, bacteria, plant or animal) millions of times in a matter of hours. It is especially useful because:

• it is highly speciﬁc; • it is easily automated; • it is capable of amplifying minute amounts of a sample.

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Before considering PCR, you should know the following:

• double-stranded DNA can be separated into single strands by heating to 90 - 95 ◦C: the DNA is then said to be ’denatured’;

• under the correct conditions, complementary base sequences will bond or ’hybridise’ with each other;

• ’primers’ are short sequences of single-stranded DNA;

• DNA polymerases are enzymes required for the synthesis of DNA;

• Taq polymerase is a special DNA polymerase - it is an enzyme that is not denatured by heating to 90 - 95◦C;

• DNA polymerases can only synthesise new DNA in one direction, starting from the 3 end of a primer.

PCR depends on special heat-stable DNA polymerase enzymes from thermophilic bacteria, which can withstand temperatures of 95◦C and have an optimal temperature for activity of 72◦C. The PCR process The whole process can be divided into four stages.

1. The components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase) are mixed together.

2. Heating (90 - 95◦C) - the ﬁrst step in the PCR is to separate (denature) the double- stranded DNA containing the target DNA sequence for ampliﬁcation.

3. Annealing (54◦C) - the temperature is then reduced to allow the primers to bind (by forming hydrogen bonds) to their complementary sequences on the separated strands. Primers are short DNA sequences or oligonucleotides (each about 20 bases long) that bind at either side of the target sequence, one on each of the complementary strands of the target DNA.

4. Extension (72◦C) - Taq polymerase now adds nucleotides to the 3 ends of the primers and extends them into new complementary strands. The result is that the amount of DNA has now doubled.

Repeated heating and cooling cycles multiply the target DNA exponentially since each new double strand separates to become two templates for further synthesis. In about one hour, 20 PCR cycles can amplify the target DNA a million-fold.

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The Polymerase Chain Reaction (PCR) process: Steps The following summarises the steps involved in the PCR process. Go online

Stage 1: Mixing of the DNA template, primers, nucleotides and Taq polymerase.

Stage 2: Heating (90-94◦C) - the double-stranded template DNA is split into single strands.

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Stage 3: Annealing (54◦C) - the primers anneal to complementary sites on the single stranded template DNA.

Stage 4: Extension (72◦C) - the DNA polymerase binds to the annealed primers and extends them in a 5 to 3 direction. Result: the amount of DNA has now doubled.

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......

Ampliﬁcation and detection of DNA sequences: Question Q10: Complete the paragraph using the words from the list.

After the of the components in the PCR process, three further stages follow. Go online

Heating to which denatures the DNA by separating its two strands. The temperature is then dropped to , which allows to bind to their

complementary sequences on the separated strands. Primers are short

sequences of about 20 bases that bind on either side of the sequence. This stage is known as . The temperature is then raised again to in the

stage. Taq adds to the 3 ends of the primers and extends them into new complementary strands. Continued cycles of heating and cooling cause

the DNA to be doubled repeatedly, so that in one hour, cycles will amplify the target DNA more than 1 million times. Word list: 20, 54◦C, 72◦C, 95◦C, annealing, DNA, extension, mixing, nucleotides, polymerase, primers, target......

7.7 DNA probes Learning objective By the end of this section, you should be able to:

• state that a DNA probe is a short length of DNA with a known nucleotide (base) sequence, which has the ability to pair (bind) with the complementary sequence on a DNA strand;

• state that DNA probes contain radioactive phosphorus, which can be detected by autoradiography on X-ray ﬁlm;

• state that DNA probes are used for locating speciﬁc pieces of DNA;

• explain that DNA probes can also be tagged with fluorescent markers which allow detection of specific DNA sequences under the microscope. To identify a gene (specific section of DNA), we can use a gene probe. A gene probe is a single strand of DNA with a base sequence complementary to that of the gene which needs to be located. It will therefore bind to it. The phosphate groups in the DNA backbone of the manufactured gene probe are made with radioactive phosphorus, 32P, so the probe can be detected when it has bound to the gene, i.e. the radioactivity is easy to detect using X-ray film. This process is called autoradiography. A fluorescent label can be added instead of a radioactive label. The probe would then be detected using laser scanning rather than autoradiography.

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DNA probes: Steps The following summarises the steps involved in DNA probes. Go online

......

DNA probes: Questions Q11: A DNA probe can be used to: Go online a) insert a gene into the bacterium E. coli. b) construct a new gene. c) identify a particular sequence of DNA. d) produce DNA.

......

Q12: Which of the following statements concerning gene probes is false?

a) A gene probe is a single strand of DNA. b) The pentose sugar ring of a gene probe is labelled with radioactive 14C. c) A gene probe has a complementary base sequence to that of the gene which needs to be located. d) Gene probes can be located by autoradiography.

......

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7.8 Medical and forensic applications Learning objective By the end of this section, you should be able to:

• explain that the diagnosis of disease status or risk of disease onset can be made by screening a cell sample from a patient for the presence or absence of a particular sequence;

• state that this information can be used to diagnose a condition or describe the likelihood of a condition developing;

• state that women with a family history of cancer can be tested for mutations in the BRCA 1 and BRCA 2 genes;

• explain that DNA proﬁling allows the identiﬁcation of individuals by means of comparison of regions of the genome with highly variable numbers of repetitive sequences of DNA;

• explain that if a tiny sample of blood is left at a crime scene, the DNA it contains can be ampliﬁed using PCR;

• explain that a probe can then be created, which binds to one of these highly variable sequences;

• explain that if a suspect is apprehended, their DNA can be tested with the same probe to look for the sequence in question. Medical application Some human disorders are inherited and are frequently associated with congenital abnormalities. Thousands of inherited disorders have been known for many years because they run in families. Many of these are severe and are due to single gene defects. Fortunately, most of these are rare disorders. Modern genetics provides a way of accurately diagnosing who is at risk of inheriting some of these conditions. It is possible to make a diagnosis of disease status or risk of disease onset. One example is that of using a cell sample from a patient to screen for the presence or absence of a particular sequence, e.g. a mutation in a gene. The information can be used to diagnose a condition or describe the likelihood of a condition developing. For example, women with a family history of cancer can be tested for mutations in the BRCA 1 (BReast CAncer gene one) and BRCA 2 (BReast CAncer gene two) genes. Mutations in these genes can increase the risk of developing breast or ovarian cancer. Forensic application The DNA of every individual is unique (except for identical twins). We will discuss how this variation is used by forensic scientists to identify individuals in criminal cases and paternity disputes. Using PCR, the forensic scientist can amplify DNA extracted from tiny fragments of skin, bloodstains or semen samples left by a criminal at the scene of a crime. DNA can even be extracted from the saliva left on a stamp! This DNA can be matched against that of any suspect.

©HERIOT-WATT UNIVERSITY 160 TOPIC 7. HUMAN GENOMICS

Dr. Alec Jeffries, a scientist at Leicester University, developed the original technique of DNA profiling in 1984. By treating DNA in a certain way, Dr. Jeffries was able to produce patterns similar to the bar codes we see on goods from the supermarket. The likelihood of DNA from two individuals producing the same pattern is so low that it can be used to identify individuals. DNA profiling depends on the use of special DNA sequences called minisatellites. If the number of repeat units within the minisatellite shows variation (polymorphism) between individuals, it is called a variable number tandem repeat, or VNTR. VNTRs can be detected by the changes they cause in the length of DNA fragments following digestion by restriction enzymes. The patterns produced are called restriction fragment length polymorphisms, or RFLPs. The steps involved in creating a DNA fingerprint using a VNTR are summarised below.

1. Extraction of DNA.

2. Digestion of DNA using a restriction enzyme.

3. Separation of the DNA fragments by size using gel electrophoresis.

4. Blotting, or transfer, of the DNA fragments onto a ﬁlter.

5. Hybridisation, or binding, of a labelled copy of a chosen VNTR to the DNA on the ﬁlter.

The accuracy of the technique is increased by using VNTRs that are multi-locus, or present at more than one site in the genome. The DNA profiling patterns produced using multi-locus VNTRs are much more complex than those produced using single-locus VNTRs. The technique can also be made more reliable by doing several fingerprints using different VNTRs. The technique relies on amplification by PCR of small segments of DNA which carry a particular microsatellite, or short tandem repeat (STR). The steps involved in creating a DNA profile using STRs are summarised below.

1. Isolation of DNA sample.

2. Ampliﬁcation of DNA by PCR using primers that bind to the unique DNA sequence at either side of the STR.

3. Repetition of the process to amplify three to four different loci.

4. Labelling of the ampliﬁed DNA product with ﬂuorescent dyes.

5. Analysis of the DNA product using an automated DNA sequencer (the chart produced by the sequencer gives peaks indicating the length of each ampliﬁed fragment).

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Medical and forensic applications: Questions Q13: Complete the ﬂow chart by using each of the six phrases to ﬁll in the empty boxes. Go online 30 min

......

Read the passage below and then answer the questions that follow. A terrible crime has taken place. A middle-aged man was murdered during a burglary. Forensic scientists collected a blood sample from the victim and a sample from the scene of the crime. Blood samples were also taken from four suspects identiﬁed by the police. Forensic scientists have prepared the autoradiograph shown from DNA extracted from the blood samples. The prosecution wishes to use it as evidence in a court case.

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DNA proﬁles

Q14: In the correct order, brieﬂy list the six steps used to produce an autoradiogram. The ﬁrst and last steps have been done for you. The technique was developed by Ed Southern at Edinburgh University.

1. DNA extracted from blood sample. 2. 3. 4. 5. 6. Autoradiography......

Q15: Who should be charged with the crime?

a) Suspect 1 b) Suspect 2 c) Suspect 3 d) Suspect 4

......

Q16: The prosecution wishes to use the forensic evidence in court, but is concerned that the autoradiograph is not conclusive proof. Why? ......

Q17: How could the accuracy of the analysis be improved? ......

......

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7.9 Learning points

Summary Sequencing DNA

• Sequencing DNA and the sequence of bases can be used to determine individual genes and entire genomes.

• Bioinformatics - enormous amounts of data produced by DNA and protein sequencing can be managed and analysed using computer technology.

• Systematics - comparison of human genome sequence data (and genomes of other species) provides information on evolutionary relationships and origins.

• The analysis of an individual’s genome may lead to personalised medicine through understanding the genetic component of risk of disease.

Polymerase chain reaction (PCR)

• This is a technique for amplifying (increasing in quantity) a sample of DNA.

• It is therefore useful in any area requiring larger samples of DNA, e.g. forensics and archaeology.

• It is a four step process, involving mixing, heating, annealing and extension:

◦ Mixing together the components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase). ◦ Heating (90 - 95◦C) separates the DNA strands. ◦ Annealing (54◦C) is the binding of the primers, which mark the start and end of the sequence to be ampliﬁed, as well as preventing reattachment of the separated DNA strands. ◦ Extension (72◦C) (of the primers to complete the complementary strand) is carried out by heat stable DNA polymerase.

• Repeated cycles of heating and cooling amplify this region of DNA.

• PCR uses the same mechanism for copying the DNA as our cells use for DNA replication.

• PCR uses the enzyme DNA polymerase to replicate, or ’amplify’, the DNA strand.

DNA probes

• A DNA probe is a short length of DNA with a known nucleotide (base) sequence, which has the ability to pair (bind) with the complementary sequence on a DNA strand.

• It has radioactive phosphorus, which can be detected by autoradiography on X-ray ﬁlm.

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Summary continued • They are therefore useful for locating speciﬁc pieces of DNA.

• DNA probes can also be tagged with ﬂuorescent markers which allow detection of speciﬁc DNA sequences under the microscope.

Medical and forensic applications

• The diagnosis of disease status or risk of disease onset can be made by screening a cell sample from a patient for the presence or absence of a particular sequence.

• This information can be used to diagnose a condition or describe the likelihood of a condition developing.

• Women with a family history of cancer can be tested for mutations in the BRCA 1 and BRCA 2 genes.

• DNA proﬁling allows the identiﬁcation of individuals through comparison of regions of the genome with highly variable numbers of repetitive sequences of DNA.

• If a tiny sample of blood is left at a crime scene, the DNA it contains can be ampliﬁed using PCR.

• A probe can then be created, which binds to one of these highly variable sequences.

• If a suspect is apprehended, their DNA can be tested with the same probe to look for the sequence in question.

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7.10 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of the PCR process before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: The PCR process Give an account of the ampliﬁcation of DNA sequences by the Polymerase Chain Reaction (PCR). (8 marks) ......

©HERIOT-WATT UNIVERSITY 166 TOPIC 7. HUMAN GENOMICS

7.11 End of topic test End of Topic 7 test Q18: Complete the sentences by matching the descriptions on the left with the terms Go online on the right. (8 marks)

The sequence of bases can be determined for individual genes by bioinformatics. using

The use of computer technology to polymerase chain reaction (PCR). identify DNA sequences is known as

Comparison of the human genome with genomes of other species is DNA probes. known as

An understanding of the genetic component of the risk of disease is DNA proﬁling. required for

A technique for the ampliﬁcation of systematics. DNA in vitro is known as

Short sequences of single-stranded DNA that are complementary to DNA sequencing. speciﬁc target sequences are known as

Short single stranded fragments of personalised medicine. DNA are known as

Comparison of highly variable repetitive sequences of DNA is primers. known as ......

Q19: Rearrange the following steps to describe the order in which the polymerase chain reaction (PCR) takes place. (6 marks)

• Temperature of the reaction adjusted to 72◦C. • Temperature of the reaction adjusted to 55◦C. • Penetration of the double-stranded DNA. • Temperature of the reaction adjusted to 95◦C. • Synthesis of DNA by the enzyme DNA polymerase. • Annealing of the primers to the single stranded DNA.

......

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Q20: The genome of an organism is: (1 mark) a) the full complement of DNA possessed by the organism. b) a description of the genes possessed by the organism. c) a physical map of the chromosomes possessed by an organism. d) a genetic map of the organism.

......

Q21: The DNA polymerase used in the polymerase chain reaction (PCR) possesses a particular characteristic that makes it ideally suited to the purpose. What is it? (1 mark) a) It contains DNA ligase. b) The enzyme is relatively stable at high temperatures. c) The enzyme is very accurate at copying DNA from a template. d) The enzyme synthesises DNA at a very rapid rate.

......

◦

Q22: During the process of PCR, occurs at 72 C. (pick from ’annealing’, ’exponential’, ’extension’ and ’heating’) (1mark) ......

Q23: During the process of PCR, DNA primers bind to DNA single strands during

. (pick from ’annealing’, ’exponential’, ’extension’ and ’heating’) (1mark) ......

Q24: During the process of PCR, separates the DNA into single strands. (pick from ’annealing’, ’exponential’, ’extension’ and ’heating’) (1mark) ......

Q25: Starting with a single molecule of DNA, the polymerase chain reaction was allowed to go through three complete cycles. How many molecules of DNA would be produced? (1 mark) a) 4 b) 8 c) 16 d) 32

......

Q26: Which of the following statements with regard to the primers involved in the polymerase chain reaction is incorrect? (1 mark) a) Primers allow DNA probes to attach to single stranded DNA. b) Primers allow DNA polymerase to attach. c) Primers prevent strands re-joining. d) Primers start addition of nucleotides.

......

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©HERIOT-WATT UNIVERSITY 169

Topic 8

Cell metabolism

Contents

8.1 Introduction ...... 171 8.2 Types of metabolic pathway ...... 172 8.2.1 Anabolic and catabolic pathways ...... 172 8.2.2 Reversible and irreversible pathways ...... 173 8.3 Control of metabolic pathways - the action of enzymes ...... 177 8.3.1 Control by regulation of enzyme production ...... 177 8.3.2 Regulation of the activity of continuously present enzymes ...... 179 8.4 The role of the active site ...... 180 8.4.1 The induced ﬁt model ...... 180 8.4.2 Reversible reactions ...... 184 8.4.3 Multi-enzyme complexes ...... 184 8.5 Control of metabolic pathways through enzyme inhibition ...... 185 8.5.1 Competitive and non-competitive enzyme inhibition ...... 185 8.5.2 Feedback inhibition ...... 187 8.6 Learning points ...... 189 8.7 Extended response question ...... 191 8.8 End of topic test ...... 191

Prerequisite knowledge You should already know that:

• enzymes are biological catalysts made by all living cells;

• enzymes are made of protein;

• enzymes lower the energy input required for chemical reactions;

• enzymes can speed up chemical reactions but remain unchanged;

• enzymes are involved in degradation and in synthesis reactions (you should be able to give examples);

• the characteristic shape of an enzyme molecule is complementary to its substrate. 170 TOPIC 8. CELL METABOLISM

You should also remember that chemical reactions in the human body involve the breakdown or synthesis of molecules to provide energy and building blocks to be used in other reactions. Learning objectives By the end of this topic, you should be able to:

• describe the role of metabolic pathways in the cell;

• explain the role of enzymes in metabolic pathways;

• describe the control of metabolic pathways.

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8.1 Introduction Metabolism is the term used to describe all the enzyme-catalysed reactions that take place within the cells of the body. A metabolic pathway is a sequence of reactions controlled by enzymes which change one metabolite into another. The metabolite produced at each stage, the product, is used as the substrate for the next stage of the pathway. For the cell, and the body of which it is part, to function coherently, these pathways must be integrated and subject to control processes.

Metabolism: Question Q1: Match each phrase on the left with the relevant term on the right. (4 marks) Go online All the chemical reactions of the body: substrate

The chemical on which an enzyme acts: metabolites

Chemical produced by an enzyme’s action: metabolism

Chemicals involved in cells reactions: product ......

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8.2 Types of metabolic pathway Learning objective By the end of this section, you should be able to:

• state that anabolic pathways are biosynthetic, building up more complex molecules from simpler ones and requiring the input of energy;

• state that catabolic pathways break down molecules, providing building blocks for other reactions and usually releasing energy;

• explain that metabolic pathways may have reversible and irreversible steps;

• explain that metabolic pathways may contain alternative routes that by-pass steps in the pathway.

8.2.1 Anabolic and catabolic pathways There are two types of enzyme-catalysed reaction:

1. biosynthetic reactions (also known as synthesis reactions), which are also referred to as anabolism;

2. breakdown reactions, which are also referred to as catabolism.

Anabolism Anabolic reactions require the input of energy to proceed, which is usually obtained from the breakdown of ATP to ADP and the transfer of the P i group (inorganic phosphate) to the substrate molecule. Example 1: glycogen synthetase is an enzyme that adds glucose residues to the storage polysaccharide, glycogen. It adds a high-energy form of glucose, uridine-diphosphate- glucose (UDP-glucose), to the existing chain of glucose residues that make up the glycogen molecule.

glycogen synthetase UDP- glycogen glycogen glucose + (n → UDP + (n+1 residues) residues)

Example 2: in the replication of DNA, the enzyme DNA polymerase adds free nucleotides to the 3 end of the newly-forming DNA strand. This results in elongation of the strand in the 5 → 3 direction.

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Catabolism Catabolic reactions involve the breakdown of complex molecules into simpler ones. This may also involve the release of energy, which is used to add an additional inorganic phosphate group to ADP to form ATP. Example: glycogen phosphorylase catalyses the removal of glucose-1-phosphate from glycogen in the liver and muscles, which can then be passed into the glycolysis pathway.

Anabolic and catabolic pathways: Question Q2: Complete the table using the words from the list. (4 marks) Go online

Anabolic Catabolic Biosynthesis Requires energy input

ATP → ADP + Pi DNA polymerase

Word list: ADP → Pi → ATP,Breakdown, Glycogen phosphorylase, May release energy......

8.2.2 Reversible and irreversible pathways Metabolic pathways are chains of enzyme catalysed reactions in which the product of one reaction is the substrate for the next. In some cases these steps can be reversed, whereas others are irreversible, so that once that reaction has taken place the metabolite must pass onto a new stage in the pathway. The following activity illustrates the start of the respiration process in all cells. It includes the first five stages in the glycolysis pathway, known as the (energy) investment, or preparatory, phase. This subject will be revisited in the next topic, but as it illustrates so many of the key points of this section, it is useful to consider it here. There are two important points to be aware of: firstly, although initially it looks very complicated, the logic of the activity and diagram should become clear with explanation; secondly, you will not asked to remember it.

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Glycolysis: Steps The following provides a summary of the steps involved in the energy investment phase Go online of glycolysis.

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General points

• The molecular diagrams show the arrangement of the carbon, hydrogen and oxygen atoms in each molecule, and the position of the phosphate groups which become attached as the metabolites pass along the chain.

• The names of the enzymes are certainly more complicated that the pepsin and amylase of earlier courses, but they are also much more informative once the code is understood. An important example in this pathway are the kinases, a family of enzymes which catalyse the addition to molecules of phosphate derived from the breakdown of ATP to ADP, in a process called phosphorylation.

• Isomerase enzymes alter the chemical structure of a molecule without changing the chemical formula.

• The ﬁrst part of the enzyme name identiﬁes the substrate, so phospho-fructo- kinase is a kinase enzyme which acts on fructose that has been phosphorylated.

• The product of the action of phosphofructokinase is fructose 1,6-biphosphate, i.e. a fructose sugar molecule which has phosphate groups attached to carbons 1 and 6.

• The second part of glycolysis pathway, the pay-off phase, which releases energy at some steps to make ATP, has as its ﬁnal product two molecules of pyruvate for every one of glucose used. This is covered in detail in the next topic.

Reversible and irreversible steps

• The steps in the chain which involve the addition of phosphate (phosphorylation) also use up energy. These steps are irreversible.

• The step from glucose 6-phosphate to fructose 6-phosphate is reversible, with the phosphoglucose isomerase enzyme re-arranging the atoms of the glucose to form fructose, its isomer, without the addition of other atoms or groups. Under normal conditions, the fructose 6-phosphate is used up in the next step, so its concentration is low and presence of glucose 6-phosphate causes the reaction to proceed from left to right. However, if the fructose 6-phospoate builds up, the reaction will run the other way, changing it back to glucose 6-phosphate.

• The ﬁnal step in this part of the chain sees the action of an aldolase enzyme which forms, or in this case breaks, carbon-carbon bonds to split fructose 1,6- biphosphate into two 3-carbon molecules which are isomers of each other.

• Of the isomers so-formed, only glyceraldehyde 3-phosphate is the substrate for the next step so the dihydroxyacetone phosphate must be converted into it by another isomerase enzyme.

• In times of excess lipid or protein energy sources, certain reactions in the glycolysis pathway may run in reverse in order to produce glucose 6-phosphate which is then used for storage as glycogen.

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Alternative routes

• An alternative route exists, called the pentose phosphate pathway, which takes glucose 6-phosphate and eventually produces NADPH, important in reductive processes such as the synthesis of fatty acids, and ribose, the 5-carbon sugar in nucleic acids.

• For those metabolic pathways that have stages that cannot be reversed, there are often alternative pathways that can overcome any blockage.

......

Reversible and irreversible pathways: Questions Q3: State two key differences between reversible and irreversible reactions in the Go online investment phase of glycolysis......

Q4: State the conditions which will determine the direction in which a reversible reaction will run......

Q5: Describe an alternative pathway to the investment phase of glycolysis......

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8.3 Control of metabolic pathways - the action of enzymes Learning objective By the end of this section, you should be able to:

• state that metabolic pathways may be controlled by the production, activation or inhibition of particular enzymes;

• state that metabolic pathways may be controlled by the regulation of the activity of key enzymes in the pathway;

• state that the rate of action of particular enzymes in a pathway may be regulated by both intra- and extracellular signal molecules;

• state that the genes for some enzymes are continuously expressed, and so these enzymes are always present in the cell;

• explain that control of such continuously present enzymes is achieved through regulation of their rate of reaction;

• state that most metabolic reactions are reversible;

• explain that the direction of reversible pathways is determined by the presence of the substrate and the relative concentration of the product of the reaction. In all cells, there is the potential for a huge range of enzyme-assisted reactions to take place. Control of the metabolic pathways of the cell is achieved by adjusting the activity of the enzymes which catalyse the reactions in the pathway. During the process of differentiation, many of the genes which code for enzymes will have been inactivated. In addition, in any particular cell, only a subset of the total possible enzymes for that cell will be required to be functioning at any one time. Of the rest, some will not be produced at all as their genes have been ’switched off’ temporarily; others will be present in the cytoplasm, but either not activated or inhibited from acting.

8.3.1 Control by regulation of enzyme production Inborn errors of metabolism The ultimate examples of the control of metabolic pathways by the production of enzymes are the inherited diseases caused by single gene mutations mentioned in Topic 6.6 - the so-called inborn errors of metabolism. One such example is phenylketonuria (PKU), which results from a mutation to the gene that codes for the conversion of the amino acid phenylalanine into tyrosine. The entire metabolic pathway is summarised as:

phenylalanine → tyrosine → LDOPA → dopamine → noradrenaline → adrenaline Phenylalanine is an ’essential’ amino acid, i.e. one that cannot be synthesised in the body and must be obtained from the diet. Tyrosine, on the other hand, can be synthesised by the above pathway and so is ’non-essential’. The last three compounds in the pathway are all important neurotransmitters.

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People with PKU lack the enzyme phenylalanine dehydroxylase and cannot carry out the conversion of phenylalanine to tyrosine. They can obtain tyrosine from their diet in the form of high protein foods and so can produce the neurotransmitters. However, a build-up of phenylalanine in the body will lead to intellectual disabilities and seizures unless its intake in proteins is controlled. Regulation of gene expression Some enzymes are only required occasionally or at specific times, and so are not continuously present in the cytoplasm. Their genes are expressed in response to particular circumstances, which are usually the presence of certain signal molecules in the tissue fluid. One example involves the hormone estrogen. It is most often associated with its production in the ovaries, triggering the development of secondary sexual characteristics in girls, and with the homeostatic control of the menstrual cycle. However, estrogen is also produced in smaller quantities in the liver, adrenal glands and the breasts, and is involved in the induction of a wide range of changes in the body. It is also produced in men, where it contributes to the maturation of sperm. The production of estrogen varies widely and it is the different levels of estrogen reaching cells that influence their activity. Because of its hydrophobic nature, estrogen is able to cross the phospholipid bilayer of the cell membrane and so enter all cells, but it will only become effective in cells which contain estrogen receptor molecules in the cytoplasm. The estrogen molecules attach to these receptors and then they move as a complex to the nucleus. This complex binds to its specific target gene and activates its transcription to form the enzymes which initiate the desired reactions. Regulation by intra- and extracellular signal molecules In cellular terms, signal molecules transmit information either between or within cells. In the example given in the text above, estrogen carries information to the cells of the body, causing them to initiate various cellular activities. Although the receptors for estrogen are located within the cell, other receptors for signal molecules are found in the cell membrane. Such receptors span the membrane and set off a sequence of protein-protein interactions within the cell when a signal molecule attaches to them. These proteins act as intracellular signal molecules which ultimately trigger the expression of the target gene in the nucleus. Such complex transduction pathways offer opportunities for:

• feedback, through the level of products affecting earlier steps on the pathway;

• signal ampliﬁcation, where one signal molecule leads to many responses in the cell; and

• interactions between pathways, whereby products from reactions in one pathway are passed to others.

Examples are the cytokines, which are involved in the inﬂammatory response. Here, the extracellar signal molecule is the cytokine molecule and the intracellular signal molecules are the cascade of interacting proteins within the cell which inﬂuence gene transcription.

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Control by regulation of enzyme production: Questions Q6: What event causes the condition phenylketonuria? ...... Go online

Q7: How does PKU affect the metabolic pathway leading to adrenaline? ......

Q8: Explain the difference between extra- and intracellular signal molecules......

Q9: Give one example of: (i) an extracellular signal molecule; (ii) an intracellular signal molecule......

8.3.2 Regulation of the activity of continuously present enzymes Enzymes that are continuously present in the cytoplasm can be regulated in a number of ways.

• As described in a later section, the concentration of the end-product of a reaction may act as a negative feedback mechanism to control a reversible reaction.

• Post-translational modiﬁcation may also regulate enzyme activity. The phosphorylation of glycogen synthase in response to the hormone insulin determines whether glycogen is built up or broken down.

• The local environment may also affect the action of enzymes, e.g. changing pH.

• Localisation within the cell. Enzymes which carry out related functions are held in groups in particular parts of the cell, e.g. in the Golgi apparatus or the mitochondrion.

Enzyme regulation: Question Q10: Match each of the enzyme regulation mechanisms on the left with the examples on the right. (4 marks) Go online

phosphorylation of glycogen Concentration of end-product: synthase

Post-translational modiﬁcation: porphyrin synthesis

Local environment: Golgi apparatus

Localisation within the cell: changing pH ......

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8.4 The role of the active site Learning objective By the end of this section, you should be able to:

• explain the induced ﬁt concept of enzyme action;

• explain that the active site orientates reactants, lowers the activation energy of the transition state, and releases products with a low afﬁnity for the active site;

• explain that most metabolic reactions are reversible;

• explain that the presence of a substrate or the relative concentration of a product will determine the direction of reversible reactions;

• explain that enzymes often act in groups or as multi-enzyme complexes.

8.4.1 The induced ﬁt model Molecules need a certain minimum energy before a reaction will take place. The energy needed to allow a reaction to occur is called the activation energy. The presence of a catalyst ensures that the initial energy requirement for a reaction is lowered, therefore more molecules have this energy and the reaction takes place faster. It is like rolling a boulder down a hill, but having to push it up a small hump ﬁrst - this initial push takes energy, but after that the boulder rolls on. Enzymes make the small hump even smaller!

Activation energy hill analogy

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Activation energy The action of an enzyme can be represented by the following equation:

E + S → ES → EP → E + P enzyme- enzyme- enzyme substrate substrate product enzyme product complex complex

This emphasises that, for a reaction to take place, the substrate must collide with the enzyme and that the reaction takes place on the surface of the enzyme. The activity of enzymes depends on their ﬂexible and dynamic shape. There is an afﬁnity of substrate molecules for the active site of an enzyme; once in position there are two proposed theories of how enzymes work.

1. The lock and key hypothesis: the substrate and enzyme ﬁt exactly.

2. The induced ﬁt hypothesis: the substrate alters the enzyme’s active site, causing a reaction and allowing the products to leave.

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The lock and key hypothesis: Steps The following provides a summary of the lock and key hypothesis - the substrate is the Go online exact ﬁt of the enzyme’s active site in the same way as a key ﬁts exactly into a lock. 5 min

Lock and key hypothesis ......

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The induced fit hypothesis: Steps The following provides a summary of the induced fit hypothesis - the substrate molecule does not fit exactly but instead induces a slight change in the shape of the active site to Go online allow the substrate molecule to fit perfectly. This conformational change in shape of the 5 min active site facilitates the reaction.

Induced ﬁt hypothesis ......

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8.4.2 Reversible reactions As mentioned above, in the glycolysis pathway there are some steps which can be reversed and others which cannot. Those which are reversible do not involve the addition of phosphate from the breakdown of ATP. For example, the step from glucose 6-phosphate to fructose 6-phosphate is reversible, the phosphoglucose isomerase enzyme re-arranging the atoms of the glucose to form fructose, its isomer, without the addition of other atoms or groups. The direction of such reversible reactions is determined by two factors. In the example mentioned in the text above, under normal conditions the fructose 6-phosphate product is used up as the substrate in the next step so its concentration is low. The presence of glucose 6-phosphate then causes the reaction to proceed from left to right. However, if the concentration of fructose 6-phosphate builds up, the reaction will run the other way, changing it back to glucose 6-phosphate.

Reversible reactions: Question Q11: State the two factors which can determine the direction of a reversible reaction. Go online ......

8.4.3 Multi-enzyme complexes Enzymes often act in groups, or as multi-enzyme complexes, which are a form of quaternary protein structure. In these complexes, the proteins are linked by weak bonds, and are held together by membrane proteins and the proteins of the cell cytoskeleton. Where the enzymes in the complex form a sequence which is a metabolic pathway; they are referred to as a metabolon. These complexes are a vital part of most cellular processes, e.g. synthesis of DNA and RNA, glycolysis, and the Krebs/Citric Acid/Tricarboxylic Acid (TCA) cycle. The presence of the enzymes in close proximity to each other means that not only is the product of the action of one enzyme immediately available as the substrate for the next enzyme in the chain, but it is also presented so that it ﬁts immediately into the active site.

Multi-enzyme complexes: Questions Q12: Give an example of a multi-enzyme complex (metabolon). Go online ......

Q13: State two ways in which the metabolon makes such complex biochemical pathways possible......

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8.5 Control of metabolic pathways through enzyme inhibition Learning objective By the end of this section, you should be able to:

• describe non-competitive inhibition or stimulation of enzyme activity by the binding of molecules that change the shape of the active site;

• describe how competitive inhibitors compete for the active site and resemble the substrate;

• state that competitive inhibition is reversible;

• explain how competitive inhibition is reduced by increasing the substrate concentration;

• explain how end products can inhibit an enzyme which catalyses a reaction early in a pathway, and therefore control the pathway. Genes for some enzymes are continuously expressed, leading to the permanent presence of these enzymes within cells. These enzymes are always present in the cell and their control involves regulation of their rate of reaction. There are many different ways in which gene expression is controlled. Enzymes may have their activity reduced or stopped, either temporarily or permanently, by different forms of inhibition. Alternatively, they may be ’switched on’ by enzyme activators.

8.5.1 Competitive and non-competitive enzyme inhibition An inhibitor is a substance which slows down, or stops, the activity of an enzyme. Two different types of inhibition are competitive and non-competitive. Competitive inhibition In competitive inhibition, a molecule that is very close in shape to the true substrate competes for the active site of the enzyme and effectively reduces the concentration of the available enzyme. The inhibitor attaches temporarily to the active site, blocking the substrate from attaching. This type of inhibition can be reversed by increasing the concentration of the substrate. At high concentration, there is a greater likelihood of a substrate molecule attaching to an active site than an inhibitor molecule.

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Competitive inhibition The following provides a summary of the steps involved in the process of competitive Go online inhibition.

...... Non-competitive inhibition An inhibitor molecule binds to the enzyme at a different area (away from the active site), changing the conformation (the shape) of the enzyme and altering the shape of the active site. As a result, the catalytic efficiency of the enzyme is reduced; indeed, the shape of the active site may be altered so much that the reaction cannot occur at all. Increasing substrate concentration will not increase the rate of reaction. The binding site of a non-competitive inhibitor molecule is called an allosteric site. On the enzyme molecule, there will actually be allosteric sites for both inhibitors and activators. This allows the enzyme to have its activity switched ’on’ and ’off’ by the presence of the relevant inhibitor and activator molecules. In the same way that an inhibitor molecule binding to an allosteric site alters the shape of the enzyme molecule so that the active site no longer fits the substrate molecule, an activator molecule binding to an activator site will cause the active site to alter so that it does fit the substrate.

Non-competitive inhibition: Steps The following provides a summary of the steps involved in the process of non- Go online competitive inhibition.

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...... In non-competitive inhibition the active site is not available to the substrate, thus the rate of reaction is not generally inﬂuenced by the substrate concentration. Increasing the substrate concentration will have little or no effect on enzyme activity because the enzyme molecules have been denatured by the non-competitive inhibitor.

Competitive and non-competitive inhibition of enzymes: Question Q14: Complete the table using the labels provided. Go online

......

8.5.2 Feedback inhibition Metabolic pathways can be controlled by feedback inhibition. When the product of a series of enzymatic reactions, for example an amino acid, starts to accumulate in the cell, it can inhibit the action of the ﬁrst enzyme involved in its synthesis. Further action of the enzyme is halted. This is called feedback inhibition. As the level of the end product drops, the inhibition is removed and the pathway will proceed. Feedback inhibition prevents a wasteful build-up of product occurring.

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Feedback inhibition: Summary The following provides a summary of the process of feedback inhibition. Go online

......

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An example of feedback inhibition in a synthetic enzyme pathway is the transamination (change in the R group) of the amino acid threonine to isoleucine. Isoleucine, the end- product in the sequence, can inhibit threonine deaminase, the initial enzyme in the sequence. The following diagram illustrates this.

Feedback inhibition example

8.6 Learning points

Summary Types of metabolic pathway

• Anabolic pathways are biosynthetic, building up more complex molecules from simpler ones and requiring the input of energy.

• Catabolic pathways break down molecules, providing building blocks for other reactions and usually releasing energy. • Metabolic pathways may have reversible and irreversible steps.

• Metabolic pathways may contain alternative routes that by-pass steps in the pathway.

Control of metabolic pathways - the action of enzymes

• Metabolic pathways may be controlled by the production, activation or inhibition of particular enzymes.

• Metabolic pathways may be controlled by the regulation of the activity of key enzymes in the pathway.

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Summary continued • The rate of action of particular enzymes in a pathway may be regulated by both intra- and extracellular signal molecules.

• The genes for some enzymes are continuously expressed and so these enzymes are always present in the cell.

• Control of such continuously present enzymes is achieved through regulation of their rate of reaction.

• Most metabolic reactions are reversible.

• The direction of reversible pathways is determined by the presence of the substrate and the relative concentration of the product of the reaction.

The role of the active site

• By the induced ﬁt concept of enzyme action, the substrate entering the active site makes the site change and ﬁt more closely, causing a reaction before allowing the products to leave.

• The active site orientates reactants, lowers the activation energy of the transition state, and releases products with a low afﬁnity for the active site.

• Most metabolic reactions are reversible.

• The presence of a substrate, or relative concentration of a product, will determine the direction of reversible reactions.

• Enzymes often act in groups or as multi-enzyme complexes.

Control of metabolic pathways through enzyme inhibition

• The binding of molecules that change the shape of the active site may cause non-competitive inhibition or stimulation of enzyme activity.

• Competitive inhibitors compete for the active site and resemble the substrate.

• Competitive inhibition is reduced by increasing the substrate concentration.

• Competitive inhibition is reversible.

• Feedback control acts through end products, inhibiting an enzyme which catalyses a reaction early in a metabolic pathway.

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8.7 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of the control of the action of enzymes before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended response question: The control of the action of enzymes Describe the control of the action of enzymes which are continuously present in the cell. (10 marks) ......

8.8 End of topic test End of Topic 8 test Q15: Complete the table using the words and phrases from the list. (6 marks) Go online

Biosynthetic means Anabolic pathways Catabolic pathways Steps which require energy are Reactions which can run both ways are Alternative routes

Word and phrase list: building up complex molecules, by-pass steps in a pathway, irreversible, require energy input, reversible, usually release energy.

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......

Q16: Complete the sentences by matching the parts on the left with the parts on the right. (7 marks)

Regulation of the activity of key enzymes substrate. enables control of

continuously Extracellular signal molecules regulate expressed genes.

Protein-protein interactions are set off by metabolic pathways.

Some enzymes are always present in cells as a product. result of

A reaction may be initiated by the presence of hormones.

intracellular signal A reaction may be reversed by the build-up of molecules......

Q17: Complete the paragraph using the words from the list. (10 marks) The site is the point on the enzyme molecule to which the molecule(s) temporarily binds. In many cases, the active site does not exactly ﬁt the substrate and it

is the substrate molecule itself which causes the necessary change in the of the

enzyme molecule. This is called ﬁt. The active site reactants, lowers energy, and releases products with low

for the active site. Many metabolic reactions are and the presence of a substrate or relative of a product will determine their direction.

Glycolysis is an example of metabolic pathway controlled by a metabolon, or . Word list: activation, active, afﬁnity, concentration, conformation, induced, orientates, multi-enzyme complex, reversible, substrate.

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......

Q18: Complete the deﬁnitions by matching the statements on the left with the descriptions and terms on the right. (6 marks)

Sites where activators may bind: competitive.

Inhibitors bind to sites away from the active site: allosteric.

Inhibitors which resemble substrate: reversible.

Increase reduces competitive inhibition: feedback control.

substrate Competitive inhibition: concentration.

End-products inhibit earlier enzymes: non-competitive......

Q19: Explain the difference between anabolic and catabolic pathways. (4 marks) ......

Q20: What controls the direction of a reversible pathway? (2 marks) ......

Q21: Explain the role of the active site. (3 marks) ......

Q22: Explain feedback control of a metabolic pathway. (2 marks) ......

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Topic 9

Cellular respiration I: Glycolysis

Contents

9.1 Introduction ...... 196 9.2 Summary of glucose breakdown ...... 197 9.3 Roles of ATP in the cell ...... 199 9.3.1 What is ATP? ...... 199 9.3.2 ATP and energy transport ...... 200 9.3.3 ATP and phosphorylation ...... 201 9.4 Glycolysis ...... 202 9.4.1 Introduction ...... 202 9.4.2 The energy investment stage ...... 203 9.4.3 The energy pay-off stage ...... 205 9.5 Regulation of the respiratory pathway ...... 208 9.5.1 Feedback inhibition and regulation of the glycolytic pathway ...... 208 9.5.2 Synchronisation of the rates of glycolysis and the citric acid cycle . . . . 208 9.6 Learning points ...... 210 9.7 Extended response question ...... 212 9.8 End of topic test ...... 212

Learning objectives By the end of this topic, you should be able to:

• give a general description of cellular respiration;

• describe the structure, formation and functions of ATP;

• describe the process of glycolysis;

• explain how the respiratory pathway is regulated. 196 TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS

In this topic, some of the material included is not in the syllabus. The purpose of this is to set the required content into its proper context so that it may be learnt through an understanding of the processes, rather than memorised by means of repetition. The questions in each section largely focus on the syllabus material. The learning points and end-of-topic test address only the knowledge required by the syllabus.

9.1 Introduction Learning objective By the end of this section, you should be able to:

• explain the central role of respiration in cell metabolism. One of the laws which govern the behaviour of energy in the universe is the Second Law of Thermodynamics, a simple interpretation of which is that complex structures will tend to break down and release their energy to other simpler structures which are part of the same system. Living things constantly oppose this tendency by building more complex molecules from simpler ones. The energy for these anabolic reactions and for other purposes, such as movement or keeping warm, is provided by respiration, which is a catabolic process. Respiration provides the fuel for all of the metabolic processes that require an energy input in the cell. Furthermore, being a long chain of reactions, the metabolic pathways of respiration are connected to many other pathways because intermediate metabolites can be passed down alternative routes.

Introduction: Questions Q1: Why is respiration central to all metabolic processes? Go online ......

Q2: What name describes reactions which build up complex molecules from simpler ones? ......

Q3: What name describes reactions which break down complex molecules to simpler ones? ......

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9.2 Summary of glucose breakdown Learning objective By the end of this section, you should be able to:

• describe in outline the process of glucose breakdown in respiration. Respiration can be divided into several linked stages. The initial one is the breakdown of glucose to pyruvate in the cytoplasm, which is called glycolysis (from ’glycose’, an earlier name for glucose, and ’lysis’, meaning ’splitting’). It releases a relatively small quantity of energy in the form of two molecules of ATP per glucose molecule and two carrier molecules with hydrogen attached. Subsequently, different pathways are available depending on whether oxygen is available or not. In anaerobic conditions, in the absence of oxygen, pyruvate enters an anaerobic pathway which has lactic acid as its ﬁnal product. In aerobic conditions, when oxygen is present, pyruvate enters the mitochondria where it reacts to form acetyl Co-enzyme A and passes into a cycle of reactions known variously as:

• the Krebs Cycle, named after Hans Adolf Krebs, the Nobel Prize-winning biochemist who first worked out the steps in the cycle; • the Citric Acid Cycle, named after the molecule which is first formed after acetyl Co-enzyme A reaches the cycle, the name being derived from the citrus fruit from which it was first identified; • the Tricarboxylic Acid (TCA) Cycle, using the modern name for citric acid.

In the course of the cycle, dehydrogenase enzymes remove hydrogen ions and electrons to carriers which transport them to the ﬁnal stage of aerobic respiration. Additionally, carbon dioxide is released, ATP (or GTP) is generated, and the initial acceptor molecule is re-established. In aerobic conditions, the last stage of respiration is the cytochrome/hydrogen transfer/electron transfer system, which is also located in the mitochondrion. The alternative names stem from:

• cytochrome, a respiratory pigment involved as a hydrogen receptor in the system (a related pigment is active in the chloroplast); • hydrogen and (high energy) electrons which are passed along a chain of receptors, giving out energy as they go. The energy released during aerobic respiration is used to generate a total of 38 molecules of ATP from each glucose molecule; • ﬁnally, the hydrogen is combined with oxygen and forms water, the other ’waste’ product of the pathway.

Glycolysis and the citric acid cycle are considered to be very ancient, probably having evolved along with the earliest organisms. Interestingly, some direct descendents of these ancient organisms, the bacteria-like Archaea which live in extreme conditions such as hot springs in volcanic regions, are able to run the TCA cycle in reverse, generating carbon compounds from carbon dioxide and water. They use hydrogen and sulﬁdes to supply the electrons to drive the process.

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Summary of glucose breakdown: Question Q4: Choose words from the list to complete the labelling of the respiration diagram. Go online

Word list: 2, 36, acetyl CoA, citric acid cycle, electron transport system, glycolysis, hydrogen, lactic acid, oxygen present, pyruvate......

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9.3 Roles of ATP in the cell Learning objective By the end of this section, you should be able to:

• describe the role of ATP in the transfer of energy within the cell;

• state that ATP is used to transfer energy to synthetic pathways and other cellular processes where energy is required;

• explain the role of ATP in phosphorylation.

9.3.1 What is ATP? ATP is adenosine triphosphate and, in common with nucleic acids, glycolysis and the citric acid cycle, it is found in virtually all living cells. Typically, although only 250g of ATP are present in the adult human body at any one time, in the course of a day the body will make its own weight of the chemical as it is being constantly created and broken down. ATP’s principal function is very short-term and local: it transports energy from the site of its release in the cell to areas where anabolic reactions are occurring in that cell. If energy is to be stored in the body, it is put into the form of carbohydrate, as glycogen, or fat; for transport between cells, it is in the form of glucose, lipids (triglycerides) or fatty acids. Chemically, ATP consists of three parts: the 5-carbon sugar ribose; the base adenine, as found in DNA and RNA, attached to the 1 carbon atom of the ribose; and three phosphate groups, attached in a chain to the 5 carbon of the ribose. Adenosine is the term for the adenine base attached to the ribose sugar. In simple form, the molecular structure of ATP is:

adenine - ribose - phosphate - phosphate - phosphate

A more accurate depiction of this structure is as follows:

The structure of ATP

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A little background information Adenosine monophosphate (AMP) is also present in the cell, both as the adenine nucleotide in nucleic acids and, in a cyclic form (cAMP), as an important messenger in intracellular signalling. Of the other nitrogenous bases found in nucleic acids, guanine has the same basic two- ring structure as adenine; together they are called purines. Cytosine and thymine/uracil are single-ring molecules and called pyrimidines. Guanine is also found acting as part of an energy transport molecule as guanosine triphosphate (GTP) but, unlike ATP, it is only involved in protein synthesis and in gluconeogenesis, the formation of glucose from non-carbohydrate substrates (vital to maintaining constant blood sugar levels in humans, but also found in all organisms). GTP is also essential to signal transduction, in which extracellular signal molecules attach to receptors on the cell membrane and secondary messengers amplify the signal into the cell.

What is ATP: Question Q5: In simple terms, describe the structure of the ATP molecule. Go online ......

9.3.2 ATP and energy transport Because the third phosphate group in ATP is furthest from the ribose sugar, the formation and breaking of its bond with the second phosphate involves more energy than either of the bonds of the other two phosphates. In the watery environment of the cell, ATP is relatively unstable, and decomposes to ADP (adenosine diphosphate) and Pi (inorganic phosphate).

It is the ready interconversion of ATP and ADP (adenosine diphosphate) + P i (inorganic phosphate) which allows ATP to act as the cell’s principal energy transporter. ATP is reconstituted by phosphorylation, i.e. the enzyme-assisted addition of a phosphate group, which reforms the third phosphate bond.

The interconversion of ATP and ADP

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ATP is generated both by glycolysis and from the electron transport/cytochrome system. It then travels to those areas of the cell which require energy to power their activity; this might be anabolic processes such as protein synthesis or movement in muscle cells, or signal transduction through the cell membrane.

ATP and energy transport: Question Q6: Explain why ATP is used as an energy transporter in the cell...... Go online

9.3.3 ATP and phosphorylation Phosphorylation is the addition of a phosphate group to a molecule. The formation of ATP from ADP, in glycolysis and the electron transport chain, is therefore a phosphorylation. However, when ATP is hydrolysed to ADP and P i, the energy released can be used to add the phosphate group to another molecule. The ﬁrst step in the glycolysis pathway involves such a phosphorylation. The enzyme hexokinase transfers the phosphate group from ATP to the 6 carbon of the glucose molecule.

The ﬁrst step in the glycolysis pathway The effect of this is twofold. Firstly, it prepares the glucose for the next step in the chain of reactions. Secondly, it reduces the concentration of glucose in the cell and, at the same time, prevents its loss from the cell as glucose 6-phosphate is much less able to cross the cell membrane. Phosphorylation is a common means of switching enzymes ’on’ and ’off’ as the addition of the phosphate group alters their structure and activity. This is a form of post- translational modiﬁcation of the proteins.

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ATP and phosphorylation: Question Q7: What is meant by the term phosphorylation? Go online ......

Q8: When glucose is phosphorylated, what is formed? ......

Q9: What effect does post-translational modiﬁcation by phosphorylation have on enzymes? ......

9.4 Glycolysis Learning objective By the end of this section, you should be able to:

• summarise the main points of the glycolysis pathway;

• explain the key aspects of the energy investment and pay-off stages;

• explain the role of phosphorylation in glycolysis;

• describe the alternative pathways connected to the glycolytic pathway;

• describe the role of phosphofructokinase;

• describe the progression pathways in the presence and absence of oxygen;

• describe the ways in which the glycolytic pathway is regulated.

9.4.1 Introduction Glycolysis takes place in the cytoplasm. It is a metabolic pathway involving ten steps, each of which is enzyme-assisted. Such a sequence would be impossible if these enzymes were randomly distributed in the cytoplasm; in fact, they are held in multi- enzyme complexes that are attached to the structural proteins of the cytoskeleton so that the product of each step is readily available as the substrate for the next enzyme in the chain. In summary, during glycolysis a 6-carbon molecule of glucose is split into two molecules of 3-carbon pyruvate. In the course of the pathway, two molecules of ATP are used up in the initial ’energy investment’ stage and four are released in the later ’energy pay-off’ stage. Two hydrogen ions are also released and, in the presence of oxygen, they are transported by carrier molecules (NADH) to the cytochrome system. In the absence of oxygen, they are used to turn pyruvate into lactic acid.

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Glycolysis: Questions Q10: Where in the cell does glycolysis take place? ...... Go online

Q11: How does the arrangement of the enzymes of glycolysis in a multi-enzyme complex improve their efﬁciency? ......

Q12: What are the ﬁnal products when one molecule of glucose is passed along the glycolytic pathway? ......

9.4.2 The energy investment stage The first part of glycolysis, the energy investment stage, is so-called because it uses up two molecules of ATP. It starts with a glucose molecule as the first substrate and finishes with two molecules of glyceraldehyde 3-phosphate as its final product. The 6-carbon glucose is split into two 3-carbon molecules. During this stage, two phosphorylations take place. In each of these, a molecule of ATP is converted to ADP; the phosphate group which is released is added to the substrate molecule. The first occurs when glucose is converted to glucose 6-phosphate by the enzyme hexokinase. This step is irreversible, but the glucose 6-phosphate can be used in a variety of metabolic pathways:

• it can progress to the next step in the glycolysis pathway;

• it can enter the pentose phosphate pathway, generating the reducing agent NADPH, which supplies the hydrogen necessary for several reactions, such as fatty acid synthesis, and the precursor of nucleotides for DNA synthesis in the form of ribose phosphate;

• it can be converted to glycogen for storage.

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The energy investment stage

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The second phosphorylation is the conversion of fructose 6-phosphate to fructose 1,6- biphosphate, which adds a second phosphate at the 1’carbon position. This step is catalysed by the enzyme phosphofructokinase, is irreversible, and can only lead on to the glycolytic pathway.

The energy investment stage: Questions Q13: Why is the energy investment stage so-called? ...... Go online

Q14: Apart from energy, how does ATP contribute to the phosphorylation reactions in this stage during glycolysis? ......

Q15: Comparing the phosphorylation reactions to the other reactions in the pathway, suggest what makes the phosphorylations irreversible......

9.4.3 The energy pay-off stage During this stage, a total of 4 ATPs are generated, two from the processing of each molecule of glyceraldehyde 6-phosphate. This gives a net gain of two molecules of ATP for each glucose molecule, as this stage runs twice for each glucose molecule entering the pathway.

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The energy pay-off stage

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Like the energy investment stage, the pay-off stage consists of several enzyme-assisted reactions, some of which are reversible and others which are not. Two of the reactions are phosphorylations, catalysed by kinase enzymes, but, in this case, the phosphate group is being transferred to ADP, forming ATP, rather than from ATP to a substrate molecule. This is known as substrate-level phosphorylation because the phosphate group comes directly from the substrate (as opposed to the oxidative phosphorylation that takes place in the cytochrome system). The ﬁrst step in the pay-off stage is catalysed by a dehydrogenase enzyme, and the hydrogen released is attached to the coenzyme carrier molecule NAD+ to form NADH. In the presence of oxygen, this enters the electron transport chain of the mitochondrion and generates ATP.In the absence of oxygen, it is used to convert pyruvate to lactic acid in the cytoplasm. The ﬁnal product of this stage is pyruvate, which has links to several metabolic pathways:

• in the presence of oxygen, pyruvate can enter a mitochondriom and be converted to an acetyl group attached to Co-enzyme A (acetyl-coA) by the loss of carbon dioxide and hydrogen. Acetyl-coA then enters the citric acid cycle;

• in the absence of oxygen, it can be converted by fermentation to lactic acid with the addition of the hydrogen released in glycolysis. No further ATP is generated;

• it can be converted to carbohydrates, by the process of gluconeogenesis, fatty acids, and the amino acid alanine.

The energy investment stage: Questions Q16: How many ATP molecules are generated from each pyruvate molecule? ...... Go online

Q17: What is the net yield of ATP molecules from each glucose molecule? ......

Q18: What type of reaction produces ATP? ......

Q19: Why is this type of ATP formation called substrate-level phosphorylation? ......

Q20: In the presence of oxygen, how may further energy be released from pyruvate? ......

Q21: In the absence of oxygen, how may pyruvate be removed from the cell? ......

Q22: Apart from glucose and lactic acid, name two other substances into which pyruvate may be converted......

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9.5 Regulation of the respiratory pathway Learning objective By the end of this section, you should be able to:

• explain what is meant by ’feedback inhibition’;

• describe the regulation of the glycolysis pathway;

• explain how the rates of glycolysis and the citric acid cycle are synchronised.

9.5.1 Feedback inhibition and regulation of the glycolytic pathway Feedback inhibition in a metabolic pathway occurs when an enzyme’s activity is reduced by an increase in the concentration of a product from a later step in the pathway. This ensures that the potentially damaging build-up of metabolites is avoided, and that resources, e.g. energy in the form of ATP, are not expended unnecessarily. In the glycolytic pathway, there is a good example of feedback inhibition. In the energy investment stage, the enzyme phosphofructokinase catalyses the phosphorylation of fructose 6-phosphate to fructose 1,6-biphosphate. This step is irreversible and can only lead on to the energy pay-off stage, serving as a key point to control the whole respiration process. Phosphofructokinase is inhibited by ATP, which binds to a part of the enzyme other than the active site (an allosteric site) and reduces its afﬁnity for its substrate, fructose 6-phosphate. Consequently, if the ATP concentration rises in the cell, phosphofructokinase will become less active and the production of ATP from both glycolysis and the cytochrome system will be reduced. As the concentration of ATP falls, the concentration of AMP (adenosine monophosphate) rises; as this reverses the effect of ATP on the enzyme, the pathway will be activated again.

Feedback inhibition and regulation of the glycolytic pathway: Questions Q23: What is meant by the term ’feedback inhibition’ in a metabolic pathway? Go online ......

Q24: Of what advantage to the cell is feedback inhibition? ......

Q25: Explain the inhibition of phosphofructokinase by ATP......

9.5.2 Synchronisation of the rates of glycolysis and the citric acid cycle The energy pay-off stage also includes steps which may be regulated. Both of the phosphorylation reactions which generate ATP are sensitive to the concentration of ADP in the cytoplasm. Low levels of ADP, which would indicate that ATP is not being used up and is accumulating, inevitably mean that these steps will be held up, controlling the production of ATP from glycolysis and the supply of pyruvate to the mitochondrion.

©HERIOT-WATT UNIVERSITY TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS 209

This, together with the feedback inhibition of phosphofructokinase by ATP, ensures that glycolysis is only supplying pyruvate to the mitochondria at a rate which is suited to the ATP availability within the cell, and hence the cell’s rate of use of energy. When the concentration of ATP is high, resources will be conserved; when the ATP concentration is low, indicating high activity in the cell, pyruvate will be made available at the maximum rate. It has been suggested that phosphofructokinase is also inhibited by high citrate levels (citrate being a salt of citric acid). Such conditions would indicate that the citric acid cycle was absorbing the Acetyl CoA produced from pyruvate but not using it up, and so it was diffusing out into the cytoplasm. This would then inhibit phosphofructokinase, slowing the glycolysis pathway and the production of pyruvate. However, this is open to debate. While it can be demonstrated in the lab (in vitro) that high levels of citrate will indeed inhibit phosphofructokinase, such high levels have not been found to occur in living cells (in vivo). In the same way, in vivo ATP concentrations do not seem to vary sufficiently for ATP to act as an effective inhibitor. These are good examples of the need to evaluate findings from experimental situations in the light of the conditions which exist within living organisms, and the difficulty of accurately simulating the conditions in the cell during experiments.

Synchronisation of the rates of glycolysis and the citric acid cycle: Questions Q26: How does the cell’s use of energy inﬂuence the pay-off stage of glycolysis? ...... Go online

Q27: Why is it important that the rates of glycolysis and the citric acid cycle are synchronised? ......

©HERIOT-WATT UNIVERSITY 210 TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS

9.6 Learning points

Summary Respiration and metabolism

• Respiration provides the energy for all of the cell’s anabolic reactions, movement and heat release.

• The respiration pathways provide intermediate compounds which may be used in several other metabolic pathways.

• Carbon dioxide and water are released as waste products.

Summary of glucose breakdown

• During the respiration process, glucose is broken down.

• Hydrogen ions and electrons are removed by dehydrogenase enzymes.

• ATP is formed and made available to the cell.

Roles of ATP

• ATP is adenosine triphosphate.

• ATP is readily hydrolysed to ADP and Pi (adenosine diphosphate and inorganic phosphate), releasing energy.

• ADP can be phosphorylated to ATP by the addition of P i and the necessary input of energy.

• ATP is generated through the glycolysis pathway and the cytochrome system.

• ATP travels from where it is formed to those areas of the cell which need energy.

• ATP provides energy for anabolic processes (e.g. protein synthesis), movement, and signal transduction across the cell membrane.

• Phosphorylation is the addition of a phosphate group to a molecule.

• The phosphorylation of glucose to glucose 1-phosphate at the ﬁrst step of glycolysis uses the phosphate released from the hydrolysis of ATP to ADP and Pi. Glycolysis

• Glycolysis takes place in the cytoplasm.

• In summary, one molecule of glucose yields two molecules of pyruvate, two molecules of ATP and two hydrogen ions as 2NADH.

• Glycolysis consists of two stages: an energy investment stage followed by an energy pay-off stage.

©HERIOT-WATT UNIVERSITY TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS 211

Summary continued • The energy investment stage includes two phosphorylations, both of which are irreversible.

• The ﬁrst phosphorylation adds phosphate to glucose to form glucose 6- phosphate.

• Glucose 6-phosphate can continue into several different metabolic pathways.

• The second phosphorylation involves the enzyme phosphofructokinase; its product can only progress further down the glycolytic pathway.

• The energy pay-off stage generates two molecules of ATP per pyruvate molecule.

• These ATP molecules are formed by the transfer of phosphate directly from the substrate to ADP, hence they are called substrate-level phosphorylations.

• The pyruvate formed from the energy pay-off stage can be used as a metabolite in several metabolic pathways.

• In the presence of oxygen, pyruvate may enter the mitochondria and the citric acid cycle.

• In the absence of oxygen, pyruvate may be converted to lactic acid.

Regulation of the respiratory pathway

• Feedback inhibition occurs when an enzyme’s activity is reduced by an increase in the concentration of a product from a later step in the pathway.

• Phosphofructokinase, in the energy investment stage of glycolysis, is inhibited by high levels of ATP.

• This will reduce the production of pyruvate, and therefore of ATP, when the cell’s need for energy is low.

• Phosphofructokinase may also be inhibited by high citrate levels in the cell.

• Glycolysis and the citric acid cycle are synchronised by the feedback inhibition of phosphofructokinase in the energy investment stage, and the supply of ADP to the phosphorylation reactions in the energy pay-off stage.

©HERIOT-WATT UNIVERSITY 212 TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS

9.7 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of feedback inhibition before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: Feedback inhibition State what is meant by the term ’feedback inhibition’, give an example of it, and explain 10 min its advantage to the cell. (7 marks) ......

9.8 End of topic test End of Topic 9 test

Q28: Complete the paragraph using the words from the list. (14 marks)

Go online Respiration provides for all the cell’s reactions, , and

release. In addition, compounds may be used in many other pathways.

During respiration, is broken down, is used up, and carbon

dioxide and are released. ions and are removed by

enzymes. As a result, is formed and made available to the cell. Word list: ATP, aerobic, anabolic, dehydrogenase, electrons, energy, glucose, heat, hydrogen, intermediate, metabolic, movement, oxygen, water.

©HERIOT-WATT UNIVERSITY TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS 213

......

Q29: Complete the table using the phrases from the list. (5 marks)

ATP is hydrolysed to ATP is formed by Glycolysis and the cytochrome system Energy for anabolic processes is Glucose to glucose 1-phosphate occurs by

Phrase list: ADP and Pi, generate ATP, phosphorylation, provided by ATP, the addition of Pi to ADP......

Q30: Complete the sentences by matching the parts on the left with the parts on the right. (10 marks)

forms two molecules of ATP per Glycolysis takes place pyruvate.

Glycolysis of one glucose molecule uses up two molecules of ATP.

pyruvate can be converted to lactic The energy investment stage acid.

The energy pay-off stage can join several metabolic pathways.

The energy investment stage yields two molecules of pyruvate. contains

Glucose 6-phosphate is formed in the cytoplasm.

Glucose 6-phosphate and pyruvate two irreversible phosphorylations.

The product of phosphofructokinase by phosphorylation of glucose.

Transfer of phosphate directly to ADP can only proceed to glycolysis.

In the absence of oxygen substrate-level phosphorylation.

©HERIOT-WATT UNIVERSITY 214 TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS

......

Q31: Complete the paragraph using the words from the list. (13 marks)

Glycolysis and the cycle are synchronised by the inhibition of

in the energy investment stage and the supply of to the phosphorylation

reactions in the energy stage. Feedback occurs when ’s activity is reduced by in the concentration of a product from step in

the pathway. Phosphofructokinase in the energy stage of glycolysis is inhibited

by levels of . This will reduce the production of , and therefore of ATP, when the cell’s need for energy is low. Word list: a later, an enzyme, an increase, ADP, ATP, citric acid, feedback, high, inhibition, investment, pay-off, phosphofructokinase, pyruvate......

Q32: Apart from providing energy, explain the other important role of the respiratory pathways. (1 mark) ......

Q33: List the products of glycolysis. (3 marks) ......

Q34: Explain why the formation of ATP in glycolysis is a substrate-level phosphorylation. (1 mark) ......

Q35: Explain how glycolysis and the citric acid cycle are synchronised. (2 marks) ......

©HERIOT-WATT UNIVERSITY 215

Topic 10

Cellular Respiration II: Citric acid cycle

Contents

10.1 Introduction ...... 216 10.2 The mitochondrion ...... 217 10.2.1 Structure of the mitochondrion ...... 217 10.2.2 Fascinating facts about mitochondria ...... 218 10.3 The citric acid cycle ...... 219 10.3.1 The steps of the citric acid cycle ...... 219 10.3.2 The enzymes and carriers of the citric acid cycle ...... 221 10.4 Alternative substrates for respiration ...... 224 10.5 Learning points ...... 226 10.6 Extended response question ...... 228 10.7 End of topic test ...... 228

Learning objectives By the end of this topic, you should be able to:

• describe the structure of the mitochondrion and the location of processes within it;

• describe the citric acid cycle, its generation of ATP, and its release of CO2 and hydrogen;

• list the different substrates available for respiration;

• explain how these different substrates are incorporated into the respiratory pathway. 216 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

10.1 Introduction In active cells in aerobic conditions, most of the pyruvate produced by glycolysis is used in the generation of ATP to provide the energy for the cell’s other functions. This is achieved by passing the pyruvate from the cytoplasm into the mitochondrion (plural: mitochondria), where it is first converted to acetyl coenzyme A and then passed round the citric acid cycle, from which carbon dioxide and hydrogen ions are released. The carbon dioxide passes out of the cell as a waste product, but the hydrogen ions (along with high energy electrons) are transferred by carriers to the electron transport system where they are passed along a series of acceptor molecules, generating the energy to form ATP as they go. The hydrogen is finally accepted by oxygen, so forming water which is the other waste product of aerobic respiration. Each of these many steps is catalysed by a different enzyme. Although we shall mainly consider aerobic respiration, which uses glucose as its substrate, there are several alternative respiratory substrates; these are dealt with after the basic process has been described. In this topic, we will consider the detail of the citric acid cycle, leaving the electron transport system/chain to the next topic. Despite the fact that both pathways are found in the mitochondrion, this has been done to reduce the quantity of material to be mastered at any one time. There is also the fact that, as with glycolysis, the processes are also fairly discrete. There is a point of possible confusion which should be clarified. Although the pathway is called the citric acid cycle, it is in fact the negatively charged ion (anion) citrate which is formed and passed into the cycle. In the same way, it is pyruvate (rather than pyruvic acid) which enters the mitochondrion. These anions are in the watery solution of the cell, and not linked to the positive hydrogen ions (cations) which would turn them into their acid equivalents. The names of the acids and the anions are often used interchangeably, and the distinction is not in the syllabus.

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 217

10.2 The mitochondrion Learning objective By the end of this section, you should be able to:

• describe the structure of the mitochondrion as consisting of an outer boundary membrane, an inner membrane, which is folded to form cristae, and the central matrix;

• explain that pyruvate enters the matrix of the mitochondrion and is broken down to an acetyl group which is attached to co-enzyme A, forming acetyl co-enzyme A (acetyl-coA);

• explain that acetyl-coA passes the acetate into the citric acid cycle in the matrix of the mitochondrion;

• explain that hydrogen and high energy electrons are released from the citric acid cycle and are taken by carrier molecules to the electron transport chain on the cristae.

10.2.1 Structure of the mitochondrion The mitochondrion is bounded by a double layer of membranes, each of which is itself composed of a phospholipid bilayer. The outer membrane surrounds the mitochondrion and acts as a boundary; although it is highly permeable to small molecules, the entry of larger molecules is strictly regulated. The inner membrane is folded, forming structures called cristae.

Structure of the mitochondrion Pyruvate crosses the double membrane and enters the inner space of the mitochondrion, which is called the matrix. Here it is oxidised and, in combination with co-enzyme A as acetyl co-enzyme A (or acetyl-coA), it enters the citric acid cycle. The co-enzymes which accept the hydrogen ions that are released from the citric acid cycle then move to the cristae of the inner membrane, where the hydrogen/electron transport chain and its enzymes is located. It is here that nearly all of the ATP is generated.

©HERIOT-WATT UNIVERSITY 218 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

10.2.2 Fascinating facts about mitochondria The most likely theory of the formation of mitochondria is that, about 2 billion years ago, one particular type of bacterium survived being engulfed into another. The evidence for this is the high degree of similarity between the mitochondria of different species of living organism, and between these mitochondria and certain bacteria which still exist today, namely the rickettsia. These similarities lie not only in their structure, but also in their DNA. Although many of the genes controlling the activity of the mitochondrion are now located in the nuclear DNA, key enzymes are still controlled by the circular DNA molecules in the mitochondria. Also, mitochondria control their own replication and respond to increased energy demands in the cell by dividing more frequently. The DNA of the mitochondrion is interesting from a more recent evolutionary perspective. At the point of fertilisation, the egg destroys the mitochondria which enter it from the sperm - we all carry only the mitochondria received from our mothers. As mitochondrial DNA mutates relatively slowly and passes down the generations without any of the genetic mixing associated with the formation of gametes, it can be used to ﬁnd relationships between individuals widely separated in time. It is also useful in considering the evolutionary history of populations. Mitochondria range in size from 0.5 to 1.0 μm, and in numbers from over a thousand in active cells, such as those in the liver, to relatively few in cells such as skin epidermis, and only one in some single-celled organisms. In addition, active cells have a much more intensively folded inner membrane, creating large numbers of cristae and hence more sites for ATP synthesis.

Mitochondria: Question Q1: Complete the pairs of items concerning mitochondria by matching the parts on Go online the left with the parts on the right.

Glycolysis citric acid cycle

Pyruvate to acetyl-coA cristae of mitochondrion

Mitochondrial matrix multi-enzyme complex in cytoplasm

Electron transport chain mitochondrial matrix ......

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 219

10.3 The citric acid cycle The citric acid cycle is located in the matrix of the mitochondrion. As will be shown below, it is classically described as a cycle with a starting point at citrate. However, this is really a false impression of what actually goes on. The acetyl-coA, which enters the cycle to form citrate, can be derived from several sources other than pyruvate. Amino acids, and certain fatty acids, can be synthesised into intermediate compounds in the cycle. Finally, several of the intermediates can be transferred to other synthetic metabolic pathways. Some of these points are explored in a later section.

10.3.1 The steps of the citric acid cycle Learning objective By the end of this section, you should be able to:

• state that pyruvate, produced by glycolysis, enters the mitochondrion and is broken down to form acetate with the release of carbon dioxide;

• explain that this acetate is combined with coenzyme A to form acetyl coenzyme A, or acetyl-coA;

• explain that coenzyme A passes the acetyl group on to oxaloacetate, forming citrate, the ﬁrst stage of the citric acid cycle;

• state that the citric acid cycle results in the direct generation of ATP, the release of carbon dioxide, the transfer of hydrogen ions to carriers, and the regeneration of oxaloacetate. For every glucose molecule, glycolysis takes place once, but the succeeding stages of aerobic respiration take place twice. Once inside the mitochondrial matrix, the pyruvate is converted to an acetyl group by the removal of hydrogen and carbon dioxide, a step known as pyruvate decarboxylation. The acetyl group is then attached to a carrier molecule, coenzyme A, to form acetyl coenzyme A (acetyl-coA) which then joins the citric acid cycle. The steps of the cycle are shown in the following diagram.

©HERIOT-WATT UNIVERSITY 220 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

The citric acid cycle The steps of the citric acid cycle are as follows:

+ 1. the 3-C pyruvate loses one carbon to CO2 and one hydrogen to NAD , forming NADH - the remaining 2-C acetyl group is attached to coenzyme A, forming acetyl- coA;

2. coenzyme A passes the 2-C acetyl group to oxaloacetate (4-C), forming citrate (6-C) and releasing coenzyme A to collect another acetyl group;

3. at successive steps in the cycle, two further carbon molecules are lost to CO2 and three hydrogen ions are transferred to NAD+, forming NADH;

4. sufﬁcient energy is released at one step in the cycle to form one molecule of ATP (or GTP);

5. a second hydrogen carrier, FAD, removes two hydrogen ions as FADH2; 6. ﬁnally, the 4-C oxaloacetate is formed and is ready to accept another 2-C acetyl group from acetyl-coA.

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 221

Although a knowledge of the numbers of carbon atoms in the various compounds in the cycle is not required, they are included here to show how the cycle ﬁnally regenerates the 4-carbon (4-C) molecule of oxaloacetate, which combines with the 2-C acetyl group from acetyl-coA to form the 6-C citrate.

The steps of the citric acid cycle: Questions Q2: For each molecule of glucose, how many times does glycolysis take place? ...... Go online

Q3: For each molecule of glucose, how many times does the citric acid cycle take place? ......

Q4: Name the molecule to which pyruvate is converted before entering the citric acid cycle......

Q5: To which molecule is the 2-C compound, formed from pyruvate, joined when it enters the citric acid cycle? ......

Q6: List the three products of the citric acid cycle......

Q7: Why is the citric acid cycle actually called a cycle? ......

10.3.2 The enzymes and carriers of the citric acid cycle Learning objective By the end of this section, you should be able to:

• state that each step in the citric acid cycle is mediated by a different enzyme;

• state that dehydrogenase enzymes remove hydrogen ions and electrons;

• state that coenzymes NAD+ and FAD accept these hydrogen ions and electrons to form NADH and FADH2; • state that NADH and FADH2 release high-energy electrons to the electron transport chain on the cristae, resulting in the generation of ATP.

©HERIOT-WATT UNIVERSITY 222 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

The citric acid cycle consists of many reactions.

The citric acid cycle Enzymes Each step in the cycle is catalysed by a different enzyme which is speciﬁc to the substrate involved. In particular, hydrogen is removed at four points in the cycle; although the substrates are different in each case, the enzymes involved all belong to a class of enzyme called dehydrogenases. In simple terms, these enzymes transfer hydrogen ions and high energy electrons to carrier molecules which then convey them to the electron transport system on the cristae.

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 223

Hydrogen Carriers The hydrogen ions and high energy electrons released by the various dehydrogenases are picked by coenzyme carrier molecules of two types. Most of the hydrogen attaches to NAD+, often simply referred to as NAD; the plus sign (’+’) indicates what is called the formal charge on the NAD molecule, and shows that it requires a negatively charged electron to balance its charge. In the citric acid cycle, it picks up one hydrogen ion and one electron to form NADH. NAD stands for nicotinamide adenine dinucleotide. When this complex name is broken down it reveals some ’old friends’: nicotinamide is a combination of nicotinic acid, or vitamin B3, and an amino acid; adenine is the base that is also found in nucleic acids; and the dinucleotide indicates that there are two nucleotides in the molecule. FAD stands for Flavin Adenine Dinucleotide, which also contains adenine and two nucleotides; in this case, the vitamin is riboﬂavin, also known as vitamin B2. FAD accepts two hydrogen ions and two electrons from the citric acid cycle to form FADH2.

NADH and FADH2 carry the hydrogen and high energy electrons from the citric acid cycle to the electron transport system on the cristae, where the electrons pass along a chain of receptors to generate the energy to form ATP.

The enzymes and carriers of the citric acid cycle: Questions Q8: By what is each step in the citric acid cycle mediated? ...... Go online

Q9: Name the enzymes which remove hydrogen ions and high energy electrons from the citric acid cycle......

Q10: Name the two coenzymes which accept the hydrogen and high energy electrons from the citric acid cycle......

Q11: When they accept hydrogen ions, what do these coenzymes form? ......

Q12: Where do these coenzymes take the hydrogen that they carry? ......

©HERIOT-WATT UNIVERSITY 224 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

10.4 Alternative substrates for respiration Learning objective By the end of this section, you should be able to:

• state that the storage carbohydrate glycogen can be broken down into glucose for use as a respiratory substrate;

• state that, in order to serve as respiratory substrates, other sugars must be converted into glucose or intermediate compounds in the glycolysis pathway;

• state that proteins can be broken down into amino acids, which can be converted into intermediate compounds in glycolysis and the citric acid cycle;

• state that fats can be broken down into glycerol and fatty acids;

• state that glycerol can be converted into glucose in the liver;

• state that fatty acids can be converted into acetyl-coA. The average diet in economically more developed countries contains carbohydrates, proteins and fats, all of which can be used to provide respiratory substrates. Under what might be called ’normal’ conditions, by which it is meant that there is no stress on the body in terms of food supply, temperature or exercise, glucose is the substrate used for respiration by all cells. Any excess of glucose beyond the body’s immediate respiratory needs is stored either as fat or glycogen. Excess fats are stored in the adipose tissue, and any amino acids which are in excess of the immediate demand for protein building are broken down in the liver. This forms the waste product urea, for excretion, and glucose for storage. The human body, like that of other animals, is adapted to store energy when there is more available in the diet than is needed to meet everyday needs. We are programmed to store fat because only a few generations ago people had to rely on that energy store to carry us through the times in the year when food was in short supply. As anyone who tries to reduce their weight by switching to a low calorie diet will discover, the body has clever mechanisms for adapting to a reduced energy intake. Without increasing physical activity alongside the diet, the kilograms will not melt away. In times of real famine, the different energy sources in the body are used in turn. First, the glycogen from the liver and the muscles is used up, then the fat from the various fat depots, and finally the protein of the muscles is used to generate respiratory substrates. The detailed picture is inevitably much more complex, but this is the general pattern. Glycogen Glucose is readily interconvertible with glycogen. After a meal, as digestion proceeds, glucose is absorbed into the blood and blood sugar levels rise. This is detected by the islets of Langerhans in the pancreas, and they increase the secretion of the hormone insulin into the blood. This, in turn, stimulates the liver and muscle cells to absorb glucose and convert it to glycogen. Once the blood glucose level has stabilised, the liver then reconverts the stored glycogen to glucose, under the influence of glucagon (also from the pancreas), so that the concentration of glucose in the blood, and hence the tissue fluid bathing the body’s cells, remains relatively constant. Glycogen is the

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 225

immediately available source of glucose which is used ﬂexibly throughout the day to maintain blood sugar levels. In endurance sporting events, such as marathon running, distance cycling or cross- country skiing, the body draws on the glycogen of the liver and muscles. As the muscles lack an essential enzyme which allows them to export glucose, their glycogen stores cannot contribute to the maintenance of the blood sugar level, but they do provide glucose to glycolysis in the muscle cells themselves. If these stores become depleted before the end of an event, the athlete suddenly experiences extreme fatigue, a condition known as ’hitting the wall’. Other sugars The diet of many in economically more developed countries contains relatively high concentrations of sucrose sugar (the sort which is put into coffee, tea, and a surprising range of other foods). In the small intestine, sucrose is digested by the enzyme sucrase into two simpler sugars: glucose and fructose. Like glucose, fructose is absorbed into the blood and is converted into an intermediate during glycolysis in the liver. This may either be used to form glucose (and then glycogen) or pyruvate. Proteins The basic units of proteins are amino acids, of which twenty are common. Protein is digested in the gut by various protease and peptidase enzymes to form these amino acids. These are then absorbed into the blood and pass round the body, with cells taking in what they require for protein synthesis. Excess amino acids are taken into the liver and are broken down, forming the waste product urea and compounds that are fed into the citric acid cycle and its associated pathways. Depending on their composition, these amino acids enter the citric acid cycle by several routes: many are converted to pyruvate, or acetyl-coA, others to various intermediates, and some even to oxaloacetate. As well as fuelling the citric acid cycle, some of these intermediates may alternatively be converted into glucose, lipids, or both. Amino acids are only used as a major respiratory substrate when the other fuels, glycogen and fat, are exhausted, i.e. in starvation conditions. Fats/lipids (more strictly, triglycerides) Fats are digested by the enzyme lipase and are converted into fatty acids and glycerol. Following absorption within the small intestine, these substances are reformed into fat which is carried as small particles, called chylomicrons, ﬁrst in the lymphatic system and then in the blood, to fat depots in the adipose tissue, to muscles, and to the liver. The processing of fat in the liver is a complex story which is covered in Unit 2 of the course. Glycerol and fatty acids also circulate freely in the blood and, along with chylomicrons, they supply cells with fats/lipids and their components. In the cytoplasm of cells, the lipids are broken down by hydrolysis to glycerol and fatty acids. Glycerol is converted to an intermediate in glycolysis, ready to enter the energy pay-off stage, whereas the fatty acids are converted into acetyl-coA. Glycerol can also be converted into glucose in the liver. In all cells, fatty acids are synthesised from citrate which is exported from the mitochondria back into the cytoplasm before entering the metabolic pathway which forms fatty acids.

©HERIOT-WATT UNIVERSITY 226 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

Alternative substrates for respiration: Questions Q13: Name the storage carbohydrate in humans. Go online ......

Q14: Why is this useful in the control of blood sugar levels? ......

Q15: State two possible fates in the cell for sugars other than glucose......

Q16: How is energy released from amino acids in the cell? ......

Q17: Where is glycerol converted to glucose? ......

Q18: How is glycerol respired in cells? ......

Q19: How do fatty acids become available for respiration in the cell? ......

10.5 Learning points

Summary The mitochondrion

• The mitochondrion consists of an outer boundary membrane, an inner membrane, which is folded to form cristae, and the central matrix.

• Pyruvate enters the matrix of the mitochondrion and is broken down to an acetyl group which is attached to co-enzyme A, forming acetyl co-enzyme A (acetyl-coA).

• Acetyl-coA passes the acetate into the citric acid cycle in the matrix of the mitochondrion.

• Hydrogen and high energy electrons are released from the citric acid cycle and are taken by carrier molecules to the electron transport chain on the cristae.

The steps of the citric acid cycle

• Pyruvate, produced by glycolysis, enters the mitochondrion and is broken down to form acetate with the release of carbon dioxide.

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 227

Summary continued • This acetate is combined with coenzyme A to form acetyl coenzyme A, or acetyl-coA.

• Coenzyme A passes the acetyl group on to oxaloacetate, forming citrate, the ﬁrst stage of the citric acid cycle.

• The citric acid cycle results in the direct generation of ATP, the release of carbon dioxide, the transfer of hydrogen ions to carriers, and the regeneration of oxaloacetate.

The enzymes and carriers of the citric acid cycle

• Each step in the citric acid cycle is mediated by a different enzyme.

• Dehydrogenase enzymes remove hydrogen ions and electrons.

• Coenzymes NAD+ and FAD accept these hydrogen ions and electrons to form NADH and FADH2. • NADH and FADH2 release high-energy electrons to the electron transport chain on the cristae, resulting in the generation of ATP.

Alternative substrates for respiration

• The storage carbohydrate glycogen can be broken down into glucose for use as a respiratory substrate.

• In order to serve as respiratory substrates, other sugars must be converted into glucose or intermediate compounds in the glycolysis pathway.

• Proteins can be broken down into amino acids, which can be converted to intermediate compounds in glycolysis and the citric acid cycle.

• Fats can be broken down into glycerol and fatty acids.

• Glycerol can be converted into glucose in the liver.

• Fatty acids can be converted into acetyl-coA.

©HERIOT-WATT UNIVERSITY 228 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

10.6 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of enzymes and coenzymes before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: Enzymes and coenzymes Give an account of the enzymes and coenzymes involved in the citric acid cycle. (6 10 min marks) ......

10.7 End of topic test End of Topic 10 test Q20: Complete the paragraph using the words from the list. (10 marks)

Go online The mitochondrion consists of an outer membrane, an inner membrane which

is folded to form , and the central . enters the matrix of the mitochondrion, is converted to , and attached to to form acetyl co-

enzyme A (). This passes the acetate into the Citric Acid cycle in the matrix of the mitochondrion. and high energy are released from the citric acid

cycle and are taken by carrier molecules to the electron chain on the cristae of the inner membrane of the mitochondrion. Word list: acetate, acetyl-coA, boundary, co-enzyme A, cristae, electrons, hydrogen, matrix, pyruvate, transport.

©HERIOT-WATT UNIVERSITY TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE 229

......

Q21: Complete the sentences by matching the parts on the left with the parts on the right. (10 marks)

Pyruvate is produced by oxaloacetate.

Pyruvate is broken down to form NAD+ and FAD.

Acetyl-coA passes the acetyl group to dehydrogenases.

Oxaloacetate combines with the ATP. acetyl group to form

The citric acid cycle directly glycolysis. generates

Hydrogen is removed from the citric cristae. acid cycle by

Hydrogen is transferred from the citric citrate. acid cycle to

High energy electrons are carried by an acetyl group.

NADH and FADH2 carry hydrogen to coenzymes. the

The electron transport chain is electron transport chain. located on the

©HERIOT-WATT UNIVERSITY 230 TOPIC 10. CELLULAR RESPIRATION II: CITRIC ACID CYCLE

......

Q22: Complete the table by matching the descriptions on the left with the terms on the right. (6 marks)

A storage carbohydrate: fructose

A sugar which can be converted into amino acids a glycolysis intermediate:

Some can be converted to citric acid glycogen cycle intermediates:

Formed from glycerol and fatty acids: fatty acids

Converted into glucose in the liver: lipids

Converted into acetyl-coA: glycerol ......

Q23: Summarise the end results of the citric acid cycle. (4 marks) ......

Q24: Explain the role of coenzymes in the citric acid cycle. (4 marks) ......

©HERIOT-WATT UNIVERSITY 231

Topic 11

Cellular Respiration III: Electron transport system

Contents

11.1 Introduction ...... 232 11.2 The electron transport chain ...... 233 11.3 ATP synthesis ...... 236 11.4 Learning points ...... 238 11.5 Extended response question ...... 239 11.6 End of topic test ...... 239

Learning objectives By the end of this topic, you should be able to:

• describe the electron transport chain;

• explain the generation of ATP by the electron transport chain. 232 TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM

11.1 Introduction The electron transport chain is the point in the process of aerobic respiration at which the bulk of the energy in the respiratory substrate is ﬁnally made available as ATP. The pathway is located on the cristae of the inner membrane of the mitochondrion. Although it has a variety of names, including the cytochrome system and the hydrogen carrier system, we shall use the term electron transport system because it is high energy electrons which drive it. There are two parts to the system which are said to be coupled, meaning that one cannot occur without the other:

• the electron transport chain, which comprises a series of enzyme-controlled electron donors and electron acceptors - each transfer of an electron provides the energy to pass a hydrogen ion (or proton) across the inner membrane into the space between the inner and outer membranes of the mitochondrion;

• the ATP synthesis stage, which involves the enzyme ATP synthase - the protons are passed back from the intermembrane space into the matrix of the mitochondrion, releasing energy which is used to form ATP.

In this, and the preceding topics, reference has been made to hydrogen ions and electrons. Hydrogen is the simplest of all atoms and is found throughout the universe, being by far the commonest element. It consists of a nucleus comprising a single particle which carries a positive electrical charge, a proton, and an orbiting negatively charged particle (an electron). The atom is therefore electrically neutral. If the proton is separated from its electron, it becomes a hydrogen ion with a positive charge, H+. Electrons are always negatively charged, but they can exist at different energy levels. They can absorb energy and be raised to a higher energy level, or give out energy and drop to a lower energy level. If a compound combines with hydrogen ions or gains electrons, it is said to be reduced. Conversely, and perhaps confusingly, the loss of hydrogen or electrons is known as oxidation, as is the combination with oxygen. In the summary equation of aerobic respiration, glucose is oxidised to carbon dioxide and oxygen is reduced to water:

C6H12O6 +6O2 → 6CO2 +6H2O

©HERIOT-WATT UNIVERSITY TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM 233

11.2 The electron transport chain Learning objective By the end of this section, you should be able to:

• state that coenzymes NAD+ and FAD collect hydrogen ions and high energy electrons from glycolysis and the citric acid cycle;

• state that, as NADH and FADH2, the reduced coenzymes travel to the inner mitochondrial membrane;

• state that, at the inner mitochondrial membrane, NADH and FADH 2 pass the hydrogen ions and high energy electrons to carriers in the electron transport chain;

• state that the electron transport chain is a collection of proteins attached to the inner mitochondrial membrane;

• state that as the high energy electrons released from NADH and FADH 2 pass along the chain, they release their energy in stages;

• state that the energy so released is used by enzymes to pump hydrogen ions (H+) across the inner mitochondrial membrane into the intermembrane space;

• state that the ﬁnal electron acceptor is oxygen, which combines with hydrogen ions to form water.

The reduced coenzymes NADH and FADH2 bring the hydrogen ions (protons) and high energy electrons to the inner membrane of the mitochondrion. Here the protons are released and the electrons are passed to a series of protein acceptors and donors which are arranged in three closely linked complexes. When the electrons are passed from donor to acceptor, energy is released. This energy is used to pump the hydrogen ions across the inner membrane into the intermembrane space, building up a concentration there that can be up to ten times that of the matrix. As the electrons pass along the chain, their energy level declines until eventually they reach their lowest energy level. At that point they are released to combine ﬁrst with oxygen, and then with two hydrogen ions to from a water molecule (H 2O).

©HERIOT-WATT UNIVERSITY 234 TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM

The following illustration shows the stages in the electron transport chain. The electrons are shown passing from their coenzymes to carriers in the transport chain. The changing height of the carriers across the diagram indicates the relative energy level of the electrons.

The electron transport chain

1. NADH carries high energy electrons and hydrogen to the inner mitochondrial membrane, where the electrons are passed to an electron carrier in the protein complex on the membrane. Each transfer is catalysed by a different enzyme.

2. FADH2 passes electrons to a carrier slightly further along the chain. For those interested in counting molecules of ATP generated, this means that FADH 2 generates less ATP than NADH.

3. As the electrons pass from carrier to carrier, the energy released is used by enzymes to pump hydrogen ions across the inner membrane into the intermembrane space. This leads to a large difference in the hydrogen ion concentration between the matrix and the intermembrane space, which can be sufﬁcient to cause a difference of 1pH unit.

4. Having released their energy, the electrons are transferred to an oxygen atom which then carries two negative charges. It combines with two hydrogen ions from the matrix to form water, the second waste product from respiration (the other being carbon dioxide).

©HERIOT-WATT UNIVERSITY TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM 235

The electron transport chain: Questions

Q1: What do NADH and FADH2 carry? ...... Go online

Q2: To where do NADH and FADH2 go? ......

Q3: Of what type of compound is the electron transport chain composed? ......

Q4: What is the function of these compounds in relation to electrons? ......

Q5: What is the function of these compounds in relation to hydrogen ions? ......

Q6: What is the ﬁnal electron acceptor in the electron transport chain? ......

Q7: Name the waste product from the electron transport chain......

©HERIOT-WATT UNIVERSITY 236 TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM

11.3 ATP synthesis Learning objective By the end of this section, you should be able to:

• state that as electrons pass along the electron transport chain, energy is released;

• explain that this energy helps enzymes pump hydrogen ions across the inner membrane of the mitochondrion into the intermembrane space;

• state that the return ﬂow of these H+ ions rotates part of the enzyme ATP synthase;

• explain that the rotation of ATP synthase catalyses the formation of ATP;

• state that this process generates most of the ATP in aerobic respiration. As the electrons pass along the electron transport chain, energy is released as the electrons pass from donor to acceptor molecule. Eventually, the electrons reach their lowest energy level and are passed to the final acceptor, which is oxygen. The energy released is used to enzymatically pump H+ ions across the inner mitochondrial membrane into the intermembrane space. As the membrane is impermeable to protons, this results in a build-up in their concentration in the intermembrane space compared to the matrix, leading to as much as a tenfold difference (equivalent to 1pH unit). This considerable difference in the concentration of positively charged hydrogen ions causes a large electrochemical gradient between the intermembrane space and the matrix. The only route back for the H+ ions is through molecules of the enzyme ATP synthase, which span the inner membrane. The ATP synthase molecule is roughly rod- shaped, with a central rotating subunit, or rotor, and three active sites in its circular base. As the H+ ions pass through the molecule, they cause the central subunit to rotate inside the base. As it does so, it alters the shape of the active sites, which have three states. The first state is ’open’, in which ADP and Pi can enter the site; the second is ’loose’, in which the site closes loosely around the two substrates; and the final state is ’tight’, during which the site forces the ADP and Pi together to form ATP. As the site returns to its ’open’ state, the newly-formed ATP molecule is ejected. It is a beautiful example of a biochemical machine!

©HERIOT-WATT UNIVERSITY TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM 237

Synthesis of ATP by ATP synthase

i. Hydrogen ions accumulate in the intermembrane space, resulting in an electrochemical gradient across the inner membrane between the intermembrane space and the matrix. ii. This, in turn, provides the energy necessary to make ATP synthase function. It causes the rotor and rod of the ATP synthase to rotate inside the circular base unit. iii. The mechanical energy from this rotation is converted into chemical energy as phosphate is added to ADP to form ATP.

Unlike the formation of ATP in the energy pay-off stage of glycolysis, in which phosphate is directly transferred from a substrate molecule to ADP (hence substrate level phosphorylation), this phosphorylation involves the removal of hydrogen and electrons from substrate molecules. Hence, it is known as oxidative phosphorylation.

ATP synthesis: Questions Q8: What is released as electrons pass along the electron transport chain? ...... Go online

+ Q9: What happens to the H ions released from the NADH and FADH2? ......

Q10: Name the enzyme which forms ATP from ADP and Pi? ...... Q11: What causes part of this molecule to rotate? ......

©HERIOT-WATT UNIVERSITY 238 TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM

Q12: What is the outcome of this rotation? ......

Q13: From where does most of the ATP generated in aerobic respiration come? ......

Once you have completed this topic, it would be a good idea to go back to Topic 9 Section 2, just to remind yourself of the overall picture of aerobic respiration. Although the total yield of energy from one molecule of glucose is theoretically sufﬁcient to generate 38 molecules of ATP (2 from direct phosphorylation during glycolysis and 36 from oxidative phosphorylation), in practice the yield is closer to 30 molecules as there are, just as in industrial power systems, losses of energy at various stages in the process.

11.4 Learning points

Summary The electron transport chain • Coenzymes NAD+ and FAD collect hydrogen ions and high energy electrons from glycolysis and the citric acid cycle. • As NADH and FADH2, they travel to the inner mitochondrial membrane. • At the inner mitochondrial membrane, NADH and FADH 2 pass the hydrogen ions and high energy electrons to carriers in the electron transport chain. • The electron transport chain is a collection of proteins attached to the inner mitochondrial membrane. • As the high energy electrons released from NADH and FADH 2 pass along the chain, they release their energy in stages. • The energy so released is used by enzymes to pump hydrogen ions (H+) across the inner mitochondrial membrane into the intermembrane space. • The ﬁnal electron acceptor is oxygen, which combines with hydrogen ions to form water. ATP synthesis • As electrons pass along the electron transport chain, energy is released. • This energy helps enzymes pump hydrogen ions across the inner membrane of the mitochondrion into the intermembrane space. • The return ﬂow of these H+ ions rotates part of the enzyme ATP synthase. • The rotation of ATP synthase catalyses the formation of ATP. • This process generates most of the ATP in aerobic respiration.

©HERIOT-WATT UNIVERSITY TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM 239

11.5 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of the electron transport chain before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: The electron transport chain Give an account of the electron transport chain. (8 marks) ...... 10 min

11.6 End of topic test End of Topic 11 test

Q14: Complete the paragraph using the words from the list. (14 marks) The electron transport chain is a collection of attached to the Go online

membrane. NAD and collect hydrogen ions and electrons from and the citric acid cycle, forming NADH and . Theytravelto

the inner mitochondrial membrane and pass the and high energy electrons to

in the electron transport chain. As the high energy electrons released from and FADH2 pass along the chain of electron acceptors and , they

release which is used by to pump hydrogen ions (H ) across the inner mitochondrial membrane into the space. The ﬁnal electron acceptor is which combines with hydrogen ions to form water.

Word list: carriers, coenzymes, donors, energy, enzymes, FAD, FADH2, glycolysis, high energy, hydrogen ions, inner mitochondrial, intermembrane, NADH, oxygen, proteins.

©HERIOT-WATT UNIVERSITY 240 TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM

......

Q15: Complete the sentences by matching the parts on the left with the parts on the right. (5 marks)

Energy is released as electrons pass part of ATP synthase. along

Enzymes pump H+ ions into the formation of ATP.

The return ﬂow of H+ ions into the the ATP from aerobic respiration. matrix rotates

ATP synthase catalyses the electron transport chain.

The electron transport chain the intermembrane space. produces most of ......

Q16: Explain the role of electrons in the electron transport chain. (6 marks) ......

©HERIOT-WATT UNIVERSITY 241

Topic 12

Energy systems in muscle cells

Contents

12.1 Introduction ...... 242 12.2 Creatine phosphate ...... 244 12.3 Lactic acid metabolism ...... 246 12.4 Types of skeletal muscle fibres ...... 248 12.4.1 Slow twitch (Type 1) fibres ...... 249 12.4.2 Fast twitch (Type 2) fibres ...... 251 12.5 Learning points ...... 252 12.6 Extended response question ...... 254 12.7 End of topic test ...... 254

Learning objectives By the end of this topic, you should be able to:

• describe the unique demands made on the respiratory pathway by muscle cells;

• explain the roles of aerobic and anaerobic respiration in muscle cells;

• describe creatine phosphate metabolism;

• describe lactic acid metabolism;

• describe the different types of muscle ﬁbres and explain their role in different sports. 242 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

12.1 Introduction Muscle tissue has as its function the exertion of force by contraction. This may be to reduce the diameter of arteries or the gut (involuntary smooth muscle), to contract the heart chambers (cardiac muscle), or to move parts of the skeleton (voluntary striped muscle). Muscle tissue is never totally at rest, being constantly stimulated to contract at a low level of intensity to maintain muscle tone. Whereas smooth muscle is not generally required to contract suddenly or violently, cardiac and skeletal muscle do both, and skeletal muscle may go from its relaxed state to strong contraction very rapidly. In this topic, we shall limit our consideration to skeletal muscle and its use in different types of sport. Skeletal muscle is required to work in a range of different ways. When we are doing anything other than lying ﬂat out, many sets of muscles must work antagonistically to maintain our posture by supporting our body weight and keeping us upright. Any movement, from picking up a pencil to sprinting for a bus, similarly requires co-ordinated activity by different muscles, all of which is controlled by the cerebellum in the brain. The way in which muscle tissue gains the energy for its activity varies with the duration of the exercise in which the body is engaged. There are four potential energy sources.

• ATP, which is by far the most important supplier of energy for cell activities, is synthesised in response to demand, and is only present in small quantities in the cell. The adult human body only contains about 250g of ATP at any one time, although during the course of a day the body will use up the equivalent of its own weight of ATP.There is no store of ATP in muscle cells so the immediately available ATP is very rapidly depleted when sudden activity takes place.

• Creatine phosphate (or phosphocreatine) is found in tissues with a high and variable energy demand, such as muscle and brain. For a brief period (2-10s), it can supply phosphate to convert ADP to ATP without the need for oxygen (anaerobically).

• Lactic acid metabolism occurs when the oxygen supply is insufﬁcient to support the electron transport chain. Pyruvate can be converted to lactic acid thus allowing glycolysis to proceed and continue to generate ATP, albeit in relatively small quantities.

• The citric acid cycle and the electron transport chain, in the presence of an adequate oxygen supply, yield at least 15 times more ATP per glucose molecule than lactic acid metabolism. However, creatine phosphate and lactic acid metabolism can provide ATP 50 times faster, and so are vital for short bursts of strenuous activity. The glucose fuel must come either from stored glycogen or from the blood.

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 243

The following diagram shows the relative importance of these different energy sources as the duration of exercise changes.

The four potential energy sources during exercise In terms of sport, we can see that events which involve an explosive burst of activity, such as weight-lifting or discus throwing, are going to depend on ATP and creatine phosphate. On the other hand, endurance events, such as triathlon or marathon, rely on aerobic respiration of glucose derived from glycogen stored in the muscles. Not only are there different energy sources for these different types of event, but also the muscle best suited to each differs as well. Explosive sports require muscle which will contract fast, but for a short time only; these are known as ’fast twitch’ muscle ﬁbres. Endurance events demand muscles which will contract repeatedly over a long period of time, even though they may contract more slowly; these are ’slow twitch’ ﬁbres.

©HERIOT-WATT UNIVERSITY 244 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

12.2 Creatine phosphate Learning objective By the end of this section, you should be able to:

• state that muscle cells do not contain any reserve of ATP which will fuel more than 2s of activity;

• state that creatine phosphate is an energy reserve that is found in cells with high and variable energy demands, such as muscle and brain;

• state that creatine phosphate breaks down anaerobically to release energy and phosphate, which is used to convert ADP and Pi to ATP at a fast rate; • state that there is only sufﬁcient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s;

• explain that when energy demands are low, creatine phosphate reserves are restored using ATP provided by aerobic respiration. Apart from the protein from which they are made, muscles contain three types of energy store.

• ATP. Although it is not a storage compound, a small quantity of ATP is held within the muscle ﬁbres. It is sufﬁcient to fuel about 2s of intense activity.

• Glycogen is the storage carbohydrate in the human body, with about 100g in the liver (used in the control of blood sugar levels) and 300g in the muscles (used as a fuel reserve). It is made from glucose that is supplied to the cells by the blood.

• Creatine phosphate, which is found in cells which intermittently have a high energy demand (e.g. muscle and brain), sufﬁcient quantities of which are stored in the muscle ﬁbres to support up to 10s of vigorous exercise. It can be obtained from two sources: ingestion of meat and internal production by the liver and kidneys. It is also taken as a supplement by some athletes, although it is only of relevance to sports requiring short bursts of energy and, more importantly, no studies have been carried out on long-term effects on the body of taking such supplements.

Creatine phosphate is reversibly converted to creatine and phosphate by another kinase enzyme, creatine kinase, which carries out a direct phosphorylation by transferring the phosphate group to ADP, so reforming ATP. This is another example of coupled reactions. They are anaerobic, meaning that this source of ATP is not dependent on the delivery of oxygen to the tissues by the blood.

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 245

Creatine phosphate energy store When muscle energy demand is reduced, ATP from aerobic respiration allows creatine kinase to run the reaction in the opposite direction, transferring phosphate from ATP to creatine to reform creatine phosphate and ADP. The dephosphorylation of ATP is catalysed by the enzyme ATPase. Notice that, like many enzymes, creatine kinase is able to catalyse the interconversion of creatine and creatine phosphate in both directions, depending on conditions such as the concentrations of reactants. Again, these are coupled reactions but, being dependent on the supply of ATP from the electron transport chain, they only occur when the cell has an oxygen supply.

Creatine phosphate: Questions Q1: For how many seconds will stored ATP sustain muscle contraction? ...... Go online

Q2: Name two tissues which have creatine phosphate as an energy store......

Q3: Why do these tissues have creatine phosphate as an energy store? ......

Q4: What are formed when creatine phosphate is anaerobically broken down? ......

Q5: What is the maximum time in seconds for which creatine phosphate can sustain strenuous muscle activity? ......

Q6: How are creatine phosphate reserves restored? ......

©HERIOT-WATT UNIVERSITY 246 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

12.3 Lactic acid metabolism Learning objective By the end of this section, you should be able to:

• explain that during vigorous exercise, muscle cells do not get enough oxygen to support the electron transport chain;

• state that under these conditions, pyruvate is converted to lactic acid;

• state that this conversion involves the transfer of hydrogen from the NADH formed during glycolysis to pyruvate, forming lactate;

• state that the use of NADH to form lactate regenerates the NAD+ that is required to maintain ATP production by glycolysis;

• state that the accumulation of lactate causes muscle fatigue;

• state that lactate is reconverted to pyruvate in the muscles and to glucose in the liver;

• state that the reconversion of lactate to pyruvate or glucose requires ATP that is formed by aerobic respiration;

• explain that this additional respiratory activity creates an oxygen debt. The demands of vigorous exercise on a muscle’s energy supply are such that aerobic respiration cannot keep up the necessary flow of ATP. The supply of oxygen from the blood limits the rate at which the electron transport chain can generate it. Whereas ATP and creatine phosphate reserves can provide fuel for some 10s, it is anaerobic respiration that must meet the need for additional supplies of ATP for up to 90s thereafter. Anaerobic respiration is also known as (lactate) fermentation. Lactate formation In the absence of oxygen as a final hydrogen acceptor, the electron transport chain cannot function so that pyruvate and NADH accumulate in the cytoplasm. Pyruvate has been shown to inhibit the enzyme phosphofructokinase, which catalyses the second irreversible reaction in the energy investment stage of glycolysis. Likewise, a lack of NAD+ limits the first reaction in the energy pay-off stage. For both of these reasons, it is vital to the continued activity of the muscle cell that pyruvate is removed and that NADH is oxidised back to NAD+ by removing hydrogen. This is achieved by converting it to lactate. Lactate is continuously formed by muscle cells whether oxygen is present or not. It is also the only source of energy for red blood cells, which lack mitochondria. However, when the energy demand in muscle cells exceeds the ability of aerobic respiration to supply all of the ATP needed and anaerobic respiration increases, lactate production exceeds the cell’s ability to use it or export it, and so it accumulates. This leads to a lowering of the cell’s pH and glycolysis is inhibited, resulting in muscle fatigue. The associated burning pain is a signal to allow the body to clear the muscles of lactate and the other metabolites which have built up in the cells. Persisting with the exercise can result in nausea and vomiting. However, build-up of lactate is not the cause of the

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 247

soreness and stiffness that may be felt in the succeeding days; that is a result of minor muscle damage. Lactate removal Lactate is excreted from muscle (and other) cells and is carried in the blood to the liver cells. Here it is ﬁrst converted to pyruvate and then passed into the mitochondria, where several of the steps in glycolysis are reversed to form glucose. Muscle cells are also capable of using up lactate and converting it to pyruvate but, instead of reconverting it to glucose, they pass it into the citric acid cycle in the mitochondria. Both of these processes require an input of ATP so can only take place to any signiﬁcant extent when the vigorous exercise has ceased, in which case aerobic respiration can provide a surplus of ATP beyond the metabolic needs of the body. The effect of this is noticed as a need to keep breathing deeply after the exercise is over in order to supply the oxygen to the cells so that they can carry out this additional respiratory activity. This gives rise to the term ’oxygen debt’ and the idea that it is being repaid by deep breathing.

Lactic acid metabolism: Questions Q7: Why do muscle cells increase lactic acid formation during vigorous activity? ...... Go online

Q8: What is converted to form lactate/lactic acid? ......

Q9: What does this conversion regenerate for glycolysis? ......

Q10: What does the accumulation of lactic acid cause in the muscle? ......

Q11: To what is lactate reconverted in the (i) liver, (ii) muscles? ......

Q12: What must aerobic respiration supply for lactate to be reconverted? ......

Q13: Why does this create an oxygen debt? ......

©HERIOT-WATT UNIVERSITY 248 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

12.4 Types of skeletal muscle ﬁbres Learning objective By the end of this section, you should be able to:

• state that there are two types of muscle ﬁbre, namely slow and fast twitch;

• state that all skeletal muscles contain a mixture of the two types of ﬁbre;

• explain that athletes show patterns of muscle fibres that reflect their events. There are two basic types of muscle fibre found in skeletal muscle, namely slow and fast twitch fibres (or Type 1 and Type 2 fibres). These are so named because fast twitch fibres contract 2-3 times faster that slow twitch. Muscles always contain a mixture of the two fibres, but the proportions differ considerably between different muscles. The actual proportions in any individual person appear to be genetically determined, although training can force one type of fibre to develop more of the characteristic of the other. Slow twitch fibres are especially suited to endurance activities, whereas fast twitch fibres are required for power sports which demand an explosive burst of activity.

Types of skeletal muscle ﬁbres: Questions Q14: Name the two types of muscle ﬁbre. Go online ......

Q15: How many of these types of ﬁbre are found in any one muscle? ......

Q16: What type of ﬁbre is relatively common in endurance athletes? ......

Q17: What type of ﬁbre is relatively common in power athletes? ......

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 249

12.4.1 Slow twitch (Type 1) ﬁbres Learning objective By the end of this section, you should be able to:

• state that slow twitch (Type 1) ﬁbres contract relatively slowly, but can sustain contractions longer;

• explain that slow twitch ﬁbres are good for endurance sporting activities, such as long distance cycling or running;

• state that slow twitch ﬁbres rely on aerobic respiration to generate ATP;

• explain that slow twitch ﬁbres have many mitochondria, a large blood supply, and a high concentration of the oxygen-storing protein myoglobin, which makes them dark red in colour;

• explain that slow twitch ﬁbres use both glycogen and fat as fuel depending on the intensity of the exercise. As we have seen in the earlier sections of this topic, a muscle cell has four sources of ATP to fuel its contraction:

1. ATP stored in small quantities in the muscle ﬁbres, able to support up to 2s of contraction; 2. creatine phosphate, anaerobically fuelling up to 10s of activity; 3. lactic acid metabolism, able to sustain up to 90s of activity anaerobically; 4. aerobic respiration, able to sustain strenuous activity until the stores of glycogen run out, which is generally after about 4hr.

From this, it is obvious that any endurance event will require muscle to be operating aerobically. The type of events which are in this category are running any distance over 800m (but especially longer races such as 10 000m and marathon), longer distance cycling, kayak, swimming or ski races, and mixed discipline distance events like triathlon. The type of muscle fibre which is suited to the demands of these events is the slow twitch (Type 1) fibre. Although they contract more slowly than fast twitch fibres, they are able to contract for much longer before fatiguing. To do this, they rely on aerobic respiration which demands a high oxygen supply. This is provided by a good blood supply through an extensive capillary network, and a high concentration of the respiratory protein myoglobin. Myoglobin is related to the haemoglobin found transporting oxygen in red blood cells; it too stores oxygen and is deep red in colour. Aerobic respiration is usually fuelled by either glucose or, less frequently, fatty acids and glycerol. Glucose may be supplied in the blood (it can be quickly absorbed from energy drinks), but essentially during endurance events it must come from stored fat and especially glycogen. Interestingly, the balance of supply between fats and glycogen varies with the intensity of the exercise. If the athlete is operating at 75% or more of their maximum capability, then glycogen is the sole source of fuel in the form of glucose.

©HERIOT-WATT UNIVERSITY 250 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

Below this level, at roughly the intensity of exercise which puts one out of breath, but not stressed, with a heart rate of roughly 130bpm (this depends on age), the balance shifts much more in favour of fat as the fuel source. At this point, fat consumption is at its maximum, making up 40-60% of the energy supply. For those trying to lose some excess fat, this is the level at which to exercise. At still less demanding levels of exercise, the balance shifts back to glucose, either derived from glycogen in the muscles or liver, or from food in the gut.

Slow twitch (Type 1) ﬁbres: Questions Q18: Describe the contraction of slow twitch muscle ﬁbres. Go online ......

Q19: To which sporting events are Type 1 muscle ﬁbres best suited? ......

Q20: On which source of ATP do slow twitch ﬁbres rely? ......

Q21: Complete the table to show the sources of fuel for slow twitch ﬁbres at different levels of exercise.

Level of sustained exercise Source of respiratory substrate Light Moderate High ......

Q22: Explain the concentrations of myoglobin, mitochondria and capillaries in muscles with a high proportion of slow twitch ﬁbres......

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 251

12.4.2 Fast twitch (Type 2) ﬁbres Learning objective By the end of this section, you should be able to:

• state that fast twitch (Type 2) ﬁbres contract relatively quickly, but for short periods;

• explain that fast twitch ﬁbres are good for power sporting activities such as sprinting or weight-lifting;

• explain that fast twitch ﬁbres generate ATP through glycolysis and the breakdown of creatine phosphate;

• state that fast twitch ﬁbres have few mitochondria and a relatively low blood supply;

• state that fast twitch ﬁbres have a low level of myoglobin, making them light pink in colour;

• explain that fast twitch fibres use both glycogen and creatine phosphate as fuel. In man’s evolutionary past, there must have been many occasions when it was necessary to contract muscles faster than could be achieved if the necessary ATP was supplied by aerobic respiration alone. One need only think of avoiding a predator or one’s enemies, running after prey, or lifting rocks and tree trunks, to imagine the sort of sudden movement required. As mentioned in a previous section, using the small reserves of ATP and creatine phosphate, followed by anaerobic respiration, can provide the sudden burst of muscular contraction needed to cope with such situations for up to 90s. These types of activity, modelled by sports activities such as sprinting and weight-lifting, demand a different type of muscle fibre to the slow twitch (Type 1) fibre required by endurance activities, namely the fast twitch (Type 2) fibre. In adapting to contracting much faster than slow twitch fibres, these Type 2 fibres have lost the ability to sustain their contraction. Also, relying largely on anaerobic sources of ATP, i.e. creatine phosphate and glycolysis, they have little myoglobin, so are pink or even white, and have a relatively poor capillary blood supply because they do not need to store or be supplied with large quantities of oxygen. Of course, fast twitch fibres do carry out aerobic respiration, both to maintain their general metabolism and, after exertion, to provide the ATP necessary to reconstitute creatine phosphate from creatine, and pyruvate from lactate. However, they do not use aerobic respiration to the same extent that slow twitch fibres do, and so they have far fewer mitochondria. Because of their greater reliance on anaerobic respiration, they have a relatively higher concentration of glycolytic enzymes. The energy stores in fast twitch fibres are creatine phosphate and glycogen, which can be rapidly broken down to glucose 6-phosphate and fed into the glycolysis pathway.

©HERIOT-WATT UNIVERSITY 252 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

Fast twitch (Type 2) ﬁbres: Questions Q23: Describe the contraction of fast twitch muscle ﬁbres. Go online ......

Q24: To which sporting events are Type 2 muscle ﬁbres best suited? ......

Q25: On which sources of ATP do slow twitch ﬁbres rely? ......

Q26: Which compounds are used as energy stores in fast twitch ﬁbres? ......

Q27: Compared to slow twitch ﬁbres, explain the relative concentrations of myoglobin, mitochondria and capillaries in fast twitch ﬁbres......

12.5 Learning points

Summary Creatine phosphate

• Muscle cells do not contain any reserve of ATP which will fuel more than 2s of contraction.

• Creatine phosphate is an energy reserve found in cells with high and variable energy demands, such as muscle and brain.

• Creatine phosphate breaks down anaerobically to release energy and phosphate, which is used to convert ADP and Pi to ATP at a fast rate. • There is only sufﬁcient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s.

• When energy demands are low, creatine phosphate reserves are restored using ATP provided by aerobic respiration.

Lactic acid metabolism

• During vigorous exercise, muscle cells do not get enough oxygen to supply all their energy needs through the electron transport chain.

• Under these conditions, pyruvate is converted to lactic acid.

• This conversion involves the transfer of hydrogen from the NADH formed during glycolysis to pyruvate, forming lactate.

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 253

Summary continued • The use of NADH to form lactate regenerates the NAD+ required to maintain ATP production by glycolysis.

• The accumulation of lactate causes muscle fatigue.

• Lactate is reconverted to pyruvate in the muscles, and to glucose in the liver.

• The reconversion of lactate to pyruvate or glucose requires ATP formed by aerobic respiration.

• This additional respiratory activity creates an oxygen debt.

Types of skeletal muscle ﬁbres

• There are two types of muscle ﬁbre, namely slow and fast twitch.

• All skeletal muscles contain a mix of the two types of ﬁbre.

• Athletes show patterns of muscle ﬁbres that reﬂect their events.

• Slow twitch (Type 1) ﬁbres:

◦ contract relatively slowly but can sustain contractions longer; ◦ are good for endurance sporting activities, such as long distance cycling or running; ◦ rely on aerobic respiration to generate ATP; ◦ have many mitochondria, a large blood supply, and a high concentration of the oxygen-storing protein myoglobin, which makes them dark red in colour; ◦ use both glycogen and fat as fuel depending on the intensity of the exercise.

• Fast twitch (Type 2) ﬁbres:

◦ contract relatively quickly, but for short periods; ◦ are good for power sporting activities, such as sprinting or weight- lifting; ◦ generate ATP through glycolysis and the breakdown of creatine phosphate; ◦ have few mitochondria and a relatively low blood supply; ◦ have a low level of myoglobin, making them light pink in colour; ◦ use both glycogen and creatine phosphate as fuel.

©HERIOT-WATT UNIVERSITY 254 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

12.6 Extended response question The activity which follows presents an extended response question similar to the style that you will encounter in the examination. You should have a good understanding of creatine phosphate in muscles before attempting the question. You should give your completed answer to your teacher or tutor for marking, or try to mark it yourself using the suggested marking scheme.

Extended answer question: Creatine phosphate in muscles Give an account of the role of creatine phosphate in muscles. (8 marks) 10 min ......

12.7 End of topic test End of Topic 12 test

Q28: Complete the paragraph using the words from the list. (10 marks) Go online Muscle cells do not contain any reserve of which will fuel more than of

activity. Creatine phosphate is an energy reserve found in cells with high and

energy demands, such as the muscles and the . There is only sufﬁcient creatine

phosphate to fuel strenuous muscle activity for a maximum of .

Creatine phosphate breaks down to release energy and which is used to convert ADP and to ATP at a fast rate. When energy demands are ,

creatine phosphate reserves are restored using ATP provided by respiration.

Word list: 2s, 10s, aerobic, aerobically, anaerobic, anaerobically, ATP, brain, low, Pi, phosphate, variable.

©HERIOT-WATT UNIVERSITY TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS 255

......

Q29: Complete the sentences by matching the descriptions on the left with the terms on the right. (10 marks)

During vigorous exercise, muscle muscles. cells cannot support

Without oxygen, pyruvate is muscle fatigue. converted to

Conversion of pyruvate involves the electron transport chain. transferring hydrogen from

The use of NADH to form lactate lactate. regenerates

NAD+ is required to maintain ATP NADH. production by

Accumulation of lactate causes oxygen debt.

Lactate is reconverted to pyruvate in AD+. the

Lactate is reconverted to glucose in glycolysis. the

Reconversion of lactate requires ATP liver. from

Reconversion of lactate causes aerobic respiration.

©HERIOT-WATT UNIVERSITY 256 TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS

......

Q30: Complete the statements by matching the parts on the left with the parts on the right. (12 marks)

There are two types of muscle ﬁbre, endurance events. namely

All skeletal muscles contain short periods only.

Athletes show patterns of muscle slow and fast twitch. ﬁbres that

Slow twitch ﬁbres contract slowly but aerobic respiration.

Slow twitch ﬁbres are good for a high concentration of myoglobin.

Slow twitch ﬁbres rely on glycogen and fat.

A large blood supply and many power events. mitochondria are linked to

Slow twitch ﬁbres use fuel stores of little myoglobin.

Fast twitch ﬁbres contract quickly but reﬂect their events. for

Fast twitch ﬁbres are good for creatine phosphate and glycolysis.

Fast twitch ﬁbres generate ATP by both types of muscle ﬁbre.

Few mitochondria and a low blood sustain contraction longer. supply are linked to ......

Q31: Explain the occurrence of creatine phosphate in the body. (2 marks) ......

Q32: Describe the supply of energy to muscle cells as they undertake a minute of intense activity. (3 marks) ......

Q33: Explain the difference between the colour of slow and fast twitch muscle ﬁbres. (6 marks) ......

©HERIOT-WATT UNIVERSITY 257

Topic 13

End of unit test 258 TOPIC 13. END OF UNIT TEST

End of Unit 1 test Q1: Complete the statements by matching the parts on the left with the parts on the Go online right. (12 marks)

Cellular differentiation means that regulate cell research in the UK. cells

Muscle cells can only maintain the chromosome number.

express genes characteristic of their Embryonic stem cells can cell type.

Tissue stem cells can develop more specialised functions.

Germline cells are able to exemplify epithelial tissue.

Mitosis provides the ability to develop into any cell type.

Mutations in germline cells produce a limited range of cell types.

Skin epidermis can be used to invade neighbouring tissues.

Stem cell research lets scientists produce gametes.

The Human Tissue Authority has a respond to normal regulatory signals. remit to

Cancer cells do not affect all cells of offspring.

understand cell growth and A tumour is benign if it does not differentiation.

©HERIOT-WATT UNIVERSITY TOPIC 13. END OF UNIT TEST 259

......

Q2: Complete the statements by matching the parts on the left with the parts on the right. (12 marks)

A nucleotide is composed of strong covalent bonds. deoxyribose,

The sugar-phosphate backbone is deoxyribose and phosphate. held together by

The base pairings of DNA are ligase. adenine and thymine,

At the 3 and 5 ends of the DNA uracil. strands are

The enzyme which joins fragments of the primary transcript. DNA together is

The formation of mRNA from a DNA guanine and cytosine. template is

The base in RNA which replaces carbohydrate groups. thymine is

The ﬁrst strand of RNA formed from a stop codon. DNA is

The protein-coding regions of mRNA phosphate and a nitrogenous base. are

The polypeptide is released when several different mRNA molecules. reaching

One primary transcript can produce exons.

Post-translational modiﬁcation may transcription. involve adding

©HERIOT-WATT UNIVERSITY 260 TOPIC 13. END OF UNIT TEST

......

Q3: Complete the statements by matching the parts on the left with the parts on the right. (12 marks)

Pathways that need an energy input pyruvate. are called

The active site lowers oxaloacetate.

Glycolysis is catalysed by activation energy.

The product of glycolysis is fructose.

The product of phosphofructokinase a multi-enzyme complex. only enters

Acetyl-coA passes the acetyl group to the intermembrane space.

A substrate that is converted into a NAD+ and FAD. glycolysis intermediate is

Enzymes pump H+ ions into anabolic.

High energy electrons are carried by glycolysis......

Q4: Complete the paragraph using the words from the list. (7 marks) Without oxygen, is converted to .

The use of NADH to form lactate regenerates .

Lactate is reconverted to glucose in the .

reserves are restored by aerobic respiration.

Slow twitch ﬁbres are good for events.

Fast twitch ﬁbres generate ATP by creatine phosphate and . Word list: creatine phosphate, endurance, glycolysis, lactate, liver, NAD+, pyruvate......

The following questions relates to different types of body cell.

The diagrams show two types of somatic cell.

Q5: Name the type of cell division by which somatic cells are formed. (1 mark) ......

Q6: As somatic cells have developed, they have differentiated. What is meant by ’differentiated’? (1 mark) ......

Q7: Name a type of somatic tissue other than those to which the cells above belong and state its function. (1 mark) ......

Q8: State two differences between cancer cells and normal tissue cells. (2 marks) ......

Q9: Compare the signiﬁcance of mutations to somatic and germline cells. (2 marks) ......

The following questions relate to nucleic acids.

DNA is made up of nucleotides.

Q10: List the components of a nucleotide. (1 mark) ......

Q11: Name the type of bond which holds DNA strands together. (1 mark) ......

Q12: Explain what is meant by the antiparallel structure of DNA. (3 marks) ......

Q13: State one feature of the diagram above which indicates that DNA replication rather than RNA synthesis is taking place. (1 mark) ......

Q14: Name the enzyme principally involved with DNA replication. (1 mark) ......

Q15: What does this enzyme need before it can start DNA replication? (1 mark) ......

Q16: The enzyme can only add nucleotides to one end of the DNA strand. How does this affect the replication of the DNA strands? (1 mark) ......

The following questions concern mutations.

Q17: What are mutations? (1 mark) ......

Q18: What effect do mutations have? (1 mark) ......

Q19: Which type of chromosome mutation involves the loss of a segment of chromosome? (1 mark) a) Deletion b) Duplication c) Translocation

......

Q20: Which type of chromosome mutation involves the repetition of a segment or section of chromosome? (1 mark) a) Deletion b) Duplication c) Translocation

......

Q21: Which type of chromosome mutation involves the rearrangement of chromosomal material involving two or more chromosomes? (1 mark) a) Deletion b) Duplication c) Translocation

......

Q22: State the most likely outcome of chromosome mutations and suggest an explanation. (2 marks) ......

The following questions relate to metabolic pathways and their control.

Q23: Metabolic pathways are either anabolic or catabolic. Choose one of these terms and explain what it means. (2 marks) ......

Q24: Control of metabolic pathways may involve the active site of enzymes. Explain the role of the active site. (2 marks) ......

Q25: Give an example of a pathway using a multi-enzyme complex and state the advantage of such an arrangement. (3 marks) ......

Q26: Feedback inhibition is very important in regulating the rate of production of metabolites. State the meaning of the term ’feedback inhibition’ and give an example. (2 marks) ......

The following questions relate to the processes of respiration.

Q27: In aerobic respiration, the product of glycolysis enters the mitochondrion. Name the product. (1 mark) ......

Q28: State what happens to this product ﬁrst in the mitochondrion. (2 marks) ......

Q29: Lactic acid metabolism is very important to muscle tissue under certain circumstances. Describe these circumstances. (1 mark) ......

Q30: When their event is over, athletes may be left gasping for breath. Explain this odd situation, as they are no longer exerting themselves. (3 marks) ......

......

Glossary Activation energy the minimum energy required by reactants to allow a reaction to occur

ADH (anti-diuretic hormone) a hormone produced by the pituitary gland which promotes water retention by the kidney

Adipose connective tissue composed mainly of cells that store fat, found beneath the skin, around internal organs, in bone marrow, and in breast tissue

ADP adenosine diphosphate

Anabolic a reaction which builds up complex molecules from simpler ones

ATP adenosine triphosphate

Bile a liquid produced by the liver which helps the digestion of fats/lipids by emulsifying these molecules

Blastocyst an embryo of about 150 cells (preimplantation) produced by cell division following fertilisation

Carcinogens any agent directly involved in causing cancer

Carrier an individual who is heterozygous for a particular recessive characteristic and shows a dominant phenotype

Catabolic a reaction which breaks down complex molecules to simpler ones cDNA stands for complementary DNA, or copy DNA, and can be single-stranded or double-stranded; it is synthesized in vitro from an mRNA template

Centromere region of the chromosome where two chromosomes are closed together

Chiasma the point where two homologous chromosomes are touching

Chromatid daughter chromosomes

Chromatin a complex of DNA, histones, and non-histone proteins found in the nucleus of a human cell. This is the material from which chromosomes are made.

Codominant of alleles, when they are both fully expressed in the phenotype of the heterozygote, e.g. in the AB blood group.

Codon set of three adjacent bases on messenger RNA which correspond to a DNA triplet of bases and an anticodon on transfer RNA

Co-enzyme organic molecule that acts as a temporary carrier of atoms being removed from or added to a substrate during a reaction. They aid the action of the enzyme to which they are loosely bound. Co-enzymes belong to a larger group of molecules called cofactors which enzymes require for their proper functioning.

Collagen a tough, ﬁbrous protein that is a major component of connective tissue.

Covalent bond strong chemical bond involving the sharing of electrons between atoms

Cytokines substances secreted by speciﬁc cells of the immune system which will stimulate speciﬁc immune responses

Cytoskeleton system of microﬁlaments and microtubules made of protein that give the cell its structure and shape

Differentiation a process, which takes place after cell division, where cells become specialised cells

Endometrium inner lining of the uterus

Eukaryote an organism in which the DNA of the cells is contained within a nucleus and enclosed in a membrane

FAD ﬂavin adenine dinucleotide

Gall bladder the organ in the body which stores bile

Gene a unit in the hereditary materials of an organism

Glucagon a hormone produced by the pancreas which raises blood glucose levels, acting in opposition to insulin

Glycolysis cell process which converts glucose to pyruvate and occurs in the cytoplasm of the cells of nearly all organisms

GTP guanosine triphosphate, which performs a similar role to ATP, but is more speciﬁc

Haploid cells contain only a single set of 23 chromosomes

Histones the basic proteins around which DNA is coiled and supercoiled. Histones and non-histone proteins are the chief protein components of chromatin.

Homologous chromosome a pair of chromosomes which share the same gene sequences, each derived from one parent

Homologous genes genes of the same type

Hydrolysis the breaking of chemical bonds by reaction with water

Hydrophobic of a molecule which is repelled by water, e.g. lipids (oils and fats)

Insulin a hormone produced by the pancreas which lowers blood glucose levels

In vitro fertilisation a technique that unites the egg and sperm in a laboratory instead of inside the female body

Isomer molecule with the same chemical formula but different molecular structure

Metabolite these are intermediates and products of metabolism

Micrometre (μm), 10-6m

Minisatellites highly repetitive DNA consisting of sequences between 10 and 100 base pairs long, repeated in tandem arrays which vary in size from 0.5 to 40 kbp

Multipotent stem cells these are true stem cells that can only differentiate into a limited number of types

Mutation occurs when a DNA gene is damaged or changed in such a way as to alter the genetic message carried by that gene

Myoglobin a conjugated protein, resembling haemoglobin, involved in oxygen transport in muscles

Negative feedback mechanism self-regulating process in which the build-up of a product inhibits earlier steps in the pathway that generates it

Neurotransmitters a range of chemicals which convey messages between neurons, or neurons and effectors

Nonsense mutation a change to a single nucleotide in DNA which results in a non-functional protein being produced

Nucleolus a small, generally spherical, body found within the nucleus of eukaryotic cells

Oxytocin a hormone produced by the pituitary gland which stimulates the uterus to contract during childbirth

Phosphofructokinase enzyme which catalyses a key regulatory step in glycolysis

Phospholipid bilayer thin membrane made up of two layers of phospholipid molecules, orientated such that their hydrophilic (’water-loving’) heads are on the outside and their hydrophobic (’water-hating’) tails are on the inside

Pituitary gland small, pea-sized gland attached to the base of the brain which secretes many hormones and is central to several homeostatic processes in the body

Pluripotent stem cells these are stem cells with the potential to make any differentiated cell in the body

Prokaryote organisms which do not have any organelles contained within membranes in their cytoplasm, e.g. bacteria

Prolactin a hormone produced by the pituitary gland which stimulates milk production and secretion

Ribosomes structures found in the cytoplasm and attached to ER where protein synthesis occurs

Rough endoplasmic reticulum extensive double-membrane sheets in the cytoplasm which are studded with ribosomes and continuous with the nuclear membrane

Scaffold organ scaffolds are organs from which all the cells have been removed. What remains is the extracellular matrix - the part of the organ that supports its shape.

Stem cells undifferentiated cells that can differentiate into specialised cells or divide by mitosis to produce more stem cells

Answers to questions and activities 1 Stem cells Introduction: Questions (page 3)

Q1: Groups of the same kind of cells with a common structure and function.

Q2: It consists of several different tissues and it carries out a speciﬁc function within a body system.

Q3: A group of organs working together to perform a particular function, e.g. any two from:

• circulatory; • digestive; • endocrine; • excretory; • immune; • integumentary; • lymphatic; • musculoskeletal; • nervous; • reproductive; • urinary; • vascular; • vestibular.

Stem cells: Questions (page 4)

Q4: The process by which an unspecialised cell develops speciﬁc functions.

Q5: They are capable of repeated cell division.

Q6: Any from:

• Muscle cell: movement; • nerve cell: carries impulses; • red blood cell: transport of oxygen.

If you have different answers, check them with your teacher as they may still be correct.

Embryonic stem cells: Questions (page 6)

Q7: How to use human embryonic stem cells to form specialised cells:

1. Early human embryo Blastocyst 2. Embryo stem cell removed 3. Stem cell cultured in the laboratory 4. Formation of mass of undifferentiated stem cells 5. Undifferentiated stem cells cultured in different culture conditions 6. Formation of specialised cells: nerve cell, muscle cell, gut cells

Q8: The hollow ball of (70-100) cells which implants into the endometrium.

Q9: In the epiblast (inner cell mass) of the blastocyst.

Q10: They are pluripotent, which means that they are able to differentiate into any type of body cell.

Q11: Any three from:

• diabetes; • Duchenne’s; • muscular dystrophy; • heart disease; • Parkinson’s disease; • traumatic spinal cord injury; • vision and hearing loss.

If you have different answers, check them with your teacher as they may still be correct.

Tissue stem cells: Questions (page 8)

Q12: Each term or statement listed below refers to the type of stem cells shown in bold:

1. Pluripotent - embryonic 2. Develop into all cell types - embryonic 3. Ability to differentiate into some cells in the body - tissue 4. Give rise to a limited range of cell types - tissue 5. Develop into cell types that are closely related to the tissue in which they are found - tissue 6. Found in developing embryo - blastocyst - embryonic 7. Have the capacity to become all cell types - embryonic 8. Multipotent - tissue 9. Found in body tissues - tissue 10. Ability to differentiate into all of the cell types - embryonic

Q13: The two types of tissue stem cells found in bone marrow are:

1. bone marrow stem cells, which can generate bone, cartilage, fat, cells that support the formation of blood, and ﬁbrous connective tissue. 2. haematopoietic stem cells, which can form all the types of blood cells in the body.

Q14: They are able to give rise to other types of cells, but are limited in their ability to differentiate.

Extended response question: Embryonic and tissue stem cells (page 9)

Suggested marking scheme Each line represents a point worth one mark. The concept may be expressed in other words. Words which are bracketed are not essential. Alternative answers are separated by a solidus (/); if both such answers are given, only a single mark is allocated. In checking the answer, the number of the point being allocated a mark should be written on the answer paper. A maximum of seven marks can be gained.

1. Stem cells are undifferentiated cells capable of repeated division to both more stem cells and cells that will later differentiate to form specialised cells.

2. Embryonic stem cells (ESCs) are found in the inner cell mass of the blastocyst.

3. ESCs are capable of repeated division...

4. ...to form more ESCs and other cell types.

5. ESCs are: capable of forming all the other cell types of the body / pluripotent

6. Tissue stem cells (TSCs) are found throughout the juvenile and adult body.

7. TSCs are capable of repeated division to form more TSCs and other cell types.

8. TSCs: can only form cells of the organ to which they belong / are multipotent.

9. e.g. bone marrow tissue cells can only give rise to bone marrow cells, red blood cells, platelets, phagocytes and lymphocytes.

End of Topic 1 test (page 10)

Q15:

Cellular differentiation: cell develops more specialised functions. the genes characteristic of that type of Muscle cells only express cell. red blood cells, platelets, phagocytes Produced from bone marrow stem cells: and lymphocytes. Cells capable of repeated division: stem cells. Stem cells: relatively unspecialised. Dividing stem cells produce more stem cells. Specialised cells are produced from dividing stem cells. Embryonic stem cells are located in the inner cell mass of the blastocyst. Through the body: adult stem cells. Tissue stem cells replenish differentiated cells. Can develop into any cell type: embryonic stem cells. Tissue stem cells can produce a limited range of cell types.

Q16: The process by which a cell develops more specialised functions.

Q17: By expressing only the genes characteristic of that type of cell.

Q18: Red blood cells, platelets, and the various forms of phagocytes and lymphocytes.

Q19: Capable of repeated division, relatively unspecialised.

Q20: More stem cells, specialised cells.

Q21: The inner cell mass of the blastocyst.

Q22: Throughout the body.

Q23: Replenish differentiated cells that need to be replaced.

Q24: Embryonic stem cells are capable of developing into any of the body’s cell types whereas tissue (adult) stem cells give rise to a limited range of cell types typical of the organ of which they are a part.

2 Differentiation in cells Mitosis: Question (page 17)

Q1:

Meiosis: Questions (page 19)

Q2:

Q3:

Q4: 46

Q5: 23

Q6: 23

Q7: E

Q8: A

Q9: D

Q10: C

Q11: B

Differentiation in somatic cells: Questions (page 21)

Q12: 46

Q13: Cells that have become specialised to carry out their various speciﬁc functions.

Epithelial tissue: Question (page 23)

Q14: Protection, secretion and absorption.

Connective tissue: Questions (page 24)

Q15: Joints between bones, the rib cage, the ear, the nose, the elbow, the knee and the ankle.

Q16: Storage of energy, protection of organs, providing a structural framework for the body, connection of body tissues and connecting epithelial to muscle tissue.

Muscle tissue: Question (page 26)

Q17: The body contains three types of muscle tissue: skeletal, smooth and cardiac. Cardiac is unique in that it is only found in one organ, the heart. Two of the muscle types are not under conscious control and are called involuntary muscle. The other type of muscle, skeletal, is under conscious control and is called voluntary muscle. The two muscle types which are striped are skeletal and cardiac. The muscle type which forces food through the small intestine is smooth. The muscle type which maintains posture is skeletal, but its control is both voluntary and involuntary.

Nervous tissue: Question (page 28)

Q18: Nervous tissue consists of neurons and glial cells. Sensory neurons conduct impulses into the CNS. Motor neurons conduct impulses out from the CNS. Interneurons connect other neurons. Glial cells maintain a constant environment for neurons.

Formation of the body organs: Question (page 29)

Q19: The gall bladder is formed from epithelial and muscle tissues, each type of tissue performing a different function in the organ.

Mutations in germline and somatic cells: Question (page 31)

Q20: Similarity:

• both involve changes to the genetic material.

Differences:

• germline mutations affect all cells in the body whereas somatic mutations only affect one cell or its descendents; • germline mutations are inherited whereas somatic mutations are not.

Extended answer question: Four types of human body tissue (page 33)

Answer scheme Each line represents a point worth one mark. The concept may be expressed in other words. Words which are bracketed are not essential. Alternative answers are separated by a solidus (/); if both such answers are given, only a single mark is allocated. In checking the answer, the number of the point being allocated a mark should be written on the answer paper. A maximum of eight marks can be gained.

1. Body tissue cells derive from somatic stem cells ...

2. ...by repeated mitosis.

3. Epithelial tissue - covers the organ surfaces.

4. Protection - skin / secretion - intestinal glands / absorption - villi.

5. Connective tissue - gives shape to organs and supports them.

6. Protection - skull bones / structural framework - ribs / storage of energy - adipose tissue / connecting body organs - blood / connecting epithelial to muscle tissue - cartilage (in tendons)

7. Muscle tissue - which causes locomotion or movement within organs.

8. Skeletal muscle - locomotion / smooth muscle - in arterioles control of access to capillary bed / cardiac muscle - contraction of the heart.

9. Nervous tissue - which transmits messages between the central nervous system and the rest of the body (and within the central nervous system).

10. Neurons - conduct impulses / glial cells - maintain a constant environment for neurons.

End of Topic 2 test (page 33)

Q21:

Somatic cells are found throughout the body.

Germline cells are gamete mother cells.

Examples of somatic cells are B-lympbocytes and motor neurons.

Dividing somatic cells produce more somatic cells.

Mitosis maintains the chromosome number.

Somatic cells form different types of differentiation. tissue by

Cells capable of producing gametes are germline.

Mutations in somatic cells affect the cell or tissue involved.

Q22:

Somatic cells divide: mitosis

Number of types of tissue in stomach: four

Produces more germline cells: mitosis

Germline cells produce gametes: meiosis

Germline chromosome complement: diploid

Gamete chromosome complement: 23

Mutations in germline cells: affect all cells of offspring

Q23: Skin epidermis is an example of epithelial tissue, the function of which is protection. Adipose tissue is used to store energy and is an example of connective tissue. Smooth muscle has the function of producing force and motion and is found in artery walls. Glial cells are found in nervous tissue where they make the myelin sheath of some neurons.

Q24: Throughout the body, apart from the gamete mother cells.

Q25: Any two from:

• skin epidermis; • B lymphocytes; • motor neurons; • any other named body cells.

Q26: More somatic cells.

Q27: Mitosis

Q28: It maintains the diploid chromosome number and the genetic complement.

Q29: Differentiation

Q30:

Tissue Function Example absorption / intestinal glands / Epithelial protection / skin epidermis / secretion villi connecting body organs / adipose tissue / connecting epithelial to blood / muscle tissue / Connective cartilage of tendons / protection / ribs / storage of energy / skull bones structural framework cardiac - heart / Muscle produce force and motion skeletal - biceps / smooth - artery walls glial cells / react to stimuli, carrying inter-neurons / Nervous impulses to various parts motor / of the body sensory

Q31: All four types.

Q32: A cell that is capable of passing on its genetic information to the next generation by giving rise to gametes.

Q33: Mitosis

Q34: Meiosis

Q35: Germline cells: diploid, which means two of each type of chromosome. Gametes: haploid, which means one of each type of chromosome.

Q36: Mutations in somatic cells only affect the cell or the cell line involved. In germline cells, mutations would be passed on to all cells of the offspring.

3 Research: Stem cells and cancer cells Stem cell research: Questions (page 43)

Q1: They are produced from a single group of embryonic stem cells and can be used for research work in stem cells.

Q2: They are formed from donated spare embryos which are at the blastocyst stage.

Q3: Tissue stem cells

Q4: It does not involve the use of embryonic stem cells, which is a contentious issue when using stem cells in research work.

Therapeutic uses of stem cells: Questions (page 47)

Q5: Embryonic stem cells are pluripotent / capable of differentiating into any body cell. Epidermal stem cells are multipotent / only capable of forming a limited range of cell types.

Q6: Reduced possibility of dehydration, infection or rejection.

Q7: They are unable to differentiate into other types of cells for transplant.

Q8: The matching of genetic types is to make sure that the patient will not reject the transplantation.

Q9: Advantages of corneal repair using stem cells:

• the cornea is repaired and the limbus is able to produce new epithelial tissues to replace further damage to the corneal epithelium. • matching is required for the donor scheme, and the limbus of the person is still damaged and unable to produce corneal epithelium.

Ethical issues and the regulation of stem cell use: Question (page 49)

Q10: For :

• surplus embryos from IVF would otherwise be destroyed without any value being put on them; • as they are pluripotent, they can be used to study a wide range of conditions affecting many different cell types.

Against:

• human life should not be artiﬁcially created, especially for research; • a zygote or blastocyst is a human being with the same right to life as anyone else.

The biology of cancer cells: Questions (page 52)

Q11: Cancer cells divide excessively and do not respond to regulatory signals.

Q12: The tumour cells fail to attach to each other and can spread throughout the body in the blood and lymphatic systems.

Extended answer question: Stem cell research work (page 54)

A) Corneal transplants (maximum of 3 marks):

1. Traditionally, the person would have to wait for donated corneal tissues to become available. 2. With the use of stem cells isolated from the patient’s healthy eye, new tissue can be cultured... 3. ...which can be transplanted back into the damaged eye, 4. and will repair the damaged corneal epithelium.

B) Skin grafts (maximum of 3 marks):

i. Traditionally, a section of the patient’s healthy skin is cultured over 3 weeks. ii. During this time, dehydration and infection are risks for the patient. iii. Skin stem cells from the patient can be cultured to produce new epidermal cells much more quickly. iv. This reduces the time during which the patient is at risk.

End of Topic 3 test (page 55)

Q13:

One reason stem cells are useful they can be cultured indeﬁnitely. for research is that

Cells which can differentiate into all pluripotent. cell types are

Stem cell research has contributed cell growth and differentiation. to understanding

The use of stem cells has improved reducing the chance of rejection. corneal repair by

The use of stem cells has improved reducing the chance of dehydration. skin graft procedures by

The development of this can be motor neuron disease. studied using stem cells

The use of stem cell cultures to test a wide range of drugs helps Alzheimer’s disease. treatment of

Ethical objections have been raised embryonic stem cells. against the use of

An ethical objection against the use a zygote is a human being. of stem cells is that

This regulates cell research in the The Human Tissue Authority. UK

Q14:

Unlike normal cells, cancer cells divide uncontrollably.

Cancer cells do not respond to normal regulatory signals.

A tumour is a mass of abnormal cells.

Tumours develop as a result of mutations.

The genes involved in tumour slower cell division and programmed development control cause cell death.

it does not invade neighbouring A tumour is benign if tissues.

its cells lose the molecules that A tumour becomes malignant if normally hold them in place.

cells detach from a primary tumour Secondary tumours develop when and spread through the body.

Q15: They are pluripotent (capable of differentiating into all cell types). They can be cultured indeﬁnitely in the laboratory.

Q16: Cell growth, differentiation and gene regulation.

Q17: Stem cells have improved the therapeutic procedures of:

i. corneal repair by speeding up the process, creating new healthy tissue and reducing the chance of rejection; ii. skin grafts by the much more rapid production of a temporary graft, reducing the risk of dehydration or infection.

Q18: Parkinson’s disease, motor neuron disease.

Q19: They can be used to produce very speciﬁc cultures for use in testing a wide range of drugs.

Q20: Embryonic stem cells

Q21: A zygote is a human being with a right to life. Human beings, in the form of embryos, should not be artiﬁcially created.

Q22: The Human Tissue Authority. The Human Fertilisation and Embryology Authority.

Q23: Cancer cells divide uncontrollably to produce a mass of abnormal cells called a tumour. Cancer cells do not respond to normal regulatory signals which would instruct them to stop dividing when necessary.

Q24: Mutations in the genes controlling cell division, programmed cell death, and differentiation.

Q25: It does not invade neighbouring tissues or spread throughout the body.

Q26: The cells lose the molecules on their surface that normally hold them in place.

Q27: Cells become detached from a primary tumour, spreading through the body in the blood or lymphatic circulation.

4 DNA structure and replication Introduction: Questions (page 60)

Q1: Amino acids joined by peptide bonds.

Q2: Catalase - enzyme; collagen/elastin/keratin - structural; insulin - cell signalling. If you have different answers, check them with your teacher as they may still be correct.

DNA structure: Questions (page 67)

Q3: Double helix

Q4: Hydrogen

Q5: Weakness

Q6:

DNA Base Complementary base Adenine Thymine Cytosine Guanine Guanine Cytosine Thymine Adenine

Q7: A-T, because it only has two hydrogen bonds whereas C-G has three.

Q8: Ribose sugar, phosphate, base.

Q9: Antiparallel

Q10: 3 end - deoxyribose; 5 end - phosphate.

Arrangement of DNA in chromosomes: Questions (page 69)

Q11: It is tightly looped and coiled.

Q12: Histones and non-histone proteins.

Q13: There are 22 homologous pairs of chromosomes, with each member of each pair carrying the same genes, and the two sex chromosomes, which are X and X in a female, and X and Y in a male. Therefore, the total number of different chromosomes is 22 + 2 = 24.

DNA replication: Question (page 73)

Q14:

DNA is unwound and two template strands. unzipped to form

DNA is unzipped by the enzyme helicase.

At the end of the 3 deoxyribose. strand is

At the end of the 5 phosphate. strand is

To start attaching nucleotides, DNA a primer. polymerase requires

DNA polymerase attaches nucleotides to 3 end of the DNA strand. the

The strand that is continuously replicated is leading 3 strand the

The strand replicated in lagging 5 strand fragments is the

The enzyme which joins ligase DNA fragment is

In addition to a DNA molecule and enzymes, ATP and free nucleotides DNA replication needs

Extended response question: The location and structure of DNA, and DNA replication (page 75)

A) Describe the location and structure of DNA. (maximum of 8 marks):

1. DNA is located on chromosomes in the nucleus. 2. A DNA molecule consists of two strands wound in a double helix. 3. Each strand consists of subunits called nucleotides. 4. A nucleotide consists of a deoxyribose sugar molecule, a phosphate group and an organic / nitrogenous base. 5. The deoxyribose and phosphate are linked to their neighbours to form a sugar-phosphate backbone. 6. There are four organic bases: adenine (A), thymine (T), guanine (G) and cytosine (C). 7. Bases are linked in complementary pairs A-T and G-C. 8. Bases link the two DNA strands by hydrogen bonds. 9. The two DNA strands have an antiparallel structure / explanation. 10. Deoxyribose is found at the 3 end of each strand and phosphate is found at the 5 end.

B) Give an account of the replication of DNA. (maximum of 7 marks):

i. DNA is unwound and unzipped... ii. ...by the enzyme helicase... iii. ...to form two template strands. iv. DNA polymerase is the enzyme which adds nucleotides to the new DNA strand. v. DNA polymerase needs a primer to start replication. vi. DNA polymerase can only add complementary nucleotides to the deoxyribose / 3 end of the DNA strand. vii. This results in one strand / the leading strand being continuously replicated, viii. and the other strand / the lagging strand being replicated in fragments, ix. which are joined together (by the enzyme ligase).

End of Topic 4 test (page 76)

Q15:

DNA is the molecule of inheritance able to direct its own replication. and is

used by cells to store genetic The base sequence of DNA is information.

DNA is stored on chromosomes and tightly coiled with associated proteins. is

The basic unit of DNA is the nucleotide.

composed of deoxyribose, phosphate A nucleotide is and a base.

identiﬁed by their position in the The carbons in deoxyribose are molecule.

held together by strong covalent The sugar-phosphate backbone is bonds.

adenine, cytosine, guanine and The bases of DNA are thymine.

adenine and thymine, guanine and The base pairings of DNA are cytosine.

The two strands of DNA are held together by hydrogen bonds.

Q16:

The shape of the DNA molecule is a double helix. The two strands of DNA are antiparallel.

deoxyribose and At the 3 and 5 ends of the DNA strands are phosphate. The enzyme that adds nucleotides to the new DNA strand is DNA polymerase. Before DNA replication can start, what is required is a primer. Complementary nucleotides are added to the 3 end of the new DNA strand. The DNA strand which is continuously replicated is the 3 or leading strand. The enzyme that joins fragments of DNA together is ligase.

Q17: It is the molecule of inheritance and can direct its own replication.

Q18: In the base sequence of DNA.

Q19: It is tightly coiled and packaged with associated proteins.

Q20: Name: nucleotide Composition: deoxyribose sugar, phosphate, and organic / nitrogenous base.

Q21: They are numbered according to their position on the molecule, e.g. 1 .

Q22: By strong (covalent) bonds between deoxyribose and phosphate groups.

Q23: Bases: adenine (A), thymine (T), guanine (G) and cytosine (C). (NB: The question asks for the names so the letters alone would not be correct) Complementary pairs: A-T and G-C (NB: Letters are acceptable here because they have been linked to the base names in the ﬁrst part of the answer)

Q24: Hydrogen bonds between the complementary base pairs.

Q25: The hydrogen bonds that hold them together are weak bonds (compared to those that hold the sugar phosphate backbone together).

Q26: Double helix

Q27: Antiparallel

Q28: 3: deoxyribose;5: phosphate.

Q29: Helicase unwinds and unzips the DNA molecule.

Q30: DNA polymerase

Q31: It allows DNA polymerase to start replication.

Q32: The deoxyribose / 3 end of the strand.

Q33: The 3 strand / leading strand.

Q34: The enzyme ligase joins them.

5 Gene expression in human cells Gene expression through protein synthesis: Questions (page 82)

Q1: a) True

Q2: Extracellular

Q3: ADP

Q4: Transcription and translation

Q5: 48

Structure and functions of RNA: Questions (page 85)

Q6:

Q7:

RNA DNA Structure not a double helix double helix Preferred form single-stranded double-stranded Number of types >1 1 the cytoplasm and the Present in the nucleus nucleus adenine, uracil, guanine, adenine, thymine, Nitrogenous bases cytosine guanine, cytosine

Q8: mRNA, tRNA and rRNA

Q9: mRNA copies the code from the DNA molecule and carries it out to the ribosomes where the proteins are synthesised. tRNA molecules are found in the cytoplasm. They attach to speciﬁc amino acids and bring them to the ribosomes where the amino acids are joined together.

Q10: c) both DNA and RNA

Q11: rRNA and proteins

Transcription of DNA into mRNA: Questions (page 92)

Q12: Translation

Q13: Helicase unwinds the DNA double helix and unzips the DNA molecule. RNA polymerase binds the RNA nucleotides together.

Q14: b) DNA → RNA

Q15: d) GACUG

Q16: The mRNA molecule must start at the 5 end.

Q17: Addition of the 5 cap, joining of the exons, and removal of introns.

Q18: RNA splicing

Translation of mRNA into a polypeptide: Questions (page 98)

Q19:

1. DNA 2. tRNA 3. Hydrogen bonds 4. Amino acid 5. Start 6. Small subunit 7. mRNA 8. Peptide bonds 9. Ribosome subunits disassociation

Q20:

Single gene, several proteins: Question (page 102)

Q21: One gene can be expressed as many proteins as a result of RNA splicing and post-translational modiﬁcation. Different mRNA molecules are produced from the same primary transcript depending on which RNA segments are treated as exons and introns. Protein structure can be altered post-translation by: cutting and combining polypeptide chains, e.g. the formation of insulin from pro- insulin; adding phosphate or carbohydrate groups to the protein.

Extended response question: Transcription and post-translational modiﬁcation (PTM) (page 105)

A) Transcription (maximum of 6 marks): 1. Transcription is the formation of a mRNA molecule on a DNA template. 2. DNA unwinds, and strands separate by the action of helicase. 3. RNA nucleotides attach to exposed bases of DNA. 4. DNA A pairs with RNA U, DNA T - RNA A, DNA G - RNA C and DNAC - RNA G. 5. RNA polymerase joins nucleotides (to each other). 6. This produces the primary transcript. 7. Which contains introns and exons. 8. Exons are protein coding sections of mRNA and introns are non-coding sections. B) Post-translational modiﬁcation (PTM) (maximum of 4 marks): i. It takes place after the polypeptide has been completed on the ribosome. ii. It allows several proteins to be formed from one gene. iii. It may involve enzyme cutting and combining of polypeptide chains. iv. e.g. insulin from pro-insulin. v. It may involve adding of phosphate or carbohydrate groups. vi. Addition of phosphate enables enzymes / receptors to be switched ’on’ and ’off’.

End of Topic 5 test (page 106)

Q22: The individual’s phenotype is determined by the proteins produced by gene expression. These proteins are generated from the DNA in the nucleus of each cell, but in any one cell only a fraction of the genome is expressed. Intracellular factors can inﬂuence gene expression, e.g. phosphofructokinase is activated by high levels of ADP. Extracellular factors may also have an effect, e.g. low blood blood sugar levels activate the production of glucagon by the alpha cells of the pancreas. Gene expression has two stages, transcription in the nucleus and translation at the ribosome, both of which may be used to control the process. The code of a gene consists of a series of triplets of bases in the DNA. A triplet of bases, of which there are four types, can represent 64 different instructions. Q23:

RNA: single strand of nucleic acid Phosphate, base, ... ribose sugar Adenine, cytosine, guanine, ... uracil G-C, ... A-U Messenger, transfer, ... ribosomal mRNA: nucleus to cytoplasm tRNA: amino acid to ribosome rRNA: reads mRNA code Anticodon: 3 exposed bases on tRNA

Q24:

Transcription: DNA triplet code to mRNA.

RNA polymerase: DNA unwound and unzipped.

RNA base uracil: DNA base adenine pairs with it. mRNA codon: DNA triplet of bases matches with it.

Primary transcript: ﬁrst RNA strand formed from DNA.

Exons: protein-coding regions of mRNA.

Introns: non-coding regions of mRNA.

RNA splicing: joining up of exons.

Mature transcript: ﬁnal form of mRNA.

Q25:

1. Translation takes place at the ribosome. 2. Ribosomes consist of rRNA and proteins. 3. At the ribosome, rRNA reads the incoming mRNA. 4. When a start codon is reached, the ribosome starts attaching tRNA bases to the mRNA. 5. mRNA has sets of three bases called codons and tRNA has matching sets called anticodons. 6. Each tRNA molecule carries a speciﬁc amino acid. 7. Enzymes in ribosome join the amino acids by peptide bonds. 8. tRNA disengages and goes off to ﬁnd another of its amino acids. 9. When a stop codon is reached, the ribosome diassociates and the polypeptide is released.

Q26:

The body contains many more genes. proteins than there are

One gene must be able to code for several proteins.

One primary transcript can produce several different mRNA molecules.

Alternative RNA splicing treats exons and introns. different RNA segments as

Post-translational modiﬁcation may carbohydrate groups. involve adding

Cutting and combining of post-translational modiﬁcation. polypeptides is another form of

Q27: Gene expression produces the proteins which determine phenotype.

Q28: Phosphofructokinase is activated / by high levels of ADP. (or other suitable)

Q29: Low blood sugar levels activate the (alpha cells of the islets of Langerhans in the) pancreas / to produce glucagon.

Q30: In any nucleotide there are 4 possible bases, so with three nucleotides there are 4x4x4=64possible combinations.

Q31: A-U, G-C, T-A

Q32: messenger RNA, transfer RNA, ribosomal RNA

Q33: The transfer of the gene information from the triplet code on the DNA template to mRNA.

Q34: Introns are removed and / the exons are joined together.

Q35: Joining of the exons together.

Q36: Base pairing between different parts of the mRNA strand fold the molecule, exposing three bases at one end (the anticodon) / and an attachment point for an amino acid at the other end.

Q37: Any one amino may be coded for by more than one anticodon.

Q38: The ribosome disassociates / and the chain of amino acids is released as a polypeptide.

Q39: Different mRNA segments being treated as exons and introns.

6 Genes and proteins in health and disease Proteins: Questions (page 114)

Q1: Carbon, hydrogen, nitrogen, oxygen.

Q2: Amino acids; 20.

Q3: Peptide

Q4: Polypeptide

Q5: The arrangement of amino acids in polypeptide chains; the association of more than one polypeptide chain in the mature protein; the inclusion of non-protein structures, e.g. phosphates, metal ions, and carbohydrates.

Different levels of protein structure: Question (page 119)

Q6:

Level of structure Description Primary: the sequence of amino acids. Secondary: held by weak hydrogen bonds between amino acids. Tertiary: held by strong (disulphide) bonds between amino acids. Quaternary: two or more polypeptide chains associated together.

Functions of proteins: Questions (page 121)

Q7:

Type of protein Example structural collagen, elastin contractile actin and myosin in muscle cells hormones insulin insulin receptor in liver cells (forming part receptors of the structure of the plasma membrane) transport proteins transporter of glucose into the cell defence proteins immunoglobulins enzymes lipase, pepsin, maltase

Treatment of bacteria with UV light: Question (page 123)

Q8: As the length of time for which the bacteria are exposed to UV increases, so the number of bacteria killed increases.

Mutations and genetic disorders: Questions (page 123)

Q9: An agent which increases the rate of mutation.

Q10: Radiation, e.g. X-rays, UV light; chemicals, e.g. benzene, arsenic.

Effects of gene mutations on amino acid sequences (page 127)

Q11: 4 amino acids have been changed; yes, a deletion mutation is a frameshift mutation.

Q12: 4 amino acids have been changed; yes, an insertion mutation is a frameshift mutation.

Q13: 1 amino acid has been changed; no, a substitution mutation is a point mutation, not a frameshift mutation.

Gene mutations: Questions (page 131)

Q14: a) one nucleotide

Q15: b) nonsense

Q16: c) splice-site

Q17: a) missense

Q18: b) sometimes

Q19: b) after it

Q20: b) after it

Q21: a) 0

Q22: d) all of these

Q23:

Q24:

Q25:

Q26: If three nucleotides are involved, only a single amino acid will be affected.

Q27: A section of a gene with a repeating series of triplets of bases.

Q28: Deletion or insertion mutations.

Q29: Huntingdon’s disease / Fragile X syndrome.

The effect of mutations on the structure and function of proteins: Question (page 136)

Q30:

Inherited disease Cause Beta thalassemia splice-site, autosomal recessive Cystic ﬁbrosis frameshift deletion, autosomal recessive Duchenne muscular dystrophy (DMD) nonsense, X-linked recessive Phenylketonuria (PKU) missense, autosomal recessive Sickle-cell disease missense, codominant Tay-Sachs syndrome frameshift insertion, autosomal recessive

Chromosome structure mutations: Question (page 139)

Q31:

structural Deletion and duplication are examples of abnormality.

Chromosome mutations occur during cell division.

Deletion is loss of a segment of chromosome meiosis I. during

The condition Cri-du-Chat is caused by a deletion.

Extra chromosomal material will be present duplication. because of

Duplication of these genes is the cause of many oncogenes. cancers

The most common effect of an inversion mutation reduced fertility. is

A gene from chromosome 22 fusing a gene on translocation. chromosome 9

A cancer of stem cells in the bone marrow myeloid leukaemia.

Extended response question: Proteins and gene mutation (page 142)

A) Give an account of the structure and function of proteins.

a) Protein structure (maximum of 4 marks): 1. Proteins consist of amino acids joined together (in chains). 2. The amino acids are joined by strong peptide bonds... 3. ...to produce the primary structure of the polypeptide. 4. Further bonding between amino acids... 5. ...produces the secondary and tertiary structures. 6. Folding of the polypeptide chains forms the three-dimensional shape of the protein. b) Protein function (maximum of 4 marks): I. Some proteins are enzymes + named example (e.g. the digestive enzyme amylase). II. Some proteins are hormones + named example (e.g. insulin). III. Some proteins are antibodies which help the body ﬁght infections. IV. Some proteins transport substances + named example (e.g. haemoglobin). V. Some proteins provide structure + named example (e.g. collagen).

B) Give an account of gene mutation. (maximum of 9 marks)

i. A gene mutation is a change in the base type or sequence in a gene. ii. In a substitution (mutation) one base is replaced by another. iii. If substitution produces a new stop codon it is a nonsense mutation. iv. If substitution affects introns and exons it is a splice-site mutation. v. If substitution changes one mRNA codon it is a missense mutation. vi. In a deletion (mutation) a base is removed. vii. In an insertion (mutation) a base is added. viii. A change to a single nucleotide / base is a point mutation. ix. Insertion and deletion (mutations) can potentially cause frameshift mutation. x. A frameshift mutation alters all the triplets following it. xi. If any of these mutations occurs in a protein-coding gene, then the protein produced may be altered (or not produced at all).

End of Topic 6 test (page 143)

Q32: Proteins are composed of chains of amino acids joined together by peptide bonds to form a polypeptide. They are held in complex three-dimensional shapes by hydrogen bonds and interactions between individual amino acids. In addition, several polypeptide chains may associate together and non-protein groups such as carbohydrates and metal ions may be bound into the ﬁnal protein structure. Proteins have many functions including acting as enzymes (e.g. phosphofructokinase), hormones (e.g insulin), transport (e.g. haemoglobin), and structural components (e.g. collagen).

Q33:

Term Deﬁnition Point mutation single base involved Substitution one base replaced Missense one codon replaced Nonsense new stop codon Splice-site new exon-intron codons Deletion removal of a base Insertion addition of a base Frameshift base sequence changed

Q34:

(all of these) addition, Of what are chromosome structural mutations the result? removal, reversal Which type of mutation results from the loss of a segment of deletion chromosome? (all of these) exons, Which codons are potentially altered by structural mutations? introns, stop codons Which type of mutation causes Cri-du-Chat? deletion Which type of mutation is associated with breast cancer? duplication Which type of mutation causes chronic myeloid leukaemia? translocation Which type of mutation usually has not noticeable effect? inversion What is the likely effect of substantial changes to the lethal effect chromosome?

Q35: Peptide

Q36: A chain of amino acids

Q37: They hold the polypeptides in complex shapes.

Q38: Carbohydrates and metal ions

7 Human genomics Sequencing DNA: Questions (page 148)

Q1: 3

Q2: DNA sequence variations that occur when a single nucleotide in the genome sequence is altered.

Q3: Coding regions

Bioinformatics: Questions (page 149)

Q4: Computer science, statistics, mathematics and engineering.

Q5: Coding sequences similar to known genes, start codons, or sequences lacking stop codons.

Q6: They identify base sequences that correspond to the amino acid sequence of the protein.

Systematics: Questions (page 150)

Q7: c) Homologous

Q8: Scientists can work out the number of differences in an amino acid sequence between two species. They can then calculate the time since the two species diverged from a common ancestor. If there are lots of differences between the maps, it can be deduced that the species diverged longer ago than if there are only a few differences.

Personalised medicine: Question (page 152)

Q9: Any four from:

• production of more ’powerful’ medicines; • production of better and safer drugs more quickly; • more accurate methods of determining drug dosages; • advanced screening for a particular disease; • allows a person to make adequate lifestyle and environmental changes; • pharmaceutical companies will be able to discover potential therapies more easily.

Ampliﬁcation and detection of DNA sequences: Question (page 157)

Q10: After the mixing of the components in the PCR process, three further stages follow. Heating to 95◦C which denatures the DNA by separating its two strands. The temperature is then dropped to 54◦C, which allows primers to bind to their complementary sequences on the separated strands. Primers are short DNA sequences of about 20 bases that bind on either side of the target sequence. This stage is known as annealing. The temperature is then raised again to 72 ◦C in the extension stage. Taq polymerase adds nucleotides to the 3 ends of the primers and extends them into new complementary strands. Continued cycles of heating and cooling cause the DNA to be doubled repeatedly, so that in one hour, 20 cycles will amplify the target DNA more than 1 million times.

DNA probes: Questions (page 158)

Q11: c) identify a particular sequence of DNA.

Q12: b) The pentose sugar ring of a gene probe is labelled with radioactive 14C.

Medical and forensic applications: Questions (page 161)

Q13:

Q14:

1. DNA extracted from blood sample. 2. DNA digested with a restriction enzyme. 3. Gel electrophoresis. 4. Blotting. 5. Hybridisation with labelled probe. 6. Autoradiography.

Q15: c) Suspect 3 Q16: There is the possibility that the suspect’s DNA ﬁngerprint matches that left at the scene of the crime purely by chance. Q17: By increasing the number of probes hybridised to the DNA blot.

Extended response question: The PCR process (page 165)

1. The components of the PCR (template DNA, primers, nucleotides and the Taq DNA polymerase) are mixed together. 2. Heating to 90 - 95◦C... 3. ...separates (denatures) the double-stranded DNA (containing the target DNA sequence). 4. Annealing at 54◦C... 5. ...allows the primers to bind (by forming hydrogen bonds) to their complementary sequences on the separated strands 6. Primers are short DNA sequences (each about 20 bases long), 7. that bind at either side of the target sequence (one on each of the complementary strands of the target DNA). 8. Extension at 72◦C. 9. Taq polymerase now adds nucleotides to the 3 ends of the primers and extends them into new complementary strands. 10. Repeated cycles multiply the target DNA exponentially... 11. ...since each new double strand separates to become two templates for further synthesis.

End of Topic 7 test (page 166)

Q18:

The sequence of bases can be determined for individual genes by DNA sequencing. using

The use of computer technology to identify DNA sequences is known bioinformatics. as

Comparison of the human genome with genomes of other species is systematics. known as

An understanding of the genetic component of the risk of disease is personalised medicine. required for

A technique for the ampliﬁcation of polymerase chain reaction (PCR). DNA in vitro is known as

Short sequences of single-stranded DNA that are primers. complementary to speciﬁc target sequences are known as

Short single stranded fragments of DNA probes. DNA are known as

Comparison of highly variable repetitive sequences of DNA is DNA proﬁling. known as

Q19:

1. Temperature of the reaction adjusted to 95◦C. 2. Penetration of the double-stranded DNA. 3. Temperature of the reaction adjusted to 55◦C. 4. Annealing of the primers to the single stranded DNA. 5. Temperature of the reaction adjusted to 72◦C. 6. Synthesis of DNA by the enzyme DNA polymerase.

Q20: a) the full complement of DNA possessed by the organism.

Q21: b) The enzyme is relatively stable at high temperatures.

Q22: During the process of PCR, extension occurs at 72◦C.

Q23: During the process of PCR, DNA primers bind to DNA single strands during annealing.

Q24: During the process of PCR, heating separates the DNA into single strands.

Q25: b) 8

Q26: a) Primers allow DNA probes to attach to single stranded DNA.

8 Cell metabolism Metabolism: Question (page 171)

Q1:

All the chemical reactions of the body: metabolism

The chemical on which an enzyme acts: substrate

Chemical produced by an enzyme’s action: product

Chemicals involved in cells reactions: metabolites

Anabolic and catabolic pathways: Question (page 173)

Q2:

Anabolic Catabolic Biosynthesis Breakdown Requires energy input May release energy

ATP → ADP + Pi ADP → Pi → ATP DNA polymerase Glycogen phosphorylase

Reversible and irreversible pathways: Questions (page 176)

Q3: Reversible reactions only change the arrangement of atoms in a molecule, whereas irreversible reactions add phosphate. Reversible reactions do not require ATP to release energy, whereas irreversible reactions require the input of energy from ATP breakdown.

Q4: The presence of the substrate will cause the reaction to run from left to right. Build-up of the substrate will cause the reaction to run from right to left.

Q5: The pentose phosphate pathway takes glucose 6-phosphate and produces NADPH and a 5-carbon (pentose) sugar.

Control by regulation of enzyme production: Questions (page 179)

Q6: Mutation of the gene (for an enzyme in a metabolic pathway).

Q7: The enzyme for converting phenylalanine to tyrosine is absent.

Q8: Extracellular come from outside the cell; intracellular are formed within the cell.

Q9:

i. An example of an extracellular signal molecule is any hormone, e.g. estrogen, cytokines; ii. An example of an intracellular signal molecule is the proteins in the cascade that are initiated by cytokines attaching to their receptors in the cell membrane. (names deﬁnitely not needed)

Enzyme regulation: Question (page 179)

Q10:

Concentration of end-product: porphyrin synthesis

phosphorylation of glycogen Post-translational modiﬁcation: synthase

Local environment: changing pH

Localisation within the cell: Golgi apparatus

Reversible reactions: Question (page 184)

Q11:

1. The presence of the substrate will cause the reaction to go from left to right. 2. Build up of the product will cause the reaction to be reversed.

Multi-enzyme complexes: Questions (page 184)

Q12: Any of:

• RNA/DNA synthesis; • glycolysis; • TCA cycle.

Q13:

1. The product of one reaction is immediately available for the next. 2. The product of one reaction is presented so that it ﬁts immediately into the active site of the next enzyme in the chain.

Competitive and non-competitive inhibition of enzymes: Question (page 187)

Q14:

Extended response question: The control of the action of enzymes (page 191)

1. Enzymes that are in constant use are continuously present.

2. The presence of the substrate may initiate a reaction.

3. The build-up of the product may reverse the direction of a reaction.

4. The active site is the part of the enzyme molecule to which the substrate (temporarily) binds.

5. Competitive inhibitors resemble the substrate and compete with it for the active site.

6. Increased substrate concentration reduces the effect of competitive inhibition.

7. Competitive inhibition is reversible.

8. Non-competitive inhibitors bind to allosteric sites...

9. ...and alter the shape of the sites, preventing the substrate from binding to them.

10. Enzyme activators bind to allosteric sites and alter the shape of the active site to ﬁt the substrate.

11. Feedback inhibition involves non-competitive inhibition.

12. Products from reactions late in metabolic pathway act as non-competitive inhibitors to enzymes earlier in the pathway.

End of Topic 8 test (page 191)

Q15:

Biosynthetic means building up complex molecules. Anabolic pathways require energy input. Catabolic pathways usually release energy. Steps which require energy are irreversible. Reactions which can run both ways are reversible. Alternative routes by-pass steps in a pathway.

Q16:

Regulation of the activity of key enzymes metabolic pathways. enables control of

Extracellular signal molecules regulate hormones.

intracellular signal Protein-protein interactions are set off by molecules.

Some enzymes are always present in cells as a continuously result of expressed genes.

A reaction may be initiated by the presence of substrate.

A reaction may be reversed by the build-up of product.

Q17: The active site is the point on the enzyme molecule to which the substrate molecule(s) temporarily binds. In many cases, the active site does not exactly fit the substrate and it is the substrate molecule itself which causes the necessary change in the conformation of the enzyme molecule. This is called induced fit. The active site orientates reactants, lowers activation energy, and releases products with low affinity for the active site. Many metabolic reactions are reversible and the presence of a substrate or relative concentration of a product will determine their direction. Glycolysis is an example of metabolic pathway controlled by a metabolon, or multi- enzyme complex.

Q18:

Sites where activators may bind: allosteric.

Inhibitors bind to sites away from the active site: non-competitive.

Inhibitors which resemble substrate: competitive.

substrate Increase reduces competitive inhibition: concentration.

Competitive inhibition: reversible.

End-products inhibit earlier enzymes: feedback control.

Q19: Anabolic pathways build up more complex molecules from simpler ones and require an input of energy. Catabolic pathways break down complex molecules to simpler ones and generally release energy.

Q20: The presence of the substrate and the relative concentration of the product of the reaction.

Q21: The active site orientates reactants, lowers the activation energy of the transition state and releases products with a low afﬁnity for the active site.

Q22: Feedback control acts through end products, inhibiting an enzyme which catalyses a reaction early in the pathway.

9 Cellular respiration I: Glycolysis Introduction: Questions (page 196)

Q1: It provides the energy to drive them.

Q2: Anabolic

Q3: Catabolic

Summary of glucose breakdown: Question (page 198)

Q4:

What is ATP: Question (page 200)

Q5: ATP consists of a ribose sugar molecule with adenine attached at 1 carbon and three phosphates attached in a chain at 5 carbon.

ATP and energy transport: Question (page 201)

Q6: ATP can be readily converted to ADP and Pi to release energy, and reconverted to store energy.

ATP and phosphorylation: Question (page 202)

Q7: The addition of a phosphate group to a molecule. Q8: Glucose 6-phosphate Q9: It switches them ’on’ or ’off’.

Glycolysis: Questions (page 203)

Q10: Cytoplasm Q11: The product of each reaction is readily available as the substrate for the next enzyme in the chain. Q12: Two molecules of pyruvate, two molecules of ATP, and two hydrogen ions on carrier molecules (as NADH).

The energy investment stage: Questions (page 205)

Q13: Two molecules of ATP are used up in it. Q14: ATP contributes the phosphate which attached to the substrate molecule. Q15: The phosphorylation reactions involve the addition of phosphate to the molecule, whereas the other reactions re-arrange the existing atoms in the molecule.

The energy investment stage: Questions (page 207)

Q16: 2 Q17: 2 Q18: Phosphorylation Q19: The phosphate comes directly from the substrate. Q20: It can enter the mitochondrion, form Acetyl CoA, and enter the Krebs Cycle / Citric Acid Cycle / Tricarboxylic Acid (TCA) Cycle.

Q21: It is converted to lactic acid.

Q22: Alanine and fatty acids.

Feedback inhibition and regulation of the glycolytic pathway: Questions (page 208)

Q23: An enzyme’s activity is reduced by an increase in the concentration of a product from a later step in the pathway. Q24: Damaging metabolites don’t build-up and resources, such as ATP,are not wasted.

Q25: ATP inhibits phosphofructokinase by binding to an allosteric site and reducing its afﬁnity for its sustrate.

Synchronisation of the rates of glycolysis and the citric acid cycle: Questions (page 209)

Q26: When the cell is not using much energy, the concentration of ADP will be low as ATP is not being broken down. Consequently, the phosphorylation steps of the pay-off stage will be slowed. And vice versa. Q27: Glycolysis supplies the pyruvate which is the fuel for the citric acid cycle. If glycolysis runs too slowly, the cell will not get the energy it needs from the citric acid cycle; if it runs to fast, energy will be wasted.

Extended answer question: Feedback inhibition (page 212)

1. Feedback inhibition in a metabolic pathway occurs when an enzyme’s activity is reduced... 2. ...by an increase in the concentration of a product from a later step in the pathway.

3. In the energy investment stage of glycolysis...

4. ...phosphofructokinase is inhibited by ATP... 5. ...which is released by glycolysis and the cytochrome system.

6. If the ATP concentration rises in the cell, phosphofructokinase will become less active...

7. ...and the production of ATP will be reduced, or vice versa.

8. The production of ATP will be matched to the cell’s need for energy...

9. ...and resources will be conserved.

End of Topic 9 test (page 212)

Q28: Respiration provides energy for all the cell’s anabolic reactions, movement, and heat release. In addition, intermediate compounds may be used in many other metabolic pathways. During aerobic respiration, glucose is broken down, oxygen is used up, and carbon dioxide and water are released. hydrogen ions and electrons are removed by dehydrogenase enzymes. As a result, ATP is formed and made available to the cell.

Q29:

ATP is hydrolysed to ADP and Pi

ATP is formed by the addition of Pi to ADP. Glycolysis and the cytochrome system generate ATP. Energy for anabolic processes is provided by ATP. Glucose to glucose 1-phosphate occurs phosphorylation. by

Q30:

Glycolysis takes place in the cytoplasm.

Glycolysis of one glucose molecule yields two molecules of pyruvate.

The energy investment stage uses up two molecules of ATP.

forms two molecules of ATP per The energy pay-off stage pyruvate.

The energy investment stage two irreversible phosphorylations. contains

Glucose 6-phosphate is formed by phosphorylation of glucose.

Glucose 6-phosphate and pyruvate can join several metabolic pathways.

The product of phosphofructokinase can only proceed to glycolysis.

Transfer of phosphate directly to ADP substrate-level phosphorylation.

pyruvate can be converted to lactic In the absence of oxygen acid.

Q31: Glycolysis and the citric acid cycle are synchronised by the feedback inhibition of phosphofructokinase in the energy investment stage and the supply of ADP to the phosphorylation reactions in the energy pay-off stage. Feedback inhibition occurs when an enzyme’s activity is reduced by an increase in the concentration of a product from a a later step in the pathway. Phosphofructokinase in the energy investment stage of glycolysis is inhibited by high levels of ATP. This will reduce the production of pyruvate, and therefore of ATP, when the cell’s need for energy is low.

Q32: The respiration pathways provide intermediate compounds which may be used in several other metabolic pathways.

Q33: One molecule of glucose yields two molecules of pyruvate, two molecules of ATP and two hydrogen ions as 2 NADH.

Q34: These ATP molecules are formed by the transfer of phosphate directly from the substrate to ADP.

Q35: They are synchronised by the feedback inhibition of phosphofructokinase in the energy investment stage, and the supply of ADP to the phosphorylation reactions in the energy pay-off stage.

10 Cellular Respiration II: Citric acid cycle Mitochondria: Question (page 218)

Q1:

Glycolysis multi-enzyme complex in cytoplasm

Pyruvate to acetyl-coA mitochondrial matrix

Mitochondrial matrix citric acid cycle

Electron transport chain cristae of mitochondrion

The steps of the citric acid cycle: Questions (page 221)

Q2: Once

Q3: Twice

Q4: Acetyl coenzyme A / acetyl-coA

Q5: Oxaloacetate

Q6: ATP, hydrogen ions, carbon dioxide.

Q7: It regenerates the oxaloacetate which joins to the acetyl group to form citrate.

The enzymes and carriers of the citric acid cycle: Questions (page 223)

Q8: A different enzyme

Q9: Dehydrogenases

Q10: NAD / NAD+ and FAD

Q11: NADH and FADH2 Q12: To the electron transport chain on the cristae (of the mitochrondrion).

Alternative substrates for respiration: Questions (page 226)

Q13: Glycogen

Q14: It is readily interconvertible with glucose.

Q15:

1. Conversion to glucose. 2. Conversion to intermediates in glycolysis.

Q16: They are converted to intermediates in glycolysis or in the citric acid cycle.

Q17: Liver

Q18: It is converted to an intermediate in glycolysis.

Q19: They are converted to acetyl-coA.

Extended answer question: Enzymes and coenzymes (page 228)

1. Each step in the citric acid cycle is mediated by a different enzyme.

2. Dehydrogenase enzymes remove hydrogen and electrons.

3. Coenzymes NAD / NAD+ and FAD accept hydrogen and electrons.

4. NAD / NAD+ accepts hydrogen to form NADH.

5. FAD accepts hydrogen to form FADH2.

6. NADH / FADH2 / reduced coenzymes carry hydrogen and electrons to the electron transport chain...

7. ...which results in the generation of ATP.

End of Topic 10 test (page 228)

Q20: The mitochondrion consists of an outer boundary membrane, an inner membrane which is folded to form cristae, and the central matrix. Pyruvate enters the matrix of the mitochondrion, is converted to acetate, and attached to co-enzyme A to form acetyl co-enzyme A (acetyl-coA). Acetyl-coA passes the acetate into the Citric Acid cycle in the matrix of the mitochondrion. Hydrogen and high energy electrons are released from the citric acid cycle and are taken by carrier molecules to the electron transport chain on the cristae of the inner membrane of the mitochondrion.

Q21:

Pyruvate is produced by glycolysis.

Pyruvate is broken down to form an acetyl group.

Acetyl-coA passes the acetyl group to oxaloacetate.

Oxaloacetate combines with the citrate. acetyl group to form

The citric acid cycle directly ATP. generates

Hydrogen is removed from the citric dehydrogenases. acid cycle by

Hydrogen is transferred from the citric coenzymes. acid cycle to

High energy electrons are carried by NAD+ and FAD.

NADH and FADH carry hydrogen to 2 electron transport chain. the

The electron transport chain is cristae. located on the

Q22:

A storage carbohydrate glycogen

A sugar which can be converted into fructose a glycolysis intermediate:

Some can be converted to citric acid amino acids cycle intermediates:

Formed from glycerol and fatty acids: lipids

Converted into glucose in the liver: glycerol

Converted into acetyl-coA: fatty acids

Q23: It results in the direct generation of ATP,the release of carbon dioxide, the transfer of hydrogen ions to carriers, and the regeneration of oxaloacetate.

Q24: Coenzymes NAD+ and FAD accept hydrogen ions and electrons to form NADH and FADH2. These then release the high-energy electrons to the electron transport chain on the cristae.

11 Cellular Respiration III: Electron transport system The electron transport chain: Questions (page 235)

Q1: Hydrogen ions and high energy electrons.

Q2: The inner mitochondrial membrane.

Q3: Protein

Q4: To act as enzymes to pass electrons from donors to acceptors.

Q5: To pump hydrogen ions across the inner membrane into the intermembrane space.

Q6: Oxygen

Q7: Water

ATP synthesis: Questions (page 237)

Q8: Energy

Q9: They are pumped across the inner membrane into the intermembrane space.

Q10: ATP synthase

Q11: The return ﬂow of H+ ions from the intermembrane space into the matrix.

Q12: ATP is formed from ADP and Pi. Q13: The electron transport chain.

Extended answer question: The electron transport chain (page 239)

+ 1. NADH and FADH2 carry H ions and high energy electrons... 2. ...from glycolysis and the citric acid cycle.

3. At the inner mitochondrial membrane, NADH and FADH 2 pass the hydrogen ions and high energy electrons...

4. ...to carriers in the electron transport chain.

5. The electron transport chain is a collection of proteins attached to the inner mitochondrial membrane.

6. As the high energy electrons released from NADH and FADH 2 pass along the chain, ...

7. ...they release their energy in stages.

8. This energy is used to pump H+ ions into the intermembrane space.

9. The ﬁnal electron acceptor is oxygen.

10. The (negatively charged) oxygen combines with two H+ ions to form water.

End of Topic 11 test (page 239)

Q14: The electron transport chain is a collection of proteins attached to the inner mitochondrial membrane. Coenzymes NAD+ and FAD collect hydrogen ions and high energy electrons from glycolysis and the citric acid cycle, forming NADH and FADH 2. They travel to the inner mitochondrial membrane and pass the hydrogen ions and high energy electrons to carriers in the electron transport chain. As the high energy electrons released from NADH and FADH2 pass along the chain of electron acceptors and donors, they release energy which is used by enzymes to pump hydrogen ions (H+) across the inner mitochondrial membrane into the intermembrane space. The ﬁnal electron acceptor is oxygen which combines with hydrogen ions to form water.

Q15:

Energy is released as electrons pass the electron transport chain. along

Enzymes pump H+ ions into the intermembrane space.

The return ﬂow of H+ ions into the part of ATP synthase. matrix rotates

ATP synthase catalyses the formation of ATP.

The electron transport chain the ATP from aerobic respiration. produces most of

Q16: The high energy electrons / released by NADH and FADH 2 / pass along a chain of transporter proteins (attached to the inner mitochondrial membrane). / As they are passed from protein to protein, they release their energy in stages. / This energy is used to pump hydrogen ions (H+) / across the inner mitochondrial membrane into the intermembrane space.

12 Energy systems in muscle cells Creatine phosphate: Questions (page 245)

Q1: 2

Q2: Muscle, brain

Q3: They have high and variable energy demands.

Q4: ATP, creatine

Q5: 10

Q6: When energy demands are low, ATP from respiration is used.

Lactic acid metabolism: Questions (page 247)

Q7: There is not sufﬁcient oxygen to provide all of the cell’s energy needs from the electron transport chain.

Q8: Pyruvate/pyruvic acid

Q9: NAD+

Q10: Fatigue

Q11:

i. Glucose ii. Pyruvate/pyruvic acid

Q12: ATP

Q13: Aerobic respiration requires an additional oxygen supply (to provide this ATP).

Types of skeletal muscle ﬁbres: Questions (page 248)

Q14: Slow and fast twitch.

Q15: 2

Q16: Slow twitch / Type 1

Q17: Fast twitch / Type 2

Slow twitch (Type 1) ﬁbres: Questions (page 250)

Q18: Slow, but sustained

Q19: Endurance

Q20: Aerobic respiration

Q21:

Level of sustained exercise Source of respiratory substrate Light Blood glucose Moderate Fat and glycogen High Glycogen

Q22: High levels of myoglobin to store oxygen. High numbers of mitochondria for aerobic respiration. High density of capillaries to give maximum oxygen supply.

Fast twitch (Type 2) ﬁbres: Questions (page 252)

Q23: Fast for short periods. Q24: Power Q25: Creatine phosphate and glycolysis / anaerobic respiration. Q26: Creatine phosphate and glycogen. Q27: Lower levels of myoglobin as oxygen does not have to be stored. Lower numbers of mitochondria as there is much less aerobic respiration. Lower density of capillaries as less oxygen has to be supplied.

Extended answer question: Creatine phosphate in muscles (page 254)

1. Muscle cells do not contain any reserves of ATP which will fuel more than 2s of activity. 2. Creatine phosphate is an energy reserve found in cells with high and variable energy demands... 3. ...such as muscle / brain. 4. Creatine phosphate is used by fast twitch muscle ﬁbres. 5. Creatine phosphate breaks down anaerobically... 6. ...to release energy and phosphate...

7. ...which is used to convert ADP and Pi to ATP at a fast rate.

8. There is only sufﬁcient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s.

9. When energy demands are low, creatine phosphate reserves are restored...

10. ...using ATP provided by aerobic respiration.

End of Topic 12 test (page 254)

Q28: Muscle cells do not contain any reserve of ATP which will fuel more than 2s of activity. Creatine phosphate is an energy reserve found in cells with high and variable energy demands, such as the muscles and the brain. There is only sufﬁcient creatine phosphate to fuel strenuous muscle activity for a maximum of 10s. Creatine phosphate breaks down anaerobically to release energy and phosphate which is used to convert ADP and Pi to ATP at a fast rate. When energy demands are low, creatine phosphate reserves are restored using ATP provided by aerobic respiration.

Q29:

During vigorous exercise, muscle the electron transport chain. cells cannot support

Without oxygen, pyruvate is lactate. converted to

Conversion of pyruvate involves NADH. transferring hydrogen from

The use of NADH to form lactate NAD+. regenerates

NAD+ is required to maintain ATP glycolysis. production by

Accumulation of lactate causes muscle fatigue.

Lactate is reconverted to pyruvate in muscles. the

Lactate is reconverted to glucose in liver. the

Reconversion of lactate requires ATP aerobic respiration. from

Reconversion of lactate causes oxygen debt.

Q30:

There are two types of muscle ﬁbre, slow and fast twitch. namely

All skeletal muscles contain both types of muscle ﬁbre.

Athletes show patterns of muscle reﬂect their events. ﬁbres that

Slow twitch ﬁbres contract slowly but sustain contraction longer.

Slow twitch ﬁbres are good for endurance events.

Slow twitch ﬁbres rely on aerobic respiration.

A large blood supply and many a high concentration of myoglobin. mitochondria are linked to

Slow twitch ﬁbres use fuel stores of glycogen and fat.

Fast twitch ﬁbres contract quickly but short periods only. for

Fast twitch ﬁbres are good for power events.

Fast twitch ﬁbres generate ATP by creatine phosphate and glycolysis.

Few mitochondria and a low blood little myoglobin. supply are linked to

Q31: Creatine phosphate is found in cells with high and variable energy demands, such are muscles and the brain.

Q32: For 2 seconds the muscle cells rely on stored ATP; for the next 10 seconds creatine phosphate; and then lactic acid metabolism.

Q33: Slow twitch ﬁbres are dark red / because they have a high concentration of myoglobin / which stores oxygen for aerobic respiration. / Fast twitch ﬁbres are pale pink / because they contain low concentrations of myoglobin / as they mainly rely on anaerobic forms of respiration / namely, creatine phosphate and lactic acid metabolism.

13 End of unit test End of Unit 1 test (page 258)

Q1:

Cellular differentiation means that develop more specialised functions. cells

express genes characteristic of their Muscle cells can only cell type.

Embryonic stem cells can develop into any cell type.

Tissue stem cells can produce a limited range of cell types.

Germline cells are able to produce gametes.

Mitosis provides the ability to maintain the chromosome number.

Mutations in germline cells affect all cells of offspring.

Skin epidermis can be used to exemplify epithelial tissue.

understand cell growth and Stem cell research lets scientists differentiation.

The Human Tissue Authority has a regulate cell research in the UK. remit to

Cancer cells do not respond to normal regulatory signals.

A tumour is benign if it does not invade neighbouring tissues.

Q2:

A nucleotide is composed of phosphate and a nitrogenous base. deoxyribose,

The sugar-phosphate backbone is strong covalent bonds. held together by

The base pairings of DNA are guanine and cytosine. adenine and thymine,

At the 3 and 5 ends of the DNA deoxyribose and phosphate. strands are

The enzyme which joins fragments of ligase. DNA together is

The formation of mRNA from a DNA transcription. template is

The base in RNA which replaces uracil. thymine is

The ﬁrst strand of RNA formed from the primary transcript. DNA is

The protein-coding regions of mRNA exons. are

The polypeptide is released when a stop codon. reaching

One primary transcript can produce several different mRNA molecules.

Post-translational modiﬁcation may carbohydrate groups. involve adding

Q3:

Pathways that need an energy input anabolic. are called

The active site lowers activation energy.

Glycolysis is catalysed by a multi-enzyme complex.

The product of glycolysis is pyruvate.

The product of phosphofructokinase glycolysis. only enters

Acetyl-coA passes the acetyl group to oxaloacetate.

A substrate that is converted into a fructose. glycolysis intermediate is

Enzymes pump H+ ions into the intermembrane space.

High energy electrons are carried by NAD+ and FAD.

Q4: Without oxygen, pyruvate is converted to lactate. The use of NADH to form lactate regenerates NAD+. Lactate is reconverted to glucose in the liver. Creatine phosphate reserves are restored by aerobic respiration. Slow twitch ﬁbres are good for endurance events. Fast twitch ﬁbres generate ATP by creatine phosphate and glycolysis.

Q5: Mitosis

Q6: They only express genes that produce proteins characteristic of that type of cell.

Q7: Epithelial - covers the organ surfaces. Connective - gives shape to organs / supports organs.

Q8: Any two from:

• divide excessively; • do not respond to regulatory signals; • fail to attach to each other.

Q9: Mutations to a somatic cell will affect only the tissue in which it occurs. Mutations to a germline cell will affect all cells in any offspring produced from that cell.

Q10: Deoxyribose sugar, phosphate, base.

Q11: Hydrogen

Q12: The two strands run in opposite directions. / One strand ends with deoxyribose at the 3 end / and the other ends with phosphate at the 5 end.

Q13: The base bonding to adenine (A) is thymine (T), not uracil (U).

Q14: DNA polymerase

Q15: A primer

Q16: One strand is replicated continuously, The other strand is replicated in fragments which are joined together by ligase.

Q17: Changes to genes or chromosomes.

Q18: Proteins are not expressed / proteins do not function correctly / proteins are faulty.

Q19: a) Deletion

Q20: b) Duplication

Q21: c) Translocation

Q22: Outcome: often lethal. Explanation: mutation results in loss of genes / breaks up groups of genes which work together.

Q23: Anabolic: biosynthetic - requires an energy input. or Catabolic: breakdown of molecules - releases energy and building blocks.

Q24: Any two from:

• orientates reactants (so that they react); • lowers the activation energy of the transition state; • release of products (with low afﬁnity for active site).

Q25: Example: glycolysis (or other suitable). Advantage: product from one enzyme is substrate for the next / presented in correct position / orientation for next active site.

Q26: Meaning: the end-product binds to an enzyme which catalyses a reaction early in the pathway. Example: ATP / citrate inhibits the action of phosphofructokinase. Threonine deaminase inhibited by isoleucine (in transamination pathway).

Q27: Pyruvate / pyruvic acid

Q28: Pyruvate is broken down to form an acetyl group. It is combined with coenzyme A to form acetyl coenzyme A / acetyl-coA. Pyruvate to acetyl-coA. (only 1 mark)

Q29: Insufﬁcient oxygen reaching muscle for aerobic respiration to supply required ATP, e.g. during a power event such as sprinting (example needed).

Q30: The athletes have an oxygen debt / lactic acid has built up in their muscles. Lactic acid must be reconverted to pyruvate / pyruvic acid in the muscles. or Lactic acid must be reconverted to glucose in the liver. Reconversion of pyruvate / pyruvic acid requires ATP from aerobic respiration.