Ib Hl Biochemistry Lab Manual

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IB HL BIOCHEMISTRY PROBLEM SETS AND LAB MANUAL

FAIRVIEW HIGH SCHOOL 2019-20 IB BIOLOGY Exam

Please do not write in this book

Written/ edited/ compiled by Brian Cox

Table of Contents

Section 1: Problem Sets

Unit 1: Macromolecules

Chemistry Review Notes 1

Optional Chemistry Review Problem Set 5

Required Chapter 5 Problem Set 9

Unit 2: Introduction to Metabolism

Optional Chemistry Review Problem Set 18

Required Unit 2 Problem Set 21

Unit 3: Cellular Respiration

Required Unit 3 Problem Set 36

Unit 4: Replication, Transcription, Translation, Regulation of Gene Expression

Watson and Crick’s Structure of DNA (from the journal Nature, April 1953) 41

Required Unit 4 Problem Set 47

Unit 5: Biotechnology

Required Unit 5 Problem Set 64

Section 2: Labs

Expt # Experiment Page

1 Hydrophobic/Hydrophilic 72

2 Levels of Protein Structure and 75 Principles of Protein Folding

3 Purification of Green and Blue Fluorescent 91 Proteins and Protein Denaturation (Adapted from Edvotek Lab #255)

4 Experimental Design: Factor that affects rate of enzyme catalyzed 94 reaction

5 Investigating effect of pH on activity of perioxidase enzyme 95

6 Measuring the Rate of Cellular Respiration of 101 Peas (Adapted from AP lab #5: Cell Respiration)

7 Molecular Details of Cellular Respiration 106 Computer Simulation Lab

8 DNA Replication Simulation 108

9 Tryp And Lac Operon Simulation Activities 111

10 Eukaryotic Gene Expression Simulation 115

11 Transformation experiment – insertion of plasmid 117 containing pGlo gene. BioRad kit lab

12 Bacterial ID Lab (NIH interactive computer simulation) 121

SAFETY CONTRACT FOR IB BIOCHEMISTRY

1) Read over safety concerns described in lab manual before each experiment; listen to discussion of safety issues in class. Concentrate during the lab period.

2) Safety glasses must be worn in the lab during any experiments involving chemicals or biological samples. Gloves should be worn during some experiments.

3) Immediately inform instructor of any accidents.

4) If chemicals or bacterial sample contact your skin, you should immediately wash the affected area thoroughly with soap and water. If chemicals contact eyes, flush the eyes for 15 minutes with water from eyewash.

5) Handling organic chemicals: - Volatile organic chemicals should be handled in a fume hood. - Organic fumes are very flammable – kept away from open flames. - Organic waste must be disposed off in the appropriate waste container. It cannot be disposed of in the sink.

5) Handling Bacterial Samples:

- Although the bacteria samples we work with in this course are believed to be completely harmless to humans, always minimize your exposure. Avoid contact with bacterial colonies. - All materials that contact bacteria directly should be disposed of in a sterilizing bleach bath and not placed directly in the trash.

6) Wash hands with soap and water at the end of each laboratory period.

I have read and understand the safety rules for IB Biochemistry. I understand that actions that endanger yourself or others in the lab may result in dismissal from the lab.

Signed,

CHEMISTRY REVIEW NOTES LEWIS DOT STRUCTURES # of electrons (e-) needed to fill valence level = # of bonds (each bond is the equivalent of gaining 1 e-) Most common 4 elements in organic/biological molecules: C – makes 4 covalent bonds; N makes 3 covalent bonds (N+ will make 4) ; O makes 2 covalent bonds (O- makes 1 covalent bond); H makes 1 covalent bond Other common elements: P makes 5 bonds, S makes 2 bonds, Ca and Mg form +2 ions and makes ionic bond; K and Na form +1 ions and makes ionic bonds ORGANIC CHEMISTRY FUNCTIONAL GROUPS- structural components of hydrocarbon molecules which are most commonly involved in chemical reactions. Memorize functional groups below: See pp. 64 & 65 in Campbell’s. Note: R represents a hydrocarbon chain. FUNCTIONAL GROUP FORMULA NAME EXAMPLE Hydroxyl R -OH Alcohol Ethanol Notes: OH is polar; can form H-bonds and act as nucleophile

Carbonyl R - C – H Aldehyde Formaldehyde

R - C – R’ Ketones Acetone

Notes: CO group is polar, can act as H-bond acceptor and act as electrophile

Carboxyl R - C-OH R - C –O- Carboxylic Acid Acetic Acid (neutral) (ionized) Notes: - COOH is polar, can act as an acid (H+ donor), - COO- is resonance stabilized

+ Amino - NH2 - NH3 amines Glycine (amino acid)

(neutral) (ionized) + + Notes: -NH2 is polar, can act as base (H ) acceptor, nonbonding pair on N attract H

Sulfhydryl -SH Thiols ethanethiol

Notes: S-H moderately polar, S is good nucleophile, S-H and S-H can form S-S bond

Methyl CH3 5 – methyl cytidine

Notes: Nonpolar, addition of methyl to DNA or molecules bound to DNA can affect gene expression

Phosphate - O – P – O Organic phosphates Glycerol phosphate

Notes: Ionized at neutral pH; phosphate groups are good “leaving groups” ; often involved in energy transfer reactions

VSEPR – PREDICTING 3-DIMENSIONAL SHAPES • Biological Activity of molecule is determined by both chemical behavior and 3-D shape of molecule • VSEPR (Valence Shell Electron Pair Repulsion Theory) – predict 3-D shape by predicting how bonding and nonbonding electron pairs will arrange themselves around central atom to minimize electrostatic repulsion • Double and Triple bonds count as single bonding region • Important Geometries for common important biochemical molecules

Bonding Nonbonding Geometry Pairs Pairs 2 0 linear 3 0 Trigonal planar 4 0 tetrahedral 3 1 Trigonal pyramidal

2 2 bent

ISOMERS- Compounds with the same molecular formula, but different structures and therefore different properties. 3 types of isomers: ➢ Structural isomers – different covalent arrangements (different connection pattern) Example:

n- butane isobutane

➢ Geometric isomers – variation in arrangement in space about a double bond

Example:

Cis – dichloroethene Trans-dichloroethen 2

Key point: Rotation is impossible around a double bond. (Would require breaking pi bond)

➢ Enantiomers (chiral molecules) – variation in arrangment in space around an asymmetric carbon (C is bonded to 4 different groups). - molecules are not identical (cannot be superimposed on each other) ; mirror images.

Example: CHBrClOH has 2 forms:

DIPOLE MOMENTS Dipole – a molecule in which one side of the molecule is partially positive and the other end is partially negative; (caused by asymmetric distribution of electron density)

Steps for determining a dipole moment: 1) Determine 3-D structure 2) Look for polar bonds (electronegativity difference between bonded atoms) 3) Sum up polar bonds to determine net dipole (if symmetric = polar bonds cancel)

4) Use to indicate overall dipole. Arrow always points to towards ∂-

INTERMOLECULAR FORCES ▪ Weak forces of attraction between molecules (much weaker than covalent or ionic bonds Important Note: Macromolecules like proteins, DNA, and carbohydrate polymers are so long that different parts of the chains can interact with itself via “intermolecular” forces (i.e. these forces can be intramolecular for large molecules)

▪ 3 Intermolecular forces – Hydrogen bonding, dipole-dipole, London Dispersion Forces (also known as Van Der Waals) forces

Dipole-dipole attraction – weak electrostatic attraction between two molecules which have net dipole moments. The molecules line up with the partial positive end of one molecules next to the partial negative end of the second molecule.

Example:

Hydrogen Bonds- an especially strong dipole-dipole attraction involving a hydrogen bridging two very electronegative atoms.

Requirements for H-bonding:. 1) H-bond donor- H covalently bonded to F,O,N ( H-F, H-O, H-N) AND 2) H-bond acceptor F,O,N with nonbonding pair of e- available

Strength of attraction between + and – charges depends on: 1) Size of + and – charge: H-X, where X = F,O,N ; X is very electronegative => very polar bond => large partial + (∂+) and – (∂-)

2) Distance between charges: (H, F, O, N are all small atoms => allows for close approach)

H-bonds are very important in biological interactions which two molecules to specifically recognize each other. Specificity – requirement that partial + surface aligns with partial – surface requires specific alignment of molecules

Examples: Connection between DNA strands (base pairing), Enzyme-substrate interactions, Receptor/ligand interactions

Bond strength- H-bonds are strong enough to hold molecules near each other but don’t require large amounts of energy to break when molecules need to separate

London Dispersion Forces – “ instantaneous” dipole attraction that occurs between 2 nonpolar molecules - Generally weakest of intermolecular attractions- strength is very dependent on size of electron cloud.

Other Intermolecular/ Intramolecular bonding important in biochemistry:

Salt Bridge - ionic bond between parts of molecule with full + and – charges Ion-dipole – electrostatic attraction between a full + or – and a partially charged (dipole) region Hydrophobic interaction – term used to describe the observation that nonpolar molecules tend to aggregate together in water

Important Properties of Water

1) polarity and ability to form Hydrogen bonds 2) excellent solvent for polar substances; poor solvent for nonpolar 3) cohesion and adhesion 4) Relatively high specific heat capacity 5) Transparent to light

CHEMISTRY REVIEW PROBLEMS (Optional)

(Optional problems = not turned in for credit:) This material is foundation knowledge you need to be successful in Unit 1. Strongly recommended for careful completion for all students who have not completed 2 years of IB Chemistry.

(Use a separate sheet of paper for written answers) Reading: Quick Skim of Chapter 2, read carefully as necessary Read: Chapter 4, pp. 58-66, for background on carbon chemistry. Careful reading of isomers (pp.62-63) and functional groups pp. 64-65).

PART 1: GENERAL CHEMISTRY REVIEW (Chapter 2) 1) What is the difference between A) an atom and an ion? B) an ion and an isotope? 2) How does the # of valence electrons relate to the reactivity and chemical bonding of an atom? 3a) Distinguish between ionic, nonpolar covalent, polar covalent bonds. 3b) What is the difference between bonds listed in 3a and intermolecular forces? 4) What is electronegativity?

PART 2: LEWIS DOT STRUCTURES –Use own paper, leave plenty of space between structures- you will be circling functional groups on your structures and labeling

Problems: Draw the correct Lewis Dot structure for each of the following: 1) H2O 2) CO2 3) CH3OH (methanol) 4) CH3CH2OH (ethanol)

5) SHC2H4OH (beta-Mercaptoethanol) 6) CH2OHCHOHCH2OH (glycerol )

7) CO(NH2)2 (urea) 8) HCOH (formaldehyde)

9) HCOOH (formic acid) 10) C6H6 (benzene) (hint: ring structure)

+ - -2 11) NH3CH2COO (glycine) 12) CH2OHCHOHCH2OPO3 (glycerol phosphate)

13) Toluene C7H8 (benzene ring with a methyl group)

PART 3: FUNCTIONAL GROUPS - Circle the functional groups in Lewis dot structures in Part 2 of HW 1-1 problems #3-9 and 11-13 and name each functional group. - Also: Book Problems: p. 67, #5,6,7 (can just write down letter of answer) Check your answers in appendix A in back of book.

For each of the aqueous reactions below, write the products of the reaction:

1) CH3CH2COOH + NaOH →

2) NH2CH2CH3 + HCl →

PART 4: ISOMERS 1) Draw 3 structural isomers for pentane, C5H12 2) Draw 2 geometric isomers for dibromoethene and indicate which is cis and which is trans. 3) Draw 2 enantiomers for CHFBrOH. 4) Book Problems, p. 67, #3 & 4 (can just write down letter of answer) Check your answers in appendix A in back of book.

PART 5: DIPOLE MOMENTS For each of the following, determine whether the molecule has a net dipole moment. If it does, indicate the direction of the net dipole. Electronegativity values: H =2.1, Cl =3.0, C=2.5 S=2.4 O =3.5 N=3.0 B =2.0 F =4.0

1) HCl 2) CSO 3) CO2 4) H2O 5) HCOH 6) BF3

7) NH3 8) CCl4 9) CHCl3 10) CH3OH 11) CH3NH2

PART 6: INTERMOLECULAR FORCES 1) Draw a correctly oriented dipole-dipole attraction between : A) 2 IF molecules B) 2 HCOH molecules C) IF and HCOH

2) For each pair of molecules, indicate whether or not an H bond can form. If an H-bond can form sketch a 2-D picture showing how the molecules would be aligned. Use dashed lines to indicate the hydrogen bonds.

A) H2O and CH3OH B) BHF2 and H2O C) CH3NH2 and CH3SH - D) HCOO and CH3OH E) CH3NH2 and CH3COCH3

3) Explain the two properties of hydrogen bonding which make it an ideal choice for holding together bases between DNA strands.

Part 7: WATER: Read Chapter 3 and answer the following: 1) Draw a picture of 3 molecules of water and the hydrogen bonds that connect them. Draw covalent bonds as solid lines and hydrogen bonds as dashed lines. For each water molecule, label atoms that have a partial negative charge by putting a  - next to the atom and label the atoms that have a partial positive by putting a + next to the atom.

2) Describe each of the following properties of water and give one example: A) cohesion B) adhesion C) surface tension

3) Importance of specific heat capacity of water A) What is specific heat capacity? B) Water has a relatively high specific capacity for its density. What property of water accounts for its unusually high specific capacity?

C) Given: specific heat capacity of watr = 4.18 J/g oC, Fe = .449 J/ g oC, Q = cmT, where q = heat in J, c = specific heat capacity, m = mass, T = Tfinal - tinitial Calculate the change in temperature, T, when i) 1 gram of H2O absorbs 100 J of heat energy ii) 1 gram of Fe absorbs 100 J of heat energy

D) Biological significance of specific heat capacity of water. One of the themes of biology in homeostasis, meaning maintaining internal conditions of a cell at a steady state. i) Referring to your answers in parts B and C, how does the specific heat capacity of water help a cell maintain a constant temperature? (In other words, what is the problem in terms of heating up or cooling down with using a substance with a relatively a low specific heat capacity such as Fe) ii) What is one reason that cells are mostly made up of water?

Part 8:

4) Another property of water is that it is mostly transparent to visible light. Explain why this property of water would be critical to the survival of an algae growing on a rock submerged a meter under the water? ( Hint: What is the major energy source of the algae?)

5) Water as a solvent A) Define hydrophobic and hydrophilic. B) What kinds of substances will dissolve in water? C) Which of the following substances would be soluble in water? i) benzene, C6H6 ii) salt, NaCl iii) glucose, C6H12O6 (see picture in your textbook, p. 71, 5.4b)

D) For any of the substances in part C which is soluble in water, draw a picture which includes 6 water molecules, showing how the water would interact with the solute to dissolve it. (Hint: See figure 3.7 on page 50 and 3.8 on p.51). E) Why do you think that cells are mostly made up of water, instead of hydrophobic lipid molecules?

6) Importance of pH A) What is pH?

B) How many more times acidic is a solution with a pH of 3 than a solution with a pH of 7?

C) Effect of changing pH on a chemical reaction: When your muscle cells release CO2 into your blood, the CO2 reacts with the water in your blood to form carbonic acid, H2CO3. . As mentioned in your textbook, H2CO3., is a weak acid that - + dissociates into HCO3 and H . - + CO2 + H2O , H2CO3  HCO3 + H .

If you exercise vigorously and therefore are increasing the concentration of CO2 in your blood, what happens to the pH of your blood? (Hint: Think about Le Chatelier’s principle: which way will the equilibrium shift if the concentration of CO2 increases?)

D) One fundamental idea in biology is that living things maintain homeostasis, meaning that they try to keep their internal conditions at a steady state. i) What is a buffer? ii) How does the carbonic acid-bicarbonate buffering system help keep the pH of your blood relatively constant?

E) Describe the damage that acid precipitation can do to the environment.

Chapter 5- Macromolecules Problem Set (Required – this will be turned in for credit) Make sure that you have memorized functional groups on pp. 64 and 65 of Campbell’s.

HW 1-1: Polymers and Carbohydrates Reading: Campbell’s, pp.68-74. Powerpoint: Carbohydrates

Polymers 1) Define polymer 2) Define monomer 3) What is a condensation reaction? 4) What is a hydrolysis reaction?

Carbohydrates 1) State two roles for carbohydrates in cells. 2) What functional groups are most commonly present in carbohydrates? 2) Monosaccharides A. State Definition: B.) What pattern in the ratios of carbon:hydrogen:oxygen is present in monosaccharides?

3) Disaccharides A. State Definition: B. How are they formed? C. State three examples

4) Polysaccharides A) State Definition B) How are they formed? C) Fill in the Blank: Starch is ______- storing molecule in plants. It consists of polymers formed from ___- ______monomers. There are two forms of these polymers called ______which represents approximately ______% of the starch structure, and ______which represents approximately ______% of the starch structure. ______is more water soluble and more easily digested than ______.

D) Label the structures below as amylose or amylpectin:

E) Identify the structures below as cellulose, glycogen or starch:

E) Which of the above molecules (cellulose, glycogen, starch) are found in animals and which in plants? Which can be digested by humans?

5) IB Memorize List – Draw the following structures: A) Ribose B) α – glucose (D form, i.e. biologically active form shone in your book) C) β – glucose (D form, i.e. biologically active form shone in your book)

6) Draw the products of the following condensation reaction. The OH groups that are involved in the condensation are circled.

7) Draw the products of the following hydrolysis reaction.

HW 1-2: Lipids Reading: Campbell’s pp. 74-77; pp. 125 -131; p.136-37 – How Ion Pumps Maintain Membrane Potential Powerpoint: Lipids

IB Requirement: Calculate Body Mass Index using formula

Practice: Calculate BMI for an individual weighing 100. kg with a height of 1.75 meters and classify the weight status. Check Ans: 32.7 → obsese

1) Describe two roles of lipids in the cell. 2) What does it mean to describe all lipids as being hydrophobic? 3) Triglycerides are made by combining one molecule of glycerol with three fatty acid molecules. Draw this reaction.

4) Saturated and Unsaturated Fats

A) What is the difference between saturated and unsaturated fatty acids?

B) How does the presence of a double bond vs. a single bond impact the ability of a bond to rotate freely? How does this impact the shape of the molecule? Which type of fat unsaturated or saturated is straight chain and which has kinks or branches?

C) Effect of kinks or branches association between molecules (example application- viscosity of membranes): Let’s consider a simple case of two of the isomers of pentane that you drew on the chemistry review problem set. Normal pentane, C5H12 is considered to be a “straight chain” hydrocarbon. Since in reality the carbon-carbon bonds are at an angle of 109.5o instead of a true straight line we can represent the pattern in a simplified two-dimensional zigzag pattern. A second isomer, isopentane contains a branch. Diagrams of pairs each type of molecule are shown below:

n- pentane, bp = 36 oC isopentane , bp = 27 oC i) What type of intermolecular force attracts two n-butane molecules together? What type of intermolecular force attracts two iso-pentane molecules together?

11 ii) N-pentane has a higher boiling point than isopentane, indicating stronger intermolecular attractions. Refer to the diagrams above to explain why the force of interaction is stronger between a pair of normal pentane molecules than between a pair of iso-pentane molecules.

4E) IB exam topic: Unsaturated fatty acids come in two structural forms, cis and trans. Diets high in the trans form have been linked to significantly higher risk of death of from heart attacks and stroke. It is observed that diets high in trans fats negatively impact to markers of cardiovascular disease: 1) Increase in “bad” cholesterol (Low Density, LDL form), while decreasing “good” cholesterol (high density, HDL) 2) Increase level of C reactive protein, CRP associated with increased inflammation with blood vessels and hence greater risk of heart attack or stroke. The mechanism of action is unclear, but the leading hypothesis is that trans-fatty bind to receptors or enzymes in pathways that control the synthesis or breakdown LDL, HDL and inflammatory response molecules.

The risk is considered so great that the recommendation is to eliminate as much as possible any consumption of trans fatty acids from the diet.

Identify the saturated fatty acid, the cis unsaturated fatty acid and the trans unsaturated fatty acids in the diagram below: Which two of these acids acid are associated with increased risk of cardiovascular disease when consumed in high amounts?

5A) Draw a diagram to show the general structure of a phospholipid. Clearly indicate which structures are hydrophillic and which are hydrophobic.

5B) Draw a diagram to show how phospholipids form a bilayer to create membranes.

HW 1-3: Nucleic Acids

Reading: Campbell’s, p. 86-89. Powerpoint: Nucleic Acids

1) Why is DNA referred to as the Double Helix?

2) What is the name for the monomer used in DNA or RNA?

3) IB Memorize List: Draw diagram that includes four total nucleotides bases (two on each separate strand). Use a pentagon to represent the sugar, a circle for the phosphate group and a rectangle to represent the bases. Use solid lines for covalent bonds and dashed lines for hydrogen bonds.

Clearly Label, using horizontal arrows the following components: i) Base ii) Sugar, including 5’ and 3’ positions iii) Phosphodiester bond iv) hydrogen bonds

Note: Diagram must clearly indicate that the two strands are anti-parallel.

4) Compare and contrast in terms of structures of chains, chemical identity of sugar and bases and single strand vs. double strand forms, the differences between DNA and RNA.

5) Campbells Concept Check 5.5, p. 89, #1 and 2.

HW 1-4: Proteins

Reading: Campbell’s pp. 77 -86. Powerpoint:

1. What are the monomers for proteins called?

2. IB Memorize List - Write the general structural formula of an amino acid. Use R to represent the side chain.

3. Show the products of this condensation reaction.

4.) Show the product of this hydrolysis reaction.

5) Describe the four levels of protein structure: primary, secondary, tertiary and quaternary.

6) IB Exam topic List: What is sickle – cell anemia? What specific mutation causes this disease? Why is this specific amino-acid substitution so catastrophic?

7) Campbell’s p. 86, Concept Check 5-4, #1-3; Inquiry Fig. 25.5 What If?

HW 1-5 Exam Review: Multiple Choice Review for Chapter 5 Campbell’s Part 1: Campbell’s p. 91, #1,2,4-7 (Answers in Appendix A in the back of the book.) Part 2: Comprehensive Multiple Choice Practice (Answers at the end) 1) Which three elements are occur most frequently in living organisms? A) carbon, nitrogen and oxygen B) carbon, nitrogen and hydrogen C) carbon, hydrogen and oxygen D) nitrogen, hydrogen and oxygen E) carbon, hydrogen and nitrogen

2) Which of the following is present in the greatest quantity of living organisms? A) Vitamins B) Carbohydrates C) Proteins D) Water

3)Match the element with its role in important biological molecules.

Answer choices: A) element present in the amino acids cysteine and methione (not C,N,O or H) B) element present in the both DNA and RNA, but not amino acids, sugars or lipids C) element present in the heme group of hemoglobin D) The movement of the +1 ion of this element is important in nerve conduction E) This element is present in teeth and bone. The movement of its +2 ion is important in muscle contraction. i) Phosphorus ____ ii) sodium ___ iii) iron _____ iv) sulfur v) calcium _____

4) Polymerization is a process that A) creates bonds between amino acids in the formation of a peptide bond B) involves the removal of a water molecule C) links the phosphate of one nucleotide with the sugar of the next D) requires a condensation or dehydration reaction E) involves all of the above

5) Dissacharides can differ from each other in all of the following ways except A) in the number of monosaccharides bonded together B) as enantiomers C) in the monomers involved D) in the location of their glycolsidic linkage E) in their structural formulas.

6) Which of the following is not true of cellulose? A) It is a polymer formed by monomers of glucose B) It differs from starch because of the configuration of glucose and the geometry of the glycosidic linkage. C) Few organisms have enzymes that hydrolyze its glycosidic linkages. D) The glycosidic linkages between glucose monomers are in the α form.

7) Plants store most of their energy as A) glucose B) glycogen C) starch D) sucrose E) cellulose

8) What happens when a protein denatures? A) It loses is primary structure. B) It loses its tertiary structure. C) It always becomes irreversibly insoluble and precipitates. D) It hydrolyzes into component amino acids. E) All bonds including covalent disulfide linkages, covalent peptide bonds, hydrogen bonds, ionic bonds, hydrophobic interactions (London dispersion forces) are disrupted.

9) The alpha helix structure of proteins is A) part of the tertiary structure and is stabilized by disulfide linkages. B) a double helix C) stabilized by hydrogen bonds along the backbone and commonly found in fibrous proteins. D) found is some regions of globular proteins and is stabilized by hydrophobic interactions E) a complementary sequence to messenger RNA

10) Three molecules of a fatty acid C16H32O2 are joined to a molecule of glycerol (C3H8O3). The resulting molecule has the formula A) C48H96O6. B) C48H98O9 C) C51H102O8 D) C51H98O6 E) C51H104O9

11) Beta pleated sheets are characterized by A) disulfide bridges between cysteine amino acids. B) parallel regions of polypeptide chain held together by hydrophobic interactions. C) folds stabilized by hydrogen bonds between segments of polypeptide chains D) membrane sheets composed of phospholipids E) hydrogen bonds between adjacent cellulose molecules.

12) Which of these molecules would provide the most energy (kcal/g) when eaten? A) glucose B) starch C) glycogen D) fat E) protein

13) Which of the following would be a hydrophobic molecule? A) cholesterol B) nucleotide C) amino acid D) chitin E) glucose

14) If the nucleotide sequence of one strand of a DNA helix is GCCTAA, what would be the sequence of the complementary strand? A) GCCTAA B) CGCATT C) CGGATT D) ATTCGG E) TAAGCC

Answers for Part 2 MC: 1) C 2) D #3) Matching: A – Sulfur B- Phosphorus C-iron D- sodium E- calcium

4) E 5) A 6) D 7) C 8) B 9) C 10) D 11) C 12) D 13) A 14) C

HW 1-6 Free Response:

1)What is the difference between an organic and inorganic compound?

2) The properties of adhesion, cohesion, and the usually high boiling point and the high specific heat capacity of water are primarily the result of what type of intermolecular interaction between water molecules?

3) Write an essay outlining the principles of protein folding in water. Include all of the words in the work bank below. (Note word banks not provided on actual tests).

Word Bank: primary structure, secondary structure, tertiary structure, hydrophobic, hydrophilic, polar, nonpolar and ionized amino acid sidechains, disulfide linkages, hydrogen bonding along backbone, weak forces of attraction between sidechains, hydrogen bonding, dipole-dipole, London Dispersion Forces, Ionic bonds

Unit 2 KINETICS/THERMODYAMICS REVIEW- (Optional ) Careful completion strongly recommended for students who have completed only one year of chemistry. Check your answers online.

Thermodynamics vs. Kinetics

1) In order to predict what will happen when chemicals are mixed under a given set of conditions, it is necessary to understand both the thermodynamics and the kinetics of the system.

A) What questions does thermodynamics address?

B) What question does kinetics address?

C) What does it mean to say that a reaction is spontaneous? Does spontaneous imply fast?

D) Sketch a reaction profile diagram and mark the realm of thermodynamics and the realm of kinetics of your diagram.

Three Laws of Thermodynamics:

4) What is energy?

5) State the 3 Laws of Thermodynamics.

Enthalpy:

6) What is enthalpy? What is taking place in chemical reactions that causes changes in enthalpy?

Entropy:

7) A) What is entropy? How is entropy related to the number of possible arrangements or microstates in a system? B) For each of the following predict the sign of S, + (entropy increasing) or – (entropy decreasing) I) gas → solid II) glucose molecules → starch III) folded protein → denatured protein

C) An important consequence of the second law is that no process can ever be 100% efficient. Why not?

D) The process of taking 100 amino acids and linking them together inside a cell nd has a very large S system = -. Explain why this does not violate the 2 Law of Thermodynamics.

Entropy and Equilibrium systems separated by a semi-permeable membrane

8) For each of the following systems below draw a 2nd picture to show what the system would look like when equilibrium is reached.

A) In the diagram below, the two compartments are initially separated from each other. The left compartment contains glucose molecules () dissolved in water. The right compartment contains only water. What will happen when the contents are allowed to mix?

  open   →   partition

B) In the diagram below, the two compartments are separated from each other by a semi- permeable membrane. Both compartments contain H+ ions (+) and Cl- ions (-) dissolved in water. Draw the final picture showing the distribution of charges after equilibrium has been established. State 1 State 2 + - + - + time elapsed - + - → i) In the original picture, is the voltage the same in both compartments? Explain.

ii) (Circle the best choice) In moving from state 1 to state 2, free energy (is absorbed or released) and work (could be obtained or would be required).

KINETICS REVIEW 1A) According to collision theory, what are the 3 requirements that must be met for a chemical reaction?

1B) Fill in the blank. The rate of a chemical reaction is determined by the height of the ______. The greater the activation energy, the (faster or slower) ______the reaction rate.

2) Free Energy Reaction Profile diagrams 2A) Draw Reaction Profile diagrams (Gibb’s Free Energy (G) on the y-axis and Rxn Progress on the x-axis) for the systems given below. Label Reactants, Products, Transition state (T.S.) for each diagram.

I) G = -25 kcal/mol, Ea = +20 kcal/mol

II) G = +15 kcal/mol, Ea = +30 kcal/mol

2B) For each graph above, indicate whether the rxn is exergonic or endergonic.

2C) For which of the above graphs (I or II) is i) a net input of energy required to drive the rxn? Briefly explain.

ii) the rxn faster? Briefly explain.

iii) For any rxns that are exergonic, what is the theoretical maximum amount of work that is available from this reaction?

2D) Explain why this amount cannot actually be obtained in a real process.

HW 2-1, Metabolism, Cellular Work and Metabolic Disequilibrium and Energy Coupling (Required)

Metabolism 1A) What is metabolism?

1B) Diagram the basic pattern of a metabolic pathway. Use S for Starting Material, I1 = Intermediate 1, I2 = Intermediate 2, I3 = Intermediate 3, P = Product E1 = Enzyme 1, E2 = Enzyme 2, E3 = Enzyme 3, E4 = Enzyme 4

2) What is the difference between anabolic and catabolic pathways?

Cellular Work 3) Book problem: p. 161 #2 (Hint: Think ethanol Cannon)

4) List 3 types of work that cells need to be able to do to sustain life.

5) In cells, chemical systems are never allowed to reach chemical equilibrium. A) What is the value of G at equilibrium? What is the significance of this value for the amount of work that can be done be the reaction?

B) In a sealed test tube, all chemical reactions will eventually go to chemical equilibrium. What is different about a cell? What does the cell do that prevents equilibrium from being reached (maintaining metabolic disequilibrium)?

ENERGY COUPLING 6) Explain the term energy coupling and provide an example to support your explanation.

7A) Which of the following reactions are endergonic?

Reaction I: A + B → C ΔG = - 4 kcal/mol

Reaction II: D + E → F ΔG = +5 kcal/mol

Reaction III: G + H → I ΔG = +12 kcal/mol

Reaction IV: L + M → N ΔG = +8 kcal/mol

7B) Which of the above reaction reactions that are endergonic could be coupled to the reaction

X + Y → Z ΔG = - 10 kcal/mol to produce an overall exergonic reaction?

7C) For each of your answers to B, write the overall reaction and calculate the overall ΔG value.

8) Explaining the Concept of Metabolic Disequilibrium and Energy Coupling Into a Single Free Energy Diagram for Reaction Pathway:

Part 1: R1<=> P1 in Isolation (closed system):

Sketch a reaction profile diagram [G (kJ/mol) vs. Rxn Progress] using Ea = 40 kJ/mol and ΔG = + 20 kJ/mol.

G kJ/mol

RXN PROGRESS

QUESTION #1: Which is favored at equilibrium R1 or P1? Justify your answer in terms of relative Gibb’s Free Energy values.

Part 2: Open and coupled system (continuously adding R2)

Consider the same reaction system R1 → P1 with a second reactant R2 that reacts with P1 according the following scheme:

R1 → P1 + R2 → P2

Where R1 = reactant #1, P1 = Product # 1, R2 = Reactant #2 and P2 = product #2 (Note: In the overall reaction of R1 → P2, P1 is called an intermediate)

In our problem we assume that R2 is continuously being added and Given: R1 → P1 : Ea = 40 kJ/mol and ΔG = + 20 kJ/mol P1 + R2 → P2 : Ea = 10 kJ/mol and ΔG = - 30 kJ/mol

(measured with respect P1 + R2)

G kJ/mol

RXN PROGRESS

QUESTION #2: Which process will be favored, the reverse reaction for the first step, R1 ← P1 or the forward reaction of P1 + R2 → P2? Justify your answer

QUESTION #3: Why is the reaction P1 → R2 no longer at equilibrium when R2 is added?

QUESTION #4: Calculate the overall ΔG for the reaction R1 → P2.

2-2 ENZYMES (Required)

1) What kinds of molecules act as biological catalysts? Fill in the blanks.

Enzymes are generally made up of ______( type of biological molecule). However Dr. Cech and his research group at CU won the Nobel Prize for demonstrating that ______( type of biological molecule) can also catalyze self-splicing reactions.

2) How much of a difference do enzymes make in the rate of reactions? Fill in the blanks.

Enzyme typically accelerate the rates of reactions by factors of ______. In some cases the rate enhancement is as much as _____. In practice this means that enzyme catalyzed reactions within the cell may happen thousands or millions of times within a second, whereas without the enzyme the reaction might takes thousands or millions of years to react to the same extent! In the case of urease a reaction the takes 5 minutes with the enzyme takes almost ______years to take place without the enzyme.

3) Enzymes have been intensively studied in biochemistry for a century. State four reasons enzymes are important.

4) IB Practice question: Define the term active site of an enzyme.

5) Sketch the catalytic cycle of enzymes. Define the following abbreviations, E, S, ES, EP, P

6) Explain why one chiral configuration of an amino acid is readily incorporated into proteins within the body, but the other form is excreted without being used. Refer the picture in your answer.

7) ACTIVE SITE/ SUBSTRATE INTERACTIONS: A) Why is hydrogen bonding frequently found in substrate binding to the active site? B) Below is a diagram of an active site and a substrate. Redraw the diagrams below showing how the substrate would bind in the active site. In your diagram, label the type of intermolecular attraction between the active site and the substrate.

8) IB Practice Question: Outline enzyme-substrate specificity.

Word bank: 1) Active site 2) Complementary 3) size and shape of substrate and distribution of charge 4) Weak intermolecular attractions, Hydrogen bonding, Dipole- dipole, ion-dipole, ionic, London Dispersion Forces

9) Compare the lock and key to the induced fit model. Which model is more currently accepted?

HW 2-3: How do Enzymes catalyze reactions? 1)What is the key idea of catalysis?

2) REACTION PROFILE DIAGRAMS: Y-AXIS = FREE ENERGY, X AXIS = RXN PROGRESS

A) UNCATALYZED: R → P Sketch a reaction profile diagram for an exergonic, uncatalyzed chemical reaction. Label the reactants, products and transition state.

B) ENZYME CATALYZED : On the same axis, sketch the same reaction catalyzed by an enzyme. Label the ES and EP intermediates. (Note the reactant in the enzyme catalyzed reactant is called a substrate, abbreviated by S) E + S → ES → EP → E + P

C) What feature of your diagram explains why the enzyme-catalyzed reaction is faster than the uncatalyzed reaction?

D) ENERGIES ENZYMES CAN AND CANNOT IMPACT Enzymes cannot change the free energy of the ______or ______, but by changing the mechanism of the reaction, it changes the free energy of the ______.

E) Can an enzyme shift the equilibrium constant for a reaction? Explain.

3) Use the following reaction to explain the concepts of orientation of collisions, stretching/ stretching bonds to be broken, stabilization of geometry/ charge buildup in Transition State:

A) Orientation of collision: Which atom on HCOOH will the Cl- ion need to collide with in order to react? (Hint: Remember chemical bonding is always about attraction of opposite charges; which atom is partially positive that the Cl- will bond with?)

B) Explain why in a reaction between two large macromolecules such as protein or nucleic acids, why a major limiting factor on the rate of reaction is the likelihood of a properly oriented collision. (A labeled diagram would be helpful)

3C) One strategy enzyme use to lower activation energy is to destabilize the reactant by stressing or stretching bonds and thereby weakening bonds to be broken. Explain + charges near the two oxygen atoms would affect the strength of the C-O bond.

Hint: Think about how the average position of

Electron density will shift.

3D) Stabilize the geometry of transition state and charge buildup in transition state.

From VSEPR theory, the geometry of HCOOH is planar or flat. We thus might expect the active site to be very flat (think of inserting a flat DVD into a drive).

During catalysis, the active site is not flat. The reaction is believed proceed via a tetrahedral transition state.

Explain why the active site is not flat during catalysis.

*Practice ESSAYS* Taken from old IB exams

3) Outline how enzymes catalyze biochemical reactions.

Word Bank for essay: 1) Biological catalyst 2) Protein 3) structure-function 4) Free energy of activation 5) substrate 6) Active site 7) enzyme-substrate complex 8) induced fit model 9) Correct orientation 10) stretch /weaken bonds 11) stabilize geometry of transition state / charge buildup in transition state 12) Release products

HW 2-4 Factors affecting rate of enzyme-catalyzed reactions (Required) 1) How is a catalyst involved in a reaction? Fill in the blanks: During the course an enzyme catalyzed reaction , the substrate concentration (decreases, increases, remains the same) ______, the product concentration (decreases, increases, remains the same) ______, and the enzyme concentration(decreases, increases, remains the same)______,

2) EFFECT OF INCREASING SUBSTRATE ON RXN RATE Given the following graphs:

Rate Rate

[Substrate] [Substrate] NO CATALSYT ENZYME

Why does the rate of the uncatalyzed reaction continue to increase as more substrate is added while eventually the enzyme catalyzed rxn rate does not change even when more substrate is added?

Effect of Temperature and pH on Enzyme Activity 3) The effect of pH and temperature were studied for an enzyme catalyzed reaction. Sketch a predicted shape of the curve for Enzyme Activity and Temperature and enzyme activity and pH.

Enzyme Enzyme Activity Activity

Temperature pH

4)If you were studying the temperature dependence of the reactions of cellular respiration, photosynthesis, DNA replication or DNA transcription, how would the basic shape of the curve change?

Impact of Temperature on Enzyme activity 5A) Below is a sketch of a enzyme with some examples of bond types shown. Draw a sketch of the enzyme after it has been heated to the point of denaturing.

5B) Which bond or bonds (choices: disulfide linkage, ionic, LDF, hydrogen bond) will still likely be still in the denatured form? What has happened to the shape of the active site after heating?

6) Write a paragraph to explain the following observation: The enzyme Rubisco catalyzes the reaction in photosynthesis in which carbon dioxide is added to 5 – RuBp. The activity of Rubisco in a pea plant drops dramatically when the temperature is increased from 25oC to 95 oC. Include the following word bank words/phrases in your answer: 1) structure of protein/function of protein 2) tertiary protein structure 3) weak intermolecular attractions – hydrogen bonding, dipole-dipole attractions, ionic bonds, london dispersion forces 4) average kinetic energy 5) distance between atoms forming weak intermolecular attraction within folded protein 6) shape of active site

Effect of pH on Enzyme Activity 7) Changing pH can change the charge on amino acid side chain.

Example equilibrium #1: R’COOH <=> R’COO- + H+ ; where R’ = hydrocarbon

At pH below 5: R’COOH (neutral form predominates)

At pH above 5: R’COO- (- ion) form predominates

+ + Example equilibrium #2: R’NH3 <=> R’NH2 + H ; where R’ = hydrocarbon

+ At pH below 9: R’NH3 (+ ion predominates)

At pH above 9: R’NH2 (neutral form predominates)

Note: predominates means present in a very high concentration relative to the other form

General observations: i) when [H+] is high and thus [OH-] is low, + or neutral forms predominate ii) when [H+] is low and thus [OH-] is high, - or neutral forms predominate

In our sketches below we will use the following conventions for side chain charges:

Below is a sketch of an enzyme, showing two sets of ionic bonds at pH 7. Draw two pictures, one for pH 1 and one for pH 14, showing how the charge distribution would change at the new pH and how the shape of the enzyme would be affected.

8) Write a paragraph to explain the following observation: The enzyme DNA polymerase III catalyzes the elongation of DNA daughter strands being copied from a parent DNA template strand. The activity of the enzyme is very high at pH 7, but very low at pH 1.

Include the following word bank words/phrases in your answer: 1) structure of protein/function of protein 2) tertiary protein structure 3) weak intermolecular attractions – hydrogen bonding, dipole-dipole attractions, ionic bonds, london dispersion forces 4) concentration of H+ ion 5) charge distribution on R groups within folded protein 6) shape of active site

HW 2-5: ENZYME INHIBITORS ( Required)

1) Drug Action and Enzyme Inhibitors (Fill in the blank)

Most drugs work by ______the action of enzymes that catalyze critical reactions in a cell. Antibiotics kill bacteria by inhibiting critical reactions that the ______cells need to sustain life but that human cells do not. For example, penicillin inhibits the formation of cell walls in certain types of bacteria. Without a cell wall, the bacterial cell will be crushed and die. Since human cells only have cell membranes they are not affected by the action of penicillin.

2)Illustrate by drawing a picture, the difference between competitive and noncompetitive inhibition. Include in the picture an enzyme, a normal substrate, a competitive inhibitor and a noncompetitive inhibiting. Use different shapes to represent the substrate and inhibitors. Label enzyme, substrate, competitive inhibitor, noncompetitive inhibitor, active site and noncompetitive binding site. State one example of a competitive inhibitor and one example of a noncompetitive inhibitor.

Which of the following could be a competitive inhibitor of an enzyme given the substrate below?

Briefly explain your answer.

4) Ibuprofen in an inhibitor of prostaglandin endoperoxide synthase (PPG). By inhibiting the synthesis of prostaglandins, ibuprofen reduces inflammation and pain.

The data below are from an experiment in which prostaglandin endoperoxide synthase (PPG) activity was measured in the presence 10 mg/mL of ibuprofen inhibitor. [substrate] mM PPG activity (%) 0.5 60

1.0 68 1.5 80 2.5 94 3.5 97

Note: mM = concentration in millimolar or M x 10-3 % activity = rate of rxn with inhibitor/ rate rxn without inhibitor x 100

Is ibuprofen a competitive or noncompetitive inhibitor of PPG? Explain.

5) The drug acetazolamide is used as a diuretic (i.e. increases urine production) and to lower excessively high fluid pressure in the eye in glaucoma patients. Acetazolamide works by inhibiting the enzyme carbonic anhydrase. (Carbonic anhydrase catalyzes the reaction of H2O + CO2  H2CO3 which helps regulate pH and bicarbonate levels and hence fluid levels in the body.) The data below are from an experiment in which carbonic anhydrase activity was measured in the presence of acetazolamide.

[substrate] mM carbonic anhydrase activity % 0.2 60% 0.4 60% 0.6 60% 0.8 60% 1.0 60% Note: mM = concentration in millimolar or M x 10-3 % activity = rate of rxn with inhibitor/ rate rxn without inhibitor x 100%

Is acetazolamide a competitive or noncompetitive inhibitor of carbonic anhydrase? Explain.

6)The top curve in the graph below depicts the rate of an enzyme catalyzed reaction as a function of substrate concentration. The two lower curves represent the rate in the presence of a competitive and noncompetitive inhibitor. Which curve represents competitive inhibition and which curve represents noncompetitive inhibition? Explain your answer.

2-6 ENZYME REGULATION (Required)

1) Importance of Enzyme in Regulating Metabolic Pathways:

The rates of Virtually ALL essential chemical reactions in the cells are regulated by ______. Cells can control the rates of a chemical reactions by regulating the concentrations of reactants available or by increasing or decreasing the amount of enzyme present in the cell or by increasing or decreasing the ______of enzyme via the binding of small molecules to allosteric sites.

2A) Draw a picture of an enzyme with an allosteric site. Show how the conformation of the enzyme changes when an allosteric inhibitor binds to the protein.

2B) One of the early surprises in studying enzymes was just how huge these molecules are given how small the active site actually is. Given what you have learned about induced fit and regulation of enzyme activity, propose an explanation to explain why enzymes are not just small rigid active site size molecules.

3) Sketch a metabolic pathway showing the conversion of reactant A to product D with intermediates B and C. Label the enzymes E1, E2, E3.

4) Campbell’s p. 161, #7

5) *IB PRACTICE ESSAY*

Explain how allosteric control of metabolic pathways by end-product inhibition includes negative feedback and non-competitive inhibition.

HW 2-7 Unit 2 Review Part 1: Campbell’s p. 161, #3-6. (Answers in back of book) 3) 4) 5) 6) Part 2:

1) IB exam question: How does the enzyme lactase aid in the production of lactose- free milk? A) Lactase breaks down lactose, which some people are unable to digest because they do not produce the enzyme in sufficient quantities. B) Lactase catalyzes the synthesis of lactose. C) Lactase stimulates the production of other enzymes that breakdown lactose. D) Lactase allosterically inhibits enzymes that produce lactose.

2) Catabolic and anabolic pathways are often coupled in a cell because A) the intermediates of a catabolic pathway are used in the anabolic pathway B) both pathways use the same enzymes C) the free energy released from one pathway is used to drive the other D) the activation energy of the catabolic pathway can be used in the anabolic pathway E) their enzymes are controlled by the same activators and inhibitors

3) The ΔG for a reaction is -25 kJ/mol. Which of the following statements are true? A) 25 kJ/mol of energy is available to do work B) The reaction is spontaneous C) The reactants have more free energy than the products D) The reaction is exergonic E) All of the above are true

4) What is meant by an induced fit? A) The binding of the substrate in an energy requiring process. B) A competitive inhibitor can outcompete the substrate for the active site. C) The binding site of the substrate changes the shape of the active site, which can stress or bend substrate bonds. D) The active site creates a microenvironment ideal for the reaction. E) Substrates are held in the active site by hydrogen and ionic bonds.

5) One way in which a cell maintains metabolic disequilibrium is to A) siphon (remove) products of a reaction off to the next step in a metabolic pathway B) provide a constant supply of enzymes for critical reactions C) use feedback inhibition to turn off pathways D) use allosteric enzymes that can bind to activators or inhibitors E) use the energy from anabolic pathways to drive catabolic pathways.

6) In an experiment measuring an enzyme’s activity, changing the pH from 7 to 6 resulted in an increase in product formation. From this result we could conclude A) the enzyme active site became saturated at pH 6. B) this enzyme’s optimal pH is 6. C) this enzyme works best at neutral pH. D) the temperature must have increased when the pH was changed to 6. E) the enzyme was in a more active conformation at pH 6.

7) When substance A was added to an enzyme-catalyzed reaction, product formation decreased. The addition of more substrate did not increase product formation. From this result we can conclude that substance A could be A) a product molecule B) a cofactor C) a noncompetitive (allosteric) inhibitor D) a competitive inhibitor

8) At equilibrium, A) no enzymes are functioning B) ΔG = 0 C) the forward and back reactions have stopped D) the products and reactants have equal values of H. E) the reaction has a + ΔG.

9) When a cell breaks down glucose, only about 40% of the energy is captured within ATP molecules produced by cellular respiration. The remaining 60% of the energy is A) used to increase the order necessary for life to exist B) lost as heat because of the second law of thermodynamics C) used to increase the entropy of system by converting kinetic energy into potential energy. D) stored in starch or glycogen for use later by the cell. E) Released when the ATP molecules are hydrolyzed.

10) In the metabolic pathway A → B → C→ D→ E, what effect will molecule E likely have on the enzyme that catalyzes A → B? A) noncompetitive (allosteric) inhibitor B) allosteric activator C) competitive inhibitor D) feedback activator E) coenzyme

Answers: 1) A 2) C 3) E 4) C 5) A 6)E 7) C 8)B 9) B 10) A

UNIT 3 CELLULAR RESPIRATION PROBLEM SET (REQUIRED)

HW 3-1 - OVERVIEW OF BIOENERGETICS AND CONTROLLED REDOX REACTIONS

1) Cellular Respiration and Photosynthesis

A) What is the major purpose of both cellular respiration and photosynthesis?

B) What is the overall reaction for cellular respiration? How is it linked to photosynthesis?

2) For questions 2A and 2B below, choose between the following: I) animals only II) plants only III) both plants and animals

2A) Cellular respiration occurs in

2B) Photosynthesis occurs in

3A) What molecule is the energy coin of the cell?

3B) Approximately how many ATP/ADP molecules are present in a typical human cell?

3C) Approximately how many times is a given ADP molecule recycled to make ATP each minute?

3D) Given your answers to 3B and 3C, what can you conclude about our bodies demand for ATP? (FYI: Your body contains approximately one hundred trillion cells).

3D) List 3 types of work that cells need to carry out in the cell.

3E) List two ways that ATP is synthesized in cells and briefly explain each process.

4) OXIDATION AND REDUCTION REACTIONS 4A) Fill in the blank Not all redox reactions involve the complete ______of electrons; some change the degree of electron ______in ______bonds. Redox reactions release energy when electrons move ______to more ______atoms.

4B) Select the correct word choice that completes the sentence. i) Oxidation often involves (gain or loss) of oxygen and (gain or loss) of hydrogen.

ii) Reduction often involves (gain or loss) of oxygen and (gain or loss) of hydrogen

5) For each pair of species below, indicate which is more oxidized and which is the more reduced form. Also indicate which species has higher energy and which has lower energy. (IMPORTANT HINT: For biological molecules, More H’s and fewer O’s = more reduced and therefore higher energy)

+ A) NAD vs. NADH B) FAD vs. FADH2

- C) CH4 vs. CO2 D) C6H12O6 (glucose) vs. CH3COCOO (pyruvate)

- E) CH3COCOO (pyruvate) vs. CH3CHOHCOOH (lactate)

6) What is the role of NADH and FADH2 in cellular respiration?

7) Thermodynamics and kinetics, combustion of glucose vs. cellular respiration.

A) How does the AMOUNT of heat energy released compare for the combustion of glucose vs. oxidation of glucose via cellular respiration?

B) How does the RATE of heat energy release compare for the combustion of glucose vs. oxidation of glucose via cellular respiration?

HW 3-2 ANAEROBIC RESPIRATION

1) List the energy rich molecules (think ADP, ATP and NAD+ NADH) consumed in the energy- investment and produced energy payoff reactions of glycolysis. What is the overall net production of ATP and NADH?

2) Why is less ATP produced in fermentation than in aerobic respiration?

3) Complete the following table:

Process Where it occurs Inputs Products Alcohol Fermentation

Lactic Acid Fermentation

4A) Is ATP produced DIRECTLY when during alcohol or lactic acid fermentation?

4B) What molecule is regenerated during alcohol or lactic acid fermentation? Why is this regeneration of this molecule critical to continued production of ATP in the cell?

HW 3-3 METABOLISM OF FATS AND PROTEINS

1A). How is energy (ATP) produced from fats and proteins?

1B) What molecule is the common key entry point into the Kreb’s cycle for catabolism of proteins, carbohydrates and fats? 1C) A fatty acid has a formula of C18H35O2 . How many acetyl CoA molecules could be produced from this fatty acid? Explain.

2A) Which enzyme is the most important control enzyme in cellular respiration of glucose? (Recall that pathways are typically regulated at first steps (entry steps) or energy requiring steps).

2B) For each of the following molecules, indicate whether the molecule acts as an allosteric inhibitor or activator of phosphofructokinase. A) ATP B) AMP C) citrate

HW 3-4 Unit 3 Review Problems IB Practice Essay:

Explain the similarities and differences in anaerobic and aerobic cellular respiration.

HOMEWORK 3-5 UNIT REVIEW BIOENERGETICS

Campbell’s Review Questions:

1) p. 167: Concept check 9.1, #1,2; 2) p. 167: Concept check 9.2:# 1,2 3) p. 172: Concept check 9.3, #1,2 (skip #3) 4) p. 177: Concept check 9.4, #1,2 5) p. 179: Concept check 9.5, #1,2 6) p. 182: Concept check 9.6, #1-3

Multiple Choice: Campbell’s p. 183-184: #2-9: Draw it: p. 184, #10.

MULTIPLE CHOICE REVIEW Supplement:

1) A substrate that is phosphorylated A) Has a stable phosphate bond. B) Has been formed by the reaction ADP + Pi → ATP C) Has an increased reactivity; it is primed to do work. D) Has been oxidized. E) Will pass its electrons to the electron transport chain.

2) Which of the following is NOT true oxidative phosphorylation? A) It uses oxygen as the initial electron donor. B) It involves the redox reactions of the electron transport chain. C) It involves an ATP synthase located in the inner mitochondrial membrane. D) It produces approximately three ATP for every NADH that is oxidized. E) It depends on chemiosmosis.

3) Substrate-level phosphorylation A) Involves the shifting of a phosphate group from ATP to substrate.

B) Can use NADH or FADH2 C) Takes place only in the cytosol D) Accounts for 10% of the ATP formed by fermentation

E) Is the energy sources for an organism under anaerobic conditions

4) The MAJOR reason that glycolysis is not as energy productive as complete aerobic respiration is A) NAD+ is regenerated by alcohol or lactate production, without the high-energy electrons passing through the electron transport chain. B) It is the pathway common to fermentation and respiration C) It does not take place in a specialized membrane-bound organelle.

D) Pyruvate is more reduced than CO2; it still contains much of the energy from glucose. 5) When electrons move closer to electronegative atoms, A) Energy is released B) Energy is consumed C) A proton gradient is established D) Water is produced E) ATP is synthesized

6) When pyruvate is converted to acetyl CoA,

A) CO2 and ATP are released. B) A multienzyme complex removes a carboxyl group, transfers electrons to NAD+, and attaches a coenzyme C) One turn of the Krebs cycle is completed D) NAD+ is regenerated so that glycolysis can continue to produce ATP by substrate-level phosphorylation. E) Phosphofructokinase is activated and glycolysis continues

7) In the chemiosmotic mechanism, A) ATP production is linked to the proton gradient established by the electron transport chain. B) The difference in pH between the intermembrane space of the mitochondria and the cytosol of the cell drives the formation of ATP. C) The flow of H+ through ATP synthases from the matrix to the intermembrane space drives the phosphorylation of ADP. D) The production of water in the matrix by the reduction of oxygen leads to a net flow of water out of mitochondrion.

8) Which of the following reactions is incorrectly paired with its location? A) ATP synthesis/inner membrane of mitochondria B) Fermentation/cell cytosol C) Glycolysis/cell cytosol D) Substrate-level phosphorylation/cytosol and matrix

E) Krebs cycle/cristae of mitochrondria

9) What do human muscle cells in oxygen deprivation gain from the conversion of pyruvate to lactate? A) ATP B) recycled NAD+

C) CO2 and lactate D) ATP and alcohol

E) ATP, lactate, and CO2

Ans: 1) C 2) A 3) E 4) D 5) A 6) B 7) A 8) E 9) B

A Structure for Deoxyribose Nucleic Acid J. D. Watson and F. H. C. Crick (1)

April 25, 1953 (2), Nature (3), 171, 737-738

We wish to suggest a structure for the salt of deoxyribose nucleic acid (D.N.A.). This structure has novel features which are of considerable biological interest.

A structure for nucleic acid has already been proposed by Pauling (4) and Corey1. They kindly made their manuscript available to us in advance of publication. Their model consists of three intertwined chains, with the phosphates near the fibre axis, and the bases on the outside. In our opinion, this structure is unsatisfactory for two reasons:

(1) We believe that the material which gives the X-ray diagrams is the salt, not the free acid. Without the acidic hydrogen atoms it is not clear what forces would hold the structure together, especially as the negatively charged phosphates near the axis will repel each other.

(2) Some of the van der Waals distances appear to be too small.

Another three-chain structure has also been suggested by Fraser (in the press). In his model the phosphates are on the outside and the bases on the inside, linked together by hydrogen bonds. This structure as described is rather ill-defined, and for this reason we shall not comment on it.

We wish to put forward a radically different structure for the salt of deoxyribose nucleic acid (5). This structure has two helical chains each coiled round the same axis (see diagram). We have made the usual chemical assumptions, namely, that each chain consists of phosphate diester groups joining beta-D-deoxyribofuranose residues with 3',5' linkages. The two chains (but not their bases) are related by a dyad perpendicular to the fibre axis. Both chains follow right-handed helices, but owing to the dyad the sequences of the atoms in the two chains run in opposite directions (6) . Each chain loosely resembles Furberg's2 model No. 1 (7); that is, the bases are on the inside of the helix and the phosphates on the outside. The configuration of the sugar and the atoms near it is close to Furberg's "standard configuration," the sugar being roughly perpendicular to the attached base. There is a residue on each every 3.4 A. in the z-direction. We have assumed an angle of 36° between adjacent residues in the same chain, so that the structure repeats after 10 residues on each chain, that is, after 34 A. The distance of a phosphorus atom from the fibre axis is 10 A. As the phosphates are on the outside, cations have easy access to them.

Figure 1 The structure is an open one, and its water content This figure is purely is rather high. At lower water contents we would diagrammatic (8). The expect the bases to tilt so that the structure could two ribbons symbolize become more compact. the two phophate-sugar chains, and the The novel feature of the structure is the manner in horizonal rods the pairs which the two chains are held together by the of bases holding the purine and pyrimidine bases. The planes of the chains together. The bases are perpendicular to the fibre axis. They are vertical line marks the joined together in pairs, a single base from one chain being hydroden-bonded to a single base fibre axis.

41 from the other chain, so that the two lie side by side with identical z-coordinates. One of the pair must be a purine and the other a pyrimidine for bonding to occur. The hydrogen bonds are made as follows: purine position 1 to pyrimidine position 1; purine position 6 to pyrimidine position 6.

If it is assumed that the bases only occur in the structure in the most plausible tautomeric forms (that is, with the keto rather than the enol configurations) it is found that only specific pairs of bases can bond together. These pairs are: adenine (purine) with thymine (pyrimidine), and guanine (purine) with cytosine (pyrimidine) (9).

In other words, if an adenine forms one member of a pair, on either chain, then on these assumptions the other member must be thymine; similarly for guanine and cytosine. The sequence of bases on a single chain does not appear to be restricted in any way. However, if only specific pairs of bases can be formed, it follows that if the sequence of bases on one chain is given, then the sequence on the other chain is automatically determined.

It has been found experimentally3,4 that the ratio of the amounts of adenine to thymine, and the ratio of guanine to cytosine, are always very close to unity for deoxyribose nucleic acid.

It is probably impossible to build this structure with a ribose sugar in place of the deoxyribose, as the extra oxygen atom would make too close a van der Waals contact.

The previously published X-ray data5,6 on deoxyribose nucleic acid are insufficient for a rigorous test of our structure. So far as we can tell, it is roughly compatible with the experimental data, but it must be regarded as unproved until it has been checked against more exact results. Some of these are given in the following communications (10). We were not aware of the details of the results presented there when we devised our structure (11), which rests mainly though not entirely on published experimental data and stereochemical arguments.

It has not escaped our notice (12) that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.

Full details of the structure, including the conditions assumed in building it, together with a set of coordinates for the atoms, will be published elsewhere (13). We are much indebted to Dr. Jerry Donohue for constant advice and criticism, especially on interatomic distances. We have also been stimulated by a knowledge of the general nature of the unpublished experimental results and ideas of Dr. M. H. F. Wilkins, Dr. R. E. Franklin and their co-workers at King’s College, London. One of us (J. D. W.) has been aided by a fellowship from the National Foundation for Infantile Paralysis.

1 Pauling, L., and Corey, R. B., Nature, 171, 346 (1953); Proc. U.S. Nat. Acad. Sci., 39, 84 (1953). 2 Furberg, S., Acta Chem. Scand., 6, 634 (1952). 3 Chargaff, E., for references see Zamenhof, S., Brawerman, G., and Chargaff, E., Biochim. et Biophys. Acta, 9, 402 (1952).

4 Wyatt, G. R., J. Gen. Physiol., 36, 201 (1952). 5 Astbury, W. T., Symp. Soc. Exp. Biol. 1, Nucleic Acid, 66 (Camb. Univ. Press, 1947). 6 Wilkins, M. H. F., and Randall, J. T., Biochim. et Biophys. Acta, 10, 192 (1953).

Annotations (1) It’s no surprise that James D. Watson and Francis H. C. Crick spoke of finding the structure of DNA within minutes of their first meeting at the Cavendish Laboratory in Cambridge, England, in 1951. Watson, a 23-year-old geneticist, and Crick, a 35-year-old former physicist studying protein structure for his doctorate in biophysics, both saw DNA’s architecture as the biggest question in biology. Knowing the structure of this molecule would be the key to understanding how genetic information is copied. In turn, this would lead to finding cures for human diseases.

Aware of these profound implications, Watson and Crick were obsessed with the problem—and, perhaps more than any other scientists, they were determined to find the answer first. Their competitive spirit drove them to work quickly, and it undoubtedly helped them succeed in their quest.

Watson and Crick’s rapport led them to speedy insights as well. They incessantly discussed the problem, bouncing ideas off one another. This was especially helpful because each one was inspired by different evidence. When the visually sensitive Watson, for example, saw a cross- shaped pattern of spots in an X-ray photograph of DNA, he knew DNA had to be a double helix. From data on the symmetry of DNA crystals, Crick, an expert in crystal structure, saw that DNA’s two chains run in opposite directions.

Since the groundbreaking double helix discovery in 1953, Watson has used the same fast, competitive approach to propel a revolution in molecular biology. As a professor at Harvard in the 1950s and 1960s, and as past director and current president of Cold Spring Harbor Laboratory, he tirelessly built intellectual arenas—groups of scientists and laboratories—to apply the knowledge gained from the double helix discovery to protein synthesis, the genetic code, and other fields of biological research. By relentlessly pushing these fields forward, he also advanced the view among biologists that solving major health problems requires research at the most fundamental level of life.

(2) On this date, Nature published the paper you are reading.

According to science historian Victor McElheny of the Massachusetts Institute of Technology, this date was a turning point in a longstanding struggle between two camps of biology, vitalism and reductionism. While vitalists studied whole organisms and viewed genetics as too complex to understand fully, reductionists saw deciphering fundamental life processes as entirely possible— and critical to curing human diseases. The discovery of DNA’s double-helix structure was a major blow to the vitalist approach and gave momentum to the reductionist field of molecular biology.

Historians wonder how the timing of the DNA race affected its outcome. Science, after years of being diverted to the war effort, was able to focus more on problems such as those affecting human health. Yet, in the United States, it was threatened by a curb on the free exchange of ideas. Some think that American researcher Linus Pauling would have beaten Watson and Crick to the punch if Pauling’s ability to travel had not been hampered in 1952 by the overzealous House Un-American Activities Committee.

(3) Nature (founded in 1869)——and hundreds of other scientific journals—help push science forward by providing a venue for researchers to publish and debate findings. Today, journals also validate the quality of this research through a rigorous evaluation called peer review. Generally at least two scientists, selected by the journal’s editors, judge the quality and originality of each paper, recommending whether or not it should be published.

Science publishing was a different game when Watson and Crick submitted this paper to Nature. With no formal review process at most journals, editors usually reached their own decisions on submissions, seeking advice informally only when they were unfamiliar with a subject.

(4) The effort to discover the structure of DNA was a race among several players. They were world-renowned chemist Linus Pauling at the California Institute of Technology, and X-ray crystallographers Maurice Wilkins and Rosalind Franklin at King’s College London, in addition to Watson and Crick at the Cavendish Laboratory, Cambridge University.

The competitive juices were flowing well before the DNA sprint was in full gear. In 1951, Pauling narrowly beat scientists at the Cavendish Lab, a top center for probing protein structure, to the discovery that certain proteins are helical. The defeat stung. When Pauling sent a paper to be published in early 1953 that proposed a three-stranded DNA structure, the head of the Cavendish gave Watson and Crick permission to work full-time on DNA’s structure. Cavendish was not about to lose twice to Pauling.

Pauling's proposed structure of DNA was a three-stranded helix with the bases facing out. While the model was wrong, Watson and Crick were sure Pauling would soon learn his error, and they estimated that he was six weeks away from the right answer. Electrified by the urgency—and by the prospect of beating a science superstar—Watson and Crick discovered the double helix after a four-week frenzy of model building.

Pauling was foiled in his attempts to see X-ray photos of DNA from King's College—crucial evidence that inspired Watson's vision of the double helix—and had to settle for inferior older photographs. In 1952, Wilkins and the head of the King's laboratory had denied Pauling's request to view their photos. Pauling was planning to attend a science meeting in London, where he most likely would have renewed his request in person, but the United States House Un-American Activities Committee halted Pauling’s trip, citing his antiwar activism. It was fitting, then, that Pauling, who won the Nobel Prize in Chemistry in 1954, also won the Nobel Peace Prize in 1962, the same year Watson and Crick won their Nobel Prize for discovering the double helix.

(5) Here, the young scientists Watson and Crick call their model “radically different” to strongly set it apart from the model proposed by science powerhouse Linus Pauling. This claim was justified. While Pauling’s model was a triple helix with the bases sticking out, the Watson-Crick model was a double helix with the bases pointing in and forming pairs of adenine (A) with thymine (T), and cytosine (C) with guanine (G).

(6) This central description of the double helix model still stands today—a monumental feat considering that the vast majority of research findings are either rejected or changed over time.

According to science historian Victor McElheny of the Massachusetts Institute of Technology, the staying power of the double helix theory puts it in a class with Newton’s laws of motion. Just as Newtonian physics has survived centuries of scientific scrutiny to become the foundation for today’s space programs, the double helix model has provided the bedrock for several research fields since 1953, including the biochemistry of DNA replication, the cracking of the genetic code, genetic engineering, and the sequencing of the human genome.

(7) Norwegian scientist Sven Furberg’s DNA model—which correctly put the bases on the inside of a helix—was one of many ideas about DNA that helped Watson and Crick to infer the molecule’s structure. To some extent, they were synthesizers of these ideas. Doing little laboratory work, they gathered clues and advice from other experts to find the answer. Watson and Crick’s extraordinary scientific preparation, passion, and collaboration made them uniquely capable of this synthesis.

(8) A visual representation of Watson and Crick’s model was crucial to show how the components of DNA fit together in a double helix. In 1953, Crick’s wife, Odile, drew the diagram used to represent DNA in this paper. Scientists use many different kinds of visual representations of DNA.

(9) The last hurdle for Watson and Crick was to figure out how DNA’s four bases paired without distorting the helix. To visualize the answer, Watson built cardboard cutouts of the bases. Early one morning, as Watson moved the cutouts around on a tabletop, he found that only one combination of base molecules made a DNA structure without bulges or strains. As Crick put it in his book What Mad Pursuit,Watson solved the puzzle “not by logic but serendipity.” Watson and Crick picked up this model-building approach from eminent chemist Linus Pauling, who had successfully used it to discover that some proteins have a helical structure.

(10) Alongside the Watson-Crick paper in the April 25, 1953, issue of Nature were separately published papers by scientists Maurice Wilkins and Rosalind Franklin of King’s College, who worked independently of each other. The Wilkins and Franklin papers described the X-ray crystallography evidence that helped Watson and Crick devise their structure. The authors of the three papers, their lab chiefs, and the editors of Nature agreed that all three would be published in the same issue.

The “following communications” that our authors are referring to are the papers by Franklin and Wilkins, published on the journal pages immediately after Watson and Crick’s paper. They (and other papers) can be downloaded as PDF files (Adobe Acrobat required) from Nature’s 50 Years of DNA website (http://www.nature.com/nature/dna50/archive.html).

Here are the direct links:

Molecular Configuration in Sodium Thymonucleate Franklin, R., and Gosling, R. G. Nature 171, 740-741 (1953) URL: http://www.nature.com/nature/dna50/franklingosling.pdf

Molecular Structure of Deoxypentose Nucleic Acids Wilkins, M. H. F., Stokes, A. R., & Wilson, H. R. Nature 171, 738-740 (1953) URL: http://www.nature.com/nature/dna50/wilkins.pdf

(11) This sentence marks what many consider to be an inexcusable failure to give proper credit to Rosalind Franklin, a King’s College scientist. Watson and Crick are saying here that they “were not aware of” Franklin’s unpublished data, yet Watson later admits in his book The Double Helix that these data were critical in solving the problem. Watson and Crick knew these data would be published in the same April 25 issue of Nature, but they did not formally acknowledge her in their paper.

What exactly were these data, and how did Watson and Crick gain access to them? While they were busy building their models, Franklin was at work on the DNA puzzle using X-ray crystallography, which involved taking X-ray photographs of DNA samples to infer their structure. By late February 1953, her analysis of these photos brought her close to the correct DNA model.

But Franklin was frustrated with an inhospitable environment at King’s, one that pitted her against her colleagues. And in an institution that barred women from the dining room and other social venues, she was denied access to the informal discourse that is essential to any scientist’s work. Seeing no chance for a tolerable professional life at King’s, Franklin decided to take another job. As she was preparing to leave, she turned her X-ray photographs over to her colleague Maurice Wilkins (a longtime friend of Crick).

Then, in perhaps the most pivotal moment in the search for DNA’s structure, Wilkins showed Watson one of Franklin’s photographs without Franklin’s permission. As Watson recalled, “The instant I saw the picture my mouth fell open and my pulse began to race.” To Watson, the cross- shaped pattern of spots in the photo meant that DNA had to be a double helix.

Was it unethical for Wilkins to reveal the photographs? Should Watson and Crick have recognized Franklin for her contribution to this paper? Why didn’t they? Would Watson and Crick have been able to make their discovery without Franklin’s data? For decades, scientists and historians have wrestled over these issues.

To read more about Rosalind Franklin and her history with Wilkins, Watson, and Crick, see the following:

“Light on a Dark Lady” by Anne Piper, a lifelong friend of Franklin’s URL: http://www.physics.ucla.edu/~cwp/articles/franklin/piper.html

“The Double Helix and the Wronged Heroine,” an essay on Nature’s “Double Helix: 50 years of DNA” Web site URL: http://www.nature.com/cgi-taf/DynaPage.taf?file=/nature/journal/v421/n6921/full/nature01399_fs.html

A review of Brenda Maddox’s recent book, Rosalind Franklin: The Dark Lady of DNA in The Guardian (UK) URL: http://books.guardian.co.uk/whitbread2002/story/0,12605,842764,00.html

(12) This phrase and the sentence it begins may be one of the biggest understatements in biology. Watson and Crick realized at the time that their work had important scientific implications beyond a “pretty structure.” In this statement, the authors are saying that the base pairing in DNA (adenine links to thymine and guanine to cytosine) provides the mechanism by which genetic information carried in the double helix can be precisely copied. Knowledge of this copying mechanism started a scientific revolution that would lead to, among other advances in molecular biology, the ability to manipulate DNA for genetic engineering and medical research, and to decode the human genome, along with those of the mouse, yeast, fruit fly, and other research organisms.

(13) This paper is short because it was intended only to announce Watson and Crick’s discovery, and because they were in a competitive situation. In January 1954, they published the “full details” of their work in a longer paper (in Proceedings of the Royal Society). This “expound later” approach was usual in science in the 1950s as it continues to be. In fact, Rosalind Franklin did the same thing, supplementing her short April 25 paper with two longer articles.

UNIT 4 CHAPTER 16 – REPLICATION HOMEWORK

HW 4-1 Evidence for DNA as Genetic Material:

1) It was known in the 1940’s that chromosomes contained the genetic material. Chromosomes were known to contain both nucleic acids and proteins. Why did most biologists assume that the genetic material was protein and not DNA? (It was incorrectly believed that nucleic acids existed to provide a structural scaffold for the proteins).

2) Summarize the Avery-McCloud evidence that DNA and not protein was the genetic material.

3) Hershey-Chase experiment A) What is a bacterial phage? Briefly explain how a bacterial phage infects a bacteria. B) Explain why Hershey and Chase used radioactive S and radioactive P in their experiments to distinguish whether the genetic material was DNA or protein. C) Using a well-labeled diagram, summarize the design and results of the Hershey – Chase experiment.

HW 4-2- Watson and Crick Nature Paper on the Structure of DNA -

1) Watson and Crick’s first Nature paper published April 25, 1953 is considered the most famous paper in biochemistry. Why? 2) Describe the Watson Crick DNA model. 3) The second to last paragraph starting “It has not escaped our notice …” lays claim to a model for what important process involving DNA? 4) Who is acknowledged in the last paragraph? In class we will discuss the controversy in science about who gets the credit for a discovery. 5) What was Rosalind Franklin’s important contribution to evidence for the double- helix structure of DNA? What is controversial about the degree of credit she received for her work?

HW 4-3 – DNA as Genetic Material and Semi-conservative Replication and Basics of DNA structure

Part 1: Basics of DNA Structure 1) IB Memorize List: Draw diagram that includes four total nucleotides bases (two on each separate strand). Use a pentagon to represent the sugar, a circle for the phosphate group and a rectangle to represent the bases. Use solid lines for covalent bonds and dashed lines for hydrogen bonds.

Clearly Label, using horizontal arrows the following components: i) Base ii) Sugar, including 5’ and 3’ positions iii) Phosphodiester bond iv) hydrogen bonds

Note: Diagram must clearly indicate that the two strands are anti-parallel.

2a) Are the sugar-phosphate groups and bases on the interior or exterior of the helix? 2b) What kind of bond connects the nucleotides WITHIN a strand?

2c) Which functional groups (i.e. type and carbon # are they attached to) are involved in the formation of the phosphodiester bond?

2d) What does it mean to say that the strands are antiparallel?

3a) What type of bond holds the two STRANDS together?

3b) Which bases are complementary? How many hydrogen bonds connect to each base pair?

Part 2: Semi-Conservative Replication 4A) Briefly explain the differences between the Conservative, Semi-Conservative and Dispersive models of DNA Replication.

4B) Outline the design and results of the Meselson-Stahl Experiment with emphasis on how the experiment distinguishes between Semi-conservative, Conservative and Dispersive Models. Use a diagram and words.

4C) ) Suppose that you had an actively dividing culture of E. coli bacteria to which radioactive thymine had been added. What would happen if a cell replicated once in the presence of this radioactive base? A) One of the daughter cells, but not the other would have radioactive DNA. B) Neither of the two daughter cells would be radioactive. C) All four bases of the DNA would become radioactive. D) Radioactive thymine would pair with nonradioactive guanine. E) DNA in both daughter cells would be radioactive.

HW 4-4 Details of Replication – 1A) What is replication?

1B) Why is accurate replication of DNA so critical to the survival of the cell?

2a) What is the origin of replication? How does the origin of replication differ in bacteria chromosomes compared to eukaryotic chromosomes?

2b) In bacteria, it has been observed that the origin of replication tends to have a high proportion of A-T base pairs compared to C-G base pairs. Speculate on why bacterial origins to be A-T rich. (Hints: What has to happen to the strands before replication can occur; Also look at the number of H-bonding contacts in each pair)

2c) What is a replication fork?

2d) Is replication unidirectional or bidirectional? 3a) Diagram out the addition of a nucleotide triphosphate base (either ATP, GTP, TTP or CTP) to an existing DNA strand. 3b) Explain the following statement: “Synthesis of a strand of DNA can ONLY occur in a 5’ to 3’ direction.”

4a) What is meant by the terms leading strand and lagging strand?

4b) What is the apparent direction (direction of growth) for the leading strand vs. the lagging strand? DRAW A DIAGRAM to support your answer.

4c) Explain why continuous synthesis of both DNA strands is not possible. Use a diagram to support your answer.

4d) Explain how the lagging strand problem of discontinuous synthesis is solved by the use of Okasaki fragments.

5A) What is priming of DNA synthesis? What property of DNA polymerases requires the use of primers?

5B) What molecule is used as a primer in place of DNA?

HW 4-5 End Replication , DNA Proofreading and Repair 1) What is the ends replication problem? How does a eukaryotic cell solve this problem? Draw diagrams to support your answer. Include the terms telomere and telomerase in your answer. 2A) What is a mismatch repair? 2B) What is excision repair? 2C) Explain the roles of DNA polymerase I, ligase and repair enzymes in DNA proofreading and repair. 2D) Explain how UV light can cause damage to DNA that can lead to skin cancer.

HW 4-6 Structure of Chromosomes - Role of DNA packing in eukaryotic cells:

1) The wrapping of DNA around histones is called supercoiling. If the DNA in a typical human chromosome were uncoiled it would be approximately 3.5 cm in length. How does this length compare to the diameter of the cell?

2) State 3 reasons that DNA must be tightly packed and highly organized within the cell.

3) How does the amount of protein associated with DNA in bacteria compare the amount in eukaryotes? 4) What are histones? What kinds of charges are found on the surface of histones? Why do histones have these charges?

5) What are the components a nucleosome? What are the “beads” and what is the “string”? How many histones are there in each nucleosome? Sketch a nucleosome and label the DNA and the histones.

Chapter 17 - Transcription and Translation HW 4-6- Read Chapter 17 pp. 325-331 (Powerpoint: Intro to Genetic Code )

1) Distinguish between replication, transcription and translation.

2) What is the central dogma of molecular biology?

3A) Compare and contrast the function and cellular locations of mRNA, tRNA and rRNA.

3B) What is the approximate % in the cell for each type of RNA? Why is rRNA so abundant?

4A) Describe where transcription and translation occur in prokaryotes and eukaryotes.

4B) What is a primary transcript in eukaryotes?

4C) Explain the significance of, transcription and translation being separated in space and time in eukaryotes.

5A) What is the sense strand of the DNA? What is the antisense strand? Is the sense strand or the antisense strand defined as the gene? Which serves as the template for the RNA, the sense strand or the antisense strand?

5B) How many amino acids must be coded for? How many different DNA bases exist? How many DNA bases are need to code for an amino acid?

5C) What is a codon?

5E) What does it mean to say the genetic code is universal? What important implication does this have for biotechnology?

6A) What does it mean to describe the genetic code as redundant and unambiguous?

6B) What does it mean to describe the genetic code as “universal?” What is the evolutionary significance of this? (remember, evolution is a major theme in biology.) What important implication does this have for biotechnology?

6C) What is the wobble effect?

HW 4-8 –Read Chapter 17, pp. 331-336. (Powerpoint: Introduction to Transcription 2010) 1) Campell’s Concept Check 17.2, p. 334 #1-4.

2A) Name and briefly describe, the 3 steps of transcription.

2B) Describe the steps for formation of the transcription initiation complex. Word bank: TATA box, promoter, transcription factors, RNA polymerase

3) Circle the correct choice in the ( ): Elongation of the mRNA by RNA polymerase always procedes in the (3’ to 5’ or 5’ to 3’) direction?

4) mRNA processing in Eukaryotes: How is eukaryotic RNA modified between transcription and translation? Where does processing take place in the cell? What are the purposes of these modifications?

HW 4-8 TRANSLATION : READ CHAPTER 17, pp. 337- 348. (Ppt: Translation)

1) Coupled transcription and translation – Campbell’s p. 347, Concept Check 17.6, #1.

2A) Describe the role of tRNA in translation.

2B) What is an anticodon?

2C) Draw the structure of tRNA and label the amino-acid attachment site and the anti-codon site.

3) Fill in the following table using figure 17.4 on p. 330 of Campbell’s DNA triplet 5’ DNA triplet 3’ to mRNA codon 5’ tRNA anticodon Amino acid to 3’ 5’ (template or to 3’ (sense strand antisense strand) or gene) ATG TAC AUG UAC methionine GGA TTC UAG

4) Describe the role of the enzyme aminoacyl tRNA synthetase in translation.

5A) Describe and draw the structure of a ribosome. Label the small and large subunits and the E, P, A sites.

5B) Describe the roles of the E, P, and A sites of the large ribosomal subunit.

6A) Describe, using diagrams and words, the process of translation initiation. What amino acid always starts an amino acid chain?

6B) ELONGATION- describe using diagrams and words how an amino acid moves through the different site of the ribosome.

6C) TERMINATION – include in your answer a discussion of the stop codon and release factor.

7) Campbell’s p. 344, Concept Check 17.4, #2.

8) Campbell’s p. 346, Concept Check 17.5, #1 and 2.

9) Describe the signal hypothesis (Campbell’s Concept Check 17.4, p.344, #3)

10) What are posttranslational modifications? What is the purpose of a posttranslational modification? Give specific examples to support your answers.

HW 4-10 Bacterial Operons – 18.1 Powerpoints: Try Operon, Lac Operon Chapter 18 – Regulation of Gene Expression – Read in Campbell’s pp. 351-356.

1) What is an operon?

2) What is an operator?

3) Why would a bacterial cell need to control the expression of a gene based on environmental conditions?

4) Explain the differences between inducible and repressible operons.

5) Campbell’s p. 356, Concept Check 18.1, #1,2,3 and p. 379, #1 & 3.

6) COMPARISON OF TRYP vs. LAC OPERONS

Tryp operon Lac operon

Repressible?

Inducible?

Default on or off?

Repressor activated for DNA Binding By metabolite product?

Repressor inactivated for DNA binding by first substrate in pathway?

7) GENE EXPRESSION OF LAC OPERON RELATED TO [GLUCOSE] AND [LACTOSE]

[cAMP] CRP Repressor Lac gene high or low? binds ? binds ? expression high or low?

[Glucose] high, [lactose] low

[Glucose] high, [lactose] high

[Glucose] low, [lactose] low

[Glucose] low, [lactose] high HW 4-11 Chapter 21 (Genomes and Their Evolution) 21.3, 21.4, pp. 432-437

1) Complete the following table by filling in the appropriate description of each property associated with prokaryotes vs. eukaryotes.

Property Prokaryote Eukaryote

Single Cell vs. Specialized Cell

Size of Genome (large or small)

% of genes expressed (20-60 % vs. almost 100%) % of DNA that codes for Genes (1.5% vs. vast majority)

2) List 3 categories of a noncoding DNA (i.e that does not code directly for proteins).

3) Which of the categories that you listed in the previous question is utilized in producing DNA fingerprints for identifying individuals? Can you utilize this same region for identifying bacteria? Explain.

4A) How does the sequence of DNA compare for a nerve cell, eye cell, liver cell and white blood cell? 4B) How can we explain the extreme differences in the structure and function of these cells if the DNA sequences are identical?

5A)How similar are chimpanzees in human in terms of anatomy, behavior and DNA sequences?

5B) How similar are the DNA sequences of human beings of different “races” / sex?

5C) How can we make sense of the answers to 5A and 5B?

6) How can we explain differences in phenotypic expression on molecular level?

7) How can a very small number of mutations change the expression of many genes? Specifically what types of genes when mutated can impact the expression of many other types of genes? What critically important \implication does this have for evolution?

HW 4-11 Chapter 18: Eukaryotic Gene Regulation 18.2 and 18.3 Reading: Campbell’s pp. 356-366

1) Is the default condition for most eukaryotic genes on or off?

2) Which step in the production of proteins (gene expression) is the most heavily regulated?

Epigenetics Problem Set

3) The dark banded areas between the telomeres and the centromeres show great variation in sequence. The light colored areas (described as orange below) consistent mainly of repeat sequences. Which region is most likely to contain genes? Explain.

Diagram 1

4) The dark region in diagram 1 above is called Euchromatin and light colored (labeled as orange) is call heterochromatain. Diagrams 2 and 3 below give a more magnified representation of these regions. What do you notice about the packaging of DNA in euchromatin vs heterochromatin? How do you think the packaging is related to gene expression?

Diagram 2

Diagram 3 5) DNA methylation and levels of gene transcription:. DNA methylation is the attachment of a CH3 group to the base in the DNA, usually cytosine, C. DNA methylation, especially in promoter regions of gene, is linked to the levels of gene transcription. In diagrams 4 and 5, white circles represent the absence of methylation,

dark circles represent the presence of methylation.. Diagram 5 depicts the promoter regions of a normal cell and a cancerous cell. (CpG islands are regions within the promoter with alternating C and G sequences. The cytosine bases, C, are often targets for methylation.) In diagram 5, a tumor suppression gene is turned OFF in the cancer cell. What is the connection between methylation and the level of gene activity (transcription)?

Diagram 4

Tumor suppressor gene IS expressed

Tumor suppressor gene NOT expressed Diagram 5

6)Histone Acetylation – recall that DNA is wrapped around proteins called histones. (Recall 8 histones with DNA wrapped around = nucleosome). A) What type of charge distribution (positive, negative or neutral) would expect to find on the surface of a histone if the histone bonds strongly to DNA? Explain.

B) The structure of a histone is shown below (diagram 6). The histone “tails” structures helping which help secure the DNA to the histone have a high proportion of the amino acid lysine. (Specifically the tail regions have strong attractions for the minor grove of the DNA and hence have the effect of pulling the DNA in closer towards the histone).

Diagram 6. It is observed that presence or absence of acetyl groups (COCH3) attached to the lysine groups changes the expression of the gene. The chemical structures of lysine and acetylated lysine is shown below in Figure 7.

Diagram 7

Which form of lysine acetylated or deacetylated do you think would have a stronger attraction to DNA? Explain.

C)What is the relationship between the presence or absence of acetyl groups and the presence or absence of gene expression? Explain.

Summary The presence or absence of the methyl groups on DNA or acetyl groups on histone are referred to as epigenetic markers. These chemical groups can be added or removed to in

56 response to environmental exposures and hence are believed to be at least a partial explanation of the molecular basis of “the nurture” aspect of who we are.

Advanced Extension Note: Acetylation of the histone is the best understand chemical modification of histones. Increased acetylation almost always results in an increase in gene activity. Other modifications such adding a phosphate group or methyl groups to histone sometimes increase and sometimes decrease activity depending on the location. It is likely that these other types of tags act as recognition binding sites for proteins that can either increase or decrease transcription. In the case of acetylated histones, the presence of acetyl groups attracts the binding of particular proteins. In one of the most well-studied cases, a general transcription factor called TFIID, which is responsible for binding to the TATA box in the promoter and critically important for recruiting the binding RNA polymerase II to the transcription initiation complex has been shown to have a much higher binding affinity to the DNA/histone complex when the histone contains acetyl groups. Thus it is likely this epigenetic marker groups such as acetyl or phosphate not only impact electrostatic attractions between histone and DNA, but also serve as recognition sites for the binding of proteins important for regulating transcription rates.

7)Epigenetic Case Study

One of the most important medical advances of the 20th century was the realization that diseases such as sickle cell anemia can be caused by mutations that cause chemical changes and therefore structural changes in proteins responsible coordinating the activities of the cell. In the 21st century our understanding of the molecular basis of disease is becoming more sophisticated and nuanced.

Twin studies are often used to determine the role of genetics on health. A woman developed breast cancer at age 60. Her identical twin sister did not. When the sequences of genes known to be linked with breast cancer were compared in the twins, NO DIFFERENCES IN GENE SEQUENCES were detected. Propose a hypothesis to account for the presence of cancer in one twin but not the other and an experiment (i.e. what would you look for) to test your hypothesis.

8) Initiation of transcription.

A) What is a transcription factor?

B) What region of the DNA do general transcription factors bind to?

C) What is an activator protein (specific transcription factor)?

D) What region of the DNA do the activator proteins bind to?

E) How is the binding of transcription factors and activator proteins impacted by the how tightly DNA is bound to histones (level of DNA packing)?

F) RNA polymerase will not bind to the DNA of the promoter region by itself. What other molecules must be present to form the transcription initiation complex?

G) How can the presence of different enhancer regions change the expression of different genes? How does the presence or absence of transcription factors in a cell change the expression of different genes? In your answer refer to figure 18-10 in your textbook, which compares the gene expression in a liver cell and the lens cell in your eye.

H) Activator proteins and some transcription factors are allosteric proteins. These proteins can have different conformations when small signal molecules such as hormones are bound or absent or when a phosphate group is present or absent. In one configuration of the protein it will tightly bind to DNA but in the other configuration its affinity for DNA is low. Explain how the rate of transcription initiation can be regulated by the small signal molecules such as hormones.

9) RNA processing

A) Concept of alternative splicing: Explain how different can proteins produced from the same “gene”.

B) Describe the relationship between the 5’ Cap and Poly A tail sequences and the lifetime of the mRNA in the cytosol.

10) Regulating Rate of Translation A) Removal of the 5’caps/3 tails results in mRNA being ______by endonucleases in the cell. How does this change the rate of translation? B) Micro-interfering mRNA’s are small noncoding segments of RNA which are complementary to an mRNA. Describe what happens when an microRNA which is complementary to mRNA is produced. What happens to the rate of translation and why? 11) Final level of regulation of gene expression – In addition to regulating the rate at which proteins are produced in the cell, the final concentration of active protein is controlled by regulating the rate at which proteins are ______by proteases. 12) Campbell’s, Concept Check 18.2, p. 364 # 1-4;

Homework 4-12 Unit Review for Chapters 17 + 18 Multiple Choice:

Part 1: Campbell’s End of Chapter Multiple Choice Self-Quiz Review Questions for Chapter 17, p. 350 #1-8. (Answers in Appendix A, back of book) . 1) 2) 3) 4) 5) 6) 7)

8) Fill in Table Below

Type of RNA Functions Messenger RNA (mRNA)

Transfer RNA (tRNA)

Plays catalytic (ribozyme) roles and structural role in ribosomes Primary Transcript

Small nuclear RNA (snRNA)

Part 2: Extra Practice MC for Chapter 17 (Answers at top of page 3)

1) Transfer RNA A) Forms hydrogen bonds between its codon and anticodon of an mRNA in the A site of the ribosome B) binds to its specific amino acid in the active site of an aminoacyl-tRNA synthetase C) uses GTP as the energy source to binds its amino acid D) is translated from mRNA

2) Translocation involves A) the hydrolysis of a GTP molecule B) the movement of the tRNA in the A site to the P site C) the movement of the mRNA strand one triplet length in the A site D) the release of the unattached tRNA from the E site E) all of the above

3) Several proteins may be produced at the same time from a single mRNA A) the action of several ribosomes in a string called a polyribosome B) several RNA polymerase molecules working sequentially C) signal peptides that associate ribosomes with rough ER D) containing several promoter regions E) the involvement of multiple splicesomes

Use the choices below for questions #4 and 5

A) RNA replicase B) RNA polymerase C) aminoacyl-tRNA synthetase D) ribosomal enzymes E) ribozymes

4) Which enzyme is responsible for the synthesis of mRNA?

5) Which enzymes is responsible for charging tRNA, e.g. attaching an amino acid to the tRNA?

6) A signal peptide A) is most likely to be found on proteins produced by bacterial cells B) directs an mRNA molecule into the cisternal space of the ER C) is a sign to help bind the small ribosomal unit at the initiation codon D) would be the first 20 or amino acids of a protein destined for secretion from the cell. E) is part of the 5’ cap

7) Which of the following is not involved in the formation of a eukaryotic transcription initiation complex? A) TATA box B) transcription factors C) snRNA D) RNA polymerase II E) promoter

8) A prokaryotic gene 600 nucleotide bases long can code for a polypeptide chain of about how many amino acids?

A) 100 B) 200 C) 300 D) 600 E) 1800

9) All of the following would be found in a prokaryotic cell except

A) mRNA B) rRNA C) simultaneous transcription and translation D) snRNA E) RNA polymerase

10) Place the following events in the synthesis of protein in the proper order. 1) A peptide bond forms. 2) A tRNA matches its anticodon to the codon in the A site. 3) A tRNA translocates from the A to the P site, and an unattached tRNA leaves the ribosome from the E site. 4) The large subunit attaches to the small subunit and the initiator tRNA fits in the P site. 5) A small subunit binds to an mRNA and an initiator tRNA

A) 4-5-3-2-1 B) 4-5-2-1-3 C) 5-4-3-2-1 D) 5-4-1-2-3 E) 5-4-2-1-3 ANSWER KEY FOR Part 2: #1-10:

1) B 2) E 3) A 4) B 5) C 6) D 7) C 8) B 9) D 10) E

REVIEW CHAPTER 18: REGULATION OF GENE EXPRESSION

Part 3: Campbell’s End of Chapter Multiple Choice Self-Quiz Review Questions for Chapter 18, p. 379 #1-4 & 6-9. (Answers in Appendix A, back of book).

1) 2) 3) 4) 6) 7) 8) 9)

Part 4: Extra Practice Multiple choice for both bacterial operons (#1-9) and eukaryotic gene expression (#10 - (Answers at end)

1) Circle the correct word: Repressible operons are always (on or off) unless a metabolite is present. Inducible operons are always (on or off) unless a substrate for one of the enzymes on the pathway is present.

For the following questions, match the terms with the appropriate phrase or description below. Each term can be used once, more than once, or not at all.

A) operon B) operator C) promoter D) repressor

2) a protein that is produced by a regulatory gene 3) a mutation is this region could change the rate at which RNA polymerase binds to DNA 4) The binding of an active repressor molecule at this site prevents transcription.

5) The role of a metabolite that controls a repressible operon is to A) bind to the promoter region and decrease the affinity of RNA polymerase for the promoter. B) bind to the operator region and block the attachment of RNA polymerase to the promoter. C) increase the production of inactive repressor proteins. D) bind to the repressor protein and inactivate it. E) bind to the repressor protein and activate it.

6) For a repressible operon to be transcribed, which of the following must be TRUE? A) Corepressor must be present. B) RNA polymerase and the active repressor must be present. C) RNA polymerase must bind to the promoter and the repressor must be inactive. D) RNA polymerase cannot be present and the repressor must be inactive. E) RNA polymerase must not occupy the promoter and the repressor must be inactive.

7) Inducible enzymes A) are usually involved in anabolic pathways

B) are produced when a small molecule inactivates the repressor protein C) are produced when activator molecule enhances the attachment of RNA polymerase with the operator D) are regulated by inherently inactive repressor molecules. E) are regulated almost entirely by feedback inhibition.

8) In E. Coli, tryptophan switches off the trp operon by A) inactivating the repressor protein B) inactivating the gene for the first enzyme in the pathway by feedback inhibition C) Binding to the repressor and increasing the latter’s affinity for the operator. D) Binding to the operator. E) Binding to the promoter.

9) A mutation that renders nonfunctional the product of a regulatory gene for an inducible operon would result in A) continuous transcription of the genes of the operon. B) complete blocking of the attachment of RNA polymerase C) irreversible binding of the repressor to the operator D) no difference in transcription rate when an activator protein E) negative control of transcription

Use the choices below to answer questions #10 and 11.

A) 1-2% B) 20 %-60% C) 95-100%

10) What % of genes in a typical eukaryotic cell are typically expressed?

11) What percentage of the DNA in a typical eukaryotic cell directly codes for protein?

12) A eukaryotic gene typically has all of the following features except A) a promoter B) an operator C) enhancers D) introns and exons E) control elements 13) Gene expression in eukaryotes may depend upon A) the position of the gene on the chromosome B) the state of the external environment C) the stage of development of the organism. D) A and C are correct. E) A,B, and C are correct.

14) DNA methylation of cytosine residues A) can be induced by drugs that reactivate genes 62

B) may contribute to long-term gene inactivation C) produces the promoter regions that specifically bind RNA polymerase D) makes satellite DNA a different density so it can be separated by centrifugation E) may be related to the transformation of protoncogenes to oncogenes.

15) Which of the following is NOT true of enhancers? A) They may be located thousands of nucleotides upstream from the genes they affect. B) When bound with activators, they interact with the promoter region and other transcription factors to increase the activity of the gene. C) They may complex with steriod-activated receptor proteins and thus selectively activate specific genes. D) They may coordinate the transcription of enzymes in the same metabolic pathway. E) They are located within the promoter, and when complexed with a steriod or other small molecule, they release an inhibitory protein and thus make DNA more accessible to RNA polymerase.

16) Which of the following is NOT an example of the control of gene expression that occurs after transcription?

A) mRNA stored in the cytoplasm needing a control signal to initiate translation B) the length of time mRNA lasts before it is degraded C) rRNA genes amplified in tandem arrays produced extra copies of DNA D) alternative RNA splicing before mRNA exits from the nucleus E) post-translational modification of a polypeptide

17) The control of gene expression is more complex in eukaryotic cells because A) chromosomes are contained in a nucleus B) gene expression differentiates specialized cells C) the chromosomes are linear and more numerous D) operons are controlled by more than one promoter region E) inhibitory or activating molecules may help regulate transcription

18) Histones are A) Small positively charged proteins that bind tightly to DNA B) Small bodies in the nucleus involved in rRNA synthesis C) Basic units of DNA packing consisting of DNA would around a protein core. D) Repeating arrays of six nucleosomes organized around an H1 molecule. E) Proteins responsible for producing repeating sequences of telomeres

ANSWER KEY FOR PART 4: 1) Repressible = on; Inducible = off

2) D 3) C 4) B 5) D 6) C 7) A 8) C 9) A 10) B 11) A (1.5%)

12) B 13) E 14) B 15) E 16) C 17) B 18) A

Biotechnology Chapter 20

Homework 5-1 – Basics of DNA Recombination and Gene Cloning Powerpoint: Gene cloning 2010. Reading: Campbell pp. 396- 403. 1) What does the term Recombinant DNA mean?

2) Summarize the arguments in favor and against genetically modified crops.

3) Petroleum-lysing bacteria are being engineered for the removal of oil spills. What is the most realistic danger of these bacteria to the environment? A) Mutations leading to the production of a strain pathogenic to humans B) Extinction of natural microbes due to the competitive advantage of the “petro- bacterium” C) Destruction of natural oil deposits D) Poisoning of the food chain

4A) What is gene cloning?

4B) What is a plasmid?

5)The role of restriction enzymes in DNA technology is to A) Provide a vector for the transfer of recombinant DNA. B) Produce cDNA from mRNA C) Produce a cut (usually staggered) at specific recognition sequences on DNA. D) Reseal “sticky ends” after base-pairing of complementary bases E) Digest DNA into single strands that can hybridize with complementary sequences.

6)Which of the following DNA sequences would be most likely to be a restriction site? (Hint: Look for palindromic sequence in choices)

a) AACCGG b) AAGG c) CTGCAG TTGGCC T T CC GACGTC

7) List the 5 steps necessary to clone a gene.

8) Inserting a Eukaryotic gene in a bacterial host: A) Why can’t a eukaryotic gene simply be inserted directly into a bacterial host? B) What does the enzyme reverse transcriptase do? What organism is the source of reverse transcriptase? C) Explain the process for preparing eukaryotic genes for insertion into a bacterial host.

9) What enzyme is needed to seal the nicks after sticky ends are combined?

10) What is meant by the term transformation? How can transformation be accomplished in the lab? 11) List 4 different methods for screening for cell colonies that contained the desired DNA insert.

12) Yeast has become important in genetic engineering because it A) Has RNA splicing machinery. B) Has plasmids that can be genetically engineered C) Allows the study of eukaryotic gene regulation and expression D) Grows steadily and rapidly in the laboratory. E) Does all of the above.

13) A plasmid has two-antibiotic-resistance genes, one for ampicillin and one for tetracycline. It is treated with a restriction enzyme that cuts in the middle of the ampicillin gene. DNA fragments containing a human globin gene were cut with the same enzyme. The plasmids and fragments are mixed, ligase, and used transform bacterial cells. Clones that have taken up the recombinant DNA are the ones that A) Are blue and can grow on plates with both antibiotics B) Can grow on plates with ampicillin but not with tetracycline C) Can grow on plates with tetracycline but not with ampicillin D) Cannot grow with any antibiotics. E) Can grow on plates on tetracycline and are not blue.

Answers to MC questions 3)B 5) C 6) C 12) E 13) C

5-2 Biotechnology Techniques Powerpoint: Biotechnology Techniques 2012 Reading: Campbell, p. 403-411, especially studying insert Research method boxes and 419 to 420.

1) Gel Electrophoresis A) Describe a gel electrophoresis experiment. What is the purpose of gel electrophoresis? B) Why does DNA move toward the positive electrode rather than the negative electrode in gel electrophoresis? C) Since DNA is invisible to the naked eye, how do we visualize the bands of DNA?

Analyzing Fragments formed by Digestion of DNA by Restriction Enzymes

2) Linear vs. Circular DNA - If you start with linear DNA, how many cut sites do there have to be to produce 2 fragments that will run separately on a gel? If you start with a circular plasmid of DNA, how many cut sites do there have to be to produce 2 fragments that will run separately on a gel?

3) A segment of DNA has restriction sites I and II which create restriction fragments a,b, and c. Which of the following gel(s) produced by electrophoresis would represent the separation and identity of these fragments?

4) Given the following restriction map

A B C D

Draw a sketch of the pattern that would be obtained if the 4 fragments were separated by gel electrophoresis.

5) Polymerase Chain Reaction (PCR) A) What is PCR and how is it used? B) List the 3 steps in one PCR cycle. C) Why is the polymerase used in automated PCR systems taken from bacteria that live in a hot springs? D) What happens to the number copies of the target DNA after each cycle?

6) DNA Fingerprinting A) What regions of our DNA are most useful in identifying individuals? Why aren’t coding regions of our DNA (genes) very useful?

B) Draw a diagram and also explain verbally why regions that contain no differences in the individual repeat DNA sequences (Short Tandem Repeats or STR’s) are so useful in distinguishing individuals. (Specifically, the sequence of an individual repeat is identical between individuals but what is different between individuals?) C) What is a DNA fingerprint? How is a DNA fingerprint made? D) Give two examples of how DNA fingerprints are used.

E) Who is the father of the baby in the DNA fingerprint below?

7) One application of DNA fingerprinting technology has been to identify stolen children and return them to their parents. Bobby Larson was taken from a supermarket parking lot in New Jersey in 1978, when he was 4 years old. In 1990, a 16-year-old boy called Ronald Scott was found in California, living with a couple named Susan and James Scott, who claimed to be his parents. Authorities suspected that Susan and James Scott might be the kidnappers and that Ronald Scott might be Bobby Larson. DNA samples were obtained from Mr. And Mrs. Larson and from Ronald, Susan, and James Scott. PCR of several VNTR loci was used to generate a DNA fingerprint of each individual with the results shown in the figure below. From the information in the figure, what can you say about the parentage of Ronald Scott? Explain.

5-2 Answers to MC and fingerprinting: Answer to MC for 5-2, #3: B; Answer to First Paternity Problem: Sample 4

Answer to Paternity Kidnapping Case: The Larsons are likely to be Ronald Scott’s parents. Almost all of his bands on the gel are found in either Mr or Mrs Larson. There is not a significant match with the Scotts.

5-3 Applications of Biotechnology – Reading Campbells, pp. 412 -420; 426-432. - Powerpoint: DNA Sequencing and the Human Genome Project Powerpoint

- 1) Sanger Sequencing Method - A) What is does the Sanger Sequencing method accomplish? - B) What is a dideoxyribonucleotide? What happens to the elongation of a DNA strand when a dideoxyribonucleotide is incorporated in place of a normal deoxyribonucleotide?

- C) Explain how is the sequence of nucleotides determined.

2) How can be a person’s DNA sequence be potentially used in diagnosing and treating diseases?

3) Discuss limitations and concerns about the use of DNA sequences to guide medical treatments? (Hints: Can we, with 100% accuracy, to predict the likelihood of a person developing a disease? What are the privacy concerns about who has access to medical records?) . 5-4 Gene Therapy,Cloning, Stem Cells Gene Therapy, Cloning Organims, Stem Cells Powerpoint 1A) What is gene therapy? 1B) What vector is commonly used to insert DNA into new host cells? 1C) What kinds of diseases are currently targets of gene therapy? 1D) What are the risks of gene therapy?

Cloning of an organism from differentiated cells 2A) Who is Dolly the Sheep? What is different about her? 2B) List the steps necessary to do cloning using differentiated cells. 2C) How efficiently does this technique work? 2D) What are the potential benefits of cloning organisms? 2E) What are some of the potential concerns of cloning organisms?

Production of differentiated tissue cells from stem cells 3A) What are stem cells? What are differentiated cells?

3B) Place Multipotent (Adult Stem Cells), Pluripotent, Totipotent and Unipotent cells in order of degree of differentiation, starting from completed undifferentiated and proceeding to completely differentiated.

3B) A totipotent stem cell has the potential to develop into a complete organism. A pluripotent stem cell has the potential to become any ______of ______cell.

3C) Induced Pluripotency - Shinya Yamanaka of Kyoto University shared the 2012 Nobel Prize in Physiology and Medicine based on a series of groundbreaking experiments published in 2007. What was the revolutionary result of Yamanaka’s experiments? Briefly describe the experimental design.

3D) Transdifferentiation – 2010 – Marius Wernig and his group at Stanford demonstrated that it was possible to converted a fibroplasts (a type of connective tissue skin cell) to ______cells by adding 3 specific ______. The was the first demonstration that one type of differentiated cell could be converted directly into another differentiated cell without going through the pluripotent stem cell stage.

3E) State two potential medical benefits of stem cell research.

5-5 Unit Review

PART 1: Multiple Choice (Answers at end). Use the following choices to answer questions #1-2.

A) Restriction enzyme B) Reverse transcriptase C) Ligase D) DNA polymerase E) RNA nuclease

1) Which is the first enzyme used for the production of cDNA

2) Which enzyme is used in the polymerase chain reaction? Part 2: Campbell’s Self-Quiz Problems: p. 424, #1,5,6,8 5-5 PART 1 MC ANSWERS:

1) B 2) D

Possible Final Exam Free Response Questions (Word banks are NOT provided on actual exam) Unit 1:

1) Describe the 4 levels of protein structure and describe the principles of protein in water. Word Bank: primary structure, secondary structure, tertiary structure, hydrophobic, hydrophilic and hydrophobic sides chains, disulfide linkages, hydrogen bonding between backbone amino acids, weak forces of attraction between sidechains, hydrogen bonding, dipole-dipole, London Dispersion Forces, Ionic bonds 2) Explain the molecular basis of sickle cell anemia: Word Bank: hemoglobin, oxygen carrying molecule, 4 subunits, alpha subunits, beta subunits, quaternary interactions, single point mutation, hydrophobic vs. hydrophilic amino acid side chains; valine vs. glutamic acid; hydrophobic patches; change in folding patterns; change in quaternary interaction formation of fibers; change in capacity to bind oxygen

Unit 2: 1) Outline how enzymes catalyze biochemical reactions. Word Bank for essay: 1) Biological catalyst 2) Protein 3) structure-function 4) Free energy of activation 5) substrate 6) Active site 7) enzyme-substrate complex 8) induced fit model 9) Correct orientation 10) stretch/weaken bonds 11) stabilize charge buildup in transition state 12) release products

2)Describe the effect of temperature on rates of a biochemical reaction. Word Bank for essay: 1) structure of protein/function of a protein 2) tertiary protein structure 3) weak intermolecular attractions-hydrogen bonding; dipole-dipole attractions, ionic bonds, London dispersion forces 4) average kinetic energy 5) distance between atom forming weak intermolecular attraction within folded proteins 6) shape of active site Unit 3: Cellular Respiration Compare and contrast aerobic and anaerobic respiration. Word Bank for essay: 1) cytosol 2) glycolysis 3) glucose 4) pyruvate 5) presence / absence of oxygen 6) lactate acid 7)Ethanol 8) ATP 9) NADH 10) substrate level phosphorylation

11) cytosol 12) mitochondria 13) link reaction 14) Krebs/ citric acid cycle 15) electron transport chain 16) pump protons 17) proton gradient 18) ATP synthease 19) ATP

Unit 4: Replication, Transcription and Translation 1) Replication- Word Bank for essay 1) Definition of Replication 2) Origin of Replication 3) Helicase 4) SSB 5) primase 6) DNA polymerase III 7) Nucleotide triphosphate bases 8) 5’ and 3’ 9) Antiparallel 10) Leading strands 11) Lagging strands 12) Okazaki fragments 13) ligase 2) Regulation of Transcription – The most heavily regulated step in the expression of gene is transcription. Write a paragraph explaining how transcription is regulated in Eurkaryotes. Word Bank: DNA packing, DNA/histone interactions, DNA methylation, histone acetylation, transcription factors and activator proteins, signal molecules such as hormones 3) Process of Transcription Word Bank: Definition of Transcription, Transcription Factors, Promoter, RNA polymerase II, enhancer region, activator proteins, promoter, Antisense/template strand, sense strand gene, initiation, elongation, polyadenylation sequence for termination, complementary base-pairing, 5’ to 3’, triphosphate bases, uracil, thymine 4) mRNA processing Work Bank: nucleus, cytosol, modified Guansine cap, Poly- A tail, splicing, introns, exons, Unit 5: Biotechnology 2002 AP EXAM 1) The human genome illustrates both continuity and change. A) Describe the essential features of two of the procedures/techniques below. For each of the procedures/techniques you describe, explain how its application contributes to understanding genetics. • The use of bacterial plasmids to clone and sequence a human gene. • Polymerase chain reaction (PCR) • Restriction fragment length polymorphism (RFLP) analysis

B) All humans are nearly identical genetically in coding sequences and have many proteins that are identical in structure and function. Nevertheless, each human has a unique DNA fingerprint. Explain this apparent contradiction.

Lab #1: LOVE WATER OR FEAR WATER? INVESTIGATING THE RELATIONSHIP BETWEEN CHEMICAL STRUCTURE AND SOLUBILITY IN WATER

Purpose: To investigate the relationship between the detailed chemical structure of a substance and whether a substance is hydrophilic or hydrophobic. The concepts from this investigation will provide a critical foundation for understanding protein folding, DNA structure and the biochemistry of lipids and membrane structure and the principles of chromatography.

Introduction The cells in our body are typically approximately 65% by weight water. The aqueous environment of within our cells in absolutely critical for the functioning of the molecules and ions that comprise the cell. Many substances are hydrophilic (water seeking or loving) and readily dissolve, while others are hydrophobic (water fearing) and are insoluble. Some large macromolecules such as proteins and nucleic acids are water soluble under certain conditions of pH or ionic strength but not others. Dissolving large molecules is very complex because of the incredibly large number of interactions required for the solvent to completely surround the solute molecules. In biology, the systems we wish to understand are very big and complex so we start by studying smaller, simpler systems (e.g. small molecules instead of complex macromolecules or single-celled bacteria instead of a human) to make it easier to identify the important principles. The results of our experiments will help us next week in developing an understanding of the structures and properties of important biological molecules. For example, folding of proteins is largely driven by the interactions hydrophobic and hydrophilic structures within the polypeptide chain and with the solvent environment. Thus proteins fold differently in a polar aqueous environment than in the nonpolar interior of a membrane. The results of this experiment will also help us to understand how the cell is able to establish partitions to separate different types of molecules and control their movements between cellular compartments. In this lab, you will investigate the relationship between chemical structure and whether a compound is hydrophilic or hydrophobic. The key to understanding the properties of a substance lies in analyzing its distribution of electrical charge in the chemical structure. The investigation will be subdivided into six parts. We will start by analyzing patterns in small compounds and work our way towards more complex structures.

SAFETY: -Wear safety glasses. - AVOID SKIN CONTACT AND AVOID BREATHING VAPORS of Heptane, xylene, methanol and iodine. The instructor will handle this chemicals for you to minimize any risk of your contacting these chemicals due to their toxicity. - NaOH is very corrosive. If NaOH is contacted on your skin, flush affected area immediately with large amounts of water. If NaOH solution gets in your eyes, flush eyes for 15 minutes with water.

WASTE DISPOSAL: - Any experiment which contains heptane must be disposed of in the organic waste jar in the fume hood.

PART 1: Investigating the Relationship between Polarity and Solubility

In part 1 you will classify a list of substances on your student answer sheet as being polar or nonpolar on the basis of chemical structure. You will observe whether or not a solution is formed by various combinations of substances and use the patterns of your results to determine the relationship between polarity and solubility.

AID TO VISUALIZATION OF LIQUID-LIQUID LAYERS: Determining whether two layers or one present when mixing two clear, colorless liquids can be difficult. In order to make the results of some of our experiments easier to see, we are going to take advantage of the fact the food coloring is very soluble in water and insoluble in heptane.

Expt Chemicals mixtured Directions: # (P= polar, NP = nonpolar 1 NaCl ( ) & Place a spatula tip of salt to ≈ 2 cm of water, swirl and H2O ( ) observe. DISPOSAL: Rinse in sink 2 NaCl ( ) & The hexane has been added to the tube with a sealed lid to Hexane ( ) prevent escape of vapors. Add a spatula tip of salt and swirl. DISPOSAL: ORGANIC WASTE JAR IN HOOD 3 MetOH ( ) & DEMO: OBSERVE SEALED TUBE. (MeOH is very H2O ( ) toxic, do not open sealed tube) ; Note: Food coloring has been added for contrast since both are clear, colorless liquids. DISPOSAL: None – return sealed tube unopened to storage jar. 4 MetOH ( ) & DEMO: OBSERVE SEALED TUBE. (MeOH is very Hexane ( ) toxic, do not open sealed tube) DISPOSAL: None – return sealed tube unopened to storage jar. 5 Hexane ( ) & DEMO: OBSERVE SEALED TUBE. (xylene is very Xylene ( ) toxic, do not open sealed tube) DISPOSAL: None – return sealed tube unopened to storage jar. 6 Hexane ( ) & The hexane has been added to the tube with a sealed lid to H2O ( ) prevent escape of vapors. Add 1-2 cm of water and reattach lid. Swirl and observe. DISPOSAL: None - Use this same tube for expt #7. 7 Tube # 6 contents + Bring tube from 6 with mixture of water and hexane to I2 instructor in fume hood. The instructor will add a small solid crystal of solid iodine to your tube. DISPOSAL: ORGANIC WASTE JAR IN HOOD

PART 2: Hydrophilic/Hydrophobic Behavior of Carbohydrates and Lipids Directions: Study structures, answer questions on student answer sheet.

Part 3: Solubility of Alcohols (Done as demo) In this experiment, you will observe the solubility of 3 alcohols in water and hexanes, and analyze your results in terms of the chemical structures.

Alcohols: Methanol (MeOH) : CH3OH

Ethanol, CH3CH2OH

Isopentyl alcohol, (CH3)2CH2CH2CH2OH

Part 4: Effect of Detergent on hexane and water interactions

Expt #1: Record observations of demo tube consisting of Water + Food Coloring + hexane (Demo tube) record

Expt #2: Water + Food Coloring + hexane + Soap Add a couple of drops of detergent to the tube used in Expt. #1; Record your observations, answer questions on student answer sheet.

Part 5: Effect of Salt on Water + isopropyl alcohol interaction (Demo Tubes) Observe the demo tubes below, record observations, answer questions on student answer sheet.

Expt #1: Water + Food Coloring + Isopropyl alcohol

Expt #2: Saturated Salt Water + Food Coloring + Isopropyl alcohol

Part 6: Effect of pH on solubility of salicyclic acid (Done as Demo)

Observe the demo tubes below, record observations, answer questions on student answer sheet. Salicylic acid + hexane, Salicylic acid + water, salicyclic acid + water + NaOH.

Lab #2: LEVELS OF PROTEIN STRUCTURE AND PRINCIPLES OF PROTEIN FOLDING LAB

Introduction Proteins are essential for a wide range of important functions in the body including reaction catalysis, immune function, nerve transmission and chemical signaling as well as structural roles in tendons, ligaments and muscle. Proteins are generally very large molecules with complex three-dimensional shapes. Proteins have a very definite shape which is essential to the function of the protein. Interestingly, although large changes in shape (denaturation) make the protein unable to carry out is its function, proteins typically have flexible structures that allow for small changes in shape in response to changing environmental conditions, such as the binding of a substrate. In this lab exercise, we will use models and computer simulations to explore the structure of proteins and the principles which guide the folding of proteins. This lab is structured as a guided inquiry, with data presented and leading questions asked to direct you to the appropriate conclusions.

Purpose: To review the four levels of protein structure and to examine the guiding principles of protein folding.

Overview of Lab: 1) Complete the prelab and minimum essential learnings on the student sheet provided. You have the option of answering the guided inquiry leading questions on your own paper or just discussing them verbally with your team.

2) 25 out of the 35 points for the lab will be earned on an individual, OPEN, LAB REPORT AND LAB MANUAL, CLOSED BOOK AND CLOSED FRIEND QUIZ given at the completion of the lab period.

INVESTIGATING THE GENERAL RULES THAT DETERMINE THE 3-D FOLDING PATTERNS (TERTIARY STRUCTURE) OF A PROTEIN Introduction The overall three-dimensional folding pattern of a protein is called the tertiary structure. The tertiary structure of the protein is primarily determined by the interactions of the amino-acid side chains. Amino acid side chains can be grouped into 3 general categories: nonpolar, polar or charged. Charged amino acid side chains may be + or -. Positively charged side chains are called basic (a neutral atom has acquired an H+ to become +). Negatively charged side chains are referred to as acidic (a neutral atom has lost an H+ to become -).

PART 1: DISCOVERING THE RULES THAT DETERMINE PROTEIN FOLDING PATTERNS IN WATER

Examine the model of a folded protein in water on the plastic laminated sheets. The amino acid side chains are color coded to represent different types of amino acids. The color code is: Nonpolar sidechains = Green Polar sidechains = Yellow + charged sidechains = Blue - Charged sidechains = Red

Your goal is to search for patterns in the locations of different types of amino acid side chains that can be used to generate folding rules.

Complete the following table Type of # of a.a. sidechains on # of a.a. sidechains on sidechain INTERIOR EXTERIOR Nonpolar (green) Polar (yellow) + (blue) - (red)

Questions: 1) How is the distribution of nonpolar vs. polar buried on the interior of the protein away from water vs. outside interacting directly with water consistent with last week’s hydrophobic/hydrophilic lab?

2) Red (-) side chains are generally near ______side chains.

3) Two ______can form a covalent disulfide linkage.

Use your observations to complete the Rules of Protein Folding Chart:

BASIC LAWS OF CHEMISTRY THAT DRIVE PROTEIN FOLDING IN WATER

1 ) NONPOLAR (HYDROPHOBIC) SIDECHAINS will be buried on the ______of a globular protein, where they are hidden from polar water molecules.

2) POLAR SIDE CHAINS – will be on the ______of the protein where they can hydrogen bond with water. If polar sidechains are in the interior of the protein they will be generally paired with another (polar or nonpolar) ______sidechain.

3) CHARGED SIDECHAINS (+ and -) will be ______each other on the ______of the protein where they often neutralize each other to form salt bridges.

4) CYSTEINE sidechains often interact with each other to form covalent ______bonds that stabilize protein structures.

Note that the presence of disulfide linkages reduced the “floppiness” of the polypeptide chain, i.e. it reduces the # of possible conformations. If multiple disulfide linkages are present, changing which cysteine are linked will change the folding pattern.

PART #2: PROTEIN FOLDING INSIDE A MEMBRANE Not all globular proteins are folded in water. A number of very important proteins, such as those involved in oxidative phosphorylation (the process the produces the majority of the ATP produced in cellular respiration and photosynthesis) are transmembrane proteins. The folding patterns of proteins inside are membrane are different from those in water. i.) Examine the model of a phospholipid bilayer. The circular caps represent the charged heads and the pipe-cleaners represent the nonpolar tails. Sketch a diagram of a lipid bilayer below and label the polar heads and nonpolar tails. ii.) The interior environment of the membrane is (circle the correct choice) polar or nonpolar.

The toober represents a transmembrane protein. iii.) Draw a diagram of your protein and clearly indicate the location (inside or outside) of each type of amino acid side chain (polar vs. nonpolar) iv.) Explain why this pattern is different than the pattern in water

PART 3: COMPUTER SIMULUATION SOFTWARE FROM CENTER FOR MOLECULAR MODELING – (Projected in the front of the room; small group discussion with instructor).

Folding website reference: www.rpc.msoe.edu/cbm/resources/jmol.php

We will first run Basic Principles of Chemistry that Drive Protein Folding, (Amino acid starter kit) Part 1. Beta-Globin structure.

From Basic Principles of Chemistry that Drive Protein Folding, (Amino acid starter kit) Part 2 for the proteins beta-globin and potassium ion channel protein. Beta-globin is a component of hemoglobin. It binds oxygen and is folded in the aqueous environment (blood). Potassium ion channel protein is a transmembrane protein that regulates the flow of potassium ion across membranes.

1) What are the advantages of using computer generated models of proteins over physical models? (Hints: Discuss rotation, color coding, backbone vs space-filling features, etc).

2) Label the hydrophilic and hydrophobic sections of beta-globin, a water soluble protein, and potassium ion channel protein, transmembrane protein. (For simplicity, we will represent the globular proteins as oval shapes;).

Interior

Beta-Globin Potassium Ion Channel Protein 3) Explain why the arrangements of hydrophilic and hydrophobic side chains are different in beta-globin vs. potassium ion channel protein.

Extension Questions: Speculate on the design features of the channel for the potassium ion. 1) What types of chain sides would you expect to find lining the interior of the channel? 2) How is it possible to exclude other +1 ions, such as Li+1, Na+1, Rb+1, Ag+1, from passing through the channel?

ACTIVITY #4: Role of disulfide linkages in stabilizing number of conformations (positions of polypeptide chain) of proteins

Part 1: Role of changing disulfide linkages in hair styling

Using the video clip and your toobers with pipe cleaners to represent disulfide linkages, explore the effect on protein conformation of changing disulfide linkage partners. USE DIAGRAMS and WORDS to support your answer.

1) Briefly discuss the role of alpha helices and intertwined chains in the structure of a hair fiber.

2) Explain how hair can be permanently changed from straight to curly by changing disulfide linkages.

Extension Part 2: Anfinsen’s Protein Folding Experiments: Connection between primary and tertiary structures and role of disulfide linkages in determining tertiary structure

In the 1960’s Anfinsen and his colleagues at the National Institutes of Health conducted a series of ground-breaking experiments on the nature of protein folding. The protein they studied was an enzyme called ribonuclease A (RNAase A), which is composed of 124 amino acids and 4 disulfide linkages. Urea was added to a solution of RNAase A to denature the enzyme. The urea could be removed by dialysis. The disulfide linkages could be broken and by the addition of a Beta-mercaptoethanol to reduce the cysteine side chains back to SH. Disulfide linkages can be restored by the addition of an oxidizing agent.

Questions asked by the experimenters included: “To what extent is the folding of the correct tertiary structure of a protein encoded in the primary structure?” “ What is the role of disulfide linkages in determining the correct folding pattern of the protein?”

3) What conclusions can be drawn from the above experiments?

4) One of the major themes of biology is the concept of structure determining function. Long polypeptide chains have billions of different positions (conformations) possible for the atoms in the chain. Using a diagram and words, explain the role of disulfide linkages in stabilizing the biologically active conformation of a polypeptide. ACTIVITY 5: EFFECT OF MUTATION: UNDERSTANDING THE RELATIONSHIP BETWEEN PRIMARY STRUCTURE AND TERTIARY STRUCTURE. (Modified from the Center for Biomolecular Modeling)

One of the major themes of biology is the concept of structure determining function. Long polypeptide chains have billions of different positions (conformations) possible for the atoms in the chain. In order to be biologically active the polypeptide has to be in the correct set of conformations. One of our most important tools in biology is studying mutations, i.e. what is the effect on normal biological function when a change is made. In our next round of simulations, you explore using flexible toober models to understand why changing the primary structure of a protein can potentially critically impact biological function.

Part 1: Folding your Protein

In your kit, you will find a Styrofoam sphere. Your protein must fold in a configuration to specifically bind your sphere in a pocket of the structure. Fold your protein, in a suitable configuration and then draw of sketch of your structure.

Part 2: Determining the Primary Sequence needed to support your folded tertiary structure

The conformation of your protein must obey the basic laws of chemistry for folding in water. In your kit you have 15 amino acids, some are hydrophobic, some are hydrophilic, some have the potential to form disulfide linkages. Place your amino acids in a correct order so that the both the laws of chemistry are obeyed and your protein remains in its biologically correct form. ( In some kits, the amino acids are magnetic and attach to metal clips; in other kits the plastic amino acids snap into place around the plastic bands).

Part 3: How does a random primary sequence impact tertiary structure and hence biological function?

RANDOM ARRANGMENT OF AMINO ACIDS

Have one team member close his or her eyes and RANDOMLY PULL OUT the 15 amino acids, one at a time and place them on your protein. Are the rules of chemistry still be obeyed? Is it likely that your protein refold into a conformation that would no longer bind the Styrofoam sphere?

1) Is it harder or easier to follow the rules of folding and end up with a biological active structure with a random primary sequence?

2) What is the potential effect on the tertiary structure to changing the primary structure as a result of a gene mutation?

PART 6: EFFECT OF MUTATION: SICKLE CELL ANEMIA ; IMPACT OF CHANGING PRIMARY STRUCTURE ON QUATERNARY STRUCTURE OF HEMOGLOBIN Red Blood cells, the cells are responsible for transporting oxygen and carbon dioxide in the blood, consist mainly of water and hemoglobin. Hemoglobin is the transport protein responsible for binding oxygen in the lungs and releasing it to tissues. Hemoglobin consists of 4 polypeptides chains. The polypeptides actually consist of two sets of two identical chains. The first set of chains are called alpha and the second set are called beta. (These chains are named for the first two letters of the Greek alphabet and are completely unrelated to secondary structure alpha helix and beta sheet – a point which is often confusing for students). The different polypeptides chains are not linked by covalent bonds, rather they are attracted by a combination of ionic, dipole- dipole, hydrogen bonding and hydrophobic attractions.

Sickle cell anemia is a genetic disease which results in abnormally shaped sickled blood cells. Individuals who live in parts of Africa or other areas in which malaria is present are more resistant to the disease if they contain 1 copy of normal hemoglobin and 1 copy of the sickle version. Sickled hemoglobin does not transport oxygen as efficiently and the shape of the red blood increases the risk of heart attacks and strokes

Look at the poster depicting a normal red blood cell.

1) Sketch the shape of a normal (wild type) red blood, containing normal wild type hemoglobin molecule in 2D, outlining the 4 subunits of the hemoglobin.

1B) Are the different hemoglobin molecules in the normal red blood cell separate or stuck together?

2A) Sketch the shape of the diseased sickle red blood cell. 2B) Are the different hemoglobin molecules in the sickled cells separate or stuck together? 2C) Which subunits (α or β) are attached together in the different hemoglobin molecules that allow for the formation of the fiber?

2D) What impact do the fibers have on the shape of the red blood cell? How does this sickle shape effect the ability of red blood cells to pass through small capillaries?

4)Sickle-cell anemia is one of approximately two hundred known diseases that are caused by a single point mutation. Recall that hemoglobin consists of 4 subunits consisting of two identical alpha subunits and two identical beta subunits. The disease is caused by a mutation in position 6 on the beta subunit. The normal amino acid in position 6 on the beta subunit is a negatively charged glutamic acid. In sickle cell anemia, position 6 on the beta subunit is mutated to an electrically neutral valine. Amino acid 6 is located on the outside surface of the beta subunit.

4)While the mutation has only very minor localized effects on the folding of the beta subunit when oxygen is bound, it has major consequences for the interactions between different hemoglobin molecules in their deoxygenated forms.

A) In the wild-type hemoglobin with negatively charged glutamic acids present, what would happen if two negatively charged glutamic acids on different hemoglobin molecules approached each other?

B) Compare valine and glutamic acid in terms of: i) size ii) shape iii) charge/distribution of charge. What is the most important difference?

C) When sickle hemoglobin is deoxygenated (O2 not bound) a hydrophobic patch is exposed to the surface. Is valine hydrophobic or hydrophilic?

D) Why does valine stick to the hydrophobic patch on another sickle cell hemoglobin?

EXTENSION:

5) Important clues about the molecular level differences between wild type (normal) hemoglobin and sickle-cell hemoglobin were first discovered in the 1950’s by clever experiments conducted by Harvey Itano, working at the time in the lab of Linus Pauling.

It was known that wild type hemoglobin changed its conformation (shape) when oxygen was bound to it. It order to keep the conformation of hemoglobin the same throughout the experiment, so that the changed shape of oxygenated vs deoxygenated would not affect how the molecules moved through an electrophoresis gel, Itano saturated the hemoglobin with carbon monoxide, CO. CO binds the same location to Fe+2 in the heme group but with much greater affinity than O2.

5A) Why does the binding of O2 to hemoglobin in cells need to be reversible?

5B) Why does CO bind some much more tightly to hemoglobin than O2? (Hint: electronegativity of C = 2.5, O = 3.5).

Itano placed hemoglobin dissolved in a neutral pH phosphate buffer solution saturated with CO gas in the center of a tube with electrodes at each end. The hemoglobin samples were subjected to an applied electric current for an hour and then the final position of the hemoglobin in the tube was noted. Experimental setup:

STARTING STARTING P

STARTING POSITION ↑

Wild Type Hemoglobin:

FINAL POSITION↑

5C) What is the overall electrical charge at neutral pH for wild type hemoglobin, positive, negative or neutral? How do you know?

Sickle Cell Hemoglobin:

FINAL POSITION↑

5D) What is the overall electrical charge at neutral pH for sickle cell hemoglobin, positive, negative or neutral? How do you know?

BONUS QUESTION 5E) Given what you know about sickle cell anemia, propose an explanation for these observations. (There is more than one possible explanation but hints for the simplest explanation are given below).

Hints: 1) If the overall charge was exactly zero (isoelectric point), what would be the final position of the protein relative to the starting position? 2) What must be the signs for overall charges for the wild type vs sickle cell 3) What is different in the primary sequences of wild type vs sickle cell that would change the charge? 4) How many beta subunits are present in each hemoglobin molecule? 5) Putting the pieces together, what must have been the overall charge and sign of charge in the wild type hemoglobin and what is the overall charge charge and sign of charge in the sickle-cell hemoglobin?

Protein Structure and principles of folding PRELAB

1A) Proteins are polymers of covalently bonded ______.

1B) Draw a general structure for an amino acid. What does R represent?

1C) Amino acids may be positively charged, neutral or negatively charge depending on the pH values. Draw the predominant equilibrium form of an amino acid in acidic, neutral and acidic conditions. Which is the zwitterion form?

2) Explain the terms hydrophobic and hydrophilic as the terms relate the R- group side chains.

Amino acid side chains can be grouped into 3 general categories: nonpolar, polar or charged. Charged amino acid side chains may be + or -. Positively charged side chains are called basic (a neutral atom has acquired an H+ to become +). Negatively charged side chains are referred to as acidic (a neutral atom has lost an H+ to become -). Use the table of R groups on page 69 of Campbell’s textbook to answer the following:

Fill in the blank: A) Hydrophobic sidechains are composed primarily of ______and ______atoms.

B) Polar R groups generally contain (besides a nonpolar hydrocarbon segment) functional groups that contain the elements ______or ______. These 2 elements are both much more electronegative than C or H, thereby creating polar bonds.

C) Acidic sidechains contain a ______functional group. This functional group always contains 2 ______atoms.

D) Basic sidechains contain an ______functional group. This functional group always contains 1 ______atom.

3) Write the complete structural formulas for the two products for the following reaction by adding the single amino acid to the end of the dipeptide reactant. Use table 5.15 on p. 69 to identify the primary sequence of your final tripeptide product.

4) Provide definitions of the different levels of protein structures:

Primary

Secondary

Tertiary

Quaternary-

5) Recopy sentence Fill in the blanks with either amino side chains or polypeptide backbone : Secondary structures are the result of hydrogen bonding between amino acids on the ______while tertiary structure is primarily determined by interactions among the ______. The two types of secondary structures are called ______helix and ______pleated sheet

6) Find and label two alpha helices and two beta pleated sheets in the structure below.

7)In the diagram below, identify type of attractive force represented for each letter. Choices are dipole-dipole, hydrogen bond, ion-dipole, LDF, ionic, disulfide linkage.

A = B =

C = D) =

8) Which picture represents a tertiary structure and which represents a quaternary structure?

MINIMUM ESSENTIAL UNDERSTANDINGS FROM LAB ACTIVITIES

Part 1: BASIC LAWS OF CHEMISTRY THAT DRIVE PROTEIN FOLDING IN WATER

1 ) NONPOLAR (HYDROPHOBIC) SIDECHAINS will be buried on the ______of a globular protein, where they are hidden from polar water molecules.

3) CHARGED SIDECHAINS (+ and -) will be ______each other on the ______of the protein where they often neutralize each other to form salt bridges.

Note: If fully charged side chains are present on the interior, they will always be paired with an oppositely charged ion in an ionic bond (salt bridge) or a highly polar side chain (ion-dipole). On the outside surface however, individual charged side-chains can exist. What is the key difference explains this observation?

4) CYSTEINE sidechains often interact with each other to form covalent ______bonds that stabilize protein structures.

PART #2: PROTEIN FOLDING INSIDE A MEMBRANE

A) Sketch a diagram of a lipid bilayer below and label the polar heads and nonpolar tails.

B) ) The interior environment of the membrane is (circle the correct choice) polar or nonpolar.

C) Draw a diagram of your protein and clearly indicate the location (inside or outside) of each type of amino acid side chain (polar vs. nonpolar)

C) Explain why this pattern is different than the pattern in water.

PART #3: COMPUTER MODELING OF BETA-GLOBIN AND POTASSIUM ION CHANNEL PROTEIN 1) Label the hydrophilic and hydrophobic sections of beta-globin, a water soluble protein, and potassium ion channel protein, transmembrane protein. (For simplicity, we will represent the globular proteins as oval shapes;).

Interior

Beta-Globin Potassium Ion Channel Protein 2) Explain why the arrangements of hydrophilic and hydrophobic side chains are different in beta-globin vs. potassium ion channel protein. PART 4: IMPORTANCE OF DISULFIDE LINKAGES IN TERTIARY STRUCTURE Long polypeptide chains have billions of different positions (conformations) possible for the atoms in the chain. One of the major themes of biology is the concept of structure determining function. Using a diagram and words, explain the role of disulfide linkages in stabilizing the biologically active conformation of a polypeptide. PART 5: EFFECT OF MUTATION: RELATIONSHP BETWEEN PRIMARY STRUCTURE AND TERTIARY STRUCTURE AND IMPACT ON BIOLOGICAL ACTIVITY 1) Is it harder or easier to follow the rules of folding and end up with a biological active structure with a random primary sequence?

2) What is the potential effect on the tertiary structure to changing the primary structure as a result of a gene mutation?

PART 6: EFFECT OF MUTATION: SICKLE CELL ANEMIA – IMPACT OF CHANGING PRIMARY STRUCTURE ON QUATERNARY STRUCTURE OF HEMOGLOBIN

Write a short essay briefly explaining sickle anemia with a focus on the molecular level causes of the disease. (IB requirement)

Lab #3: PURIFICATION OF GREEN AND BLUE FLUORESCENT PROTEINS AND EXPLORING FACTORS THAT CAUSE DENATURATION OF PROTEINS LAB (Adapted from Edvotek Lab #255)

Introduction In this experiment you will isolate a protein from an extract using a technique called column chromatography. Once you have isolated your protein, you will investigate factors that can cause the protein to denature or unfold. In order to study the properties of a protein, it must first be obtained in purified form. The source of a protein is generally either tissue or microbial cells. The first step in any protein purification is to break open these cells, releasing their proteins into a solution called a crude extract.1 In this experiment, you will be provided with a crude extract of proteins and will isolate either a green or blue fluorescent protein by using a technique called column chromatography. In our experiment, you will prepare a column made of a material called Sephadex. Sephadex beads act as molecular sieves. Each bead contains small holes that will trap or slow down smaller molecules while allowing larger molecules such as protein to pass by. As a result, large proteins will move through faster than smaller molecules. (This is the opposite of gel electrophoresis in which the larger molecules pass through more slowly). The rate at which each protein molecule passes through the column depends on its size, shape and whether it has smaller nonprotein molecules such as carbohydrates attached to it.2 Since proteins dissolved in water are generally clear and colorless, we need a method for detecting which of the fractions of solution that you collect from the column actually contain the proteins. In this experiment, we will use a special class of fluorescent proteins that were originally isolated from the bioluminescent jelly fish Aquorea Victoria. When these proteins are exposed to long-wave ultraviolet light they fluoresce, emitting in a green or blue light. The green fluorescent protein (gfp) is made up of 238 amino acids. The region primarily responsible for light emission is a cyclic structure formed by the interaction of Ser-Tyr-Gly at position 65-67 (see figure below). If the interaction of these amino acids is disrupted by denaturation (unfolding of the tertiary structure), the fluorescence is quenched. If the Tyr in position 66 is substituted to His and a Tyr at position 145 is substituted to Phe-145, the protein will emit a blue color instead of a green.3 The experimental protocol is divided into 3 parts. In part 1, you will prepare your column. In part 2, you will purify your protein by running your sample through the column. In the final part of the experiment, you will investigate what happens to the fluorescence of the purified protein when it is exposed to high temperature, and very high or low pH.

1) From Lehninger, Principles of Biochemistry, 4th edition, Freeman and Company, p. 89. 2, 3) From Edvotek, Background information for experiment #255.

PROCEDURE:

SAFETY: 1) Wear safety glasses at all times. 2) In part 3, avoid contact with HCl and NaOH solutions. These solutions are corrosive and will burn skin and eye tissue. If you get some on yourself, immediately rinse affected area with water.

PART 1: PACKING AND EQUILIBRATING THE COLUMN (from Edvotek Kit #255) KEY OBJECTIVE: To make a column that contains about a 3 cm packed height of white Sephadex material, suspended in buffer solution (column must not be allowed to dry).

2 mm buffer

3 cm sephadex

Detailed Directions:

1) Vertically mount the column on a ring stand. Make sure it is straight.

2) Slide the cap onto the spout at the bottom of the column. Fill about one-third of the column with the elution buffer.

3) Mix the Sephadex suspension (white gel column material) thoroughly by swirling or gently stirring.

4) Carefully pipet the mixed Sephadex slurry into the column by letting it stream down the inside walls of the column.

If the flow is stopped by an air pocket, stop adding the slurry and firmly tap the column until the air is removed and the slurry continues to flow down the side of the column.

5) Place an empty beaker under the column to collect wash buffer.

6) Remove the cap from the bottom of the column and allow the matrix to pack into the column.

7) Wash the packed column with 5 mL of 1x elution buffer. DO NOT ALLOW THE COLUMN TO DRY!

PART 2: PURIFYING PROTEIN EXTRACT USING A SEPHADEX COLUMN OBJECTIVE: TO OBTAIN A PURE PROTEIN SAMPLE

1) Label four micro tubes 1-4.

2) Slowly load the column with 0.2mL (200 microliters) of gfp extract OR bfp extract. Allow the extract to completely enter the column.

Useful hint: If you shine long U.V. light (black light) on the column, you will see gfp or bfp migrating through the column and you can predict the peak tubes that will contain protein.

3) Elute the column with 1x elution buffer. ∙ Add buffer slowly (several drops at a time) to avoid diluting the protein sample. ∙ using the graduated marks on the sides of the tubes, collect 0.5 mL fractions in the labeled microcentrifuge tubes.

4) Check all fractions by using long wave U.V. light to identify tubes that contain the fluorescent gfp or bfp proteins.

SAMPLE PREPARATION: You will need 3 separate samples of gfp or bfp for part 3. If your protein is concentrated in one fraction transfer about 1/3 to two separate microtest tubes. If it is divided between 3 fractions, you can use the 3 separate fractions.

PART 3: INVESTIGATING DENATURATION OF PROTEINS

OBJECTIVE: To examine the effect of temperature and pH on protein structure.

The gfp and bfp protein lose their ability to fluoresce when they are unfolded (denatured).

PART A: EFFECT OF DENATURING SOLUTION AND BOILING

After making sure that the sample tubes are tightly capped, place the microtube in an empty well in the hot bath. Monitor the fluorescence with your UV light.

PART B: EFFECT OF ADDING ACID (WITH INSTRUCTOR IN FUME)

CAUTION: 6 M HCl is very corrosive. Wear safety glasses and avoid contact with skin or eyes. If contacted immediately, wash affected area and inform instructor.

1) Assign one partner to monitor fluorescence by holding the UV light near the tube.

2) The instructor will add 6 M HCl dropwise until the fluorescence disappears.

PART C: EFFECT OF ADDING BASE- same as above, including caution warning, except use 6 M NaOH.

CLEANUP:

RINSE OUT COLUMNS AND MICROTEST TUBES WITH WATER, AND THEN DISPOSE OF IN TRASH.

EXPERIMENT 4: PRACTICING EXPERIMENTAL DESIGN DESIGNING AN EXPERIMENT TO MEASURE THE EFFECT OF A VARIABLE THAT CHANGES THE RATE OF THE DECARBOXYLATION OF PYRUVATE CATALYZED BY PYRUVATE DEHYDROGENASE

Purpose: The AP and IB programs and college level science courses all require students to design experiments. Working in a team, you will design an experiment to determine the effect of a variable that changes the rate of enzyme catalyzed reaction.

Overview This is a research proposal that will not be actually carried out, but your proposal should be feasible. The enzyme pyruvate decarboxylase is an important enzyme in the anaerobic pathway in yeast respiration. It catalyzes the conversion of pyruvate to acetaldehyde. (In a separate reaction the enzyme alcohol dehydrogenase converts acetaldehyde to ethanol, thus regenerating NAD+ for glycolysis).

Notes: TTP = Thiamine Triphosphate and Mg+2 are cofactors required for the reaction. Product molecules: CO2(g) is produced as a gas and Acetaldehyde is aqueous (aq) Acetaldehyde is a substrate for alcohol dehydrogenase, a reaction for which NADH is a cofactor. Design an experiment to measure the effect of a 1 variable (not pH) on the rate of conversion of pyruvate to acetaldehyde catalyzed by the enzyme pyruvate decarboxylase. This reaction takes place in an aqueous solution.

Report must be typed and should include: - Statement of question to be studied. - Brief abstract statement outlining general experimental approach. - Identify the independent and dependent variables. - A sample data table that illustrates independent variable values selected and number of trials. (Leave dependent variables values blank) - List any variables held constant during experiment. You must clearly explain how your procedure insures that variables are in fact constant during the experiment. - List any experimental controls. - Hypothesis statement (e.g. based the model of enzyme action, why would you expect the independent variable you selected might reasonably be expected to impact enzyme activity?)

- Prediction: What would you expect the general results (i.e., general trend without predicting specific values) to be?

- Materials: List all equipment and chemicals including concentrations of any solutions varied during the experiment, e.g. if concentration of substance is independent variable. - Procedure: Should sufficiently detailed that someone else carry out your procedure. (Can be numbered steps or narrative style).

You may research literature sources that provide techniques or information for measuring something but you may not merely recopy an exact experiment design from the literature that someone else has already conducted. You must cite any literature sources used.

Lab #5: Measuring the Rate of an Enzyme Catalyzed Reaction (Adapted from Advanced Placement Biology Advanced Inquiry Lab Investigation #13; Procedure adapted from Flinn Scientific Publication No. 11141).)

Purpose: To experimentally measure the rate of the enzyme catalyzed decomposition of hydrogen peroxide at room temperature using absorption spectroscopy.

Background Enzymes are biological catalysts that accelerate and regulate the rate of virtually all essential chemical reactions in the cell. Enzymes are proteins and their activity is depends on the reaction conditions and the ability of the enzyme to maintain its correctly folded conformation under the reaction conditions. Measuring enzyme activity (i.e. how quickly the enzyme catalyzes a particular reaction) is crucial step in understanding how metabolic pathways such as glycolysis or the Calvin cycle (part of photosynthesis) function. The reaction we will be studying in the reduction of hydrogen peroxide catalyzed by the enzyme turnup peroxidase. Peroxidase catalyzed reactions follow the general pattern of

H2O2 + 2 AH2 → 4 H2O + A2

In which AH2 represent organic reducing agent (electron donor).

In our experiment we will use the Guaiacol, C7H8O2, as the organic reducing agent. Guaiacol is colorless but its oxidized form, tetraguaiacol, is a dark orange color. This difference in color can be used to measure the rate of the reaction. The rate of a chemical reaction can be measured by either monitoring the appearance of a product or the disappearance of a reactant as a function of time. Measuring a reaction rate requires a method for determining the concentration of the reactant or product at different times during the course of the reaction. In our experiment we will use one of the most commonly utilized techniques for measuring concentration called absorption spectroscopy. The theoretical basis of this technique is the observation that at a particular wavelength light absorbance is

95 directly proportional to concentration. Hence by measuring the change of light absorbance we can determine the change in concentration. The equation that precisely defines the relationship between absorbance and concentration is Beer’s Law

A = ε c l Where A = absorbance of light at particular wavelength ε = extinction coefficient constant (measuring how strongly a given chemical absorbs light at a particular wavelength) c = concentration in moles/ liter l = pathlength (distance light travels through cuvette; the typical pathlength for most cuvettes is 1 cm)

Importance of an Experimental Control in Measuring the Catalyzed Rate:

Overall Measured Rate = Rate of reaction without catalyst + Rate with catalyst

Note: You cannot assume that the rate of reaction without catalyst is zero, especially under extreme reaction conditions. The substrate may be unstable.

Materials: pH 3-9 Buffer Solutions pH 7 Phosphate Buffer Guaiacol, C7H8O2, 0.2% solution in isopropyl alcohol Hydrogen Peroxide, H2O2, 0.02% solution Peroxidase enzyme extracted in phosphate buffer, 6 mL Serological pipets 1, 2 and 5 mL Pipet pump Spectrophotometer 8 Test Tubes, 13 x 100 mm Test Tube Rack Timer (can use your cell phone)

Safety:

Wear Safety Glasses Guaiacol is toxic by ingestion. The guaiacol solution is prepared in isopropyl alcohol and had an aromatic, creosote-like odor that may be irritating to the nose and throat. Isopropyl rubbing alcohol (70%) is a flammable liquid. Keep away from flames heat and other sources of ignition. Dilute hydrogen peroxide solution may be irritating to the eyes and skin. Avoid contact of all chemicals with eyes and skin and wash hands thoroughly with soap and water before leaving laboratory. If you do get any chemicals on your skin wash hands immediately. If you get chemicals in your eyes, flush eyes for 15 minutes in eye wash.

Procedure:

1) Be sure to read the entire procedure before starting. Once you have mixed the enzyme and substrate solutions you are committed and will have to start completely over if you are not immediately prepared for the subsequent steps.

2) Turn on the spectrophotometer, adjust the wavelength setting to 500 nm, and allow the instrument to warm up for 15-20 minutes.

Labeling Test Tubes and Pipettes for Experiment

Label tape at top to avoid blocking passage of light through the bottom during measurements

Handle the test tube only at the top near the tape to avoid getting your finger prints on the area of the tube where the absorbance is measured. Wipe tubes with Kim wipe to make sure tubes are clean. Glass thickness may vary on different sides of tube, thus it is important to have the test tube in the same orientation each time. Many tubes have arrows already marked so that the direction will be the same.

Be sure to label each pipet with a flag of masking tape placed just above the 0 mark on the pipet .

Attach Label tape flag here

Solution Volume in mL pH X Buffer 6.0 mL 0.2 % Guaiacol 1.0 mL

3) Zero the spectrophotometer (zero absorbance, 100% transmittance) at 500 nm using the blank solution.

4) Set up a data table for the control, trial 1 and trial 2 with two columns: Time (seconds) and Absorbance.Starting with 0 seconds, record time every 20 seconds out to 300 seconds,( e.g. 0, 20, 40, 60, etc).

MAKE SURE YOU HAVE READ THE PROCEDURE AND ARE READY BEFORE MIXING.

5) Prepare control tube and 3 sets of Substrate Tube and 3 sets of Enzyme Tube

Control Tube: (At least once; repeat 3X if possible)

Solution Volume in mL pH X Buffer 4.0 mL 0.02% Hydrogen peroxide 2.0 mL solution 0.2 % Guaiacol 1.0 mL

Reaction Tubes: (Repeat 3 Times)

Substrate Tube

SOLUTION VOLUME (mL) pH X Buffer 1.0 mL 0.02 % H202 2.0 mL 0.02 % Guaiacol 1.0 mL

Enzyme Tube

SOLUTION VOLUME (mL) pH X buffer 2.0 mL Enzyme Extract 1.0 mL

Record absorbance every 20 seconds to 300 seconds at 500 nm. Clock starts 0 secs as soon as enzyme and substrate are combined.

6) When you are ready to begin a trial for kinetics run, (1) carefully pour the contents of tube S1 and into E1 and immediately start timing. (2) Pour the combined contents back into tube S1 and immediately insert tube into spectrophometer. 7)

S1 E1 99

Tube S1 into spec 20

8) Leave the tube inside the spectrophotometer and measure and record the absorbance every 20 seconds. Since the initial rate data in which the change is approximately linear is more accurate than the longer term data, it is crucial to obtain accurate measurements as early as possible. The elapsed time between the initial tube mixing and recording the first absorbance measurement should be no greater than 20-40 seconds.

9) Estimating the Experimental Uncertainties.

pH of buffers = +/- 0.02

Vender’s uncertainty claims on Spectrophotometer: Wavelength + /- 2 nm Absorbance + / - 0.005 Drift on absorbance is likely greater than 0.005 in practice. Based on amount of drift you observe with a sample whose absorbance should be constant, estimate the uncertainty.

Time Measurement Uncertainty: Estimate the best you can.

Consider that realistically the limit on the measurement is most likely not due to your cell phone timer that measures probably to the nearest 0.1 seconds, but rather your reflex time in responding to the time and looking the measurement. Make your best estimate.

10) Rate can be obtained from plotting an absorbance (proportional to concentration) vs. time graph

Rate = slope of linear portion of curve = Δ [absorbance] [time]

Since absorbance is directly proportional to the concentration of tetraguaiacol, one of the reaction products, the rate you calculated is a measure of the rate of the reaction.

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Lab #6: Measuring the Rate of Cellular Respiration of Peas (Adapted from AP lab #5: Cell Respiration)

Purpose: In this investigation you will design an experiment to answer the following question: Is the Rate of Cellular Respiration of Dormant Peas different from the rate of germinating peas? Introduction Cellular Respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. The chemical reactions in cellular respiration are all catalyzed by enzymes. In this experiment you will work with seeds that are living but dormant. A seed contains an embryo plant and food supply surrounded by a seed coat. When necessary conditions are met, germination occurs. The fundamental question that you will ask in this experiment is: Does a pea seed that is germinating have an increased metabolic rate (i.e. more rapid cellular respiration) compared to a dormant seed? In order to answer this question you will have to plan the details of the experiment using your biology and chemistry knowledge (and with the help of hints).

Planning Goal: Write up a Team Plan: Using the Lab introduction/procedure sheet and the answers to the prelab questions below to guide your thinking, your lab team of 3-4 members should write a plan and assign tasks to each team in order to ensure that lab is completed. The written plan is for your group -you don’t have to turn in a copy to the instructor. It needs to only be as organized, long or detailed as you need to accomplish your goals. Each individual needs to complete a student report sheet. Discussion of the report questions within the group is encouraged, however your final answers in your report should be your own.

Materials List: - Germinating Peas - Dormant Peas - Plastic pea size beads - 100 mL graduated cylinder - Respirometer: consisting of 30 mL glass cylinder, 1 – holed rubber stopper, 1 mL pipet with graduation markings - A large rectangular tub that can hold water. - Food coloring - Cotton - KOH - Grease/parafilm to help seal respirometer leaks, if necessary

Supplemental Procedure Details 1) Fill a rectangular plastic tub approximately 2/3 full with tap water and allow set it aside to allow the temperature to equilibrate while you setting up the rest of the experiment.

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2) Look at the diagrams below for hints on how to build respirometers and to setup the experiment. Then follow the guided instructions prelab to design the experiment. Some additional hints about experimental details are provided below:

Experimental technique Hint #1: ➢ Assemblying respirometer. Place 1 cotton ball or a small wad of absorbent cotton in the bottom of each vial and use a small dropper to saturate the cotton with 15% KOH. Do not get KOH on the sides on the respirometer (make sure that the respirometer vials are dry on the inside). Place a small a wad of dry cotton on the top of the KOH-soaked absorbent cotton. It is important that the amounts of cotton and KOH be the same for each respirometer.

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Respirometers in the Water Bath 3) Make a sling of masking tape attached to each side of the water baths to hold the pipettes out of the water during an equilibration period of seven minutes.

4) After the equilibration of seven minutes, immerse all 3 respirometers entirely in their water baths. Water will enter the pipettes for a short distance and then stop. If the water continues to move into a pipette, check for leaks in the respirometer. Work swiftly and arrange the pipettes so that they can be read through the water at the beginning of the experiment. They should not be shifted during the experiment. Hands should be kept out of the water bath after the experiment has started. Make sure that a constant temperature is maintained.

5) Allow the respirometers to equilibrate for three more minutes than record, to the nearest 0.01 mL, the initial position of water in each pipette (time 0). Every 5 minutes for 20 minutes take readings of the water’s position in each pipette.

Experimental technique Hint #2:

Helpful suggestion for measuring volume: Add a drop of food coloring to the tip of pipette using a small micropipet; the food coloring can provide a color contrast that makes it easier to measure changes.

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Prelab Planning Questions (For discussion by team members; written answers for the team as part of the planning process is optional)

Part 1: Determining how to measure the rate: The overall equation for cellular respiration is:

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

To measure the rate of a reaction, you can either measure the rate at which reactants are consumed or the rate at which products appear. In this experiment, you will measure the rate of disappearance of oxygen, O2. Rate = - Δ[O2] Δ tjme

The plan for this experiment is to measure the change in volume of trapped air in the respirometer during the experiment. The change in volume will be the result of a change in pressure as oxygen is consumed.

1A) As oxygen is consumed, what will happen to the pressure of the trapped air inside the respirometer (increase, decrease or remain constant)?

1B) Since the pressure of air trapped inside the respirometer must equal the atmospheric pressure, what will happen to the volume of trapped air during the experiment (increase, decrease or remain constant)?

1C) Look at the phases of matter in the above equation for each reactant and each product. Which product (due to its phase of matter) will also affect the pressure of gas trapped inside the respirometer?

1D) CONTROL #1: One experimental control is to include cotton balls soaked with KOH inside the respirator. CO2 gas reacts with KOH by the following equation.

CO2(g) + KOH(s) → K2CO3(s)

Explain why KOH –soaked cotton balls act as a control:

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Part 2: Determining how to setup constant initial volume conditions 2A) The goal of the experiment is to compare the rate of respiration of dormant vs. respiring peas. If you use 25 dormant peas in one respirometer, how many respiring peas should be used in the other respirometer?

2B) The experimental approach is to measure the change in the volume of trapped air in the respirometer during the experiment. In designing the experiment, what should be true about the INITIAL (STARTING) VOLUMES of trapped air inside the respirometer for both dormant peas experiment and for the respiring peas experiment?

2C) What should be true about the # of KOH soaked cotton balls (and hence the volume occupied by cotton balls) in the respirometer with dormant peas vs. respirometer with germinating peas?

2D) Dormant peas are small and shriveled. Germinating peas are larger and typically more green in color and have a round smooth appearance. Since the # of dormant peas and germinating peas in their respective respirometers needs to be the same but the volume of each type of pea and the volume of each individual is different, you need to have a method for making the volume of trapped air the same in each container. i) How could you experimentally measure the volume your 25 peas using the experiment given in the materials list? ii) Assuming that the volume of the cotton balls is essentially equal in both respirometers, what could you add as a control material that would not respire to help make the volumes of trapped air equal? How could you determine the volume to add?

Part 3: Controlling for other factors that might impact volume of trapped air in respirometer during experiment:

3A) Since your measurement of rate depends upon measuring the volume change of trapped air, you need to also consider factors that could impact the volume besides the respiration of the peas and control for these factors. What is the ideal gas equation?

3B) If # of moles of trapped air is varying, which in turn is determining the volume of trapped air, which other two variables in the equation must be held constant?

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3C) Describe how you would assemble a third, separate respirometer with no peas but with the same initial volume of trapped air to control for factors that might change the volume of trapped air besides respiration of the peas. (Hint: what material could replace the peas in the vial?)

3D) Why is it important to insure that no outside air enters or leaves the respirometer once the experiment is underway?

Part 4: Data Collection: Build your respirometers, implement your design plan and collect data. You will need initial values for dormant peas, germinating peas and control and data collection for each category every 5 minutes up to 20 minutes total time. (See data tables on student report sheet).

Lab #7: Molecular Details of Cellular Respiration Computer Simulation

Introduction Cellular Respiration is the process by which cells oxidize energy rich molecules to obtain chemical energy stored in an easily utilized form, ATP. In this course we will focus primarily on how cells metabolize glucose and oxygen to carbon dioxide and water. Aerobic (in the presence of oxygen) cellular respiration is organized into four parts: 1) Glycolysis 2) The link or transition reaction 3) Krebs or citric acid cycle and 4) the electron transport chain/ATP synthesis. In this lab, you will work in small teams of 2-3 students exploring computer programs that cover each of the major parts of cellular respiration. You may run the programs at your own pace and you are encouraged to discuss with your group, however each student is responsible to for completing his or her own report.

Procedure: Read the directions below. Fill in your answers on the student answer sheets provided in class. .

➢ Best overview Starting Program: http://www.sumanasinc.com/webcontent/animations/content/cellularrespiration.h tml ➢ Good Overall Site for Biology Links: http://nhscience.lonestar.edu/biol/bio1int.htm

➢ At the top menu bar, select Cellular Respiration.

➢ Start by selecting and running Overall Reaction by Dr. Meyers at CUNY to get an overview of aerobic cellular respiration. http://www.qcc.cuny.edu/BiologicalSciences/Faculty/DMeyer/respiration.html

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➢ Next select Inquiry into Biology - Metabolism McGraw-Hill to obtain a menu of programs http://highered.mcgrawhill.com/sites/0070960526/student_view0/chapter5/animations.html

➢ Run the programs listed below and fill in the missing information on the student information sheet.

Program #1: How Glycolysis works: http://www.mcgrawhill.ca/school/applets/abbio/quiz/ch05/how_glycolysis_works.swf

Program #2: How the Krebs cycle works: (note: this includes the link or transition rxn) http://www.mcgrawhill.ca/school/applets/abbio/quiz/ch05/how_the_krebs_cycle_wor.swf

Program #3: Electron transport and ATP synthesis (use this program for diagrams and fill in blanks ) http://www.mcgrawhill.ca/school/applets/abbio/ch05/electrontrans_electron.swf

➢ Additional Programs for more review:

Electron transport system and the formation of ATP http://www.mcgrawhill.ca/school/applets/abbio/quiz/ch05/electron_transport_syst38.swf

Oxidative Phosphorylation – Campbell’s Interactive Resources http://www.brookscole.com/chemistry_d/templates/student_resources/shared_resources/a nimations/oxidative/oxidativephosphorylation.html

Pyruvate Dehydrogenase Complex – Campbell’s Interactive http://www.brookscole.com/chemistry_d/templates/student_resources/shared_resources/a nimations/pdc/pdc.html

Boyer’s Oxidative Phosphorylation http://www.wiley.com/legacy/college/boyer/0470003790/animations/electron_transport/e lectron_transport.htm

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Lab #8 DNA REPLICATION SIMULATION

Introduction Replication is the process of copying DNA before cell division can take place. The concept of semi-conservative replication (using parent strands as templates for the daughter strands) is straightforward but the details are complex and involve many different proteins.

Procedure: You will work in teams of 3 to 5 (depending on class size), however each student is responsible for his or her own final report. In Part 1 of this lab you will review these details by demonstrating the process to the instructor using models and then drawing pictures to represent the process. Be prepared to explain the solutions to both the lagging strand and ends replication problems. In part 2, your team will discuss and interpret the experimental work of an important replication researcher.

Part 1: Follow Directions on Student Answers sheet for modeling and drawing section

Part 2: DNA Replication Lab Interpretation Question In 1968, researchers were investigating the details of DNA replication. One of the most perplexing issues was to understand how synthesis occurred in the 3’ to 5’ direction. The only DNA polymerases (enzymes that can synthesize DNA from a DNA template) that had been discovered only added nucleotides in the 5’ to 3’ direction.

Three possibilities were considered: 1) There was another, as yet undiscovered DNA polymerase that worked only in the 3’ to 5’ direction. 2) There was another, as yet undiscovered DNA polymerase that worked in both the 5’ to 3’ and 3’ to 5’ direction. 3) DNA polymerases can only proceed in the 5’ to 3’ direction and there is a different detailed mechanism for synthesizing strands in the 3’ to 5’ direction as opposed to the 3’ to 5’ direction.

Background information for Question #1: A very talented young Japanese researcher named Reijji Okazaki, (who survived the Hiroshima atomic bombing but tragically died at age 42 from leukemia), set out to design an experiment to distinguish between the various possibilities. Okazaki’s experimental plan required a method to radioactively label specifically DNA in a cell extract that contained all the components necessary to support replication. Okazaki chose to use a radioactively labeled thymine triphosphate, TTP, a technique first developed by Arthur Kornberg. Kornberg earned a share of the 1959 Nobel Prize in Physiology or Medicine for discovery of DNA polymerase.

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1) Why do you think that Okazaki chose to use radioactively labeled thymine triphosphate, TTP instead of one of the other nucleic acid bases, ATP, CTP, GTP or UTP(uracil triphosphate) to monitor to the synthesis of DNA? (Hint: The researchers were not working with purified DNA; they were working with cell extracts which contained many different types of molecules mixed all together)

Background information for question #2: One of the great challenges in studying the details of DNA replication is the amazing speed of the reaction. At room temperature, cells incorporate new DNA nucleotides at approximately 1000 base pairs per second. Okazaki needed to slow down the process so that he could potentially trap strands in the process of being formed (intermediates) in sufficient amounts that he could easily detect them. He therefore ran the assays at very low temperatures. He utilized a sedimentation technique (DNA strands separated by density using a centrifuge) that enabled him to separate the DNA into approximate lengths. (Note: The technique of gel electrophoresis had not yet been fully developed) Okazaki carried out two types of experiments:

Experiment #1: Short Pulse – • Thymine provided is radioactive; any new strands being synthesized will incorporate radioactivity • Very short replication times; the replication reactions were only allowed to proceed for extremely short time intervals.

The DNA that was collected in these short increments revealed the following pattern:

Short fragments Long fragments

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Experiment #2: Long Chase – In the second experiment, Okazaki: • After a short initial phase of using radioactive thymine (referred to as the pulse phase), a hugh excess of nonradioactive thymine is added to mixture (called the chase phase). Any new strands synthesized after the chase will not incorporate radioactivity and therefore cannot be visualized in the experiment; any extensions of existing DNA strands will not incorporate new radioactivity • Longer replication times

The DNA that was collected in the longer chase experiments revealed the following pattern:

Long strands only

2) Interpretation of results:

2A) Based on our current model of how replication occurs, explain why are there a mixture of short and long fragments when replication is only allowed to proceed for a very short time interval. Draw a diagram of the replication fork in which the short and long fragments are labeled.

2B) In the second experiment, why are there only long strands after the “chase”? What happened to the short strands that were present initially from the short pulse experiment?

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Lab #9: TRYP and LAC OPERON SIMULATION

Introduction Operons are groups of genes that encode proteins for a metabolic pathway. The transcription of these genes share the same regulatory system, thus allowing the genes to be turned on or off as a group. Operons can be classified as inducible or repressible. For inducible operons the default condition is off and the gene can be turned on in response to a signal. In repressible operons, the default condition is off and the gene can be switched on in response to a signal. Two of the first operon systems studied were bacterial operons for controlling the uptake and catabolism (breakdown) of lactose (referred to as the lac operon) and the synthesis of tryptophan (referred to as the tryp operon).

Purpose: To reinforce our understanding of the lac and tryp operons by building and drawing symbolic models.

Procedure: In groups, build models and/or run computer simulations of the tryp and lac operons. Complete the drawing for the simulations described below and answer the questions on provided student answer sheets. You are encouraged to discuss the lab within your group, but each individual is responsible for his or her own answer sheet. You will then take an individual quiz.

Part 1: TRYP OPERON SIMULATION LAB Draw a diagram of the tryp operon using the symbols provided below. For each the listed conditions, indicate whether gene activity would be high or low, and briefly explain why:

Repressor Protein Repressor protein (Tryp not bound) Tryp bound

tryptophan RNA Polymerase

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Note that the shape of the DNA binding site is complementary to the active form of the repressor protein.

Part 2: LAC OPERON SIMULATION DRAWINGS

For each set of conditions given, fill in on the diagram using the following symbols, indicate whether gene activity would be high or low, and briefly explain why:

Repressor protein (active) Repressor protein inactive

allolactose

RNA POLYMERASE

cAMP CRP INACTIVE

CRP ACTIVE

REPRESENTATION OF LAC OPERON DNA

PROMOTER OPERATOR LAC Z LAC Y LAC A

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CRP DNA Binding Site REPRESSOR DNA BINDING SITE

Extension: Genetic Studies of the Lac Operon – (Discuss with your lab team, final answers individual, not a part of operon lab quiz)

The first model of gene regulation in bacterial systems was the lac operon system in E. Coli. The development of this model and the experiments that demonstrated the soundness of the concept won the 1965 Nobel Prize for Physiology or Medicine for its discoverers the French biochemists, Francois Jacob, Jacques Monot and Andre Lwoff. Some of the most critical experiments involved the use of genetics, specifically the use of mutants to test key ideas. One such experiment involved the use of constitutive mutants, mutants in which the lac genes are always expressed instead of being expressed only under specific conditions. The level of expression of the lac operon genes was determined by measuring the enzyme activity of beta- galactosidase. ( Recall that beta-galactosidase key enzyme in the metabolism of lactose – it breaks down lactose into monosaccharide products).

Table 1: Concept of Constitutive mutants.

Lactose present? Beta-galactosidase Activity

Wild Type (Normal) Lactose Present 50 units Lactose Absent 2 units

Constitutive Mutant Lactose Present 50 units Lactose Absent 50 units

Note: glucose is absent in the growth media for all experiments.

The wild type and constitutive mutants in the above experiment were typical haploid E. Coli, meaning that they had only one chromosome with one copy of the lac operon genes.

Figure 1: Typical Haploid E. Coli

Lac operon

E. Coli Cell

Chromosome

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Monod + Jacob isolated haploid mutants and induced the mutants to take in a second segment of DNA called a plasmid that contained a second copy of a wild type lac operon. The resulting mutants had been changed into diploid cells, meaning that they contained two copies of the lac operon, one mutated and one normal.

Figure 2: Diploid Bacteria with 1 mutant and 1 normal copy of lac operon.

Chromosome with mutant copy of lac operon

E. Coli Cell

Inserted plasmid containing normal copy of lac operon

Chromosome

The key question the researchers asked was does the addition of a second, normal copy of the lac operon cause the E. coli cell to now behave as a normal, wild type cell?

The answer turned out to be sometimes. In other words, the researchers identified two different types of mutants that we will call Class I and Class II mutants. Class I mutants were converted back to wild type when a normal copy of the lac operon genes were added; Class II mutants remained constitutive mutants even when a normal a normal copy of the lac genes were added.

Table 2: Lactose Present? Beta-galactosidase Activity

Class I mutants Lactose present 100 units Lactose absent 2 units

Class II mutants Lactose present 100 units Lactose absent 50 units

Jacob and Monot were delighted by these results because they believed the data supported their lac operon model. Answer questions about these expts on your student answer sheet.

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Lab #10: EUKARYOTIC GENE EXPRESSION SIMULATION

Introduction Expression of a eukaryotic gene is a complex process involving transcription, mRNA processing and translation. In this lab you will simulate the various parts of the process using models.

Purpose: To model the process of creating a short amino acid chain in a eukaryote by simulating transcription, mRNA processing and translation.

Materials:

Part 1: Modeling the Transcription Initiation Complex

PART PLAYED BY DNA WEBBING GENE, PROMOTER, ENHANCER MAGNET STRIP AND COLORED LETTERS TRANSCRIPTION FACTORS RECTANGULAR MAGNETS labeled TF ACTIVATOR PROTEINS Circular MAGNETS labeled ACT RNA POLYMERASE Large magnetic clips

Also in Part 1, Poster showing how signals such as hormones can impact trancription by binding to transcription factors and activator proteins.

Part 2: Production and Processing of mRNA RNA: blue web with plastic slots for mRNA RNA bases: small cardboard letters Intron (webbing with velco)

Part 3: Modeling Translation

PART PLAYED BY Small ribosomal subunit Poster board cutout Large ribosomal subunit Poster board cutout mRNA Blue webbing from Part 2 t-RNA with amino acid Poster board codon linked to key holder labels Amino acids Key holder labels Anticodon bases Cardboard cutouts

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Procedure: You will work in groups (group sizes will vary based on class sizes) for the hands-on modeling of each step of gene expression: Transcription, RNA processing and Translation. You will need to be able to demonstrate each step to the instructor. You will also need to complete the questions and diagrams on the student answer sheet. While you are encouraged to discuss within your group, the question answers and drawings must be your own work. Instructions Part 1: Using the materials provided, model the steps needed to build the transcription initiation complex.

Part 2: Using the materials provided, model the transcription and processing steps necessary to produce a mature mRNA.

Part 3: Using the materials provided, model the steps for translation of the message.

PRELAB QUESTIONS:

Helpful diagrams from Campbell’s to review before class: p.365, Fig. 19.5 A eukaryotic gene and its transcript p.366, Fig. 19.6 A model for the action of enhancers and transcription activators p. 319, Fig. 17.11 The roles of snRNP’s and spliceosomes in pre-mRNA splicing. p. 321, Fig. 17.14 The structure of transfer RNA p. 323, Fig. 17.17 The initiation of translation p. 324, Fig. 17.18 The elongation cycle of translation

Complete the Prelab Questions on your student answer sheet BEFORE class.

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LAB #11 : pGLOTM Transformation Experiment

(Adapted from Biorad Experiment #166-0003EDU)

Introduction In this experiment you will perform a procedure known as transformation on E. coli cells. Transformation is the process of inducing a bacterial cell to take in a small circular segment of DNA called a plasmid. The addition of the plasmid changes or transforms the genetic makeup of the cell. In addition to one large chromosome, bacteria naturally contain one or more plasmids. Plasmids generally contain “extra” genes such as resistance to an antibiotic that may give the cell a survival advantage under specific environmental conditions.

We have purchased a plasmid from Bio-Rad Corporation that contains genes coding for resistance to the antibiotic ampicillin and for the production of the Green Fluorescent Protein that we isolated earlier in the semester. [Recall that Green Fluorescent Protein (GFP) fluoresces green when exposed to black light (long wave UV)]. In order to induce the E. coli cells to take in our plasmid, you will first mix the cells with plasmid together at ice cold temperature in a calcium chloride solution. The Ca+2 ions neutralize the negative charges on the phosphate groups DNA backbone and the surface of the cell membrane. Cooling the solution reduces thermal motion and allows for the Ca+2 ions to strongly attract the negative charges on the DNA backbone and cell membrane. Next you will heat shock the cells for 50 seconds. The temperature difference between the inside and outside of the cells creates a gradient that pulls the plasmid through the pores in the membrane into the interior of the cell. Finally you provide the cells with nutrients and temperature that will ensure rapid growth and expression of the newly acquired genes. The next class period you will check your plates for growth and expression of the pGLO gene.

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Materials per lab group - E. coli starter plate - 4 poured agar plates (1 LB only ; 2 LB/amp, 1 LB/amp/ara) (abbreviations used: LB = Luria Nutrient Broth – contains nutrients needed for bacterial growth; amp = ampicillin- an antibiotic; ara = arabinose- a sugar).

- Foam microtube holder/float with 4 microtest tubes labeled: Tube 1 = + pGLO Tube 2 = -pGLO Tube 3 = CaCl2 (transformation solution) Tube 4 = LB (Luria Broth)

- 5 sterile pipets - 1 package of sterile inoculation loops

Procedure – see next two pages for detailed instructions.

Lab Safety: - Wear safety glasses. - The E. coli cell line used in this experiment is believed to be completely harmless however taking basic precautions is always recommended. - Avoid contact with any E. coli or surfaces that have contacted E.coli. If you accidentally touch E. coli, immediately wash your hands thoroughly with soap and water. Wash hands with soap and water at the end of the lab.

IMPORTANT DISPOSAL PROCEDURES: Any item that has come in contact with E. coli during the experiment should be placed in the bleach bath in the fume hood. This includes:

- all agar plates - All inoculation loops - pipets used to transfer bacterial solutions

Data Collection

Observe the results you obtained from the transformation lab under normal room lighting. Then turn out the lights and hold the ultraviolet light over the plates.

1) Carefully observe and draw what you see on each of the four plates. Put your drawings in the data in the column to the right. Record your data to allow you to compare observations of the “+ p GLO” cells with your observations for the non-transformed E. coli. Write down the following observations for each plate.

2) For each plate comment on the following: ii) What color are the bacteria? iii) How many bacterial colonies are on each plate (count the spots you see).

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Lab #12 Bacterial Identification Virtual Lab

Introduction In this lab you will see how many of the different biotechnology techniques that we have been discussing in class can be combined to solve a real-life medical problem. You will run a simulation program developed by the National Institutes of Health (NIH) to identify a bacteria species from its DNA sequence. In the simulation, a patient in the hospital has a suspected bacterial infection. For some infections it is important to know the identity of the bacteria in order to select an antibiotic that will be most effective against that particular strain. A blood or tissue sample believed to contain the bacteria of interest is sent to the lab for analysis. You (the lab technician) will identify the bacteria by determining the DNA sequence for an rRNA gene and comparing the results of your sequence analysis to a computerized data base. The 16S rRNA gene was selected because it an essential gene present in all bacteria and the sequence is slightly different in different bacterial species. The entire sequence of the human genome is known for a handful of individuals as well as the sequence of many different commonly used species of lab organisms (bacteria, yeast, nematodes, mice, etc). The ability to search computer data-bases for sequence matches is considered a basic skill for molecular biologists. The search engine you will use in this simulation is one of the commonly utilized programs and is referred to as BLAST.

Purpose: To simulate the application of biotechnology techniques to identifying a bacteria species based on DNA sequence.

Procedure: Run the program on the disk provided in class and answer the questions below and ON YOUR OWN PAPER.

Makeup Lab: This lab is own the internet at the Howard Hughes Medical Center site: Go To Biology I Animations Homepage, Under Over All Collections select Howard Hughes Medical Institute, then under the Virtual Labs menu, Bacterial ID. http://www.hhmi.org/biointeractive/vlabs/bacterial_id/index.html

Hints for using BLAST: Highlight the sequence in the virtual lab and copy it. (Either use edit menu or Control + C command). Open a second window to the BLAST web page. Paste the sequence in the BLAST box.(Either use edit menu or Control + V command). Don’t change anything else – the data is already in the correct format. Submit your data. Ignore the next screen about access numbers. On the next screen, scroll down the page to see the similarity scores.

Questions (on your own paper, don’t have to recopy questions):

1) Construct a flowchart that lists the different procedures that will be carried out in the experiment in order.

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2) PCR (Polymerase Chain Reaction) A) What is PCR used for? B) Explain how PCR works.

3) DNA Sequencing A) Explain how the Sanger method of DNA sequencing works. B) Briefly explain how the process can be automated.

4) In part 6 of the lab (sequence analysis), refer to the sections on “The science behind sequence mapping” and “Interpreting BLAST results” to answer the following questions: A) What reasons might a scientist have for doing a search of a sequence database? (What might he or she hope to learn?)

B) Do high or low similarity scores indicate a good match between sequences?

5) What is your conclusion about the identity of the sample?

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