GENETICS

Chapter 15

Memorise Understand Importance

* Define: phenotype, genotype, gene, locus, * Importance of meiosis; compare/contrast with High level: 10% of GAMSAT Biology allele (single and multiple) mitosis * Homo/heterozygosity, wild type, * Segregation of genes, assortment, linkage, questions released by ACER are related to recessiveness, complete/co-dominance recombination content in this chapter (in our estimation). * Incomplete dominance, gene pool * Single/double crossovers * Note that approximately 75% of the * Sex-linked characteristics, sex determination * Hardy-Weinberg Principle questions in GAMSAT Biology are related * Test cross: back cross, concepts of parental, to just 7 chapters: 1, 2, 3, 4, 7, 12, and 15. F1 and F2 generations; evaluating a pedigree

Introduction

Genetics is the study of heredity and variation in organisms. The observations of Gregor Mendel in the mid- nineteenth century gave birth to the science which would reveal the physical basis for his conclusions – DNA – about 100 years later.

This is the last of the 3 ‘high-level importance’ chapters. Approximately 45% of GAMSAT Biology questions among ACER’s official practice materials contain content related to 3 chapters: 1, 7 and 15, this chapter. Also, be forewarned: despite the fact that a dihybrid cross (BIO 15.3) is time consuming, it appears in 3 separate ACER GAMSAT practice booklets.

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THE BIOLOGICAL SCIENCES BIO-267 15.1 Background Information

Genetics is a branch of biology which environment. Consider a heterozygote that deals with the principles and mechanics of expressed one gene (dark hair) but not the heredity; in other words, the means by which other (light hair). The expressed gene would traits are passed from parents to offspring. be called dominant while the other unex- To begin, we will first examine some relevant pressed allele would be called recessive. The definitions - a few of which we have already individual would have dark hair as their phe- discussed. notype, yet their genotype would be hetero- zygous for that trait. The dominant allele is Chromosomes are a complex of DNA expressed in the phenotype. This is known as and proteins (incl. histones; BIO 1.2.2). A Mendel’s Law of Dominance.

High-level ImportanceHigh-level gene is that sequence of DNA that codes for a protein or polypeptide. A locus is the posi- It is common to symbolise dominant tion of the gene on the DNA molecule. Recall genes with capital letters (A) and recessive that humans inherit 46 chromosomes - 23 genes with small letters (a). From the pre- from maternal origin and 23 from paternal ceding paragraphs, we can conclude that origin (BIO 14.2). A given chromosome from with two alleles, three genotypes are pos- maternal origin has a counterpart from pater- sible: homozygous dominant (AA), heterozy- nal origin which codes for the same products. gous (Aa), and homozygous recessive (aa). This is called a homologous pair of chromo- Note that this only results in two phenotypes somes. since both AA and Aa express the dominant gene, while only aa expresses the recessive Any homologous pair of chromosomes gene. have a pair of genes which codes for the same product (i.e. hair colour). Such pairs of Each individual carries two alleles genes are called alleles. Thus for one gene while populations may have many or mul- product, a nucleus contains one allele from tiple alleles. Sometimes these genes are maternal origin and one allele from paternal not strictly dominant or recessive. There origin. If both alleles are identical (i.e. they may be degrees of blending (= incomplete code for the same hair colour), then the indi- dominance) or sometimes two alleles may vidual is called homozygous for that trait. If be equally dominant (= codominance). ABO the two alleles differ (i.e. one codes for dark blood types are an important example of mul- hair while the other codes for light hair), then tiple alleles with codominance. the individual is called heterozygous for that trait. Incomplete dominance occurs when the phenotype of the heterozygote is an inter- The set of genes possessed by a particu- mediate of the phenotypes of the homozy- lar organism is its genotype. The appearance gotes. A classic example is flower colour in or phenotype of an individual is expressed snapdragon: the snapdragon flower colour is as a consequence of the genotype and the red for homozygous dominant and white for

BIO-268 Chapter 15: GENETICS GAMSAT MASTERS SERIES homozygous recessive. When the red homo- snapdragon flower. The pink snapdragon is zygous flower is crossed with the white homo- the result of the combined effect of both domi- zygous flower, the result yields a 100% pink nant and recessive genes. High-level Importance

Figure IV.A.15.1: Genotype vs. phenotype; dominant allele W, recessive allele w, in a fly. (Courtesy: National Human Genome Research Institute)

15.2 ABO Blood Types

Codominance occurs when multiple is A; if they have only antigen B, then the blood alleles exist for a particular gene and more type is B; if they have both antigens, the blood than one is dominant. When a dominant allele type is AB; if neither antigen is present, the is combined with a recessive allele, the phe- blood type is O. There are three allelic genes notype of the recessive allele is completely in the population (IA, IB, iO). Two are codominant masked. But when two dominant alleles are (IA, IB) and one is recessive (iO). Thus in a given present, the contributions of both alleles do population, there are six possible genotypes not overpower each other and the phenotype which result in four possible phenotypes: is the result of the expression of both alleles. A classic example of codominance is the ABO blood type in humans. Genotype Phenotype IAIA, IAiO blood type A Red blood cells can have various antigens B B B O I I , I i blood type B (BIO 8.1) or agglutinogens on their plasma A B membranes which aid in blood typing. The I I blood type AB O O important two are antigens A and B. If the red i i blood type O blood cells have only antigen A, the blood type

THE BIOLOGICAL SCIENCES BIO-269 High-level ImportanceHigh-level

Figure IV.A.15.2: Diagram of the ABO blood groups, the antigens and the antibodies (IgM; BIO 8.2) present in each.

Blood typing is critical before doing a The only other antigens which have blood transfusion. This is because people some importance are the Rh factors which are with blood type A have anti-B antibodies, coded by different genes at different loci from those with type B have anti-A, those with type the A and B antigens. Rh factors are either AB have neither antibody, while type O has there (Rh+) or they are not there (Rh-). 85% of both anti-A and anti-B antibodies. If a person the population are Rh+. The problem occurs with type O blood is given types A, B, or AB, when a woman is Rh- and has been exposed the clumping of the red blood cells will occur to Rh+ blood and then forms anti-Rh+ antibod- (= agglutination). Though type O can only ies (note: unlike the previous case, exposure receive from type O, it can give to the other is necessary to produce these antibodies). If blood types since its red blood cells have no this woman is pregnant with an Rh+ fetus her antigens {type O = universal donor}. Type AB antibodies may cross the placenta and cause has neither antibody to react against A or B the fetus’ red blood cells to agglutinate (eryth- antigens so it can receive blood from all blood roblastosis fetalis). This condition is fatal if left types {type AB = universal recipient}. untreated.

BIO-270 Chapter 15: GENETICS GAMSAT MASTERS SERIES

15.3 Mendelian Genetics High-level Importance Recall that in gametogenesis homolo- predict that the results of a genetic cross gous chromosomes separate during the first should be the same regardless of which par- meiotic division. Thus alleles that code for the ent introduces the allele. However, it can be same trait are segregated: this is Mendel's shown that some traits follow the inheritance First Law of Segregation. Mendel's Sec- of the sex chromosomes. Humans have one ond Law of Independent Assortment states pair of sex chromosomes (XX = female, XY = that different chromosomes (or factors which male), and the remaining 22 pairs of homolo- carry different traits) separate independently gous chromosomes are called autosomes. of each other. For example, consider a pri- mary spermatocyte (2N) undergoing its first Since females have two X chromosomes meiotic division. It is not the case that all 23 and males have only one, a single recessive chromosomes of paternal origin will end up in allele carried on an X chromosome could be one secondary spermatocyte while the other expressed in a male since there is no sec- 23 chromosomes of maternal origin ends up ond allele present to mask it. When males in the other. Rather, each chromosome in a inherit one copy of the recessive allele from homologous pair separates independently of an X chromosome, they will express the trait. any other chromosome in other homologous In contrast, females must inherit two copies pairs. to express the trait. Therefore, an X-linked recessive phenotype is much more frequently However, it has been noted experi- found in males than females. In fact, a typi- mentally that sometimes traits on the same cal pattern of sex linkage is when a mother chromosome assort independently! This passes her phenotype to all her sons but none non-Mendelian concept is a result of cross- of her daughters. Her daughters become car- ing over (recall that this is when homologous riers for the recessive allele. Certain forms of chromosomes exchange parts, BIO 14.2). In hemophilia, colourblindness, and one kind of fact, it has been shown that two traits located muscular dystrophy are well-known recessive far apart on a chromosome are more likely to sex-linked traits. {In what was once known as cross over and thus assort independently, as Lyon’s Hypothesis, it has been shown that compared to two traits that are close. The pro- every female has a condensed, inactivated pensity for some traits to refrain from assort- X chromosome in her body or somatic cells ing independently is called linkage. Double called a Barr body.} crossovers occur when two crossovers hap- pen in a chromosomal region being studied. Let us examine the predictions of Men- del’s First Law. Consider two parents, one Another exception to Mendel's laws homozygous dominant (AA) and the other involves sex linkage. Mendel's laws would homozygous recessive (aa). Each parent can

THE BIOLOGICAL SCIENCES BIO-271 only form one type of gamete with respect to be aa. In other words the genotypic ratio of that trait (either A or a, respectively). The next homozygous dominant to heterozygous to generation (called first filial or F1) must then homozygous recessive is 1:2:1. However, be uniformly heterozygotes or hybrids (Aa). since AA and Aa demonstrate the same phe- Now the F1 hybrids can produce gametes notype (i.e. dominant) the ratio of dominant that can be either A half the time or a half the phenotype to recessive phenotype is 3:1. time. When the F1 generation is self-crossed, i.e. Aa X Aa, the F2 generation will be more Now we will consider the predictions genotypically and phenotypically diverse and of Mendel’s Second Law. To examine inde- we can predict the outcome in the next gen- pendent assortment, we will have to con-

High-level ImportanceHigh-level eration (F2) using a Punnett square: sider a case with two traits (usu. on different chromosomes) or a dihybrid cross. Imagine a parent which is homozygous dominant for two traits (AABB) while the other is homo- zygous recessive (aabb). Each parent can only form one type of gamete with respect to those traits (either AB or ab, respectively). The F1 generation will be uniform for the dominant trait (i.e. the genotypes would all Here is an example as to how you derive be AaBb). In the gametes of the F1 genera- the information within the square: when you tion, the alleles will assort independently. cross A with A you get AA (i.e. 1/2 A × 1/2 A Consequently, an equal amount of all the = 1/4 AA). Thus by doing a simple monohy- possible gametes will form: 1/4 AB, 1/4 Ab, brid cross (Aa × Aa) with random mating, the 1/4 aB, and 1/4 ab. When the F1 genera- Punnett square indicates that in the F2 gen- tion is self-crossed, i.e. AaBb X AaBb, we eration, 1/4 of the population would be AA, can predict the outcome in the F2 generation 1/2 would be Aa (1/4 + 1/4), and 1/4 would using the Punnett square:

1/4 AB 1/4 Ab 1/4 aB 1/4 ab 1/4 AB 1/16 AABB 1/16 AABb 1/16 AaBB 1/16 AaBb 1/4 Ab 1/16 AABb 1/16 AAbb 1/16 AaBb 1/16 Aabb 1/4 aB 1/16 AaBB 1/16 AaBb 1/16 aaBB 1/16 aaBb 1/4 ab 1/16 AaBb 1/16 Aabb 1/16 aaBb 1/16 aabb

BIO-272 Chapter 15: GENETICS GAMSAT MASTERS SERIES

Thus by doing a dihybrid cross with ran- AABb + AaBb + AaBB = 9), the expression dom mating, the Punnett square indicates of only the first trait (AAbb + Aabb = 3), the

that there are nine possible genotypes (the expression of only the second trait (aaBB + High-level Importance frequency is given in brackets): AABB (1), aaBb = 3), and the expression of neither trait AABb (2), AaBb (4), AaBB (2), Aabb (2), aaBb (aabb = 1). Now we know, for example, that (2), AAbb (1), aaBB (1), and aabb (1). Since 9/16 represents that fraction of the popula- A and B are dominant, there are only four tion which will have the phenotype of both phenotypic classes in the ratio 9:3:3:1 which dominant traits. are: the expression of both traits (AABB +

15.3.1 A Word about Probability

If you were to flip a quarter, the prob- kind of sperm, occur independently, the gen- ability of getting "heads" is 50% (p = 0.5). If otype of one child has no effect on the geno- you flipped the quarter ten times and each types of other children produced by a set of time it came up heads, the probability of get- parents. Thus in the previous example of the ting heads on the next trial is still 50%. After dihybrid cross, the chance of producing the all, previous trials have no effect on the next genotype AaBb is 4/16 (25%) irrespective trial. of the genotypes which have already been produced. For more about probability, see Since chance events, such as fertilisa- GM 6.1. tion of a particular kind of egg by a particular

15.4 The Hardy-Weinberg Law

The Hardy-Weinberg Law deals with evolution is the changing of alleles in a gene population genetics. A population includes pool from one generation to the next. all the members of a species which occupy a more or less well defined geographical area Evolution can be viewed as a changing and have demonstrated the ability to repro- of gene frequencies within a population over duce from generation to generation. A gene successive generations. The Hardy-Wein- pool is the sum of all the unique alleles for berg Law or equilibrium predicts the outcome a given population. A central component to of a randomly mating population of sexually

THE BIOLOGICAL SCIENCES BIO-273 reproducing diploid organisms who are not quencies are given. This can be summarised undergoing evolution. by the following:

For the Hardy-Weinberg Law to be applied, the idealised population must meet pA qa the following conditions: i) random mating: pA p2AA pqAa the members of the population must have no 2 mating preferences; ii) no mutations: there qa pqAa q aa must be no errors in replication nor similar event resulting in a change in the genome;

High-level ImportanceHigh-level iii) isolation: there must be no exchange of The Punnett square illustrates the genes between the population being con- expected frequencies of the three genotypes sidered and any other population; iv) large in the next generation: AA = p2, Aa = 2pq, and population: since the law is based on statisti- aa= q2. cal probabilities, to avoid sampling errors, the population cannot be small; v) no selection For example, let us calculate the per- pressures: there must be no reproductive centage of heterozygous individuals in a advantage of one allele over the other. population where the recessive allele q has a frequency of 0.2. Since p + q = 1, then p = To illustrate a use of the law, consider 0.8. Using the Hardy-Weinberg equation and an idealised population that abides by the squaring p and q we get: preceding conditions and has a gene locus occupied by either A or a. Let p = the fre- 0.64 + 2pq + 0.04 = 1 quency of allele A in the population and let q = the frequency of allele a. Since they are the 2pq = 1 - 0.68 = 0.32 only alleles, p + q = 1. Squaring both sides we get: Thus the percentage of heterozygous (2pq) individuals is 32%. (p + q)2 = (1)2 OR A practical application of the Hardy- Weinberg equation is the prediction of how p2 + 2pq + q2 = 1 many people in a generation are carriers for a particular recessive allele. The values would have to be recalculated for every gen- The preceding equation (= the Hardy- eration since humans do not abide by all the Weinberg equation) can be used to calculate conditions of the Hardy-Weinberg Law (i.e. genotype frequencies once the allelic fre- humans continually evolve).

BIO-274 Chapter 15: GENETICS GAMSAT MASTERS SERIES

15.4.1 Back Cross, Test Cross High-level Importance A back cross is the cross of an individual for that trait so the unknown genotype can (F1) with one of its parents (P) or an organism be determined from that of the offspring. For with the same genotype as a parent. Back example, for P: AA x aa and F1: Aa, we get: crosses can be used to help identify the genotypes of the individual in a specific Backcross #1: Aa x AA type of back cross called a test cross. A test Progeny #1: 1/2 Aa and 1/2 AA cross is a cross between an organism whose genotype for a certain trait is unknown and Backcross #2: Aa x aa an organism that is homozygous recessive Progeny #2: 1/2 Aa and 1/2 aa

15.5 Genetic Variability

Meiosis and mutations are sources of tive (i.e. cancer) than positive for an organ- genetic variability. During meiosis I, crossing ism’s survival. Nonetheless, such a change over occurs between the paternal and mater- in the genome increases genetic variability. nal genes which leads to a recombination of Only mutations of gametes, and not somatic parental genes yielding unique haploid gam- cells, are passed on to offspring. etes. Thus recombination can result in alleles of linked traits separating into different gam- The following are some forms of muta- etes. However, the closer two traits are on a tions: chromosome, the more likely they will be linked • Point mutation is a change affecting a and thus remain together, and vice versa. single in a gene

Further recombination occurs during the • Deletion is the removal of a sequence random fusion of gametes during fertilisation. of DNA, the regions on either side being Consequently, taking Mendel’s two laws and joined together recombination together, we can predict that • Inversion is the reversal of a segment parents can give their offspring combinations of DNA of alleles which the parents never had. This • Translocation is when one chromo- leads to genetic variability. some breaks and attaches to another

Mutations are rare, inheritable, random • Duplication is when a sequence of changes in the genetic material (DNA) of a DNA is repeated. cell. Mutations are much more likely to be • Frame shift mutations occur when either neutral (esp. silent mutations) or nega- bases are added or deleted in numbers other

THE BIOLOGICAL SCIENCES BIO-275 than multiples of three. Such deletions or are not. The Ames test is a widely used test additions cause the rest of the sequence to to screen chemicals used in foods or medica- be shifted such that each triplet reading frame tions for mutagenic potential. is altered. Mutations can produce many types A mutagen is any substance or agent of genetic diseases including inborn errors that can cause a mutation. A mutagen is not of metabolism. These disorders in normal the same as a carcinogen. Carcinogens are metabolism are usually due to defects of a agents that cause cancer. While many muta- single gene that codes for one enzyme. gens are carcinogens as well, many others High-level ImportanceHigh-level

15.6 Genetics and Heredity: A Closer Look

The rest of this chapter begins to push of seeds also influences the expression of into more advanced topics in genetics. How- starch metabolism. For example, in wrinkled ever, these topics continue to represent legiti- seeds there is more unconverted glucose mate exam material. which leads to an increase of the osmotic gradient. These seeds will subsequently con- Epistasis occurs when one gene masks tain more water than round seeds. When they the phenotype of a second gene. This is often mature they will dehydrate and produce the the case in pigmentation where one gene turns wrinkled appearance. on (or off) the production of pigment, while a second gene controls the amount of pigment Polygenic inheritance refers to traits that produced. Such is the case in mice fur where cannot be expressed in just a few types but one gene codes for the presence or absence rather as a range of varieties. The most pop- of pigmentation and the other codes for the ular example would be human height which colour. If C and c represent the alleles for the ranges from very short to very tall. This phe- presence or absence of colour and B and b nomenon (many genes shaping one pheno- represent black and brown then a phenotype type) is the opposite of pleiotropy. of CCbb and Ccbb would both correspond to a brown phenotype. Whenever cc is inherited Penetrance refers to the proportion of the fur will be white. individuals carrying a particular variant of a gene (allele or genotype) that also express Pleiotropy occurs when a single gene the associated phenotype. Alleles which are has more than one phenotypic expression. highly penetrant are more likely to be noticed. This is often seen in pea plants where the Penetrance only considers whether individu- gene that expresses round or wrinkled texture als express the trait or not. Expressivity refers

BIO-276 Chapter 15: GENETICS GAMSAT MASTERS SERIES to the variation in the degree of expression of can be ascertained. For example, say you a given trait. have a fly with genotype BBTTYY and the

crossover frequency between B and T is High-level Importance Nondisjunction occurs when the chro- 26%, between Y and T is 18% and between mosomes do not separate properly and do B and Y is 8%. Greater recombination fre- not migrate to opposite poles as in normal quencies mean greater distances so you anaphase of meiosis (BIO 14.2). This could know that B and T are the furthest apart. arise from a failure of homologous chro- This corresponds to a gene order of B-Y-T mosomes to separate in meiosis I, or the and since frequencies are a direct measure failure of sister chromatids to separate dur- of distance you know exactly how far apart ing meiosis II or mitosis. Most of the time, each allele is and can easily calculate the gametes produced after nondisjunction are map distances. sterile; however, certain imbalances can be fertile and lead to genetic defects. Down Twin studies (nature vs. nurture) help Syndrome (Trisomy 21 = 3 copies of chro- to gauge the relative importance of environ- mosome 21 due to its nondisjunction, thus mental and genetic influences on individuals the person has an extra chromosome mak- in a sample. Twins can either be monozy- ing a total of 47 chromosomes); Turner and gotic (“identical”), meaning that they develop Klinefelter Syndrome (nondisjunction of sex from one zygote (BIO 14.5) that splits and chromosomes); and Cri du Chat (deletion forms two embryos, or dizygotic (“fraternal”), in chromosome 5) are well known genetic meaning that they develop from two separate disorders. Hemophilia and red-green colour eggs, each fertilised by separate sperm cells. blindness are common sex-linked disorders Thus fraternal twins are like any 2 siblings and are recessive. from a genetic point of view, but they may share the same environment as they grow up Phenylketonuria, sickle-cell anemia and together. Tay-Sachs disease are common autosomal recessive disorders. To control for environment, the classical twin study design compares the similarity of Gene linkage refers to genes that monozygotic and dizygotic twins. If identical reside on the same chromosome and are twins are considerably more similar than fra- unable to display independent assortment ternal twins (which is found for most traits), because they are physically connected this implies that genes play an important (BIO 15.3). The further away the two genes role for those specific traits. By comparing are on the chromosome the higher prob- hundreds of families of twins, researchers ability there is that they will crossover dur- can then understand more about the roles ing synapsis. In these cases recombination of genetic effects, shared environment, and frequencies are used to provide a linkage unique environment in shaping behavior or in map where the arrangement of the genes the development of disease.

THE BIOLOGICAL SCIENCES BIO-277 15.6.1 Mitochondrial DNA

Mitochondrial DNA (mtDNA or mDNA) mosome. In most species, including humans, has become increasingly popular as a tool to mtDNA is inherited solely from the mother. determine how closely populations are related The DNA sequence of mtDNA has been as well as to clarify the evolutionary relation- determined from a large number of organisms ships among species (= phylogenetics). Mito- and individuals (including some organisms chondrial DNA is circular (BIO 1.2.2, 16.6.3) that are extinct). and can be regarded as the smallest chro- High-level ImportanceHigh-level

Figure IV.A.15.3: Pedigree ("Family tree"): Maternal inheritance pattern of mtDNA for 3 generations with the grandparents in the top row. As is the standard, circles represent females, squares represent males. Colours show the inheritance of the same mt genome (from mother to offspring, i.e. children). Don’t be surprised to see a pedigree of some sort in your practice questions and/or on the real exam.

15.7 DNA Recombination and Genetic Technology

DNA recombination involves DNA that nition sites correspond to different contains segments or genes from different sequences and produce sticky and blunt ends sources. The foreign DNA can come from when a double stranded DNA segment is cut. another DNA molecule, a chromosome or from a complete organism. Most DNA transferred The sticky end is the unpaired part of is done artificially using DNA recombination the DNA that is ready to bind with a comple- techniques which use restriction enzymes mentary codon (sequence of three adjacent to cut pieces of DNA. These enzymes origi- ; BIO 3.0). These cut pieces or nate from bacteria and are extremely specific restriction fragments are often inserted into because they only cut DNA at specific recogni- plasmids (circular piece of DNA that is able to tion sequences along the strand. These recog- replicate independently of the chromosomal

BIO-278 Chapter 15: GENETICS GAMSAT MASTERS SERIES

Bacterium and Vector Plasmid

Bacterial DNA Plasmids High-level Importance

DNA) which are then able to be introduced into through a gel which is under the influence the bacteria via transformation (see BIO 2.2). of an electric field. Since DNA is negatively charged it will move towards the cathode Treating the plasmid, or replicon, with (positive electrode). The shorter fragments the same restriction enzymes used on the move faster than the longer ones and can be original fragment produces the same sticky visualised as a banding pattern using autora- ends in both pieces allowing base pairing to diography techniques. occur when they are mixed together. This attachment is stabilised by DNA ligase. After SDS-PAGE, sodium dodecyl sulfate the ends are joined and the recombinant polyacrylamide gel electrophoresis (ORG plasmid is incorporated into bacteria, the 13), also separates proteins according to their bacteria become capable of producing copi- electrophoretic mobility. SDS is an anionic ous amounts of a specific protein that was detergent (i.e. negatively charged) which has not native to its species (i.e. bacteria with the following effect: (1) linearise proteins and recombinant DNA producing insulin to treat (2) give an additional negative charge to the diabetes). linearised proteins. In most proteins, the bind- ing of SDS to the polypeptide chain gives an Gel electrophoresis is a method of even distribution of charge per unit mass, thus separating restriction fragments of differing fractionation will approximate size during elec- lengths based on their size (as described in trophoresis (i.e. not dependent on charge). the previous section, a restriction fragment is a fragment of DNA cleaved by a restriction Restriction fragment length polymor- enzyme). The DNA fragments are passed phisms or RFLP is a technique that exploits

THE BIOLOGICAL SCIENCES BIO-279 variations in restriction fragments from one Sometimes it is necessary to obtain individual to another that differ in length due to the DNA fragment bearing the required gene polymorphisms, or slight differences in DNA directly from the mRNA that codes for the sequences. The process involves digest- polypeptide in question. This is due to the ing DNA sequences with different restriction presence of introns (non-coding regions on a enzymes, detecting the resulting restriction DNA molecule; BIO 3.0) which prevent tran- fragments by gel electrophoresis, and com- scription of foreign genes in the genome of paring their lengths. In DNA fingerprinting, bacteria, a common problem in recombinant commonly used to analyse DNA left at crime technology. To carry this out one can use scenes, RFLP’s are produced and compared reverse transcriptase producing complemen-

High-level ImportanceHigh-level to RFLP’s of known suspects in order to catch tary DNA (cDNA) which lack the problematic the perpetrator. introns.

- Electrode Bu er

Well

Larger Samples fragments

Gel

Smaller fragments

Plastic frame Electrode +

Bu er

Figure IV.A.15.4: Gel Electrophoresis.

BIO-280 Chapter 15: GENETICS GAMSAT MASTERS SERIES

Rather than using a bacterium to clone DNA fragments, sometimes DNA is copied SNOW DROP Southern DNA directly using the polymerase chain reac- High-level Importance tion (PCR). This method allows us to rapidly Northern RNA amplify the DNA content using synthetic prim- O O ers that initiate replication at specific nucleo- Western Protein tide sequences. This method relies on ther- mal cycling (repeated heating and cooling) of DNA microarray technology (= DNA chip the DNA primers and can lead to thousands or biochip or “laboratory-on-a-chip”) helps to and even millions of copies in relatively short determine which genes are active and which periods of time. are inactive in different cell types. This tech- nology evolved from Southern blotting and Southern blotting, named after Dr. E. can also be used to genotype multiple regions Southern, is the process of transferring of a genome. DNA microarrays are created DNA fragments from the electrophoresis by robotic machines that arrange incredibly agarose gel onto filter paper where they small amounts of hundreds or thousands of are identified with probes. The procedure gene sequences on a single microscope slide. begins by digesting DNA in a mixture with These sequences can be a short section of restriction endonucleases to cut out spe- a gene or other DNA element that is used to cific pieces of DNA. The DNA fragments hybridise a cDNA or cRNA (also called anti- are then subjected to gel electrophoresis. sense RNA) sample. The hybridisation is usu- The now separated fragments are bathed ally observed and quantified by the detection in an alkaline solution where they immedi- of fluorescent tag. ately begin to denature. These fragments are then placed (or blotted) onto nitrocel- NB: The techniques lulose paper and then incubated with a of FRAP (BIO 1.5) and ELISA (BIO 8.4) were specific probe whose location can be visu- described earlier in this book. Electropho- alised with autoradiography. resis and chromatography are discussed in Organic Chemistry Chapter 13. Northern blotting is adapted from the Southern blot to detect specific sequences of RNA by hybridisation with cDNA. Similarly, Western blotting is used to identify specific amino-acid sequences in proteins. Since this is not required for you to memorise for the GAMSAT, you may want to set aside this :) mnemonic for when you are attending medi- cal school: SNOW DROP.

THE BIOLOGICAL SCIENCES BIO-281 CHAPTER 15: Genetics GOLD STANDARD FOUNDATIONAL GAMSAT PRACTICE QUESTIONS

1) Although the cloned sheep Dolly contains the 3) Von Willebrand’s disease is an autosomal exact DNA as her genetic mother, there are dominant bleeding disorder. A man who does a few visible and behavioral differences in not have the disease has two children with Dolly. This is most probably due to: a woman who is heterozygous for the condi- tion. If the first child expresses the bleeding A. the sheep in which the embryo was implanted. disorder, what is the probability that the sec- ond child will have the disease?

High-level ImportanceHigh-level B. the induction of specific genes not expressed in the mother. A. 0.25 C. mutations caused by incubation in the B. 0.50 nutrient-deficient solution. C. 0.75 D. environmental factors. D. 1.00

2) In fruit flies, males have XY sex chromo- 4) Over 10 million North Americans are treated somes, females have XX, and white eye for thyroid diseases and, overwhelmingly, colour is sex linked. If red-eyed (heterozy- women are much more likely than men to gous) females are crossed with white-eyed succumb to these conditions. Is it reason- males, what would be the expected eye able to conclude that thyroid disease is sex- colours and sexes of the progeny? linked? A. ¾ white-eyed female and ¼ red-eyed A. No, because thyroid disease appears to male. be caused by a defect of the immune sys- B. ½ red-eyed female and ½ white-eyed tem and not a defective DNA sequence. male. B. No, because if the disease was sex- C. All red eyed, half male and half female. linked, there would be a high incidence in D. ¼ red-eyed female, ¼ white-eyed female, the male, rather than the female, popula- ¼ red-eyed male and ¼ white-eyed male. tion. C. Yes, because the high incidence of the disease in women suggests that a gene found on the X chromosome codes for the disease. D. Yes, because the same factor increases the risk of women getting the disease, regardless of familial background.

BIO-282 Chapter 15: GENETICS GAMSAT MASTERS SERIES

GOLD STANDARD GAMSAT-LEVEL PRACTICE QUESTIONS

Questions 5–7 7) Which of the following ratios best represents High-level Importance the likelihood that the parents previously Sickle cell disease (SCD), also known as sickle described would have offspring with blood cell anemia, describes a group of inherited red type O and sickle cell trait? blood cell disorders. People with SCD have abnormal hemoglobin, called hemoglobin S or A. 9/16 sickle hemoglobin, in their red blood cells, while B. 3/16 the normal variant is hemoglobin A. C. 2/16 D. 1/16 SCD is as an autosomal recessive disorder. Hav- ing one copy of HbA produces enough normal hemoglobin to prevent anemia. Thus the hetero- 8) The genes in the intestinal bacterium E. Coli, zygous condition, AS, is a carrier and is described in alphabetical order: ade, ilv, rad, and trp, all as ‘sickle cell trait’. map onto the same chromosome. Pairwise crosses between single mutants give the fre- In humans, blood type is a result of multiple quencies of wild type recombinants provided A B O alleles: I , I , and i . A few simple rules of blood- in Table 1. Which of the following represents type genetics are that: the most likely correct order in which the four genes lie along the chromosome? • IA is dominant over iO, • IB is dominant over iO, and rad ade ilv trp • IAIB are codominant. rad 0 10.2 7.8 3.4 ade 10.3 0 3.3 4.8 5) Consider the case in which two parents are ilv 7.6 4.0 0 2.8 heterozygous for type A blood and have sickle cell trait. What is the genotype of the trp 3.1 7.0 2.8 0 parents? Table 1 A O A. I i AS A. ade-trp-ilv-rad A A B. I i AS B. ilv-ade-trp-rad A O C. I i SS C. trp-rad-ilv-ade A O D. I i AA D. rad-trp-ilv-ade 6) Which of the following does NOT represent a genotype likely to be present in the gam- etes of the parents described in the previous question? A. IAA B. iOA C. iOS D. All of the above are likely.

THE BIOLOGICAL SCIENCES BIO-283 Questions 9–12 is the sticking together of complementary single strands which, of course, involves the formation The polymerase chain reaction (PCR) is a power- of hydrogen bonds between the base pairs; BIO ful biological tool that allows the rapid amplifica- 1.2.2). Once the primer has annealed, the temper- tion of any fragment of DNA without purification. ature is elevated to 72 °C to allow optimal activ- In PCR, RNA primers are made to flank the spe- ity of the DNA polymerase. The polymerase will cific DNA sequence to be amplified. These RNA continue to add nucleotides until the entire com- primers are then extended to the end of the DNA plimentary strand of the template is completed at molecule with the use of a heat-resistant DNA which point the cycle is repeated (Figure 1). polymerase. The newly synthesised DNA strand is then used as the template to undergo another One of the uses of PCR is sex determination, round of replication. which requires amplification of intron 1 of the High-level ImportanceHigh-level amelogenin gene. This gene found on the X-Y The 1st step in PCR is the melting of the target homologous chromosomes has a 184 base pair DNA into 2 single strands by heating the reaction deletion on the Y homologue. Therefore, by ampli- mixture to approximately 94 °C, and then rapidly fying intron 1, females can be distinguished from cooling the mixture to allow annealing of the RNA males by the fact that males will have 2 different primers to their specific locations (note: annealing sizes of the amplified DNA while females will only have 1 unique fragment size. 5’ 3’ 3’ 5’ 9) The polymerase chain reaction resembles target DNA to be amplified which of the following cellular process? Separate strands by heating; A. Transcription of DNA cool and anneal; B. Protein synthesis

RNA primers ( ) C. DNA replication 5’ 3’ D. Translation 5’ 5’ 10) Why is a heat resistant DNA polymerase required for successive replication in the 5’ 3’ polymerase chain reaction, rather than sim- Nucleotides added; ply a human DNA polymerase? primers extended by DNA polymerase A. The high temperatures required to melt 5’ 3’ the DNA double strand may denature a 5’ normal human cellular DNA polymerase. B. The high temperatures required to melt the 5’ DNA would cause human DNA polymerase 3’ 5’ to remain bound to the DNA strand. amplified target DNA C. Heat resistant DNA polymerase increases the rate of the polymerase chain reaction repeat cycle at high temperatures whereas human DNA polymerase lowers the rate. D. Heat resistant DNA polymerase recog- Figure 1 nises RNA primers whereas human DNA polymerase does not. BIO-284 Chapter 15: GENETICS GAMSAT MASTERS SERIES

11) The use of PCR for sex determination relies 13) The autosomal dominant gene Curly (CY) of on the fact that: the fruit fly Drosophila melanogaster results in adult flies with abnormal wings that curl up A. the amelogenin gene is responsible for High-level Importance an autosomal recessive trait. when the inactive form (the pupae) are main- tained at 25°C, but remain uncurled when the X and Y homologous chromosomes B. the pupae are kept at 19°C. Consider the F have different sizes of intron 1 of the 1 generation of a cross between two flies het- amelogenin gene. erozygous for curly wings. What would be the C. females have an X and Y chromosome expected percentage of F1 flies with normal and males have two X chromosomes. (uncurled) wings when the pupae are main- D. intron 1 has a different nucleotide length tained at either 19°C or 25°C? than intron 2. A. 100% uncurled at 19 °C; 75% uncurled at 25 °C 12) What would PCR amplification of an individu- al’s intron 1 of the amelogenin gene reveal if B. 100% uncurled at 19 °C; 25% uncurled at the individual were male? 25 °C C. 75% uncurled at 19 °C; 25% uncurled at A. One type of intron 1 since the individual 25 °C has one X chromosome and one Y chro- mosome. D. 25% uncurled at 19 °C; 75% uncurled at 25 °C B. Two types of intron 1 since the individual has only one X chromosome. C. One type of intron 1 since the individual has only one X chromosome. D. Two types of intron 1 since the individual has one X chromosome and one Y chro- mosome.

Questions 14–15 Sickle cell anemia in humans is an example of a mutation affecting a base in one of the genes involved in the production of hemoglobin. Hemoglobin, in adults, is made up of four polypeptide chains attached to the prosthetic group heme. The polypeptide chains influence the oxygen carrying capacity of the molecule. A change in the base sequence of the triplet coding for one amino acid out of the normal 146 amino acids in the beta chains gives rise to the production of sickle cell hemoglobin. The physiologi- cal effect of this is to lower the oxygen-carrying capacity of these red blood cells. In the heterozygous condition individuals show the sickle-cell trait. The red blood cells appear normal and only about 40% of the hemoglobin is abnormal. A gene represented by two or more alleles within a population is said to be polymorphic. Allelic dif- ferences are caused by sequence differences, that is, the DNA itself is polymorphic and results in sequence differences between homologous regions of DNA in different individuals. These differences can be detected even when no other differences can be found and knowledge of this polymorphism could be used for the pre- or postnatal diagnosis for the sickle-cell gene.

THE BIOLOGICAL SCIENCES BIO-285 Rather than directly sequencing a gene, differences called restriction fragment length polymorphisms (RFLPs) can be used to highlight the regions of sequence differences between individuals. The tech- nique depends on a making a cut at a site on the gene for normal hemoglobin, not present on sickle-cell hemoglobin, causing fragments cut by the given enzyme to be smaller than it is in individuals not possessing the site. High-level ImportanceHigh-level

Figure 1: Inheritance pattern of an RFLP associated with sickle-cell disease. Males are represented by squares and females by circles. An open square or circle indicates an individual who is homozygous normal. A half-filled square or circle indicates an individual with sickle-cell trait. A filled square or circle indicates a homozygous individual with sickle-cell disease. The direction of electrophoresis is indicated by an arrow.

14) Figure 1 indicates that the fragment cut by 15) If the heterozygous male in the first genera- the restriction enzyme in an individual who is tion in Figure 1 was substituted with a homo- homozygous for the sickle-cell condition is: zygous male for the normal condition, the phenotypes of the third generation would be A. 7.6 kb long. expected to be in the ratio of: B. 13 kb long. C. 20.6 kb long. A. 1 normal : 2 sickle-cell trait : 1 sickle-cell disease. D. 5.4 kb long. B. 1 normal : 1 sickle-cell trait. C. 2 normal : 1 sickle-cell trait : 1 sickle-cell disease. D. 1 sickle-cell trait : 1 sickle-cell disease. BIO-286 Chapter 15: GENETICS GAMSAT MASTERS SERIES

Bonus Question!

16) The Hardy–Weinberg principle, or law, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. High-level Importance For example, consider an idealised population that abides by certain conditions and has a gene locus occupied by either the dominant gene ‘A’ or the recessive gene ‘a’. Let p = the frequency of allele ‘A’ in the population, and let q = the frequency of allele ‘a’. Since they are the only alleles, then p + q = 1. Squaring both sides we get: (p + q)2 = (1)2 and thus p2 + 2pq + q2 = 1. Consider Figure 1. According to Figure 1, what is the frequency of the heterozygous state? A. 0.36 B. 0.4 C. 0.48 D. 0.6

Figure 1: Allele frequencies in a model population

Gold Standard has cross-referenced the content in this chapter to examples from ACER’s official GAMSAT practice materials. It is for you to decide when you want to explore these questions since you may want to preserve some of ACER’s materials for timed mock-exam practice.

Examples – Dihybrid cross: Q16-17 of 1; monohybrid cross: Q23-26 of 2; genotype/phenotype: Q1-4 of 3; pedigree with dihybrid cross: Q24-26 of 4; dihybrid cross Q54-56 of 5. Note that “Q” is followed by the question number, and, for example, “of 1” refers to booklet number 1 which is referenced in the Spoiler Alert table at the end of Chapter 1. The 10 full-length HEAPS GAMSAT practice tests (by Gold Standard and MediRed), exams 1 through 10, contain specific cross-references to this chapter within the worked solu- tions. Note that the dihybrid-cross, sickle-cell unit is a rare treat from HEAPS-1; the pedigree analysis is from HEAPS-7. Side note: HEAPS-1 also has a heavy meiosis/crossing over unit with recurring concepts.

THE BIOLOGICAL SCIENCES BIO-287 Chapter Checklist Access your online account to view answers, worked solutions and discussion boards. Reassess your ‘learning objectives’ for this chapter: Go back to the first page of this chapter and re-evaluate the top 3 boxes and the Introduction. Complete a maximum of 1 page of notes using symbols/abbreviations to represent the entire chapter based on your learning objectives. These are your Gold Notes. Consider your multimedia options based on your optimal way of learning: Download the free Gold Standard GAMSAT app for your Android device or iPhone. Create your own, tangible study cards or try the free app: Anki. Record your voice reading your Gold Notes onto your smartphone (MP3s) and listen during exercise, High-level ImportanceHigh-level transportation, etc. Try out the Gold Standard GAMSAT online videos at gamsat-prep.com, or you can try other options on YouTube like Khan Academy or Crash Course Biology. Reassess your schedule for your full-length GAMSAT practice tests: ACER and/or HEAPS exams. Ensure that you have scheduled one full day to complete a practice test and 1-2 days for a thorough assessment of worked solutions while adding to your abbreviated Gold Notes. Reassess your progress in scheduling and/or evaluating stress reduction techniques such as regular exer- cise (sports), yoga, meditation and/or mindfulness exercises (see YouTube for suggestions).

BIO-288 Chapter 15: GENETICS