University of York Department of Biology B. Sc Stage 2 Degree Examinations

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1 Examination Candidate Number: Desk Number: University of York Department of Biology B. Sc Stage 2 Degree Examinations Evolutionary and Population Genetics Time allowed: 1 hour and 30 minutes Total marks available for this paper: 80 This paper has two parts: Section A: Short answer and problem questions (50 marks) Answer all questions in the spaces provided on the examination paper Section B: Long answer question (marked out of 100, weighted 30 marks) Answer either question A or question B Write your answer on the separate paper provided and attach it to the back of the question paper using the treasury tag provided The marks available for each question are indicated on the paper A calculator will be provided For marker use only: For office use only: Total as % page 1 of 10

2 SECTION A: Short answer and problem questions Answer all questions in the spaces provided Mark total for this section: At a crime scene, a hair is found and used to obtain a DNA profile. The forensic team provides DNA fingerprints, shown in the table below. Explain how population genetic principles can be used to assess whether the hair actually belonged to suspect 1. (6 marks) Genetic locus Hair at crime scene AA AB BB BC AB AB AA AC AB BB Suspect 1 AA AB BB BC AB AB AA AC AB BB Suspect 2 AB AA AB AB CC BB BC AB AA AB Even though the DNA profile of the hair and suspect 1 match, there is a possibility that the hair came from another person (1 mark). Assessing the probability that matching DNA profiles belong to a particular person requires knowledge of the allele frequencies at the genetic loci used (1 mark). For each locus, the probability of having a particular genotype can be calculated assuming that the population is in Hardy-Weinberg equilibrium (1 mark). Caution needs to be taken if there is population structure (1 mark) or other assumptions of the H-W equilibrium are not met (1 mark). Close relatives are also more likely to share DNA fingerprints (1 mark). Credit will also be given if it is mentioned that the probabilities for the ten loci can be combined. Learning objectives addressed: 2 know that under simplified conditions allele frequencies in populations are stable, be aware that this is described by the Hardy-Weinberg equilibrium and be able to apply the Hardy-Weinberg equilibrium to simple population level problem; 8 be able to interpret and analyse simple data sets, graphs and tables that illustrate the population genetic principles listed above page 2 of 10

3 2. A poultry farmer is planning to breed chickens, aiming for an increased amount of breast meat. He has an estimate of the narrow sense heritability based on similar populations at other farms. Discuss what the farmer needs to consider before he decides which chickens to choose for breeding. (6 marks) The farmer can use the breeders equation to estimate the response to selection (1 mark). Higher heritability of the trait and a greater selection differential will lead to a stronger response to selection (2 marks), so choosing parents with a more extreme phenotype will achieve the response more quickly (1 mark). However, the farmer faces a trade-off between a quicker response to selection and potential problems of inbreeding and a loss of genetic variation in other traits (1 mark). The farmer should also be aware that the heritability of the trait might be different in his chicken population if his chickens were more or less genetically variable than those at other farms (1 mark). Learning objectives addressed: 1 know that continuous variation has the same genetic origin as discontinuous variation, be able to apply some techniques for studying continuous variation and be able to interpret its evolutionary significance 3. a) Describe two ways of estimating the genome-wide mutation rate. (4 marks) Compare sequences from different species across a phylogeny. Count number of nucleotide differences. Obtain an estimate of divergence time. Mutation accumulation lines. Maintain multiple selfing lines for multiple generations. Sequence and count the number of new mutations. Count mutations in a pedigree. Sequence genomes of parents and offspring. Count the number of new mutations. A brief explanation of each method is required to get full marks. (2 marks each) page 3 of 10

4 b) Typical mutation rates are of the order of 10-9 per base pair per generation. Explain how such a rare event as a mutation could provide sufficient material for selection to produce the diversity of life on the planet. (2 marks) The genome of each individual has several million to billion bases, and there are many individuals in each population. Therefore, huge numbers of mutations are produced each generation (1 mark). Additionally, evolution can take place over a very long time; ie over many generations (1 mark). Learning outcomes addressed: 3 be able to explain how allele frequencies in populations change due to the four microevolutionary processes: mutation, genetic drift, selection or gene flow, 4 be able to evaluate which of the microevolutionary processes is likely to be important under given circumstances and be able to explain how these processes might interact with each other 4. A sample of snow geese consists of 110 geese with blue plumage and 195 geese with white plumage. The blue allele (B) is dominant to the white allele (b). a) Assuming that the population is in Hardy-Weinberg equilibrium, what are the expected allele frequencies? (3 marks) Let the p be the frequency of the blue allele (B) and q the frequency of the white allele (b). The frequency of white geese is q 2 : q 2 = 195/( ) = (1 mark) The allele frequency of b is therefore q = ( q 2 ) = (1 mark) The frequency of B is p = 1 q = = (1 mark) page 4 of 10

5 b) What are the expected frequencies and numbers of homozygous and heterozygous blue geese? (4 marks) expected frequency of BB: p 2 = (1 mark) expected number of BB: N * p 2 = 305 * = (1 mark) expected frequency of Bb: 2pq = (1 mark) expected number of Bb: N * 2 pq = 305 * = (1 mark) c) Later these geese were genotyped at the colour locus, and it was found that there are 70 homozygous and 40 heterozygous blue geese. Comment on these new results with reference to your calculations in parts a) and b). Detailed calculations are not required. (4 marks) There are far more homozygous blue geese than expected (and fewer heterozygous blue geese) (1 mark). This lack of heterozygotes might be due to population substructure (Wahlund effect) (1 mark) or assortative mating between homozygotes (1 mark). These observed numbers also show that it was incorrect to assume that the population was in Hardy-Weinberg equilibrium, and therefore the estimates of allele and genotype frequencies are incorrect (1 mark). (Credit will be given for more detailed comments on how the allele frequencies are incorrect (p must be higher than estimated, q lower), up to a maximum of 4 marks.) Learning objectives addressed: 2 know that under simplified conditions allele frequencies in populations are stable, be aware that this is described by the Hardy-Weinberg equilibrium and be able to apply the Hardy-Weinberg equilibrium to simple population level problem; 8 be able to interpret and analyse simple data sets, graphs and tables that illustrate the population genetic principles listed above 5. In a population of wolves ( Canis lupus ), the K locus affects coat colour. page 5 of 10

6 Wolves that have the BB genotype at this locus have grey coats, whereas wolves that carry the alternative allele (genotypes AA and AB) have black coats. The mean lifetime reproductive success of the three genotypes is given in the following table: Genotype AA AB BB Mean lifetime reproductive success a) Which mechanism is likely to maintain this polymorphism? Explain why. (2 marks) Heterozygote advantage (1 mark), because the heterozygote has higher fitness than either homozygote (1 mark). b) What is the relative fitness of each genotype, assuming that there are no other effects on fitness? (3 marks) The heterozygote has the highest fitness, so the relative fitness is by definition 1. AB: 1 The relative fitness for the other genotypes is calculated relative to the fittest genotype: AA 0.031/2.36 = BB 0.62/2.36 = c) What are the selection coefficients against each genotype? (3 marks) AA =0.987 = s AB 1 1 = 0 BB = = t (1 mark each) d) Calculate the equilibrium frequencies of the A and the B allele. (3 marks) q = s/(s+t) (1 mark) q=0.987/( )=0.572 (frequency of the B allele) (1 mark) page 6 of 10

7 p= 1-q = (frequency of the A allele) (1 mark) e) Why are not all individuals in this population heterozygous? (1 mark) The alleles segregate during reproduction, half of the offspring produced by heterozygous parents will be homozygous (1 mark). Learning objectives addressed: 3 be able to explain how allele frequencies in populations change due to the four microevolutionary processes: mutation, genetic drift, selection or gene flow; 5 be able to explain how simple mathematical models can predict the direction and speed of allele frequency changes; 6 be able to apply some of these mathematical models to simple population genetic problems; 7 be able describe how genetic variation can be maintained 6. a) According to the neutral theory, changes in DNA sequence are mainly due to genetic drift. Describe with examples two pieces of evidence in support of this theory. (4 marks) Molecular clock; DNA evolves at a steady rate over time. Divergence of Hawaiian Drosophila species (2 marks). Rate of substitution is inversely proportional to the functional constraint on the gene. Histone proteins evolve very slowly relative to fibrinopeptides (2 marks). b) The figure above shows the phylogeny of the FOXP2 gene in page 7 of 10

8 primates. This gene is known to be relevant to human speech development. Vertical bars indicate nucleotide changes. The figures show the number of non-synonymous/synonymous changes that have occurred along each branch. How are the non-synonymous/synonymous mutations distributed across the different taxa in the phylogeny? (2 marks) A total of 156 synonymous mutations have occurred, 25 of these within the primates (1 mark). There are only 4 non-synonymous mutations, 2 of which are in the lineage leading to humans (1 mark). c) What can you conclude from the distribution of mutations across the phylogeny? (3 marks) The amino acid sequence is strongly conserved across a very long evolutionary time (1 mark). Two of the four non-synonymous mutations have occurred in the human lineage in a short evolutionary time, indicating that there has been selection to change the amino acid sequence (1 mark). This is likely a result of strong selection for the evolution of language in humans (1 mark). Learning objectives assessed: 3 be able to explain how allele frequencies in populations change due to the four microevolutionary processes: mutation, genetic drift, selection or gene flow. 4 be able to evaluate which of the microevolutionary processes is likely to be important under given circumstances and be able to explain how these processes might interact with each other. 8 be able to interpret and analyse simple data sets, graphs and tables that illustrate the population genetic principles listed above page 8 of 10

9 SECTION B: Long answer question Answer one question on the separate paper provided Remember to write your candidate number at the top of the page and indicate whether you have answered question A or B Mark total for this section: 30 EITHER A. Compare how genetics and the environment affect evolution in subdivided and single homogeneous populations. The key difference between a subdivided and a homogeneous population is that the individuals are more likely to be mating randomly in the single population whereas there is limited migration and gene flow between subpopulations. Processes such as genetic drift and selection are operating separately in the subpopulations, which leads to different allele frequencies in these populations. It is likely that the subpopulations are exposed to disruptive selection because the environment varies between these populations. Different genotypes will be selected for in this environments (if there are genotype x environment interactions), which may lead to local adaptation and phenotypic divergence of the subpopulations. However, each subdivided population is also likely to be smaller, so that genetic drift will play a larger role here and selection will be less efficient. The response to directional selection is likely to be stronger in the homogeneous population. The heritability of a trait is the phenotypic variation in a population that is due to genetic variation and is specific to a population. It is likely that the heritability estimates will vary between the subpopulations either due to differences in genetic variation or the stability of the environment. This will also affect the strength of the response to selection. Learning outcomes addressed: 1 know that continuous variation has the same genetic origin as discontinuous variation, be able to apply some techniques for studying continuous variation and be able to interpret its evolutionary significance; 3 be able to explain how allele frequencies in populations change due to the four microevolutionary processes: mutation, genetic drift, selection or gene flow; 4 be able to evaluate which of the microevolutionary processes is likely to be important under given circumstances and be able to explain how these processes might interact with each other; 7 be able describe how genetic variation can be maintained page 9 of 10

10 OR B. The endangered Siberian tiger ( Panthera tigris altaica ) has a fragmented habitat and its population size is estimated to be about 500. Describe a conservation strategy based on conservation genetics principles that could be used to reduce the likelihood of extinction. The smaller a population is, the stronger the effects of genetic drift. Genetic drift in small populations will result in the rapid loss of genetic diversity. While the census population is 500, the effective population size will be much lower than this, especially as the population is fragmented. Genetic drift will act faster in each of the population fragments. Genetic diversity is in general thought to be important for maintaining the adaptability and evolutionary potential of a species. Therefore, the main aim should be to increase population size quickly, and to allow the mixing of fragmented populations, perhaps by creating wildlife corridors or translocating individuals. The small populations are also at risk of suffering from inbreeding depression. The population structure of the species could be assessed using variable markers such as a panel microsatellites to determine. This would allow one to assess the connectivity of populations, and also establish which populations are particularly genetically depauperate (and most at risk), which are reservoirs of high genetic diversity (and could act as sources of genetically diverse individuals), and whether any populations are more genetically distinct (and therefore merit greater protection). It may be necessary to genetically rescue the population by introducing tigers from a different subspecies. If so, it would be important to choose a population that is most similar to the Siberian tiger to avoid introducing maladaptive alleles. Despite these measures, it is important to understanding why the Siberian tiger has become endangered in the first place (hunting? Habitat degradation? Lack of prey?). Unless these issues are tackled the genetic measures alone are unlikely to be effective. Learning objectives assessed: 3 be able to explain how allele frequencies in populations change due to the four microevolutionary processes: mutation, genetic drift, selection or gene flow. 4 be able to evaluate which of the microevolutionary processes is likely to be important under given circumstances and be able to explain how these processes might interact with each other. 5 be able to explain how simple mathematical models can predict the direction and speed of allele frequency changes. 6 be able to apply some of these mathematical models to simple population genetic problems. 7 be able describe how genetic variation can be maintained. page 10 of 10

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