1) (15 points) Next to each term in the left-hand column place the number from the right-hand column that best corresponds:

Transcription

1 1) (15 points) Next to each term in the left-hand column place the number from the right-hand column that best corresponds: natural selection 21 1) the component of phenotypic variance not explained by genotypic variation 2) measures successful DNA replication genetic drift 17 3) an environmental disease caused by homozygosity for a defective allele responding to a particular diet 4) the portion of phenotypic variance due to genotypic variation gene flow 20 5) associated with expression of sickling in the fetal stage 6) a haplotype can be found far from the geographical center of its ancestral haplotype heritability 19 7) increases genetic variability within demes, and increases genetic differences between demes 8) shows epistasis with the beta globin gene for the sickling trait environmental variance 1 9) coalesceses more rapidly than autosomal DNA 10) the component of phenotypic variance caused by variation in environmental factors phenotypic correlation 15 11) a recessive allele for malarial resistance between sibs 12) coalesceses more rapidly than mitochondrial DNA candidate locus 28 13) measures viability and mating success 14) an example of female specific gene flow hemoglobin C 11 15) greater than or equal to half the heritability 16) favored by natural selection in most populations in malarial areas scurvy 3 17) a force that causes all genetically variable populations to evolve 18) a haplotype is found within the range of its ancestral haplotype fetal hemoglobin gene 8 19) the portion of phenotypic variance due to variation in breeding values 20) increases genetic variability within demes, and decreases genetic differences between demes founder effect 27 21) a force that can prevent evolution in a genetically variable population 22) survival of the fittest range expansion 6 23) decreases genetic variability within demes, and decreases genetic differences between demes 24) a genetic disease caused by homozygotes for a defective allele responding to a particular diet X-linked DNA 9 25) influences only neutral alleles 26) less than the parent-offspring phenotypic correlation fitness 2 27) used in the successful positional cloning of the gene for Huntington s disease 28) a gene whose known function is related to a specific phenotype Lemba 29 29) an example of male specific gene flow 30) QTL

2 2) (6 points) List the three basic properties of DNA that form the basis of the neodarwinian theory of evolution. DNA can replicate DNA can mutate and recombine DNA encodes information that interacts with the environment to influence phenotype

3 3) (8 points) The frequency of the aa genotype in a population is 0.6. The frequency of individuals exposed to environmental factor X is 0.1. When an individual of genotype aa is exposed to factor X, that individual develops a severe disease. Assume genotypes and environments are independent. What is the incidence of the disease in the population? (0.6)X(0.1) = 0.06 or 6% What is the incidence of the disease in those individuals with genotype aa? Freq. Of factor X = 0.1 or 10% What is the incidence of the disease in those individuals who have been exposed to factor X? Freq. Of the Genotype = 0.6 or 60% Which is the major cause of variation for risk for this disease in this population: genetic or environmental factors. Explain your answer. The environment is the major cause of variation. Knowing environment X occurred results in a 10 fold increase in risk for the disease over the general population mean, whereas knowing an individual is aa results in only a slight increase in risk.

4 4. (6 points). A population is surveyed for genetic variation at a single locus with two alleles, A and a, with the following results: Genotype: AA Aa aa Sum Number: Test the hypothesis that this population is in Hardy-Weinberg with the frequency of the allele a = 3/4. First generated the expected values: Expected AA = (1/4) = 16 Expected Aa = 2(1/4)(3/4)256 = 96 Expected aa = (3/4) = 144 Then: Genotype AA Aa aa Sum Obs-Exp (Obs-Exp) 2 /Exp 64/16= 4 64/96= There are 2 degrees of freedom (no allele frequency was estimated), so from the Table on the title sheet, this value is not significant at the 5% level. Therefore, we fail to reject the hypothesis of HW with p = 0.75

5 5. (6 points) Species X consists of two local demes of equal size. At a particular locus, these two demes collectively have two alleles, A and a, in their combined gene pool. The frequency of the A allele is 0.5 in the combined gene pool. Within each local deme, mating is random. a). If F st =0, what are the allele frequencies of A in each of the two local gene pools, and what are the frequencies of heterozygotes within each local deme? F st =0 implies both demes have identical gene pools that are the same as the combined gene pool. Hence, the frequency of the A allele is 0.5 in each of the local gene pools (2 points). Because mating is random, and both demes have the same allele frequencies, the expected heterozygosity within a deme = 2pq = 2(.5)(.5)= 0.5 (1 point) a). If F st =1, what are the allele frequencies of A in each of the two local gene pools, and what are the frequencies of heterozygotes within each local deme? F st =1 implies that both demes have no genetic variation, so the alleles frequencies must be 0 for one deme and 1 for the other, which averages to 0.5 (2 points). Because there is no genetic variation in either deme, the heterozygosity frequencies in both demes = 0 (1 point).

6 6). (3 points). Fill in the genotypic deviations from the following data: Pop. Mean Frequency Mean Phenotype Genotypic Deviations = = = = = = (5 points). Calculate the genetic variance from the following data. Frequency Genotypic Deviations s g 2 = (-2) 2 (0.09)+(-3) 2 (0.12)+(2) 2 (0.04)+(4) 2 (0.3)+(0) 2 (0.2)+(-3) 2 (0.25) = 8.65

7 8. (6 points) Calculate the average excesses for the A, B and C alleles from the following data: Genotypic Deviations Gene Pool: Freq. Of A = 0.4, Freq. Of B = 0.3, Freq. Of C = 0.3. Average excess of A = 0.4(-15)+0.3(16)+0.3(5) = -0.3 Average excess of B = 0.4(16)+0.3(-5)+0.3(-10) = 1.9 Average excess of C = 0.4(5)+0.3(-10)+0.3(-4) = (3 points) Given that the average excesses of three alleles at a single, autosomal locus in a random mating population are a A = 1, a B = 2, and a C = -3, calculate all the breeding values. Additive Genotypic Deviations 1+1=2 1+2=3 2+2=4 1+(-3)=-2 2+(-3)=-1-3-3=-6

8 10. (3 points) Given the following data, calculate the average excesses for the three alleles, assuming random mating. Additive Genotypic Deviations The additive genotypic deviation for a homozygote = twice the average excess for the homozygous allele. Therefore, aa = 1 / 2 (40) = 20 a B = 1 / 2 (2) = 1 a C = 1 / 2 (-30) = (5 points). Calculate the additive genetic variance from the following data. Frequency Additive Genotypic Deviations s g 2 = (2) 2 (0.04)+(0.8) 2 (0.16)+(0) 2 (0.16)+(0.4) 2 (0.16)+(-0.4) 2 (0.32)+(-1) 2 (0.16) = 0.5 (0.4992)

9 12. (4 points). The environmental variance for a trait is 145, its additive genetic variance is 35, and its dominance variance is 20. Assuming a single locus model, calculate the broad and narrow-sense heritabilties. For a one locus model, the genetic variance = additive variance + dominance variance = = 55 The phenotypic variance = genetic variance + environmental variance = = 200 The broad sense heritability = genetic variance/phenotypic variance = 55/200 = The narrow sense heritability = additive genetic variance/phenotypic variance = 35/200 = (3 points). A protein coding gene is sequenced in several different species to determine its rate of molecular evolution. Order the following types of nucleotides in this gene region from fastest to slowest rates of evolution according to the neutral theory: A. first and second codon positions for codons coding for amino acids in the active site B. first and second codon positions for codons coding for amino acids in those positions other than the active site. C. Third codon positions According to the neutral theory, the rate of evolution = neutral mutation rate. A mutation is more likely to be neutral when it has fewer functional constraints. Since most third codon position mutations are synonymous, C should evolve most rapidly. Since the active site is usually the most constrained part of a protein, A should evolve the slowest. Therefore, the order is: C > B > A

10 14. (9 points). Consider the following data: Fitness Suppose a random mating population has an average fitness of 0.84 and the following allele frequencies: Freq. of A = 0.1 Freq. of B = 0.7 Freq. of C = 0.2 Show mathematically which alleles are increasing, decreasing or remaining at the same frequency under natural selection. To do this, we need to calculate the average excesses of fitness. The first step is to subtract the mean phenotype to get the genotypic deviations for fitness, which are: Fitness Deviations Then: Average excess of A = 0.1(0.06)+0.7(0.16)+0.2(0.06) = 0.13 > 0 fi selection increases the frequency of this allele. Average excess of B = 0.1(0.16)+0.7(-0.14)+0.2(0.16) = < 0 fi selection decreases the frequency of this allele. Average excess of C = 0.1(0.06)+0.7(0.16)+0.2(-0.04) = 0.11 > 0 fi selection increases the frequency of this allele.

11 15. (8 points) Consider the following haplotype tree of a protein coding gene studied within and between two species, X and Y. Haplotypes are indicated by the species letter followed by a number. Two numbers are written by each branch in the tree: the first is the number of nucleotide substitutions that occurred on that branch that did not alter the amino acid sequence of the protein, and the second number (in bold) is the number of nucleotide substitutions that occurred on that branch that did alter the amino acid sequence of the protein. Species X Species Y X1 X2 X3 Y1 Y2 Y3 Y4 6, 1 2, 0 3, 0 2, 0 1, 1 X4 Y5 4, 2 8, 1 2, 1 Y6 X5 6, 1 5, 0 Y7 11, 3 9, 2 Outgroup Set up a contingency chi-square table to test the null hypothesis of no selection on this gene. Be sure to label each row and column in your table. Then enter the appropriate number into each cell. You do not have to perform the actual test, just prepare the table. Synonymous (or Silent) Non-synonymous (or Replacement) Tip 27 3 Intraspecific Interior 12 4 Fixed ( or Interspecific) 20 5

12 16. (10 points) PCR and Human Genetic Variation Use Figure 1 (right) and Table instructed. Note that the alleles based upon size to simplify the 1 (below) when were grouped experiment. Size (bp) Allele Table 1 Allele groups based upon fragment for the Human Genetic experiment. size of DNA Variation Figure 1 An illustration of the electrophoresis gel obtained from the Human Genetic Variation experiment. a. What was the locus that was examined in lab for the Human Genetic Variation experiment? [0.5 point] D1S80 b. What is the genetic basis for the differences in the alleles? [0.5 point] Variable Number of 16 bp (Tandem) Repeats or VNTR c. What cells were the source for the template DNA? [0.5 point] Cheek cells d. What did we use to remove taq-inhibiting metal ions during the DNA isolation procedure? [0.5 point] Chelex beads e. Using Figure 1 (gel) and Table 1 (allele groups), identify the lanes for the samples that depict a HOMOZYGOUS individual? [1.5 point] 5, 9, and 11

13 f. In Figure 1 (gel), lane #6, no visible band was detected. This observation suggests that the PCR for that sample may have been unsuccessful. List 4 necessary components for a successful PCR reaction? [2.0 points] Primers Taq polymerase dntps DNA template Mg ++ Buffers g. Using Figure 1 (gel) and Table 1 (allele groups), complete the table (below) giving the frequency for each allele. [3.0 points] Allele frequency h. Assuming that the population meets all of the conditions for the Hardy-Weinberg equillibrium (including population size), predict the genotype frequencies from the allelic frequencies from the previous question (#7). [1.5 points] A. Heterozygous for allele 1 and 3: 2pq = 2 (0.278) (0.222) = B. Heterozygous for allele 2 and 5: 2pq = 2 (0.278) (0.056) = C. Homozygous for allele 4: p 2 = (0.111) 2 = 0.012