FA23 week 9 problem set

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Arizona State University *

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Biology

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Apr 27, 2024

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1 BICD100 FA2023 Week 9 Problem Set Color coded to show which ones are based Monday vs. Wednesday lecture material 1). This image shows results for 3 microsatellite markers (D8S1179, D12S91 and D19S433) that are part of a DNA fingerprint analysis of a crime scene DNA sample. Assuming random mating, use the frequency information for each allele detected here to calculate the probability that a person in the general population will have the genotype shown here. Give one answer for the entire 3 marker genotype, not separate probabilities for each marker. 7.55 × 10 −4 2). The ∆32 allele of the human gene CCR5 (encoding chemokine receptor 5) confers resistance to HIV infection. The “normal” allele of this gene (anything that is not ∆32) is referred to as “allele 1”. In a randomly mating population of 500 individuals, it was determined that 325 were 1/1 (HIV-sensitive) homozygotes. If mating is random in this population with respect to CCR5 alleles, how many 1/∆32 heterozygotes and ∆32/∆32 homozygotes do you expect there were? 156 for heterozygotes and 19 for homozygotes. 3.) Most black bears ( Ursus americanus ) are black or brown in color. However, occasional white bears of this species appear in some populations along the coast of British Columbia. Geneticists have determined that white coat color in these bears results from a recessive allele (with the nucleotide G at a certain position in the gene) in the melanocortin-1 receptor (MC1R) gene. The dominant, wild-type allele has the nucleotide A at this position in the gene and determines black or brown coat color. Genotypes for a population of bears living on an island in British Columbia are genotyped for A vs. G alleles at MC1R, with the following results. a. What are f(A) and f(G)? Do not assume random mating here. F(A) = (84+24)/174 = 0.62 f(G) = (42+24)/174 = 0.38

2 b. If the population meets the random mating condition of Hardy-Weinberg equilibrium, what number of AA, AG and GG bears is expected? AA = (0.62)(0.62) = 0.384 AG = 2(0.62)(0.34) = 0.471 GG = (0.38)(0.38) = 0.144 c. Use chi squared analysis to test whether this bear population meets the random mating expectations of Hardy-Weinberg equilibrium. Include in your answer the value of Chi squared, your p-value, and your conclusion. Use the chi squared table in the lecture slides from Nov. 27 to look up the p-value. Genotype Observed Expected O-E (O-E)^2 (O-E)^2/E AA 42 33 9 81 2.45 AG 24 41 17 289 7.05 GG 21 13 8 64 4.92 Chi square = 14.42 Degrees of freedom = 1 p-value less than 0.05 so reject the null hypothesis d. Is there evidence of inbreeding and if so, what is the inbreeding co-efficient? No inbreeding 4). In an isolated population of 50 desert bighorn sheep, homozygosity for a mutant recessive allele c causes curled coats. The wild-type dominant allele C produces straight coats. A biologist studying these sheep counts 4 with curled coats. She also takes blood samples from all of the sheep in the population for DNA analysis, which reveals that 17 of the sheep are heterozygous carriers of the c allele. What is the coefficient of inbreeding (F) for this population? CC = 29 Cc = 17. cc = 4 F(CC) = 29 50 = 0.58 F(cc) = 4 50 = 0.08 Inbreeding Coefficient: 1 − 0.34 2(0.58)(0.08) = 0.093 5). As discussed earlier in the quarter, people with xeroderma pigmentosum (XP) get skin cancers at a high rate because they lack genes necessary to repair UV-induced thymine dimers so acquire lots of mutations in skin cells, including loss of function mutations in tumor suppressor genes. This recessive trait is rare: in the US, the frequency of affected individuals is about 1/250,000 (4 X 10 -6 ).

3 a. If we consider the US population to be in Hardy-Weinberg equilibrium for this trait, what is the frequency of mutant alleles? What is the frequency of carriers? Q^2 = 0.002 1-0.002 = 0.998 2pq = 2(0.998)(4*10^-6) = 7.98*10^-6 b. What is the probability that a child resulting from a mating between first cousins (belonging to a family with no history of XP) will have the disease? Note: we calculated the inbreeding co-efficient for a first cousin mating in class on Wed. Nov. 29. 4*10^-6 c. What is the probability that a child resulting from the consanguineous (inbred) mating illustrated in this pedigree (where there is no family history of XP) will have XP? Trace loops on this pedigree to illustrate your calculation of the inbreeding co-efficient and include that value in your answer. ( 1 2 ) 5 = 0.03125 = 3.12% . There are 5 ancestors that could pass down this trait starting from first generation then to the second- generation offspring, then to 3rd generation offspring, and finally 4 th generation. 6). Bai Yun and Gao Gao are adult pandas who resided for many years at the San Diego Zoo (they have been returned to China now). While at the SD Zoo they had a few cubs. As it happens, they both came from the same remote part of China where there is only a small panda population, and are distantly related as shown in the pedigree. The black and white spotted coat typical of pandas is due to a dominant piebald spotting allele S P . Individuals who are homozygous for the (most) recessive allele of this gene have all black fur ( ss ). The frequency of the recessive s allele is 0.1 in the panda population that Gao Gao and Bai Yun come from. Assume that S p and s are the only alleles in this population. What is the probability that Gao Gao and Bai Yun's cub will be black? Trace loops on this pedigree to illustrate your calculation of the inbreeding co-efficient and include that value in your calculations (show key steps in your calculations).

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