Question

Name: QUESTIONS 38-40: You perform a genome-wide association study (GWAS) to identify genes affecting variation for sun sensitivity and the associated risk of skin cancer, using a sample of 1000 Iceland natives with no evidence of mixed ancestry. The genotype-phenotype data for SNP rs 1805007 (a C/T single-nucleotide polymorphism) in the melanocortin-1-receptor (MCIR) gene are shown in the following table. e rar Number of individuals by genotype Phenotype Burning Tanning TOTALS T/T 155 460 615 C/T 110 200 310 C/C 35 40 75 TOTALS 300 700 1000 38. What is the frequency of the T allele in the individuals with the "burning" phenotype? a. 0.517 b. 0.700 c. 0.719 d. 0.252 39. What is the frequency of the T allele in the individuals with the "tanning" phenotype? a. 0.800 b. 0.657 c. 0.748 d. 0.811 40. In the question above, assuming the differences in frequency are statistically significant, which of the following is the most likely explanation for the data? a. T nucleotides at SNP rs1805007 directly increase the risk of burning b. C nucleotides at SNP rs1805007 directly increase the risk of burning. c. MCIR alleles (haplotypes) containing a T at SNP rs1805007 increase the risk of burning d. MCIR alleles (haplotypes) containing a C at SNP rs 1805007 increase the risk of burning. e. MCIR has no effect on the risk of burning. QUESTIONS 41-43: Hardy-Weinberg (H-W) frequencies can be estimated for loci with more than two alleles, using the same principles as those for biallelic loci. In a study of the population of Iceland, you find that the frequency of alleles at the gene for the ABO blood types is as follows: 18: 0.20 1A: 0.30 41. What is the expected H-W frequency of the /I genotype? c. 0.30 d. 0.03 a. 0.09 b 0.18 42. What is the expected H-W frequency of the AP genotype? c. 0.12 d. 0.06 a. 0.50 b. 1.00 43. What is the expected H-W frequency of the /i genotype? c. 0.30 d. 0.075 a. 0.80 b. 0.15 P. 8

Answer 1

Answer:

38). B). 0.700

Explanation:

Total number of T alleles = 2*T/T+C/T = 2*155+110 = 420

Total alleles = Total no. of individuals * 2 = 300*2 = 600

Fr(T) = 420/600 = 0.7

39). a). 0.800

Explanation:

Total number of T alleles = 2*T/T+C/T = 2*460+200 = 1120

Total alleles = Total no. of individuals * 2 = 700*2 = 1400

Fr(T) = 1120/1400 = 0.8

40)..------

41). A) .0.09

Explanation:

IAIA =0.30 * 0.30 = 0.09

42). C). 0.12

Explanation:

IAIB = 2 * 0.30 * 0.20 = 0.12

43). C). 0.30

Explanation:

IAi = 2 * 0.30 * 0.50 = 0.3