Question

7. The table below shows the frequencies of the PMCT118 alleles in several different human populations. Allele Frequency (n = number of alleles typed) Allele Allele U.S. Caucasians (n = 200) African Americans (n=214) U.S. Caucasians (n = 200) African Americans (n=214) Hispanic Americans (n=210) 30 0.008 0.008 0.055 0.08 0.058 0.013 0.003 0.048 0.005 0.005 0.073 Hispanic Americans (n=210) 0.003 0.003 0.013 0.263 0.005 0.02 0.025 0.028 0.003 0.318 0.055 0.008 0.048 0.098 0.003 0.033 0.115 0.088 0.023 0.193 0.023 0.008 0.013 0.153 0.055 0.238 0.01 0.04 0.018 0.03 0.008 0.348 0.04 0.015 0.013 0.063 0.053 0.003 0.003 0.003 0.005 0.008 0.01 0.008 0.05 >41 0.005 0.01 0.005 0.055 Table adapted from: Klitz W., Reynolds R., Chen J., Erlich H.A. (2000). Analysis of genotype frequencies and interlocus association for the PM.DQA1, and PMCT118 loci in four populations. Forensic Sci. 45(5):1009-15. If we assume that each population is in Hardy-Weinberg equilibrium then the following formula is used to calculate genotype frequencies in a population with two alleles: p2 + 2pq + q2 = 1. "p" is the frequency of one allele and "q" is the frequency for the other allele. "p2" is the expected frequency of homozygotes for the first allele and "2pq" is the expected frequency of heterozygotes and q2" is the expected frequency of homozygotes for the second allele. It is therefore possible to calculate the expected frequency of a genotype from the allele frequencies, such that the frequency of homozygotes is the square of the frequency of the homozygous allele, , the frequency is two times the product of the frequency of the alleles for heterozygotes. a. Using the allele frequency data from the table above, calculate the expected frequency of the 18, 24 genotvne in U.S. Caucasians. (4.5 pts) b. Using the allele frequency data from the table above, calculate the expected frequency of the 24, 24 genotype in Hispanic Americans. (4.5 pts) c. Using the allele frequency data from the table above, calculate the expected frequency of the 21, 28 genotype in African Americans. (4.5 pts)

Answer 1

Hardy - Weinberg equations:

1 = p2 + 2pq + q2

1 = p + q

Here p is the frequency of one allele

q is the frequency of 2nd allele

p2 is the expected frequency of homozygotes for the 1st allele

q2 is the expected frequency of homozygotes for the 2nd allele

2pq is the frequency of heterozygotes

a) Expected frequency of 18,24 genotype of US Caucasians

Frequency of allele = No of occurence of the allele / total number of alleles in a population

frequency of allele 18 (p2) = 0.238

p = .0.48

frequency of allele 24 (q2) = 0.348

q = 0.58

2pq = .55

therefore expected frequencies for 18,24 genotype is .55

b) expected frequency of 24,24 genotype of Hispanic Americans

frequency of allele 24 (p2) = 0.318

p = 0.56

frequency of allele 24 (q2) = 0.318

q = 0.56

2pq = 0.62

Expected frequency of 24,24 genotype is 0.62

c) Expected frequency of 21,28 genotype of African Americans

Frequency of allele 21 (p2) = 0.115

p = 0.33

Frequency of allele 28 (q2) = 0.153

q = 0.39

2pq = .25

Expected frequency of 21,28 genotype is .25