Assume those attending our class are known to be at Hardy-Weinberg for a SNP in
ID: 161880 • Letter: A
Question
Assume those attending our class are known to be at Hardy-Weinberg for a SNP in the ABCC11 gene that determines human earwax type. There are only two alleles at this SNP-- A and G. The AA genotype at this SNP produces dry earwax, whereas GA and GG produce wet earwax. Let's assume that, in our class (150 attending-- everyone came!), 24 individuals have dry earwax. Approximately how many people in the class should be heterozygous?
A. 85
A. 85
B. 72
C. 54
D. 22
E. Can't tell with the information given
Explanation / Answer
B. 72
Given that the population is in Hardy-Weinberg equilibrium. Given that the frequency of people producing the dry ear wax, AA is = 24/150 = 0.16
Thus, the frequency of A is = 0.4
As per the Hardy-Weinberg equilibrium, the frequency of A + the frequency of G is = 1
Thus, the frequency of G is = 1- 0.6 = 0.4
The frequency of heterozygotes people is = 2*G*A = 2*0.4*0.6 = 0.48
Thus, the population of heterozygotes in the population is = 0.48*150 = 72