Question

Consider a population in which two different loci (A and B loci) each have two alleles (A & a for the A locus, B & b for the B locus). The observed frequency of the AB haplotype is 0.521, while the frequency of the A allele is 0.940 and the B allele is 0.501. What is the coefficient of linkage disequilibrium in this case?

A. 0.05
B. 0.154
C. 0.079
D. 0.425

Answer 1

The coefficient of linkage disequilibrium is calculated using the formula D = f(AB) - (f(A) * f(B)). Given the frequencies provided, D = 0.521 - (0.940 * 0.501) = 0.050, matching option A.

Step-by-step explanation:

We are tasked with calculating the coefficient of linkage disequilibrium for two loci, A and B, each with two alleles (A & a, and B & b respectively) in a population. To find the coefficient of linkage disequilibrium, we use the formula D = f(AB) - (f(A) * f(B)), where f(AB) is the frequency of the AB haplotype, f(A) is the frequency of the A allele, and f(B) is the frequency of the B allele.

Given the observed frequency of the AB haplotype is 0.521, the frequency of the A allele is 0.940, and the frequency of the B allele is 0.501, we calculate:

D = 0.521 - (0.940 * 0.501) = 0.521 - 0.471 = 0.050

Thus, the coefficient of linkage disequilibrium (D) is 0.050, which corresponds to option A.