You observe that the frequency of the AB haplotype at the A and B loci is 0.558 in a population. You also observe that the frequency of the A allele at the A locus is 0.887 while the frequency of the B allele is 0.396. What is the coefficient of linkage disequilibrium (give your answer to three decimal places)

D = 0.207

Explanation:

Coefficient of the linkage disequilibrium

D = hAB - fAfB

Where,

D is the difference between the actual frequency of the AB haplotype (hAB) and the expected frequency of AB (given independent) => (fA * fB)

D = 0 means that the alleles are not associated

=> Varies from -0.25 to 0.25

Given,

Two haplotypes

hAB = 0.558

Frequency of allele A (fA) = 0.887

Frequency of allele B (fB) = 0.396

D = hAB - fAfB

= (0.558) - (0.887 * 0.396)

= 0.558 - 0.351252

= 0.206748

D = 0.207 ( approx. 3 decimal places).

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