Question

2. Two polymorphic loci have alleles D/d and 5/6. In a sample of the population, the haplotypes are found as follows: D5, 396; D6, 198; d5, 405; d6, 1.

1. First, find the frequencies of each allele.

2. Then, determine if the loci are in linkage equilibrium.

Answer 1

1. Haplotype No. of individuals

D5 396

D6 198

d5 405

d6 1

Allele frequency of D = individuals with (D5+D6) / total no. of individuals

= (396+ 196)/ 1000 = 594/1000 = 0.594

Allele frequency of d= individuals with (d5+d6)/total no. of individuals

= (405+1)/ 1000= 406/1000 = 0.406

Allele frequency of 5= individuals with (D5+d5)/1000

= (396+405)/1000 = 801/1000 = 0.801

Allele frequency of 6 = individuals with (D6+d6)/1000

=(198+1)/1000 = 199/1000 = 0.199

2. Linkage disequilibrium can be calculated as, LD= frequency of haplotype D5 - (Frequency of allele D frequency of allele 5)

= (396/1000 ) - (0.5940.801)

= 0.396- 0.476

= -0.08

As D is not equal to zero, alleles are not in linkage quilibrium.

Haplotype frequency of D5 observed in population is not equal to product of individual allele frequencies in the population, so the loci are said to be not in linkage equilibrium.