Solutions For An Introduction to Genetic Analysis Chapter 18 Problem 2P

a) Gene diversity can be calculated using the following formula:

When we observe the figure, we notice that out of the 7 chromosomes, only two chromosomes are affected in the indel. The chromosomes 5 and 6 are affected.

The gene diversity, for the indel can be calculated in the following way:

Since, there are four differences; the gene diversity for the microsatellite locus can be calculated in the following way:

Since there is only one difference, the gene diversity for the SNP locus at position 3 would be:

b) If the given sequence is shortened and the position 24 is present at the end, the number of haplotypes would be five instead of six. The five haplotypes are: A, B, C, D, and E.

In addition, haplotype D would contain IIa and IIc and the haplotype E would contain both IIb haplotypes.

c) The linkage equilibrium parameter, D between SNPs at positions 29 and 33 stands for the difference between the expected and observed frequency of a haplotype.

At position 29, the nucleotide G is present in 6 out of the 7 haplotypes. The frequency value of 6/7 is 0.857. The nucleotide A is present at position 33. This nucleotide is present in 3 out of 7 haplotypes. The frequency value of 3/7 is 0.429.

A random combination of both the nucleotides of A and G would be the multiplication of the individual frequencies.

The observed frequency of the AG haplotype is 3 out of 7 haplotypes. The frequency value is 0.429. This is the value of . Substituting the values in the equation:

Since the value of D is not 0, so the population is in linkage disequilibrium for the haplotypes of A and G at positions 29 and 33.