Question

An investigation of 2400 births on a hospital in UK 6 deaths were observed among the new born babies caused by a recessive autosomal allele for a colon defect (co). Assume random mating and answer the following questions. a) What is the frequency of co among newborns? b) What is the frequency of heterozygotes among newborn and adults? c) Calculate the mutation rate for the normal allele to co assuming mutation-selection equilibrium.

Answer 1

Representation:

Genotype for individuals with colon defects = cc = recessive homozygous condition.

Normal individuals, will have genotypes:

CC = Homozygous dominant

Cc = Heterozygous dominant

Since 6 out of 2400 babies die due to colon defect = cc genotype

Genotype frequency for cc = q2 = 6/ 2400 = 0.0025

Allele frequency for q = 0.05 = frequency of c allele

a. Frequency of co among newborn = 0.05

As per Hardy-Weinberg equilibrium p+q = 1

Thus, p = 1- 0.05 = 0.95 = frequency of C allele

b. Heterozygous genotype frequency (Cc) = 2 x p x q =2 x 0.05x 0.95 =0.095

c.

Forward mutation rate (C-c) = µ

Backward mutation rate = (c-C) = ν

p = allele frequency of C

q = allele frequency of c

Considering equilibrium (no other change)

Rate of mutation from normal to mutant Co allele (C-c)

p= ν / (µ + ν)

Since backward mutation is not considered:

Rate of forward mutation = p= 0.95 = 9.5 x 10-1