Answer to Question #198476 in Evolution for lulu

Question

There are two alleles in the population of 7500 individuals: the wild-type allele A and the mutation allele a with frequencies of 0.98 and 0.02.

If  population is in Hardy-Weinberg equilibrium, how many individuals would you see with the disease if it was fully recessive?

 

Answer 1

Frequency of allele A = 0.98

p=0.98

Frequency of allele a= 0.02

q=0.02

Individuals with allele A;

$0.98×7500 =7350$

Individuals with allele a;

$0.02×7500 =150$

q² =$0.02×0.02=0.0004$

p²= 0.98² = $0.9604$

P²+ 2pq + q² = 1

Individuals with the disease will be obtained by;

q²×7500

$0.0004×7500= 3$