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?
Frequency of allele A = 0.98
p=0.98
Frequency of allele a= 0.02
q=0.02
Individuals with allele A;
Individuals with allele a;
q2 =
p2= 0.982 =
P2+ 2pq + q2 = 1
Individuals with the disease will be obtained by;
q2×7500
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