Question

A recessive allele in turtles (n/n) results in an exceptionally long nose. Four percent lf the turtle population have the long nose phenotype. Assuming Hardy-Weinberg Equillibrium:

a) What is the frequency of the recessive allele?

b) What is the frequency of the dominant allele?

c) What is rhe frequency of the heterozygous (N/n)?

Answer 1

Basic formulas:   

where p=frequency of dominent allele in population

q=frequency of recessive allele in population

2pq= frequency of heterozygous individuals

a)Since we believe that homozygous recessive for this gene () represents 4% (i.e=0.04),the squareroot(q) is 0.2(20%). Therefore frequency of recessive allele is 20%.

b)Since q=0.2, p+q=1 then p=0.8(80%). Therefore frequeny of dominent allele is 80%.

c)The frequency of heterozygous individuls is equal to 2pq.In this case,2pq =0.32,which means frequency for this gene is equal to 32% (i.e.2(0.8)(0.2)=0.32).