Answer :
9%
Why?
The frequency for allele a is 30% or 0.3. Let's call this frequency p.
p^2 would then represent the frequency of genotype aa (this is is derived from the formula p^2+2pq+q^2=1 where p and q represent alleles a and A respectively. So p^2=aa, pq=Aa and q^2=AA)
0.3^2=0.09 or 9%
Why?
The frequency for allele a is 30% or 0.3. Let's call this frequency p.
p^2 would then represent the frequency of genotype aa (this is is derived from the formula p^2+2pq+q^2=1 where p and q represent alleles a and A respectively. So p^2=aa, pq=Aa and q^2=AA)
0.3^2=0.09 or 9%
Answer:
The correct answer would be 9%.
As the population is in genetic equilibrium, then the population must be following the Hardy-Weinberg equations:
p + q = 1 and
p² + q² + 2pq = 1
The frequency of A (dominant allele) in a population is 70 percent.
Thus, the value of p would be 70/100 which comes out be 0.7
Similarly, the value of q would be equal to 0.3
The frequency of individual with homozygous recessive genotype, that is, aa would be equal to q².
So, q² = [tex](0.3)^{2}[/tex] ⇒ 0.09
Thus, the frequency of individual with genotype aa would be equal to 9 percent.
