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In one generation, the frequency of the A allele is 70 percent and the frequency of the a allele is 30 percent. What are the chances of an individual in the next generation having the genotype aa if the population is in genetic equilibrium? 49% 42% 30% 9%

Sagot :

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.

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