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Luden [163]
3 years ago
5

In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of the allele a is 0.70. What i

s the percentage of the population that is homozygous for this allele? (Your answer should have two decimal places following an initial 0. For example, if your answer was 56%, you should enter 0.56)
Social Studies
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer: 0.49

Explanation: In Hardy-weinberg, the population is in equilibrium, which means the expected proportions for the genotype is expressed as:

(p+q)² = 1 or p²+2pq+q² = 1, where

p is the frequency of the dominat allele;

q is the frequency of the recessive allele;

If the frequency of the allele a is 0.70, it means q=0.70.

To find the percentage  who is recessive homozygous:

q² = (0.70)² = 0.49

Therefore, the percentage of population homozygous for allele <em><u>a</u></em> is 0.49

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