Question

In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of the allele a is 0. 3. What is the frequency of individuals that are homozygous for this allele?.

Answer 1

Here, given-

homozygous alleles 'a' have a frequency of 0.3.

Also the alleles are in equilibrium in a Hardy-Weinberg population. The frequency of individuals that are homozygous for this allele are= 0.49.

The Hardy-Weinberg equilibrium can be defined as the principle which states that the variation in the genetic makeup of a population remains constant and unchanged till there are no external interferences, influencing the population.

Calculation-

$p + q = 1\\0.3 + q= 1\\q= 1- 0.3\\\\q= 0.7$

Then to find the frequency of the individuals homozygous for this allele the following formula needs to be used-

$p^{2} + 2pq + q^{2} =1$

$p^{2} = dominant homozygous frequency\\2pq= heterozygous frequency\\q^{2} = recessive homozygous frequency$

Thus, the individuals homozygous for the allele can be calculated by $q^{2} = (0.7)^{2} \\= 0.49$

Learn more about the Hardy-Weinberg equilibrium here-

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