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Chapter 25: Problem 9

In a population that meets the Hardy-Weinberg equilibrium assumptions, \(81 \%\) of the individuals are homozygous for a recessive allele. What percentage of the individuals would be expected to be heterozygous for this locus in the next generation?

Short Answer

Expert verified
Answer: In the next generation, 18% of individuals are expected to be heterozygous.

Step by step solution

01

Determine the frequency of homozygous recessive individuals (\(q^2\))

We are given that 81% of the population are homozygous recessive. To express this as a decimal we divide by 100: \(q^2 = \frac{81}{100} = 0.81\)
02

Find the frequency of the recessive allele (\(q\))

To find the frequency of the recessive allele (\(q\)), we need to take the square root of the frequency of homozygous recessive individuals (\(q^2\)): \(q = \sqrt{q^2} = \sqrt{0.81} = 0.9\)
03

Calculate the frequency of the dominant allele (\(p\))

Since there are only two alleles in this case (dominant and recessive), the frequency of the dominant allele (\(p\)) is equal to 1 minus the frequency of the recessive allele (\(q\)): \(p = 1 - q = 1 - 0.9 = 0.1\)
04

Calculate the frequency of heterozygous individuals (\(2pq\))

Now that we have the frequencies of the dominant (\(p\)) and recessive (\(q\)) alleles, we can use the Hardy-Weinberg equilibrium equation to find the frequency of heterozygous individuals (\(2pq\)): \(2pq = 2 \times 0.1 \times 0.9 = 0.18\)
05

Convert the frequency of heterozygous individuals to a percentage

To find the percentage of heterozygous individuals, we need to multiply the frequency of heterozygous individuals (\(2pq\)) by 100: Heterozygous percentage = \(0.18 \times 100 = 18\%\) The expected percentage of individuals that would be heterozygous for this locus in the next generation is 18%.

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