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Chapter 25: Problem 15

A certain form of albinism in humans is recessive and autosomal. Assume that \(1 \%\) of the individuals in a given population are albino. Assuming that the population is in HardyWeinberg equilibrium, what percentage of the individuals in this population is expected to be heterozygous?

Short Answer

Expert verified
Answer: 18%

Step by step solution

01

Determine the frequency of the recessive allele (q)

Since 1% of individuals in the population are albino, we are given the frequency of homozygous recessive individuals (aa). Therefore, $$q^2 = 0.01$$. Now we will find the frequency of the recessive allele (q) by taking the square root of $$q^2$$: $$q = \sqrt{0.01} = 0.1$$
02

Determine the frequency of the dominant allele (p)

Now that we have the frequency of the recessive allele (q), we can find the frequency of the dominant allele (p). Since we know that p + q = 1, we can calculate p like this: $$p = 1 - q = 1 - 0.1 = 0.9$$
03

Calculate the frequency of heterozygous individuals (Aa)

Now that we have both p and q, we can use the Hardy-Weinberg equation to find the frequency of heterozygous individuals (Aa). This can be calculated using the term $$2pq$$: $$2pq = 2 * 0.9 * 0.1 = 0.18$$ This means that 18% of the individuals in the population are expected to be heterozygous.

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Most popular questions from this chapter

The genetic difference between two Drosophila species, \(D\). heteroneura and \(D .\) sylvestris, as measured by nucleotide diversity, is about 1.8 percent. The difference between chimpanzees \((P\) troglodytes and humans (H. sapiens) is about the same, yet the latter species are classified in different genera. In your opinion, is this valid? Explain why.

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