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 Hardy-Weinberg equilibrium, what percentage of the individuals in this population is expected to be heterozygous?
To answer this question, we can use the Hardy-Weinberg equilibrium equation. According to this equation, in a population at equilibrium, the frequency of alleles remains constant from generation to generation.
Let's define the relevant variables first:
- p: frequency of the dominant allele in the population
- q: frequency of the recessive allele in the population
- p^2: frequency of homozygous dominant individuals (AA)
- q^2: frequency of homozygous recessive individuals (aa)
- 2pq: frequency of heterozygous individuals (Aa)
Given that 1% of individuals in the population are albino (aa), we know that q^2 = 0.01. Since albinism is a recessive trait, q^2 represents the frequency of homozygous recessive individuals.
Using these facts, we can set up the following equation:
p^2 + 2pq + q^2 = 1
Since q^2 = 0.01, we can substitute this value into the formula:
p^2 + 2pq + 0.01 = 1
Now, we need to solve for p and q.
We know that p + q = 1 since the sum of all allele frequencies must be 1. Rearranging this equation, we have q = 1 - p.
Substituting this value into the previous equation:
p^2 + 2p(1 - p) + 0.01 = 1
Simplifying:
p^2 + 2p - 2p^2 + 0.01 = 1
Combining like terms:
-p^2 + 2p + 0.01 = 0
Now we have a quadratic equation:
p^2 - 2p - 0.01 = 0
We can solve this equation using the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -2, and c = -0.01.
Using the quadratic formula, we find two values for p:
p ≈ 1.0102 and p ≈ 0.9898
Since p represents the frequency of the dominant allele, and q = 1 - p, we take the value of p that is less than 1 to be the correct one, which is p ≈ 0.9898.
Now that we have p, we can find q = 1 - p:
q ≈ 1 - 0.9898
q ≈ 0.0102
Therefore, the frequency of the heterozygous individuals (Aa) in the population is approximately 2pq:
2pq ≈ 2 * 0.9898 * 0.0102
2pq ≈ 0.0202
So, approximately 2% of individuals in this population are expected to be heterozygous.