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.