Question

A human autosomal recessive trait appears in 1 in 100 births. What percent of people are homozygous dominant for this trait?

1) 0.01
2) 0.10
3) 0.18
4) 0.81
5) 0.90

Answer 1

The percentage of people who are homozygous dominant for an autosomal recessive trait that appears in 1 in 100 births is 81%. This is calculated using the Hardy-Weinberg principle, where p² + 2pq + q² = 1 and the frequency of the recessive trait, q², is 0.01.

Step-by-step explanation:

To determine the percentage of people who are homozygous dominant for an autosomal recessive trait that appears in 1 out of 100 births, we can employ the Hardy-Weinberg principle. According to this principle, the frequency of genotypes in a population can be represented as p² + 2pq + q² = 1, where p and q are the frequencies of the dominant and recessive alleles, respectively, and p + q = 1.

If the trait appears in 1 out of 100 births, that means q² = 0.01, because q² represents the frequency of homozygous recessive individuals in the population. To find q, we take the square root of 0.01, which gives us 0.1. Then, we can calculate p as 1 - q, which equals 0.9. p represents the frequency of the dominant allele.

The percentage of homozygous dominant individuals is represented by p². Therefore, we calculate p² as 0.9², which is 0.81 or 81%. Thus, 81% of the population would be homozygous dominant for this trait, making option (4) the correct answer.