Question

If the frequency of an autosomal recessive trait in humans is 1 out of 100 births, what would be the expected frequency of heterozygote carriers for the trait if we assume that the gene is in Hardy-Weinberg equilibrium?

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

Answer 1

The expected frequency of heterozygote carriers for an autosomal recessive trait with a frequency of 1 out of 100 births, assuming Hardy-Weinberg equilibrium, is 0.18 or option 3).

Step-by-step explanation:

If the frequency of an autosomal recessive trait in humans is 1 out of 100 births, we can denote this as q² (where q is the frequency of the recessive allele). Hence, q² = 0.01, and q = 0.1. To find the expected frequency of heterozygote carriers for the trait, we use the Hardy-Weinberg equation p² + 2pq + q² = 1, where p is the frequency of the dominant allele and 2pq represents the frequency of heterozygotes. Since p + q = 1, p = 1 - q = 0.9. Thus, the frequency of heterozygote carriers, 2pq, can be calculated as 2 × 0.9 × 0.1, which equals 0.18. Therefore, the expected frequency of heterozygote carriers is 3) 0.18.