Question

Which of the following genotype frequencies of AA,Aa and aa respectively,satisfy the Hardy-Weinberg principle?
A. 0.25, 0.50, 0.25
B. 0.36, 0.55, 0.09
C. 0.49, 0.42, 0.09
D. 0.64, 0.27, 0.09
E. 0.29, 0.42, 0.29

Answer 1

To determine which genotype frequencies satisfy the Hardy-Weinberg principle, we need to check if they adhere to the equation p^2 + 2pq + q^2 = 1, where p^2 represents the frequency of homozygous dominant individuals (AA), 2pq represents the frequency of heterozygous individuals (Aa), and q^2 represents the frequency of homozygous recessive individuals (aa).

Let's calculate the values for each option:

A. 0.25, 0.50, 0.25
(0.25)^2 + 2(0.25)(0.50) + (0.50)^2 = 0.25 + 0.50 + 0.25 = 1
This option satisfies the Hardy-Weinberg principle.

B. 0.36, 0.55, 0.09
(0.36)^2 + 2(0.36)(0.55) + (0.55)^2 = 0.1296 + 0.396 + 0.3025 ≈ 0.8271
This option does not satisfy the Hardy-Weinberg principle.

C. 0.49, 0.42, 0.09
(0.49)^2 + 2(0.49)(0.42) + (0.42)^2 = 0.2401 + 0.4116 + 0.1764 ≈ 0.8281
This option does not satisfy the Hardy-Weinberg principle.

D. 0.64, 0.27, 0.09
(0.64)^2 + 2(0.64)(0.27) + (0.27)^2 = 0.4096 + 0.3456 + 0.0729 ≈ 0.8281
This option does not satisfy the Hardy-Weinberg principle.

E. 0.29, 0.42, 0.29
(0.29)^2 + 2(0.29)(0.42) + (0.42)^2 = 0.0841 + 0.2436 + 0.1764 ≈ 0.5041
This option does not satisfy the Hardy-Weinberg principle.

Therefore, the genotype frequencies that satisfy the Hardy-Weinberg principle are option A. 0.25, 0.50, 0.25.