Calculating Genotype Frequencies After One Generation of Random Mating: Hardy-Weinberg Application

The Hardy-Weinberg equilibrium is a fundamental concept in population genetics, allowing us to predict genotype frequencies from allele frequencies in a population that mates randomly and is not subject to evolutionary forces. Let’s see how this principle is applied to calculate the expected genotype frequencies after one generation of random mating.

Given Data

• Frequency of allele A1 ($p$) = 0.75

• Frequency of allele A2 ($q$) = 0.25

Hardy-Weinberg Formula

The Hardy-Weinberg equation for two alleles is:

p2+2pq+q2=1p2+2pq+q2=1

Where:

• $p^2$ = frequency of homozygous dominant genotype (A1A1)

• $2pq$ = frequency of heterozygous genotype (A1A2)

• $q^2$ = frequency of homozygous recessive genotype (A2A2)

Step-by-Step Calculation

1. Calculate $p^2$ (A1A1):

   p2=(0.75)2=0.5625p2=(0.75)2=0.5625

2. Calculate $2pq$ (A1A2):

   $$2pq=2\times 0.75\times 0.25=2\times 0.1875=0.375$$

3. Calculate $q^2$ (A2A2):

   q2=(0.25)2=0.0625q2=(0.25)2=0.0625

Resulting Genotype Frequencies

• A1A1: 0.5625

• A1A2: 0.375

• A2A2: 0.0625

Correct Option

From the options provided:

• (1) 0.5625; 0.375; 0.0625

• (2) 0.5625; 0.0625; 0.375

• (3) 0.750; 0.250; 0.350

• (4) 0.5625; 0.1525; 0.0625

The correct answer is (1) 0.5625; 0.375; 0.0625.

Why Is This Important?

• Predicting Genetic Structure: This calculation predicts the genetic structure of a population under random mating.

• Detecting Evolution: Deviations from these expected frequencies can indicate evolutionary forces at work.

• Genetic Counseling: Helps estimate carrier and disease frequencies in human populations.

Conclusion

After one generation of random mating, the genotype frequencies for A1A1, A1A2, and A2A2 will be 0.5625, 0.375, and 0.0625, respectively, according to the Hardy-Weinberg equilibrium.

Correct answer: (1) 0.5625; 0.375; 0.0625