Given:
Population: 1000
q²= .46
Answers: Homozygous Recessive Individuals // q²= .46
Frequency of Recessive allele // q= .67
Homozygous Dominant Individuals // p²= .10
Frequency of Dominant allele // p= .33
Heterozygous Individual // 2pq= .44
Steps to solving a Hardy Weinberg Problem:
- Step 1: Find either q² or p². In this case I was already given the q² value (q²= .46) which represents the homozygous recessive individuals.
- Step 2: Since we have q², we can now find q (frequency of the recessive allele) by taking the square root of .46, which then gives us q= .67 .
- Step 3: The next thing we do is find p (frequency of the dominant allele) by subtracting 1 - q . Once we subtract, our p value is now .33 .
- Step 4: To find p² (homozygous dominant individual), all we have to do is square the p value (.33) which then turns to .10.
- Step 5: To find the 2pq (heterozygous individual), you just have to multiply 2 x p x q. Once we plug in the values and after we multiply, 2 x (.33) x (.67), the frequency for heterozygous individuals is .44
- Step 6: Now to figure out the number of individuals that hold each genotype (homozygous dominant, heterozygous, and homozygous recessive) we use our given number of the population, 1000, and multiply it to q², p², and 2pq.
Homozygous Dominant individuals: 1000 x .10 = 100 individuals
Homozygous Recessive Individuals: 1000 x .46 = 460 individuals
Heterozygous Individual: 1000 x .44 = 440 individuals
- Step 7: The last step is to find the frequencies of the dominant and recessive allele. All there is to do is simply turn our p and q values into percentages.
p = .33 → 33%
q = .67 → 67%
