Hardy-Weinberg

Given that q2 = 0.04, I was able to calculate the rest of the variables by:

1. Taking the square root of 0.04 which was 0.2. This gave me q
2. Because p + q = 1, this means that p=0.8.
3. To get p2, I squared p=0.8 which got me p2=0.64
4. Finally, to get 2pq, I calculated (2)(0.8)(0.2) which gave me 0.32

From this, the equation p2 + 2pq + q2 = 1, I can check if the numbers I got were correct by substituting the variables with the numbers that I got so it would become 0.64 + 0.32 + 0.04 = 1.

Given that these numbers were taken from a population of 1000, it means that;

• Since q2=0.04 or 4% of the population, 4 individuals in total, are homozygous recessive individuals and show the recessive phenotype
• Since q=0.2 or 20%, this gave me the frequency of the recessive allele. This means that 20% of the population, 200 individuals in total, have the recessive allele.
• Since p2=0.64 or 64% of the population, 640 individuals in total, are homozygous dominant individuals and show the dominant phenotype.
• Since p=0.8 or 80% of the population, this gave me the frequency of the dominant allele. This means that 80% of the population, 800 individuals in total, have the dominant allele.
• Since 2pq=0.32 or 32% of the population, 320 individuals in total, are heterozygous dominant individuals who carry both alleles but show the dominant phenotype.