Hardy Weinberg Equilliburim Question

Question

Assume there are 200 individuals in a population. If q²=0.2 and 2pq=0.5, what is p²?

Answer choices are 0.75 , 0.03 , 0.09 , 1, and 0.55

Answer 1

p²+2pq+q²=1

q² = 0.2 ----> q = √0.2 ------> q = 0.447 or 0.45

we also know that p + q = 1 so ------> p + 0.45 = 1

p = 1 - 0.45 -------> p = 0.55

...square both sides...

p² = 0.30

p² + 2pq + q² = 1

Plug it into equation (with known variables) ------> 0.30 + 0.5 + 0.2 = 1 --------> this works!

Answer 2

If q², which is the proportion of the homozygous recessive individuals, is 0.2, and the population is 200 individuals, then 40 of the individuals are homozygous recessive. (.2 x 200 = 40)

We also know that 2pq is 0.5, so 100 of the individuals are heterozygous and have both a dominant and a recessive allele.

So, if there are 200 individuals in a population and each has 2 alleles, there are 400 total alleles in this population. 40 individuals have 2 recessive alleles each, so that’s 80 recessive alleles, and 100 have one recessive allele each, so that’s 100 recessive alleles. So, we can use this info to find q- the frequency of the recessive allele. q= 2(40)+100 / 2(200) = 0.45.

since p+q=1, p+0.45=1 so p= 0.55. P² is 0.30 ... the answer choices must be incorrect!