Question

For two alleles at frequencies p and q in a population at Hardy-Weinberg equilibrium, which of the following statements is most likely to be TRUE?

A. When p = q, all individuals in the population are heterozygous.

B. When p = q, no individuals in the population are heterozygous.

C. When p = 1, half of the individuals in the population are heterozygous.

D. When p = 0, all individuals in the population are heterozygous

E. When p = 0.5, half of the individuals in the population are heterozygous.

Please show all work and explanations. Thank you.

Answer 1

A and B

If p = q

Then, p^2 + 2pq + q^2 = 1, can be written as,

p^2 + 2pp + p^2 = 1

p^2 + 2p^2 + p^2 = 1

4p^2 = 1

p^2 = 1/4

p = 1/2 = 0.5

So, p + q = 1

q = 1 - 0.5 = 0.5

So, in the population,

p^2 = 0.5×0.5 = 0.25 = q^2

2pq = 2×0.5×0.5 = 0.5

It means half individuals in the population will be heterozygotes and one fourth will be each homozygous dominant and homozygous recessive.

So, A and B are false.

C.

If p is 1, then q will be 0.

It means 2pq and q^2 = 0

All individuals will be p^2, homozygous dominant.

So, this is also false.

D.

If p is 0, it means q is 1.

2pq and p^2 = 0

All individuals will be q^2, homozygous recessive.

So, this is also false.

E.

If p is 0.5, then q will also be 0.5.

It means p^2 = 0.5×0.5 = 0.25

q^2 = 0.5×0.5 = 0.25

2pq = 2×0.5×0.5 = 0.5

So, half will be heterozygous, and one fourth will be each homozygous dominant and recessive.

Note -

Use,

p + q = 1

p^2 + 2pq + q^2 = 1

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