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**Hardy Weinberg problem plz help**

If 4% of a population in Hardy Weinberg equilibrium expresses a recessive trait, what is the probability that the offspring of 2 individuals who do not express the trait will express it?

I did what I could and got an answer. My work is below. Could you please check my work and comment? Thanks.

rr = expresses recessive trait = 4% = 4/100

Rr, RR = doesn't express recessive trait = 100-4= 96% = 96/100

q = 2(4) + 96 = 104/200 = 0.52

The punnett square Rr X Rr can only result in offspring who will express the recessive trait so

rr would equal 0.52*o.52= 0.270