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colton4286
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Hardy Weinberg problem please 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
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