Hardy-Weinberg

[Awuah29]

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Hey guys ,

need help on this . Why is the answer C ?
According to the Hardy-Weinberg Principle, if there are only two eye colors: brown and blue and 25% of the population have recessive blue eyes, what percentage of the population is heterozygous for brown eyes?

A. 12.5%
B. 25%
C. 50%
D. 75%
E. 100%


Thanks

 

[yakuza]

Hey guys ,

need help on this . Why is the answer C ?
According to the Hardy-Weinberg Principle, if there are only two eye colors: brown and blue and 25% of the population have recessive blue eyes, what percentage of the population is heterozygous for brown eyes?

A. 12.5%
B. 25%
C. 50%
D. 75%
E. 100%


Thanks

p^2 + 2pq + q^2 = 100%

p^2= dominant
q^2= recessive
2pq= heterozygous

q^2 = 0.25 therefore q= 0.5 <- do that math

and since p + q = 1, then p=0.5 <-- do math again

SO 2pq = ? <- just plug into 2pq since 2pq stands for heterzygous

2(0.5 x 0.5) = 0.5

and there you have it

 

[AndyK]

So hady weinberg is p^2 + 2pq + q^2 = 1 and p + q = 1

q^2 = .25, t/f q = .5
p + .5 = 1 t/f p = .5

So the frequency of heterozygotes is 2pq = 2(.5)(.5) = .5 = 50%