Hardy-Weinberg Equilibrium

[RHONDAROBINSON]

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Hey,

Cannot anyone explain this question to me?

In a population in Hardy-Weinberg equilibrium, the frequency of the dominant allele D is three times that of the recessive allele (d). What is the frequency of heterzygotes in the population?

A. 6.25%
B. 25%
C. 37.5%
D. 56.25%
E. 75%

I thought that I understood the Hardy-Weinbery theory until I came across this question on a practice DAT. Thanks in advance.

 

[DrBowtie]

Hint: D=.75 d=.25

 

[Dion]

I think the answer is 37.5 %. If the Dominant allele is three times more, one ratio, I could think of is 3:1 dominant:recessive (3 of 4, or 75%, are dominant). The reccessive allele is thus 1 of 4, 25%. Because you know d is .25 and D is .75, you multiply them by each other to find the probability of the homozygote (Dd), which is 0.1875. If you do a Punnent square involving the two alleles, then the homozygote (Dd) will appear twice and therefore you take the product of Dd, 0.1875, and multipy it two times and get .375.

I hope I helped.

 

[Streetwolf]

System of equations to start it out. This will work for any setup.

D = dominant, R = recessive

D + R = 1 (frequency of alleles adds up to 1)
D = 3R (frequency of dominant is 3x that of recessive)

So 3R + R = 1 ==> (implies) 4R = 1 ==> R = 0.25 ==> D = 3(0.25), D = 0.75. Then you need to know the following:

Homozygous dominant = D^2
Heterozygous = 2DR
Homozygous recessive = R^2

Answer is 37.5%

 

[RHONDAROBINSON]

Thanks to all you for the explanations.


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