## Hardy-Weinberg equation?

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### Hardy-Weinberg equation?

I'm having some trouble with a question about calculating the frequency of alleles.

I understand that :

q= the frequency of recessive alleles (a)

p= the frequency of dominant alleles (A)

2pq= the frequency of Aa

The question asks us to find q2, q, p and 2pq for a population of 16 pigs (four of which are black and homozygous ressive aa, and the rest are white.)

This is all the information i have been given and i am completely confused.

I assume that q2=aa, and therefore q2=4 (because there are four black pigs, right?). But thats as much as i know. Can anyone help me find out what q, p and 2pq are?

This may seem really easy, but i just dont get it. Am i supposed to get a negative answer for p?

I understand that :

q= the frequency of recessive alleles (a)

p= the frequency of dominant alleles (A)

2pq= the frequency of Aa

The question asks us to find q2, q, p and 2pq for a population of 16 pigs (four of which are black and homozygous ressive aa, and the rest are white.)

This is all the information i have been given and i am completely confused.

I assume that q2=aa, and therefore q2=4 (because there are four black pigs, right?). But thats as much as i know. Can anyone help me find out what q, p and 2pq are?

This may seem really easy, but i just dont get it. Am i supposed to get a negative answer for p?

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### hardy-weinberg equation

Umm.. how about like this:

from the question we got:

there are 16 pigs and 4 of them are black (recessive aa).

it means that : 16 pigs are 100%

4 pigs are 25% (aa 25%)

q2=0.25 -------> q=0.5

p + q = 1 ---------->p + 0.5 = 1 -----> p = 0.5

AA (homozygotus) are p2----> 0.5 quadrat = 0.25 = 25%

2Aa (heterozygotus) are 2pq----> 2 x (0.5) x (0.5) = 0.5 = 50%

so the equition is: AA + 2Aa + aa = 1

25% + 50% +25% = 1

from the question we got:

there are 16 pigs and 4 of them are black (recessive aa).

it means that : 16 pigs are 100%

4 pigs are 25% (aa 25%)

q2=0.25 -------> q=0.5

p + q = 1 ---------->p + 0.5 = 1 -----> p = 0.5

AA (homozygotus) are p2----> 0.5 quadrat = 0.25 = 25%

2Aa (heterozygotus) are 2pq----> 2 x (0.5) x (0.5) = 0.5 = 50%

so the equition is: AA + 2Aa + aa = 1

25% + 50% +25% = 1

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