## hardy-weinberg HELP

Genetics as it applies to evolution, molecular biology, and medical aspects.

karenkay
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Posts: 2
Joined: Sun Jun 07, 2009 6:55 pm

### hardy-weinberg HELP

Hello,

I am having trouble understanding how to use the hardy-weinberg equilibrium equation.

I understand that the allelic frequencies must always add to 100% (i.e. p+q = 1) but I don't understand how to do a few questions.

If AA = 0.50, Aa = 0.25 and aa = 0.25, what are p and q? Is this population in equilibrium? The answer key says that p = 0.625 and q = 0.375. How is this?

I am working through this worksheet: (in particular #3)

HenryHR
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Posts: 2
Joined: Wed Jun 10, 2009 2:44 pm
let the whole population be N ,thus the total number of alleles should be 2N. so p=(0.5×2N+0.25N)/2N=0.625,and the same to q. actually,the N is not neccessary here,I use it here just for easier understanding~

anitapangz
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Posts: 4
Joined: Thu Jun 18, 2009 10:07 am
Location: jakarta ~ id
do you know the formula :
p^2 + 2pq + q^2 = 1

AA = p^2
2pq = Aa
aa = q^2

p is the allele for dominant one (A) while q is the allele for recessive one (a).
to know whether a population is in equilibrium state first you have to calculate for "expected" then compared it to "observed" one using chi-square test.
it's a test used to prove whether an experiment goes on well.

anitapangz
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Posts: 4
Joined: Thu Jun 18, 2009 10:07 am
Location: jakarta ~ id
oops...i mean p is the FREQUENCY of dominant allele while q is also the FREQUENCY of recessive allele. sorry about that =]

Jesse2504
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Posts: 47
Joined: Mon Jun 01, 2009 4:28 pm

### Re:

anitapangz wrote:do you know the formula :
p^2 + 2pq + q^2 = 1

AA = p^2
2pq = Aa
aa = q^2

p is the allele for dominant one (A) while q is the allele for recessive one (a).
to know whether a population is in equilibrium state first you have to calculate for "expected" then compared it to "observed" one using chi-square test.
it's a test used to prove whether an experiment goes on well.