hardy-weinberg HELP

[karenkay]

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)
http://life.bio.sunysb.edu/bio350/Answerkey.htm

[HenryHR]

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]

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]

oops...i mean p is the FREQUENCY of dominant allele while q is also the FREQUENCY of recessive allele. sorry about that =]

[Jesse2504]

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.

This can be misleading, since
AA = p^2 = 0.5
Using algebra you can make the assumption that
p = root 0.5 = 0.707 which is false