Hardy Weinberg HELP!
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 Garter
 Posts: 2
 Joined: Sun Feb 01, 2009 2:45 am
Hardy Weinberg HELP!
So I am in a bio class, and i am so lost on everything! If anyone has any tips for catching up quickly on all of this that would be great!
Here is the problem...
In a class, 27% of the students are PTC tasters (a dominant trait), whereas 63% are nontasters. (note: the square root of 0.27 = 0.52, the square root of 0.63 = 0.79)
I need to find the frequency of the dominant allele and the recessive allele. I know the formula, but I don't know where to plug things in. HELP!
Here is the problem...
In a class, 27% of the students are PTC tasters (a dominant trait), whereas 63% are nontasters. (note: the square root of 0.27 = 0.52, the square root of 0.63 = 0.79)
I need to find the frequency of the dominant allele and the recessive allele. I know the formula, but I don't know where to plug things in. HELP!

 Garter
 Posts: 2
 Joined: Sun Feb 01, 2009 2:45 am
I have, but i still don't understand.
I know that...
p= frequency of dominant allele
q= frequency of recessive allele
p^2 = % of homozygous dominant individuals
q^2 = % of homozygous recessive individuals
2pq = % of heterozygous individuals
and... p^2 + 2pq + q^2 = 1 and p + q = 1
I just don't understand how to use the formula.
So.. would the frequency by 27 % for the dominant allele? ( That seems to easy, and I don't understand it...)
I know that...
p= frequency of dominant allele
q= frequency of recessive allele
p^2 = % of homozygous dominant individuals
q^2 = % of homozygous recessive individuals
2pq = % of heterozygous individuals
and... p^2 + 2pq + q^2 = 1 and p + q = 1
I just don't understand how to use the formula.
So.. would the frequency by 27 % for the dominant allele? ( That seems to easy, and I don't understand it...)
 alextemplet
 King Cobra
 Posts: 5599
 Joined: Fri Dec 23, 2005 4:50 pm
 Location: South Louisiana (aka Cajun Country)
 Contact:
Re: Hardy Weinberg HELP!
hi,
I have a little problem with the question u have posted. are you sure those are the right percentage values. because the square of each, eg. the square of 0.27 = 0.52 (rounded up) and the sqaure of 0.63 = 0.79 (rounded) however both these values add up to give 1.31 which in hardyweinberg should be 1. so your percentages must be wrong to start with. if you need any more help give us a shout and i will happily lend a hand.
Bob
P.S. if u want i can take you through step by step of the calculation.
I have a little problem with the question u have posted. are you sure those are the right percentage values. because the square of each, eg. the square of 0.27 = 0.52 (rounded up) and the sqaure of 0.63 = 0.79 (rounded) however both these values add up to give 1.31 which in hardyweinberg should be 1. so your percentages must be wrong to start with. if you need any more help give us a shout and i will happily lend a hand.
Bob
P.S. if u want i can take you through step by step of the calculation.
 alextemplet
 King Cobra
 Posts: 5599
 Joined: Fri Dec 23, 2005 4:50 pm
 Location: South Louisiana (aka Cajun Country)
 Contact:
Re: Hardy Weinberg HELP!
sorry my bad. was late last night when i was doing that and i hit the wrong buttons on my calculator lol yeah the above post is correct. so just to clarift:
0.27*0.27 or 0.27^2 = 0.0729
0.63*0.63 or 0.63^2 = 0.3969
now you find the square of the above values:
square of 0.0729 = 0.27
square of 0.3969 = 0.63
0.027 + 0.63 = 0.9 (close enough to 1) lol
if the is wrong someone say, but from my understanding this give you the p value and q value. (q = recessive, p = dominant)
P.S. Hope this is right, if not ignore everything lol
0.27*0.27 or 0.27^2 = 0.0729
0.63*0.63 or 0.63^2 = 0.3969
now you find the square of the above values:
square of 0.0729 = 0.27
square of 0.3969 = 0.63
0.027 + 0.63 = 0.9 (close enough to 1) lol
if the is wrong someone say, but from my understanding this give you the p value and q value. (q = recessive, p = dominant)
P.S. Hope this is right, if not ignore everything lol
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