Hardy weinburg equation please HELP!!!!!!!!!!!!!!!!!!!!!!!!!

[confuzd441]

if 48% of a population carry dominant H gene using hardy weinberg what % have Hh??
thats all the info i have
please resond asap

[mith]

p=48
q=52
pq=?

[confuzd441]

yeah i know stupid huh??
i get that q= 0.52 but what nxt??

[mith]

pq=?
Read the post.

[victor]

48% population are dominant H (homozygotic)=0.48= p (quadrat)
p=about 0.7
p+q=1
(0.7) + q = 1
q=0.3
then how many Hh?
According to the equation= p(quadrat) + 2pq + q(quadrat)
so, Hh is 2pq
=2x(0.7)x(0.3)
=0.42
=about 42%

[MrMistery]

Since this was a homework related question i think mithril was trying to make him get the answer himself... He wasn't succeding much

[mith]

whoops, just realized I forgot the 2 infront of the pq.
@victor, he said that 48% carry the gene not that they're homozygous for it. i.e. it's allele frequency

[MrMistery]

@mithril
From what i can tell victor is right. Hrady weinberg 101
HH- p*p
Hh-2pq
hh-q*q
So to find out what p is you just need to make the sqare root of 48
Regards
ANdrew

[mith]

if 48% of a population carry dominant H gene using hardy weinberg

that just means there are 48% with the H gene, not that 48% are p^2.

[zami'87.]

I think that here p=0.48 alel freq cause p^2 is freq of diseased.
p=0.48,q=1-p,2pq=2*0.48*0.52
correct me if i'm wrong

[MrMistery]

Ok guys, here it is:
48% carry H gene. That means that the other 52% do not have the H dominant gene, which means that they are homozygotus recessive. From here we can find out the value of q(the frequence of h recessive gene) since hh=q^2. Using windows calculator we get q=0.7211. Since q+p=1 then p=0.2789.
Now the frequence of Hh heterozygotus is: 2pq=2*0.7211*0.2789=0.4022. So 40,22% of your individuals are heterozygotus.
Sorry i missread it the first time
PS: I hate this populational genetics things, never could stand them!!
Regards,
Andrew

[zami'87.]

nice job Andrew!
I got 0.4992