## Hardy Weinberg eqilibrium-HELP

Discussion of everything related to the Theory of Evolution.

Juanita1
Garter
Posts: 11
Joined: Sat Jun 13, 2009 1:18 pm

### Hardy Weinberg eqilibrium-HELP

In a human population, a mild form of stuttering is caused by an autosomal recessive allele and 1 in 400 people are affected. This condition does not carry a selective disadvantage.

How can you find out the proportion of people in the population who are carriers of the stuttering allele but do not show the condition from just the infor given above?

I really cant get my head round this. I hope someone can see the clue there.

Thank you.

AstusAleator
King Cobra
Posts: 1039
Joined: Tue Jan 24, 2006 8:51 pm
Location: Oregon, USA
if it's at equilibrium then p^2 + 2pq + q^2 = 1 and p + q = 1

where q^2 = (1/400) so q = (1/20), therefore p = 1 - (1/20) = (19/20)

Which means 2pq = (2)(19/20)(1/20) = (19/200)

check this with the original equation:
(19/20)^2 + (19/200) + (1/400) = 1

It's been a couple of years since I've done these, but I think that works.
What did the parasitic Candiru fish say when it finally found a host? - - "Urethra!!"

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