Hardy Weinberg eqilibrium-HELP

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Hardy Weinberg eqilibrium-HELP

Post by Juanita1 » Tue Jun 16, 2009 4:37 pm

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.

Please help.

Thank you.

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Post by AstusAleator » Wed Jun 17, 2009 6:11 am

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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