# Hardy Weinberg problem?

Discussion in 'DAT Discussions' started by teefRcool, Nov 24, 2005.

1. ### teefRcool Senior Member 10+ Year Member

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I'm having problem with this question.

Using Hardy-Weinberg principle, they gave me p and q outright, saying p=.7, q=.3, what is the percent of heterozygotes?

Could someone explain how this works and what the answer is and also some other scenarios that you might think they would other types of problems that deals with hardy weinberg.

Thanks guys, I just get lost when it comes to these problems.

2. ### mitshi Membership Revoked Removed 2+ Year Member

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p and q are simply allelic frequencies. They need to appear in pairs in order to constitute individuals [ i.e., homozygous (p^2 or q^2) or heterozygous (pq) ]

p + q = 1 ----- (1)

Squaring both sides of (1),

(p + q)^2 = 1^2

p^2 + 2pq + q^2 = 1 ----- (2)

From (2),

2pq = 2(0.7)(0.3) = 0.42 ==> 42%

If you understand what I meant, post your answers here for percent of each homozygote...

(Note: Questions of this type in the real DAT seldom involve calculation. They will usually be phrased with some other unrelated stuffs to test your basic understanding on Hardy-Weinberg principle. I remembered a repeated DAT question on the cheetah population has been well explained in DAT Achiever, a great program that had helped me a lot.)

Good luck to you then ...

3. OP

### teefRcool Senior Member 10+ Year Member

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Thanks mitshi, i understand a little. So if we were given pq and q we could find p by itself by using the p^2 + 2pq + q^2 = 1 right?

4. ### mitshi Membership Revoked Removed 2+ Year Member

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That's right. However, thorough understanding to the principle is of utmost importance. The algebraic stuff is just simple math not to be confused with. For instance, they can give you the percent of the dominant homozygotes as 49 % and asking you to determine the recessive allele frequency. Then:

p^2 = 0.49

Square-rooting both sides,

p = 0.7 ==> q = 0.3

5. OP

### teefRcool Senior Member 10+ Year Member

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Thank a ton mitshi!!

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