# How to solve Hardy-Weingburg problems on MCAT??

Discussion in 'MCAT Study Question Q&A' started by mytoechondriac, Jun 19, 2008.

1. ### mytoechondriac

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Hi!!

Is there an intuitive way to solve for the p and q values when you just have the p^2 or q^2 values? I can do it when the squared values are easy(i.e. 0.36 is .6 squared) but how can you do it when it's a value like .15 or .35? (these are actual values I had to use when I went through an AAMC CBT this morning).

Please help... I don't wanna be screwed on the MCAT because of this stupid mathematical ******ation issue of mine.!!!

3. ### ThePandaFactor

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well 0.15 is just under 0.16 and 0.35 is just under 0.36 so you could estimate those. Might be close enough.

4. ### mytoechondriac

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OMG, I cannot believe I did not think of that. Thank you!!

5. ### WinterLights

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well p^2+2pq+q^2=1; using p^2=.09 and q^2=.49

.09+2pq+.49=1; rearrange so that 2pq=0.42; pq=.21

at least one of the square root values (p or q) should be easy to figure out. After you figure that value out, you can plug it into pq=0.21 and get the other value.

q=(0.21/0.3)=0.7; so

p = 0.3 and q=0.7

another example

0.36+2pq+0.16=1
2pq=0.48
pq=0.24
p=0.6 (you should be able to tell that square root of 0.36 is 0.6)
q=0.4

I doubt they will give you square roots that are hard to derive (square root of 0.6 for example is too difficult to find without a calculator).

EDIT: if they gave you 0.15 and 0.35, p+q would not equal 1. Shady. If they do not give you at least one easy to find value then just do what the other poster said and estimate.

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