How to solve Hardy-Weingburg problems on MCAT??

[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.!!!

 

[ThePandaFactor]

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.

 

[mytoechondriac]

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.

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

 

[WinterLights]

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.!!!

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.

 

[mytoechondriac]

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.

Thank you winterlight!!