Hardy Weinberg question
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 Garter
 Posts: 7
 Joined: Fri Jan 25, 2008 4:33 am
Hardy Weinberg question
If you are to estimate frequencies and use HardyWeinberg rule to determine proportions of genotypes/phenotypes which method do you use?
I have quantitative information for dominant and recessive traits. Of course, we don't know what the frequency of hetero. traits are.
So, using the information I have I can determine q^2 (recessive) by dividing the recessive individuals by the total # of people. To get q, take the square root and then to get p(dominant) use the 1q=p equation. To find the frequency of hetero. then do 2pq. To check the values are correct, use the HardyWeinberg equation by plugging the values into: p^2 + 2pq + q^2 = 1
OR, I found online another method to determine if the values follow Hardy Weinberg. Using the observed and expected data I can estimate p:
2 X (# observed dominant) + (# observed recessive) / 2 X total individuals. This gives p, so to get q use the 1p=q equation.
HOWEVER, expected values are then calculated and used in a Chi Square test. No problem getting the expected values ** BUT** what do you use for the observed values when you don't know what the observed heteroz. value is?? When you do the Chi Square, you do the (OE)^2 / E for p, q and 2pq values but we only really know what the observed values for p and q are?!
How do you do the above calculations when you don't know what the observed heter. value is?
I have quantitative information for dominant and recessive traits. Of course, we don't know what the frequency of hetero. traits are.
So, using the information I have I can determine q^2 (recessive) by dividing the recessive individuals by the total # of people. To get q, take the square root and then to get p(dominant) use the 1q=p equation. To find the frequency of hetero. then do 2pq. To check the values are correct, use the HardyWeinberg equation by plugging the values into: p^2 + 2pq + q^2 = 1
OR, I found online another method to determine if the values follow Hardy Weinberg. Using the observed and expected data I can estimate p:
2 X (# observed dominant) + (# observed recessive) / 2 X total individuals. This gives p, so to get q use the 1p=q equation.
HOWEVER, expected values are then calculated and used in a Chi Square test. No problem getting the expected values ** BUT** what do you use for the observed values when you don't know what the observed heteroz. value is?? When you do the Chi Square, you do the (OE)^2 / E for p, q and 2pq values but we only really know what the observed values for p and q are?!
How do you do the above calculations when you don't know what the observed heter. value is?

 Garter
 Posts: 7
 Joined: Fri Jan 25, 2008 4:33 am
Re: Hardy Weinberg question
I agree, the Chi square is probably the better option but I don't have values for the heterozygous trait to use. I have only Recessive and Dominant values.
For example, let's say the trait is blue eyes. If 10 people have blue eyes (recessive) and 24 have brown (dominant), how would I do a punnett to get heterozygous individuals? I don't think I can.
Is it even possible to do the calculation with only Recessive and Dominant traits??
Or, is my option just the p^2 + 2pq + q^2 = 1 equation
For example, let's say the trait is blue eyes. If 10 people have blue eyes (recessive) and 24 have brown (dominant), how would I do a punnett to get heterozygous individuals? I don't think I can.
Is it even possible to do the calculation with only Recessive and Dominant traits??
Or, is my option just the p^2 + 2pq + q^2 = 1 equation
I provided an answer for your question on my web. Please, visit
http://scienceed.fullsubject.com/1on ... t11.htm#14
http://scienceed.fullsubject.com/1on ... t11.htm#14
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