Hi everyone
I just wanted to give you some help on the Hardy Weinburg homework problems.
1. In a population of 100 individuals 49% are of the NN blood type. What
percentage is expected to be MN assuming Hardy-Weinberg equilibrium?
A. 9%
B. 21%
C. 42%
D. 51%
E. insufficient information to answer question
OK, this one was a bit difficult because it is an example of codominance.
In this cause NN represents q squared.
So if q squared = .49, then if you take the square root of each side, q = .7
Remember that p + q = 1
So if q = .7 then p = .3
Now to figure out the heterozygous condition, you need to multiply using 2pq
So, 2(.7)(.3) = .42 or 42%
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2 comments:
i dont see thats question what page is it on?
I found my sheet i see the question now
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