Help with HardyWeinberg Equation?
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 Garter
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 Joined: Sun May 13, 2012 8:07 pm
Help with HardyWeinberg Equation?
If about 35% of the human population has type "O" blood,
a) What must the frequency of the type O allele?
b)what proportion of the populaton must be heterozygous carriers for type O?
a) What must the frequency of the type O allele?
b)what proportion of the populaton must be heterozygous carriers for type O?
Re: Help with HardyWeinberg Equation?
We are using the equation p2 + 2pq + q2 = 1, where
p2 = homozygous dominant frequency
2pq = heterozygous frequency
q2 = homozygous recessive frequency
You should know that blood type O is recessive. Therefore, q2 = .35. Taking the square root of .35 will give you the frequency of the type O allele (q):
sqrt (.35) = .592
Therefore, q = .592, and p = .408 (1  .592)
In the HardyWeinberg equation, the frequency of heterozygotes is equal to 2pq. Therefore:
2 (.408) (.592) = .483
So, the proportion of the population that are heterozygous carriers is equal to .483 or 48.3%.
Sorry for the late reply! I am new to the site!
p2 = homozygous dominant frequency
2pq = heterozygous frequency
q2 = homozygous recessive frequency
You should know that blood type O is recessive. Therefore, q2 = .35. Taking the square root of .35 will give you the frequency of the type O allele (q):
sqrt (.35) = .592
Therefore, q = .592, and p = .408 (1  .592)
In the HardyWeinberg equation, the frequency of heterozygotes is equal to 2pq. Therefore:
2 (.408) (.592) = .483
So, the proportion of the population that are heterozygous carriers is equal to .483 or 48.3%.
Sorry for the late reply! I am new to the site!
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