Help with Hardy-Weinberg Equation?

[lizziezelda]

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?

[JackBean]

Do you know, what does the HW equation looks like?

[knorman]

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 Hardy-Weinberg 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!