Please Help: Biology Questions?

Question

1- The frequency of recessive homozygous genotype is 0.09 in a population.What's the expected frequency of individuals homozygous for the Dominant trait?

2- Suppose that albinism (autosomal recessive trait; genotype aa) occurs among newborns in a certain population of 10,000,000 @ the frequency of 1 in 14,400. How many individuals in the population are expected to be homozygous normal (AA) for this trait?

3-In a certain human population in genetic equilibrium for the typical X-linked red-green color blindness locus,6 males in 100 are color blind. What is the expected frequency of color blind females?
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1- ? Idk if it's right but I got: .0081 are Homozygous Dominant
2-??
3-??

Did I do #1 correct? Can someone please help explain or maybe show me how to do these? They are more complicated then what I learned, I'm confused because the populations are larger, so I'm having trouble figuring out how to find the frequencies. I would really appreciate the help. Thank You! = )

Answer 1

1. qq = 0.09
q = 0.3
p = 0.7 (p + q = 1)
pp = 0.49 are homozygous dominant

2. Several steps involved here:
-- find out the frequency of a
aa= 1/14,400 = 0.0000694
a = sqrt 0.0000694 = 0.0083

-- find out the frequency of A
p + q = 1
p + 0.0083 = 1
p = 0.9917 = frequency of A

-- find out the frequency of AA
p^2 = 0.9917^2 = 0.983

-- find out how many people make up 0.983 of 10,000,000
0.983 * 10,000,000 = 9,830,000 people who are homozygous normal

3. Sorry, I haven't even thought about how Hardy-Weinberg works for X-linked traits. The best logic I can muster is this:
-- 6 males out of 100 are color blind = 0.06 = q Males only have one X chromosome, so this must be the frequency.
-- p = 0.94
-- color-blind females must have the color-blind allele on both of their X-chromosomes
-- color-blind female is qq = 0.06*0.06 = 0.0036

You might check this one with your teacher. Good luck. Don't let the bigger numbers confuse you. Just trust your calculator!