Please Help Me: Bio Questions?

Question

1- If the frequency of two alleles "A" & "a" in a gene pool is 80% & 20% respectively, what will be the expected frequency of heterozygote individuals in the population?

2- If the frequency of a recessive allele is 30% in a population of 1,000 individuals, how many individuals would you predict would be heterozygotes?

3-If gene locus has two alleles & the frequency of one of the homozygotes is 0.81, what are the frequencies of the two alleles?

4- If the frequency if gene "L" in a population is 0.6 & this gene is completely dominant to its single allele " l ", what is the % of the population expected to show the dominant phenotype?
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1- ? based on pp+2pq+qq=1, I got Aa= .32 (I did 2 x.16)
2- ??
3- ??
4- ? If I'm doing this correctly, I got 84%.
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I know it's a lot, but can someone please see if I'm on the right track & also explain how to do these. This is confusing me, Idk if I'm using the formulas correctly. I would really appreciate it. Thank You! = )

Answer 1

1. I agree with 0.32
2. The recessive allele is 0.3, so the dominant allele is 0.7.
Frequency of heterozygotes is 2pq = 2 * 0.3 * 0.7 = 0.42
0.42 * 1000 = 420 heterozygotes.

3. One of the homozygotes is 0.81. It doesn't matter if you figure it for p or q. pp = p^2 = 0.81 so p = 0.9 p+q=1 so q = 0.1

4. pp + 2pq will show the dominant phenotype.
0.6 * 0.6 + 2 * 0.6 * 0.4 = 0.36 + 0.48 = 0.84 = 84% I agree.

Answer 2

1) A is at frequency of .8 right? so the probability of two A alleles uniting is .8 x .8 = .8 squared = .64. the probability of two a alleles uniting is .2 x .2 = .2 squared = .04. the probability of a heterozygote is the prob of an A uniting with an a = .2 x .8 plus the probability of an a uniting with an A = .8 x .2 i.e. 2(.2 x .8) = .32. so answer: .32 or 32% of the individuals in the pop should be heterozygotes...under the standard hardy-weingerg assumptions.