Ap bio lab 8 [hardy weinberg].. please help!? - are ap bio labs same everywhere
Not really understand how the laboratory! Can someone please help me and explain if possible? Thank you .. best answer = 10 points:]
original class frequencies: AA = 0.25, Aa = 0.5, AA = 0.25
My first genotype = Aa
F1 = Aa genotype
F2 = AA genotype
F3 = AA genotype
Genotype F4 = Aa
F5 = AA genotype
Frequencies final class = 10 = AA, AA = 13, AA = 11
p = q = ____, ____
[I need help in finding p and q in the first place]
asks me to the A-gene alleles found # 5:
the number of offspring with genotype AA is ____ x 2 = _____ alleles
the number of offspring of genotype AA x 1 = ____ ______ alleles
= _____ Total alleles
p = total number of alleles / total alleles in the population = ______
Number of alleles of a gene, in the 5th.:
Number of children with AA genotype x 2 = _____ ______ alleles
The number of offspring of genotype Aa: x 1 = _____ _____ alleles
= Total number of alleles _____
q = total number of alleles / total alleles in the population = _____. \\ \\ \\ \\ \\ \\ \\ \\ U0026lt;br>
Please help me with that! I have to do a few pages of it, and I must be first obtained this first part, so that you can do the rest alone. Thanks for the help in advance! :]
Monday, February 1, 2010
Are Ap Bio Labs Same Everywhere Ap Bio Lab 8 [hardy Weinberg].. Please Help!?
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If their first frequencies AA, 0.25, a, 0.5, AA 0.25, while the frequency of the homozygous dominant AA p ^ 2 = 0.25, so p = 0.5 q = 0.5.
If your goal is the frequency of the 5th Generation, then the frequencies of two alleles can be calculated from the numbers of individual genotypes in the population:
AA 10
13a
AA 11
Total: 34
Frequency of A in the "frequency of the last classes:" The number of alleles in AA individuals, the number of alleles among Aa individuals by the total number of alleles shared in the community:
p = ((10 x 2) + 13) / (34 x 2) = (20 + 13) / 68 = 33/68 = 0.485
The frequency of an allele in the "frequency of the last classes:" The number of alleles in AA individuals compared to the number of alleles among Aa individuals, divided by the total number of alleles in the population:
q = ((11 x 2) + 13) / (34 x 2) = (22 + 13) / 68 = 35/68 = 0.515
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