allele frequencies?

Refine.J

2007-12-05 3:38 am

1) the allele frequencies in a population are A=.6 and a=.4
a) predict the genotype frequencies for the next generation.
AA:______ Aa_________ aa:______
b) What would the allele frequencies be for the generation you predicted above for part a?
A:______ a:______

2) suppose that one gene locus determines stripe pattern in skunks. SS skunks have two broad stripes; Ss skunks have two narrow stripes; ss skunks have white speckles down thier backs. A sampling of a population of skunks found 65 broad- striped skunks, 14 narrow- striped and 1 speckled. Determine the allele frequencies.

S:_________________ s:_______________

3) In a hypothetical population of 1000 people, tests of blood type genes show that 100 have the genotype AA, 600 have the genotype AB, and 300 have the genotype BB.
a) what is the frequency of the A allele?
b)what is the frequency ofthe B allele?
c)if there are 4000 children in this generation, how many would have AB blood?

回答 (1)

[匿名]

2007-12-05 4:22 am

✔ 最佳答案

1.
a)
p (frequency of A allele) = 0.6
q (frequency of a allele) = 0.4
and since these are the only alleles and frequencies in the population, you can use these to predict the frequencies of alleles and genotypes in the next generation:

p^2 (frequency of AA) = (0.6)^2 = 0.36
q^2 (frequency of aa) = (0.4)^2 = 0.16
2pq (frequency of Aa) = 2 * 0.6 * 0.4 = 0.48

b)
p = 0.6 (frequency of A)
q = 0.4 (frequency of a)
(note that the frequencies did not change from the previous generation, so the population is at equilibrium)

2.
65 SS, 14 Ss, 1 ss
65 + 14 + 1 = 80
So, the frequency of the S allele is:
(65*2 + 14)/160 = 144/160 = 0.9

The frequency of the s allele is 0.1
(14 + 1 * 2)/160 = 16/160 = 0.1

3.
100 AA, 600 AB, 300 BB
100 + 600 + 300 = 1000
a) frequency of the A allele
(100 * 2 + 600)/2000 = 800/2000 = 0.4

b) frequency of the B allele
(600 + 300*2)/2000 = 1200/2000 = 0.6

c) if have 4000 children, then the genotype frequencies in the children would be:

AA (p^2) 0.4 * 0.4 = 0.16
BB (q^2) 0.6 * 0.6 = 0.36
AB (2pq) 2 * 0.4 * 0.6 = 0.48

So, out of 4000 children, 48% or 1920 would have AB

👍 0 👎 0