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A sample of 100 indiv is taken from a breeding population (population x) of M&Ms that is in Hardy-Weinberg equilibrium for candy color. A=blue, A' = yellow. The heterozygote (AA') phenotype is green.

1) In the sample of 100 indiv, 30 were blue, 50 were green, and 20 were yellow. In this sample, what is the frequency of the A' allele?
2) In the original population x, p(A) =.6. If the same has the same allele frequency, how many indiv of the 100 candies in the sample would be expected to be green?

2007-02-14 05:04:16 · 2 answers · asked by Anonymous in Science & Mathematics Biology

A sample of 100 indiv is taken from a breeding population (population x) of M&Ms that is in Hardy-Weinberg equ
1) In the original breeding population, p(A) = 0.6. If the sample has the same allele frequency, what percent of the sample would be expected to be the color yellow?

2) What is the (approximate) probability that if you randomly selected 2 individuals from Population X they would both be yellow? (A) 0.03; (B) 0.16; (C) 0.32; (D) 0.36; (E) 0.50; (F) 0.60; (G) 0.72

The answer is 1) 16% and 2) A [.3=(.16)(.16)] But I dont know how you get it. Plz show me how . thx

2007-02-14 06:09:44 · update #1

2 answers

The Hardy Weinberg equilibrium is expressed as p^2+2pq+q^2=1 where p and q are the allele frequencies of the 2 alleles and (p+q)=1.
In the above example p^2 represents the number of homozygous AA (i.e. 30%). 2pq is the number of heterozygotes AA' (50%) and q^2 is the frequency of homozygote A'A' (20%)
The frequency of A' is the root of 0.2 so p(A') is 0.447.
For part 2) p is 0.6 in the above equation - this makes q=0.4 so the number of green (heterozygote) M&Ms is 2pq which is 0.48 or 48%.

For the extra bits you added:
if p(A)=0.6 then p(A')=0.4 so the number of yellows is p(A'A') which is q^2 or 0.4x0.4 which is 0.16 or 16%
If you pick 2 M&Ms at random you need the first one to be yellow with a probability of 16% or 0.16 and the second has to be yellow as well with a probability of 16% or 0.16. The chance of this occurring is 0.16x0.16 which is 0.0256 or 2.6% (3% when rounded up)

2007-02-14 06:16:45 · answer #1 · answered by Mal 2 · 0 0

1. Hardy-Weinberg says that p^2 + 2pq + q^2 = 1
Since A' is yellow and there were 20 out of 100 yellow,
q^2 = 0.20
q = 0.447 = frequency of A' allele (square root of 0.2)

2. H-W says that p + q = 1
If p = 0.6, then q = 0.4
Green candies are heterozygous, represented in H-W as 2pq.
2pq = 2 * 0.6 * 0.4 = 0.48
0.48 of 100 = 48 green candies in the sample

2007-02-14 13:34:46 · answer #2 · answered by ecolink 7 · 0 0

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