Hardy Weinberg Practice Answer Key
Answers:
1. Estimate the frequency of the recessive allele in the world population as well as the frequency of carriers (heterozygotes) that do not have the disease.
aa=.01% aa=.0001 a=square root of .0001 q=0.01 p=0.99
2xpxq = 2 x 0.01x 0.99 = Aa = 0.0198
2. Estimate the frequency of the disease carried as an autosomal dominant trait in the US. What information do you need to figure out the allelic frequencies?
½ of 3.5= 1.75 1.25million /281.42190million = 0.006218 is the frequency of "tremor with a dominant trait" in the population. Since this is a dominant trait this frequency includes both the homozyotes and the heterozygotes. In order to determine the allelic frequencies you would need to know the breakdown of this.
3. a. Which of the populations are in Hardy Weinberg equilibrium?
Populations in HW are in bold below. To find p=AA+1/2Aa.
Hint: If the population is in Hardy Weinberg the square root of Aa will also be equal to p.
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Question 3 |
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|
AA |
Aa |
aa |
p |
q |
2pq |
|
1 |
1 |
0 |
0 |
1 |
0 |
0 |
|
2 |
0 |
1 |
0 |
0.5 |
0.5 |
0.5 |
|
3 |
0 |
0 |
1 |
0 |
1 |
0 |
|
4 |
0.5 |
0.25 |
0.25 |
0.625 |
0.375 |
0.4688 |
|
5 |
0.25 |
0.25 |
0.5 |
0.375 |
0.625 |
0.4688 |
|
6 |
0.25 |
0.5 |
0.25 |
0.5 |
0.5 |
0.5 |
|
7 |
0.33 |
0.33 |
0.33 |
0.495 |
0.495 |
0.4901 |
|
8 |
0.04 |
0.32 |
0.64 |
0.2 |
0.8 |
0.32 |
|
9 |
0.64 |
0.32 |
0.04 |
0.8 |
0.2 |
0.32 |
|
10 |
0.986049 |
0.0139 |
0.000049 |
0.993 |
0.007 |
0.0139 |
4. What is the frequency of p and q, the homozygotes and heterozygotes for the next 5 generations assuming that these populations are in Hardy Weinberg equilibrium..
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Pop1 AA |
Pop2 AA |
Pop1 aa |
Pop2 aa |
Pop1 Aa |
Pop2 Aa |
Pop 1 p (A) |
Pop 2 p (A) |
Pop 1 q (a) |
Pop 2 q (a) |
|
Generation 1 |
0.3 |
0.2 |
0.6 |
0.4 |
0.1 |
0.4 |
0.35 |
0.4 |
0.65 |
0.6 |
|
Generation 2 |
0.1225 |
0.16 |
0.4225 |
.36 |
.455 |
0.48 |
0.35 |
0.4 |
0.65 |
0.6 |
|
Generation 3 |
0.1225 |
0.16 |
0.4225 |
.36 |
.455 |
0.48 |
0.35 |
0.4 |
0.65 |
0.6 |
|
Generation 4 |
0.1225 |
0.16 |
0.4225 |
.36 |
.455 |
0.48 |
0.35 |
0.4 |
0.65 |
0.6 |
|
Generation 5 |
0.1225 |
0.16 |
0.4225 |
.36 |
.455 |
0.48 |
0.35 |
0.4 |
0.65 |
0.6 |
Step 1: To find the frequency of Heterozygotes AA+aa+Aa=1 For example: 0.3+0.6+Aa=1
Step 2: To find the frequency of "A" or "p" = AA+1/2 Aa. For example: 0.3+0.5=0.35
Step 3: To find the frequency of the homozygotes in the next generation Square the allelic frequency of the previous generation. For example: (0.35)2=0.1225.
Step 4: To find the frequency of the heterozygotes in the next generation2pq. For example:
2 x 0.35 x 0.65= 0.455