In a population of Mendel's garden peas, the frequency of green-flowered plants (genotype aa) is 49%. The population is in Hardy-Weinberg equilibrium. What are the frequencies of the AA and Aa genotypes?
I got 9% AA, 42% Aa as my answer, is this correct?
To find the frequencies of the AA and Aa genotypes, we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Where:
- p^2 represents the frequency of the AA genotype
- 2pq represents the frequency of the Aa genotype
- q^2 represents the frequency of the aa genotype
Given that the frequency of aa genotype (green-flowered plants, genotype aa) is 49%, we can determine q^2.
q^2 = 0.49
Now, we know that the sum of the frequencies of all genotypes must equal 1. So, we can deduce that:
p^2 + 2pq + q^2 = 1
In this case, since p^2 represents the frequency of the AA genotype, we can substitute q^2 with 0.49 to get:
p^2 + 2pq + 0.49 = 1
Simplifying, we get:
p^2 + 2pq = 1 - 0.49
p^2 + 2pq = 0.51
We also know that p + q = 1. Solving this equation for q, we get:
q = 1 - p
Substituting q in terms of p, we can rewrite the simplified equation as:
p^2 + 2p(1 - p) = 0.51
Expanding and simplifying further:
p^2 + 2p - 2p^2 = 0.51
-p^2 + 2p = 0.51
p^2 - 2p + 0.51 = 0
Now, we can solve this quadratic equation using the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 1, b = -2, and c = 0.51. Substituting these values in the quadratic formula:
p = (-(-2) ± √((-2)^2 - 4(1)(0.51))) / (2(1))
p = (2 ± √(4 - 2.04)) / 2
p = (2 ± √1.96) / 2
Calculating further:
p = (2 ± 1.4) / 2
There are two possible solutions for p:
p1 = (2 + 1.4) / 2 = 3.4 / 2 = 1.7
p2 = (2 - 1.4) / 2 = 0.6 / 2 = 0.3
Since the frequency of a particular genotype cannot be greater than 1, we discard p1 = 1.7 as an invalid solution. Therefore, p = p2 = 0.3.
Now, we can calculate q:
q = 1 - p = 1 - 0.3 = 0.7
Finally, we can calculate the frequencies of the AA and Aa genotypes:
Frequency of AA genotype (p^2) = (0.3)^2 = 0.09 or 9%
Frequency of Aa genotype (2pq) = 2(0.3)(0.7) = 0.42 or 42%
So, the correct frequencies are 9% for AA and 42% for Aa. Your answer of 9% AA and 42% Aa is correct.