Hardy Weinberg Practice Problem



Suppose a plant population consists of 84% red flowered plants (red is dominant and represented by “R”) and 16% white flowered plants (white is recessive and represented by “r”).

1. What is the allele frequency of r?

2. What is the allele frequency of R?

3. What is the genotype frequency of RR?

4. What is the genotype frequency of Rr?

5. What is the genotype of rr?

6. Does this make sense with the second Hardy-Weinberg equation?



Answers:

1. q2 = rr
=
q = .4 This is the allele frequency of q, which is the allele frequency of r.

2.
One of the two Hardy-Weinberg equations is, p + q = 1.
It is always true as long as the Five Hardy-Weinberg Conditions are met.
If, p + q = 1
Then, p = 1 - q
and, p = 1 - .4
Therefore, p = .6
This is the allele frequency of p, which is the allele frequency of R.

3. RR = p2 = (.6)2 = .36 = 36%

4. Rr = 2pq = 2(.6)(.4) = .48 = 48%

5. rr = q2 = .16 = 16% (this was given in the question!)

6. Yes it does! The second Hardy-Weinberg Equation is p2 + 2pq + q2 = 1
It is always true as long as the Five Hardy-Weinberg Conditions are met.
If we plug in the numbers we have just calculated it works!

p2 = .36
2pq = .48
q2 = .16
p2 + 2pq + q2 = 1
.36 + .48 + .16 = 1