calculating genotype frequencies

Question

A population of snapdragons has 157 plants with white flowers (aa), 298 plants with pink flowers (Aa), and 844 plants with red flowers (AA). What are G(aa), G(Aa), and G(AA)?

And calculate the allele frequencies p (for A) and q (for a) for this population of snapdragons.

Answer 1

Hello!

For this answer, I will be assuming that your snapdragon population is in Hardy-Weinburg [HW] equilibrium.

Knowing that a population is in HW, I can use the following two equations to calculate allele frequencies and the frequesncies of the different genotypes.

1.) p + q = 1

2.) p² + 2pq + q² = 1

In the above equations:

p = frequency of the dominant (uppercase) allele (letter)

q = frequency of the recessive (lowercase) allele (letter)

p² = frequency of the dominant genotype (AA)

q² = frequency of the recessive genotype (aa)

2pq = frequency of the heterozygous genotype (Aa)

Lets start by finding the frequency of the individual alleles, A and a.

Since there are 157 plants with white flowers (aa) and we know that each plant has two recessive alleles, we can safely say that there are 314 recessive alleles being provided by the white flowers alone.

The pink flowers only provide one of each kind of allele, one dominant and one recessive, so we can safely say that from the pink flowers, we gain 298 dominant alleles and 298 recessive alleles.

The red flowers (AA), similar to the white flowers, carry two of the same allele (only this time its dominant!) so we can say that we gain 1688 dominant alleles from the red flowers alone.

It can also be noted the total number of alleles by adding up all of the aforementioned alleles:

314+298+298+1688 = 2598 total alleles across all plants.

To find the frequency for each allele, divide the total amount of dominant/recessive alleles by the total.

314+298 = 612 recessive alleles --> 612/2598 = 0.23 = q

1688+ 298 =1986 dominant alleles --> 1986 / 2598 = 0.77 = p

These are your p and q values, notice how when added together 0.23+0.77=1 (and we know this happens from p + q = 1).

Using this information, we can then "plug and chug" into the HW equation p² + 2pq + q² = 1 and solve for the genotypic frequencies.

p² = (0.77)² = 0.59 = frequency of the dominant genotype (AA)

q² = (0.23)² = 0.05 = frequency of the recessive genotype (aa)

2pq = 2(0.77)(0.23) = 0.35 = frequency of the heterozygous genotype (Aa)

To double check, add the three final values and they should equal about 1 :)

If your instructor would like these in percents, no worries! Just multiply all final answers by 100 and add a percent sign.

Ex: 0.59 * 100 = 59%

To make sure you converted right, all percents should add up to 100%

Hope this was helpful!!

~ Kayla