Question

a population of rock pocket mice were observed on a black substrate. the population contained individuals with both black and white fur. The black allele (B), is dominant over the white allele (b). A scientist in the field counts 640 black mice and 360 white mice out of 1000 total mice. She is interested in tracking evolution in the population. Please show your work for below questions.

A)What is the frequency of the white allele?

B)What is the frequency of the homozygous recessive genotype?

C)What is the frequency of the heterozygous genotype?

D)What is the frequency of the homozygous dominant genotype?

E)How will we know if the allele frequencies in the population have changed, and thus evolution has occured? Explain how you would set up a problem to figure this out.

Answer 1

The Black allele (B) is dominant over white allele (b)

So, BB and Bb both types of mice are black while bb mice are white

BB/Bb = 640

bb = 360

Total = 1000

According to Hardy Weinberg equation, if the allele frequencies of dominant (B) and recessive allele (b) are p and q respectively, the frequency of homozygous dominant is p², the frequency of heterozygous dominant is 2pq and the frequency of homozygous recessive is q².

p + q = 1.0

p² + 2pq + q² = 1.0

1. Frequency of homozygous recessive (bb) is = q² = bb individual / total population

= 360/100

= 0.36

Thus q = Frequency of white allele is = √0.36 = 0.6

2. Frequency of the homozygous recessive genotype (bb) is = q² = bb individual / total population

= 360/100

= 0.36

3. Frequency of black allele is = p = 1 – q = 1 – 0.06 = 0.4

Frequency of the heterozygous dominant genotype (Bb) is = 2pq

= 2*0.4*0.6

= 0.48

4. Frequency of the homozygous dominant genotype (BB) is = p²

= 0.4²

= 0.16

5. The problem set up is described in steps wise manner:

1. Calculate the number of homozygous dominant and heterozygous dominant mice.
2. Calculate the allele frequency using this formula:

Allele frequency of B = (2 * BB + Bb) / 2*total population

Allele frequency of b = (2 * bb + Bb) / 2*total population

Each individual with the genotype BB has two copies of the B allele; therefore the BB individuals have a count of 2*BB alleles. Heterozygote individuals (Bb) have one of each allele, the number of B allele or b allele among them is equal to the number of Bb. Individuals with the bb genotype have two copies of the b allele, so bb individuals contribute 2*bb alleles to the population. In other words, among the total individuals, there are (2*BB + Bb) number of B alleles and (2*bb + Bb) number b alleles. To calculate the allelic frequencies simply divide the number of B or b alleles by the total number of alleles 2*(BB+Bb+bb).

3. Use a statistical test (see CHI-SQUARE module) to compare the expected and observed counts.
4. If the observed allele frequency and expected allele frequency are not significantly different from one another then no evolution has occurred.
5. If the observed allele frequency and expected allele frequency are significantly different from one another then evolution has occurred in this population.

Kindly revert for any queries and concerns.