Question

In a park there is a population of 750 mice that are either brown or white in color. Of this population, in which brown is dominant to white, approximately 305 individuals are white. Assuming that the population is at equilibrium for this gene/locus, answer the following.

(a) Calculate the allele frequencies and genotype frequencies. Show your work below using proper symbols (p,2pq, etc.). Perform all calculations to three decimal places. (Note: you will not be scored on part A, but you need to do the work in part A first to help you with the bonus.)

BONUS: After the previous predictions were made, actual samples were obtained from all individuals in the population described above. Genetic analyses were done, and it was found that of those that were brown, 80 are BB and 365 are Bb. Is the population above at equilibrium? Use the chi-square test to determine this. a. (Make sure to show your work.) b. State whether the population is at equilibrium and (c) explain your answer. (4)

Answer 1

From the bonus part, it is clear that B allele causes brown colour and b allele causes white colour and that B is dominant to b.

(a) According to Hardy Weinberg Equilibrium, the sum of allelic frequencies is always equal to 1.

i.e., for a diallelic gene, (p + q )= 1

p is the allelic frequency of dominant allele ( in this case, allele B) and q is the allelic frequency of recessive allele (in this case, allele b).

Also, to find the genotype expected frequencies, we can use a relation according to Hardy Weinberg Equilibrium:

(p + q)^2 = 1

i.e., p^2 + 2pq + q^2 = 1, where p^2 is the frequency of genotype BB, 2pq is the frequency of genotype Bb and q^2 is the frequency of genotype bb.

It is given that number of white individuals is 305 and the total number of individuals is 750.

That is, q^2 = 305÷750 = 0.4066666667

Frequency of allele b = q = square root of 0.4066666667

q = 0.6377042157

We know that (p + q) = 1 , According to Hardy Weinberg Equilibrium.

So, p = 1 - 0.6377042157 =  0.3622957843

Frequency of BB genotype = p^2 = (0.3622957843)^2 = 0.1312582353

Frequency of Bb genotype = 2pq = 0.6377042157 * 0.3622957843 * 2 = 0.462075098

Frequency of bb genotype is already found out as : q^2 = 0.4066666667

Rounding to 3 decimal places:

p = 0.362

q = 0.638

p^2 = 0.131

2pq = 0.462

q^2 = 0.407

BONUS PART:

Observed number of BB genotype = 80

Observed number of Bb genotype = 365

Observed number of bb genotype = 750-(365+80) = 305

Expected number of BB individuals = 750*p^2 = 98.283

Expected number of Bb individuals = 750* 2pq = 346.434

Expected number of bb individuals = 750 * q^2 = 305

Chi square = sum of [[(O-E)^2]÷E ] of all three genotypes, where O is the observed number of individuals, E is the expected number of individuals in the respective genotypes.

If Chi square is greater than the table value, then there is no equilibrium.

Chi square = [(80-98.238)^2]÷98.238 + [(365-346.434)^2]÷346.434 + [(305-305)^2]÷305

=  3.3859061056 + 0.994984199 + 0 = 4.3808903046

Chi square = 4.3808903046

When Chi square is within the table value, observed values are not significantly different from the table value. That is, The population is at equilibrium. But, if chi square value is more than table value, the population is not at Hardy Weinberg Equilibrium.

To find table value, we need degrees of freedom and level of significance.

The level of significance here is 0.05 ( since it is a biological experiment).

The degree of freedom = number of genotypes - 1 = 3-1 = 2.

The table value for chi square at 0.05 level significance and 2 fegrees of freedom = 5.991 (from table for chi square)

Since the calculated value of Chi square is less than table value, there is no significant difference in the observed numbers in each genotype from the expected numbers.

Hence, the population is at Hardy Weinberg Equilibrium.