Determining Allele Frequencies in a Population ALLELE FREQUENCIES Suppose we hav

Question

Determining Allele Frequencies in a Population

ALLELE FREQUENCIES

Suppose we have a population of 100 snapdragons, and the phenotypes are found in the following numbers: 44 Red (RR), 30 pink (=Rr), and 26 white (rr). p = ((2*the number of RR since each has two R alleles) + the number of Rr since each has one R allele)) all divided by the total number of alleles, which equal 2 * the number of individuals in the population. = ((44*2) + 30))/200 = 0.59 What does q equal? Now try this sample problem (it is not in your lab manual): Hair color in mice shows complete dominance. Individuals that are BB and Bb have black hair; individuals that are bb have red hair. In your population of 100 mice, you find 16 individuals with red hair. Calculate the following (assuming the population is in Hardy-Weinberg equilibrium): q = (Hint: remember that 16/100 = 0.16 is the frequency of recessive homozygotes = q2) p = (Hint: What does p + q equal?) The expected frequency of the AA genotype = The expected frequency of the Aa genotype

Explanation / Answer

a) The frequency of two alleles R and r must add upto unity:

p+q = 1

p= 0.59

putting the value of p in (1) we get,

q = 1-p

q = 1- 0.59 = 0.41

Therefore, q equals 0.41

b) Number of bb individuals = 16,

therefore the frequency of the genotype bb (q2) is 16/100 = 0.16

From this, we can estimate q as:

q = square root of q2 = square root of 0.16 = 0.4.

The allele frequency of B is:

p = 1-q = 1- 0.4= 0.6.

Assuming that the population is in Hardy Weinberg Equilibrium, the expected frequencies of genotypes are:

BB = p2 = 0.36

Bb = 2pq = .48

Note: To check if the solution is correct, we place the calculated values in hardy weinberg equation:

p2+2pq+q2 = 1

.36 + .48 + .16 = 1

hence, the values are correct.

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