Question

5. Examine the following fabricated data for a population of oak trees. Genotype Height Number of individuals with this genotype in generation 1 Number of individuals with this genotype in generation 2 A1A1 5 meters 100 200 A1A2 10 meters 100 A2A2 15 meters 100 200 TOTAL: 300 450 a. (2 pts) What is the allele frequency of A1 in the first generation? b. (2 pts) is this population in Hardy-Weinberg Equilibrium? Show your work.

Answer 1

1. To calculate allele frequency we count the number of A1 alleles and dividing by the total number of alleles:

Here in A1A1 = 100 individual = 200 allele of A1

Here in A1A2 = 100 individual = 100 allele of A1

Here in A2A2 = 100 individual = 0 allele of A1

so A1 allele frequency = 300 allele of A1/600 total allele= 0.5

so A2 allele frequency = 300 allele of A1/600 total allele= 0.5

2. To find if a population is in Hardy Weinberg equilibrium. We will compare the observed and expected genotype frequency.

For observed.

For A1A1 = Total individual of A1A1/ Total individuals = 100/300 = 0.33

For A1A2 = Total individual of A1A1/ Total individuals = 100/300 = 0.33

For A2A2 = Total individual of A1A1/ Total individuals = 100/300 = 0.33

For Expected according to Hardy Weinberg equilibrium.

p² 2pq q²

A1A1 A1A2 A2A2

In this formula (p² 2pq  q² ) we have to put values of our calculated allele frequncies and see if these match our observed genotype frequency.

so p² = A1A1 = o.5 * 0.5 = 0.25

2pq = A1A2 = 2 * 0.5 * 0.5 = 0.50

q² = A2A2= 0.5 * 0.5 = 0.25

These are different from our onserved frequencies. So population is not in Hardy Weinberg equilibrium.