Question

Drosophila melanogaster populations have two alleles associated with the enzyme alcohol dehydrogenase (Adh), F (fast) and f (slow). A study conducted in Portland, Maine sampled Drosophila to test whether wild populations are undergoing evolutionary changes using the genotypes for this trait. They found 45 flies with homozygous dominant genotypes (FF), 174 heterozygotes (Ff), and 88 homozygous recessive (ff).

1. What is the frequency of the dominant allele in this population?

2. Calculate Chi-square value to test whether this population is in Hardy-Weinberg Equilibrium

Answer 1

1. Total number of flies= 45+174+88= 307

The genotype frequencies are as follows:

f(FF)= Total number of FF flies/Total number of flies= 45/307= 0.147

f (Ff)= Total number of Ff flies/Total number of flies= 174/307= 0.567

f (ff)= Total number of ff flies/Total number of flies= 88/307= 0.287

Frequency of dominant allele= 2 X Number of FF flies + Number of Ff flies/ (2 X Total number of flies)

F (F)= 2 X 45 + 174/ 2 X 307= 90+174/614= 0.43

Frequency of dominant allele= p= 0.43

2. Frequency of recessive allele= q= 1-p =1-0.43= 0.57

Null Hypothesis: This population of flies is as per Hardy Weinberg equation.

Expected Number of flies:

Expected number of FF flies= p² X Total number of flies= 0.43 X 0.43 X 307= 56.76

Expected number of Ff flies= 2pq X Total number of flies= (2 X 0.43 X 0.57) X 307= 150.5

Expected number of FF flies=q² X Total number of flies = (0.57 X 0.57) X 307 =99.74

Chi-Square test:

	Observed O	Expected E	O-E	(O-E)^2	(O-E)^2/E
FF	45	56.8	-11.8	139.24	2.451408
Ff	174	150.5	23.5	552.25	3.669435
ff	88	99.4	-11.4	129.96	1.307445
Total	307	307	0.3	821.45	7.428288

Degree of freedom= Number of phenotypes-1= 3-1= 2

The P value is 0.2435544. This p Value is greater than p=0.05.

The calculated Chi square value is greater than the chi square critical value of 5.99 for p=0.05 from Chi square table (p=0.05, df=2). Hence the null hypothesis is rejected. The Population of flies do not follow Hardy Weinberg equilibrium.