Question

You discover a cheetah population with the following abundances for the fast allele:

FF individuals = 30

Ff individuals = 15

ff individuals = 55

a) Calculate the allelic frequencies:

b) Calculate the expected genotype frequencies and expected genotype abundances for the next generation under the Hardy-Weinberg equilibrium null model

Expected genotype frequency: FF =, Ff = , ff =

Expected genotype frequency: FF =, Ff =, ff =

c) What test statistic would you use to test if the population was in Hardy-Weinberg equilibrium?

d) You carry out your test and find that your calculated test statistic is larger than the critical value. What can you conclude:

Statistically:

Biologically:

Answer 1

a).

FF = 30

Ff = 15

ff = 55

Total F alleles = 2*30+15 = 75

Total f alleles = 2*55+15 = 125

Total alleles = 200

Frequency of F allele = 75/200 = 0.375

Frequency of f allele = 125/200 = 0.625

b & c).

Expected frequency of each genotype:

FF = 0.375 * 0.375 = 0.1406

Ff = 2*0.375*0.625 = 0.4688

Ff = 0.625 * 0.625 = 0.3906

Expected number of each genotype:

Total progeny = 100

FF = 0.1406 * 100 = 14.06

Ff = 0.4688 * 100 = 46.88

Ff = 0.3906*100 = 39.06

c).

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
FF	30	14.06	15.94	254.0836	18.0714
Ff	15	46.88	-31.88	1016.3344	21.6795
ff	55	39.06	15.94	254.0836	6.5050
Total	100	100			46.2558

Chi-square value = 46.26

Degrees of freedom = no. of catergories – 1

Df = 3-1= 2

Critical value = 5.99

The chi-square value of 46.26 is greater than the critical value of 5.99. Hence, the null hypothesis is rejected and the population is not Hardy-Weinber equilibrium.

Statically: Not fit

Biologically: Not fit.