Question

 

In a bird species, the shape of the beak is encoded by ONE SINGLE locus with TWO co-dominant alleles N^a and Nz. N^a gives long beak in double dose, while individuals who are homozygous for N^z have short beak.

In a population, the following genotype frequencies were observed:

N^a N^a = 0.5
N^a N^z = 0.3
N^z N^z = 0.2

a) Assume a total of 600 individuals in the population. Set up a hypothesis, will the frequency be in Hardy-Weinberg equilibrium?

b) Use an x2 test to find out if the allele frequency is in Hardy-Weinberg equilibrium or not.

According to what I´ve heard we can´t use the frequencies listed, but instead the allel frequence:

p = 0.5 + ½ * 0.3 = 0.65

0.652 = 0.4225

(2x0.65x (1-0.65)0.35= 0.455

(1-0.652) = 0.1225

Answer 1

To calculate genotype frequency, the formula is -

Genotype freq. = # of individuals with that genotype / Total population

a)

Genotype freq. of N^a N^a = 0.5

No. of individuals having N^a N^a genotype = 0.5 * 600 = 300

Similarly, individuals having N^a N^z genotype = 0.3*600 = 180

and individuals having N^z N^z genotype = 0.2*600 = 120

Allele frequency = No. of homozgyotes carrying that allele + 1/2*no. of heterozygotes / Total population

We multiply the heterozygote freq. by 1/2 as they have only 1 copy for that allele and homozygotes carry 2 copies.

Allele frequency of "a" allele = (300 + 1/2*180 ) / 600 = 0.65

Allele frequency of "z" allele = (120 + 1/2*180) / 600 = 0.35

The sum of allele frequencies add upto 1 (0.65 + 0.35 = 1), we hypothesize that the population is in Hardy-Weinberg equilibrium and any deviation in the population of different alleles is a chance event.

 

b)

Chi square test to check if the population is in HW equilibrium :-

Genotype	Observed (o)	Expected (e)	o - e	(o - e)^2 / e
Na Na	300	(0.65)2 * 600 = 253.5 254	46	8.330
Na Nz	180	2*0.65*0.35*600 = 273	-93	31.681
Nz Nz	120	(0.35)2 * 600 = 73.5 73	47	30.26
TOTAL	600			70.271

  ² = (o - e)² / e

There are 3 phenotypic classes because alleles "a" and "z" are codominant, therefore, degree of freedom = 2

Since value of ² (70.271) is greater than critical ²_2df i.e. 5.99, null hypothesis is rejected and Population is NOT in Hardy - Weinberg equilibrium.