Question

Imagine that you have discovered a new population of curly-tailed lizards established on an island after immigrant lizards have arrived from another island during a hurricane. No lizards inhabited the island previously. You collect and tabulate the genotype data below for the lactate dehydrogenase gene (Ldh) for each of the individual lizards, numbered 1-10. Use the table to answer the questions below. The allele frequency of the A is 0.45 and the allele frequency of a is 0.55

INDIVIDUAL NUMBER INDIVIDUAL GENOTYPE FOR Ldh Аа AA 3 АА aa 5 aa Male Male Female Male Female Female Male Male Female Male AA aa aa аа АА 9 10

Is the population in Hardy-Weinberg equilibrium? Test using chi-square with an alpha value of 0.05. State your null hypothesis, your p-value and whether you reject or fail to reject your null hypothesis

Answer 1

1.Hardy -weinberg equilibrium :

Given that allele frequency of A=0.45 and allele frequency of a=0.55.

Also P=frequency of A and q=frequency of a.

Hardy-weinberg equilibrium equation =p^2+2*p*q+q^2

Where p^2=Homozygous dominant,

2*p*q=Heterozygous,

q^2=Homozygous recessive.

P^2=0.45*0.45=0.2025,

2*p*q=2*0.45*0.55=0.495,

q^2=0.55*0.55=0.3025

Applying to the formula=0.2025+0.495+0.3025=1

Therefore the population is in hardy -weinberg equilibrium

2.Chi-square test or (x^2):

X^2=€[(O-E)^2]/E

Lets calculate a table with alpha value=0.05

Observed expected Observed -expected ( O-E)^2/E

AA 2 2 0 0

Aa 5 4 1 0.25

aa 3 3 0 0

X^2=0+0.25+0=0.25

3.Here the value of P=0.45.Therefore the null hypothesis is false. If the null hypothesis want to be true p<0.05.But here p>0.05.

4.Here we fail to reject null hypothesis. Because p>0.05