The Hardy-Weinberg equilibrium is a theoretical model that predicts the frequency of different genotypes in a population that is not evolving. A scientist studying the population genetics of the rainbow fish finds that the Hardy-Weinberg model predicts that the frequencies of the three different genotypes for the albino gene in a given lake should be as follow: AA: 49%, Aa: 42%, and aa: 9%. However, the scientist obtained a sample of 320 individuals of this species and found that the numbers of individuals for the different genotypes are: AA: 180, Aa: 120, and aa: 20.

Complete question:

The Hardy-Weinberg equilibrium is a theoretical model that predicts the frequency of different genotypes in a population that is not evolving. A scientist studying the population genetics of the rainbow fish finds that the Hardy-Weinberg model predicts that the frequencies of the three different genotypes for the albino gene in a given lake should be as follow: AA: 49%, Aa: 42%, and aa: 9%. However, the scientist obtained a sample of 320 individuals of this species and found that the numbers of individuals for the different genotypes are: AA: 180, Aa: 120, and aa: 20. Use the appropriate statistical technique to test (use alpha = 0.05) whether the observed genotype frequencies are consistent with eh Hardy-Weinberg model. State your conclusion.

There is enough evidence to reject the null hypothesis. The genotypes are not in equilibrium.  The genotype frequencies are consistent with the Hardy-Weinberg model.

Explanation:

Expected Number of individuals with each genotype:  Each genotype´s frequency multiplied by the total number of individuals in the sample.

AA= 0.49 x 320 = 156.8

Aa= 0.42 x 320 = 134.4

aa= 0.09 x 320 = 28.8

Observed Number of individuals with each genotype:

AA= 180

Aa= 120

aa= 20

Chi square= ∑ ((O-E)²/E)  

AA= (O-E)² /E

AA= (156.8 - 180) ² / 156.8

AA= 538.24/156.8

AA= 3.433

Aa= (O-E)² /E        

Aa= (134.4-120)² / 134.4

Aa=207.36/134.4

Aa= 1.543

aa= (O-E)² /E

aa= (28.8-20)²/28.8

aa= 2.69

X² = ∑ ((O-E)²/E) =  3.433 + 1.543 + 2.69 = 7.666

Freedom degrees =  genotypes - alleles = 3 - 2 = 1  

Significance level, 5% = 0.05

p value less than 0.05

Table value/Critical value = 3.841

7.666 > 3.84 meaning that the difference between the observed individuals and the expected individuals in each chamber is statistically significant.

There is enough evidence to reject the null hypothesis. The genotypes are not in equilibrium.

with pass cost = $25

without pass cost = 3x number of days going the the class