Question

Consider 2 alleles in a population of seahorses: A and a. There are 2 sea horses with genotype AA, 96 seahorses with genotype Aa and 2 seahorses with genotype aa. What are the frequencies of the alleles in this population. Please show all work. Is this population in the Hardy Weinberg Equilibrium? Explain why or why not. THANKS IN ADVANCE!! :)

Answer 1

Genotype frequency in a population is the number of individuals with a given genotype divided by the total number of individuals in the population.

Here genotype frequency is given, so to find the number of individuals of each genotype we will multiply the genotype frequency of each genotype with total individuals.

Genotype frequency for AA=2/100 = 0.02

Genotype frequency for Aa= 96/100 = 0.96

Genotype frequency for aa= 2/100 = 0.02

Allele frequency is determined by counting how many times the allele appears in the population then dividing by the total number of copies of the gene.

Total population = 96+2+2 = 100

Total allele =200

Total A allele = 4 + 96 = 100

Total a allele = 4 + 96 = 100

Allele frequency of A = 100/200 = 0.5

Allele frequency of a = 100/200 = 0.5

For hardy winberg, genotype frequency will be

P² =AA= 0.5*0.5 = 0.25

q² = aa=0.5*0.5 = 0.25

2pq = Aa= 2*0.5*0.5= 0.50

	Expected Frequency	Observed Frequency
AA	0.25	0.02
Aa	0.50	0.96
aa	0.25	0.02

As we can see observed and expected frequency are different. So the population is not in Hardy Weinberg equilibrium.