In: Biology
In a sample of 200 individuals, the frequency of the BB genotype is 0.4, the Bb genotype is 0.4, and the bb genotype is 0.2. Perform a Chi-square test to test whether this population is in Hardy-Weinberg equilibrium. For full credit, be sure to include the null hypothesis, Chi-square number, degrees of freedom, p-value, and conclusion.
Ans. Part A: Calculate expected allelic frequencies
Given,
Observed Genotype frequency of BB, f(BB) = 0.4
Observed Genotype frequency of Bb, f(Bb) = 0.4
Observed Genotype frequency of bb, f(bb) = 0.2
Observed Population size = 200
Now,
Observed No. of BB individuals = f(BB) x population size = 0.4 x 200 = 80
Observed No. of Bb individuals = f(Bb) x population size = 0.4 x 200 = 80
Observed No. of bb individuals = f(bb) x population size = 0.2 x 200 = 40
# Part B: Expected Allelic frequency =
(No. of specified allele / Total number of alleles in population)
Being diploid each individual has 2 alleles of the gene. The homozygotes have two copies of the respective allele, whereas the heterozygotes have 1 copy of both the alleles. That is, BB individuals have 2 copies of allele B, whereas Bb individuals have one copy each of allele B and b.
#I. Expected Allelic frequency of B
Expected allelic frequency of B, f(B) =
(2 x No. of BB individuals + 1 x No. of Bb individuals) / (2 x population size)
Or, f(B) = (2 x 80 + 1 x 80) / [ 2 x ( 200)]
Or, f(B) = (160 + 80) / 400
Hence, f(B) = 0.60
#II. Expected Allelic frequency of b
Observed Allelic frequency of b, f(b) =
(2 x No. of bb individuals + 1 x No. of Bb individuals) / (2 x population size)
Or, f(b) = (2 x 40 + 1 x 80) / [ 2 x (200)]
Or, f(b) = 160/ 400
Hence, f(b) = 0.40
# Part C: The allelic frequency that we calculate above in #I and #II are EXPECTED allelic frequency because we expected them to be observed in the population.
We use the expected allelic frequencies to calculate the expected number of individuals of each phenotype in the population using HW equilibrium.
# Hardy- Weinberg Equation for 2 allele system (for a population at HW equilibrium) is given by-
p + q = 1 - equation 1
(p + q)2 = p2 + q2 + 2pq = 1 - equation 2
Where,
p = allelic frequency of allele dominant allele
q = allelic frequency of allele recessive allele
p2 = genotypic frequency of homozygous dominant
q2 = genotypic frequency of homozygous recessive
2pq = genotypic frequency of heterozygote
# Calculate EXPECTED number of individuals in population:
#I. Expected number of BB individuals
Expected frequency of BB individuals, f(BB)= p2 = (0.60)2 = 0.36
Now,
Expected number of BB individuals = Expected f(BB) x Actual Population size
= 0.36 x 200
= 72 (nearest whole number)
#II. Expected number of Bb individuals
Expected frequency of Bb individuals, f(Bb)= 2pq = 2 x 0.60 x 0.40 = 0.48
Now,
Expected number of Bb individuals = 0.48 x 200 = 96
#III. Expected number of bb individuals
Expected frequency of bb individuals, f(bb)= q2 = (0.40)2 = 0.16
Now,
Expected number of bb individuals = 0.16 x 200 = 32
Part D: Compare Observed and Expected number of phenotypes (individuals)
The observed number of phenotypes (individuals) are calculated using observed genotype frequencies in Part A.
The expected number of phenotypes (individuals) is calculated in #Part C.
# Compare the Observed and Expected numbers using Chi-square test as shown in figure:
Null Hypothesis: The population is at HW equilibrium.
Alternate hypothesis: The population is NOT at HW equilibrium.
# p-value = 0.2510
Since p-value is greater than 0.05 (# CI of 95%), accept the null hypothesis and reject the alternate hypothesis.
Therefore, the population is AT HW equilibrium.