A population of 5545 people is assumed to be in Hardy-Weinberg equilibrium. Of t

Question

A population of 5545 people is assumed to be in Hardy-Weinberg equilibrium. Of this population, 3607 have the dominant trait of lactase persistence. The recessive form of this trait is lactose intolerance. Showing all calculations, determine           (10 marks)

A. the frequency of the homozygous recessive genotype in the population.

B. the frequency of the recessive allele.

C. the frequency of the dominant allele.

D. the number of homozygous dominant people.

E. the number of heterozygous people.

PLEASE DO NOT ATTEMPT IF YOU DO NOT SPECIALIZE IN GENETICS!! THANK YOU!!

Explanation / Answer

Ans. 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

Let the healthy, dominant allele be “A”, and the recessive allele be “a”.

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, AA individuals have 2 copies of allele A, whereas Aa individuals have one copy each of allele A and a.

Given,

            Population size = 5545

So, total number of all alleles of the gene = 2 x population size = 2 x 5545 = 11090

#A. Number of homozygous recessive (aa) genotype or individuals =

Population size – No. of dominant genotypes

                                                = 5545 – 3607

                                                = 1938

Frequency of aa, q2= No. of aa genotype / population size

                                    = 1938 / 5545

                                    = 0.3495

#B. Frequency of recessive allele, q = (frequency of homozygous recessive genotype)½

            That is, q = (q2)½                                                                         - see equation 2

            Or, q = (0.3495)½ = 0.5912

Therefore, frequency of recessive allele, q = 0.5912

#C. Frequency of dominant allele = p = 1 – q                 - see equation 1

            Or, p = 1.0000 – 0.5912 = 0.4088

Thus, frequency of dominant allele = p = 0.4088

#D. The frequency of homozygous dominant individuals = p2            - see equation 2

                                                                        = (0.4088)2

                                                                        = 0.16711744

Number of homozygous dominant (AA) individuals = p2 x population size

                                                                        = 0.16711744 x 5545

                                                                        = 926.6662048

                                                                        = 927

The number of organisms must be a whole number. So, 926.66 is rounded off to nearest whole number.

#E. Method 1: No. of heterozygous (Aa) people =

Population size – (No. of AA + No. of aa individuals)

                                                = 5545 – (927 + 1938)

                                                = 2680

Method 2. Frequency of Aa genotype = 2 p q = 2 x 0.4088 x 0.5912 = 0.4834

Number of heterozygote (Aa) individuals = 2pq x population size

                                                                        = 0.4834 x 5545

                                                                        = 2680.453

                                                                        = 2680

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