Which of the following would be sufficient for the Hardy-Weinberg equation to ac

Question

Which of the following would be sufficient for the Hardy-Weinberg equation to accurately predict genotype frequencies from allele frequencies? p + q = 1 the population is not evolving due to natural selection. the population is not evolving due to any of the conditions that disrupt Hardy-Weinberg equilibrium. the population is infinitely large. in a village, if the proportion of individuals who have sickle-cell disease is 0.10, and the population is assumed to be at Hardy-Weinberg equilibrium, what is the expected frequency of the HbS allele? 0.32 0.18 0.90 0.01 in a village where the proportion of individuals who are susceptible to malaria (genotype HbA/HbA) is 0.31, and the population is assumed to be at Hardy-Weinberg equilibrium, what proportion of the population should be heterozygous HbA/HbS?

Explanation / Answer

Ans. 5.5. Correct option- C

Option A. Incorrect. A population NOT at Hardy-Weinberg (HW) equilibrium also has p + q= 1. Since the frequencies p, and q are calculated considering their relative abundance, their sum of always 1.0. However, the value of p and q changes over generations when population is evolving. The value of p and q remains constant over generations if the population is at HW equilibrium.      

Option B. Incorrect. Factors other than natural selection, like genetic drift, bottle-neck effect, etc. also disrupt HW equilibrium. So, even if natural selection does not occur, the population may evolve through genetic drift, etc.

Option C. Correct. HW equation holds true when no factor causes evolution of the population.

Option D. Incorrect. HW equation holds true when no factor causes evolution of the population. Being a large population do not exclude the possibilities of evolution.

Ans. 5.6.

Normal, dominant allele = HbA

Mutant, recessive allele = Hbs

Normal, homozygous genotype = HbA HbA               - susceptible to malaria

Normal, heterozygous genotype = HbA Hbs               - it is a carrier, but unaffected by disease

Recessive, affected genotype = Hbs Hbs                    - affected with sickle-cell disease

Now,

Hardy- Weinberg Equation for 2 allele system (for a population at HW equilibrium)

p + q = 1                                  - equation 1

(p + q)2 = p2 + q2 + 2pq = 1                   - equation 2

Where,

p = allelic frequency of dominant allele HbA

q = allelic frequency of recessive allele HbS        

p2 = genotypic frequency of HbA HbA

q2 = genotypic frequency of Hbs Hbs

2pq = genotypic frequency of HbA Hbs

Given,

            Frequency of Hbs Hbsindividuals = q2 = 0.10

Or, q2 = 0.10

Or, q = (0.10)1/2 = 0.316 = 0.32                                  ; [Note: (0.10)1/2 is under root (.010)]

Hence, frequency of allele HbS = q = 0.32

Correct option. A.

Ans. 5.7

Given,

            Frequency of HbA HbAindividuals = p2 = 0.31

Or, p = (0.31)1/2 = 0.55677

Hence, frequency of HbA allele = 0.55677

Putting the value of p in equation 1-

            p + q = 1

            or, q = 1 – 0.55677 = 0.44323

Hence, q = 0.44323

Now,

Frequency of the heterozygotes (HbA Hbs) = 2 pq = 2 x (0.55677) x (0.44323) = 0.493

Hence, frequency of heterozygotes = 0.493

            % of heterozygotes = 49.3 %

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