Question

In: Biology

A gene in a population of sexually reproducing organisms with the following genotype frequencies is at Hardy-Weinberg Equilibrium: BB 100, Bb 800, bb 100


A gene in a population of sexually reproducing organisms with the following genotype frequencies is at Hardy-Weinberg Equilibrium: BB 100, Bb 800, bb 100

A) true B)false C) can not be determined

Which of the following evolutionary forces would be the best explanation for the genotype frequencies in the population in the above question?

A) genetic drift B) there are no evolutionary forces acting on this gene population C)assortive mating D) Overdominance E)Directional selection

If a population is at Hardy-Weinberg equilibrium, which of the following allele frequencies would yield the smallest number of heterozygotes? A)

A=0.25, B=0.75 B)   A=0.75, B=0.25

C)A=0.50, B=0.50

D)A=0.10, B=0.90

Fixation occurs when one allele becomes the only allele for a gene in a population, meaning that all other alleles for that gene have been lost. True or false

Solutions

Expert Solution

Hardy- Weinberg Equation for 3 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 allele P

q = allelic frequency of allele Q

p2 = genotypic frequency of PP

q2 = genotypic frequency of QQ

2pq = genotypic frequency of PQ

Genotype

Number of individuals

Observed frequency

Expected frequency

(using HW equation)

Expected no. of Individuals

(expected f x population size)

BB

100

f(B) = (2BB+ 1Bb)/ 1000 = 0.5

f(BB) = (0.5)2 = .025

250

Bb

800

f(Bb) = 2 x 0.5x 0.5 = 0.50

500

bb

100

F(b) = (2bb+ 1Bb)/ 1000 = 0.5

f(bb) = (0.5)2 = .025

200

Total= 100

X2 Test, p value= 000

The population is at HW equilibrium.

Ans. 2. E. Direction selection - the process of reduction of genetic variation of the population by selecting for or against a specific allele, thus shifting predominant traits at either of one extreme.

Rest of the options are not process of natural selection.

Ans. 3. See equation 2 [Assuming A and B be allele]-

Heterozygote frequency = 2 pq

            a. 2x 0.25 x 0.75= 0.375              b. 2x 0.75 x 0.25= 0.375

            c. . 2x 0.5 x 0.5= 0.5                    d. 2x 0.1 x 0.9= 0.18

correct option d.

Ans. 3. True


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