Question

Question 1: Antibiotics are effective against a population of bacteria. However, you observe a new resistance allele arises in the bacteria population. You observe the fitness of the mutant is 1% higher than the wild type and the population size (N) of the bacteria is 10⁶.

A) How long does it take for the resistance allele to reach fixation? Hint: The fixation time is defined as the time from a single copy of the resistance allele (1/N) to near fixation (1 - 1/N).

B) If the population size is 10⁸ instead of 10⁶, what is the fixation time? (1 point)

Question 2: In an outbreeding diploid population, an autosomal locus has two alleles, A₁ and A_2. You observe the allele frequencies of A₁ and A₂ are both equal to 0.5, and the relative fitnesses of the three genotypes (A₁A₁, A₁A₂, and A₂A₂) are equal to 2, 1.5, and 1, respectively.

A) Which allele will become fixed in this population? (1 point)

B) What is the allele frequency of A₁ in the next generation? (2 points)

C) What is the allele frequency of A₁ in the next generation, if the population is inbreeding and the inbreeding coefficient is 0.5? (2 points)

Answer 1

Ans:- DA) From the given problem we know that the population size of the bacteria N=100 An fitness of the mutant is ito higher than the wild type. Now the possible coples where a new mutant could orginiale is an. Then probability of a new mutant is anu. Therefore total number of mutation will be fixed per generation Nu * 1/Neu. Finally the time taken for the resistance allele to reach fixation is = 1 unit (since u=1). B) The fixation time does not depends on population size, hence fixation time remain same..

- 2) Frequery of allele A, and Az is Equal, 0.5. we Thus, p=0.5 q=0.5 I Genotype IAIAI A, A2 ALAR I Relative titners 1.2 11.5 T T Relative fitness is used to measure how bit a genotype is as compared to that of least bit genotype. Here the least bit ; A₂ A₂ (since it has lowest relative titness value). fitness reflects reproductive Capacity of a genotype. Also, AA, individual are most tit, twice as that of A2 A2 individual and AtAz individuals have a tituess of 1.5 times that of AzAz individuals A) Eventually allele Ai should get tired, as it mas monimum relative titness, B) Allele Hequeny of A, in next generation.

Average titness world - PAWAIAI) + 2pq (W wanas) + q* (WA2A2) (0.5X0.5) (2)+ (2x6.5x0.5)(15)+(0.5X05)(1) Та , учни е и =p² ( WALA)/w + pq (WalA2)/W' - (0.500.5) (2)/1,5+ (0.5 X6.5) (1.57/1-5 +0.58) (c) Inbreeding coefficient, f=0.5. Allele Jequery of Ai iu nent generation, =p²+ fpq + pq (1-4) = (0-5)={(0.5)X0.5X0.5}+{(05) (015)) (8.–15) = 0.25 +0.125 +0.125 +10.50 please rate my Answer