# Answer on Other Biology Question for maithri

Question #52870

Two strains of E. coli, A and B, are inoculated into a chemostat in equal proportions. After one generation, the frequency of strain B is 20%. Calculate the fitness of strain B relative to strain A. Show your work.

Expert's answer

Fitness (

The frequencies of

strain A = 80%/50% = 1.6

strain B = 20%/50% = 0.4

The fitness of strain B relative to strain A:

*w*) is a survival rate of a phenotype (bacterial strain 1) relative to the maximum survival rate of another phenotype (bacterial strain 2) in population.The frequencies of

*E.coli*strains (A and B) in the input population are 50% for each strain. After one generation the frequencies of strains are 80% (strain A) and 20% (strain B). Relative changes of the frequency after one generation are:strain A = 80%/50% = 1.6

strain B = 20%/50% = 0.4

The fitness of strain B relative to strain A:

*w*= 0.4/1.6 = 0.25**Answer: 0.25**Need a fast expert's response?

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