Question

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 71 cells. (a) Find the relative growth rate. (Assume t is measured in hours.) k =

Answer 1

Answer:

Step-by-step explanation:

Given that a common inhabitant of human intestines is the bacterium Escherichia coli.

A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes.

Initial population = 71 cells

Doubles in 1/3 of an hour

$P = P_o (2)^{t/3}$, where t is the no of hours.

$P = 71 (2)^{t/3}$

Rate of growth = derivative of P

= $71(2)^{t/3} *(1/3)ln 2\\= \frac{71}{3} (2)^{t/3} ln 2$