Question

1. a population of yeast cells can double in 2 h. assume an initial population of 120 cells. a) what is the growth rate, in percent per hour, of this colony of yeast cells? 1a b) write an equation that can be used to determine the population p of cells at t hours. 1a c) use your equation to determine the population after 210 min. 3a

Explanation:

Step1: Find the growth - rate formula

The general formula for exponential growth is $P = P_0(1 + r)^t$, where $P_0$ is the initial population, $r$ is the growth rate as a decimal, and $t$ is the time. Since the population doubles in 2 hours, if $P_0$ is the initial population and $P = 2P_0$ when $t = 2$. Substituting into the formula $2P_0=P_0(1 + r)^2$. Divide both sides by $P_0$ (since $P_0
eq0$), we get $2=(1 + r)^2$. Taking the square - root of both sides, $1 + r=\sqrt{2}$ (we take the positive root since $r>0$ for growth), so $r=\sqrt{2}-1\approx0.414$. To convert to a percentage, multiply by 100. So the growth rate is $41.4\%$ per hour.

Step2: Write the population equation

The general exponential - growth equation is $P = P_0(1 + r)^t$, where $P_0 = 120$ and $r = 0.414$. So the equation is $P=120(1 + 0.414)^t=120\times1.414^t$.

Step3: Convert time for part c

210 minutes is $\frac{210}{60}=3.5$ hours.

Step4: Calculate population at $t = 3.5$

Substitute $P_0 = 120$, $r = 0.414$, and $t = 3.5$ into the equation $P = P_0(1 + r)^t$. So $P=120\times(1.414)^{3.5}$. First, $(1.414)^{3.5}=1.414^3\times1.414^{0.5}$. $1.414^3=1.414\times1.414\times1.414\approx2.828$ and $1.414^{0.5}=\sqrt{1.414}\approx1.19$. Then $(1.414)^{3.5}\approx2.828\times1.19 = 3.365$. So $P=120\times3.365 = 403.8\approx404$.

Answer:

a) The growth rate is $41.4\%$ per hour.
b) The equation is $P = 120\times1.414^t$.
c) The population after 210 minutes is approximately 404 cells.