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 $P(t)=P_0(1 + r)^t$, where $P_0$ is the initial population, $r$ is the growth rate per time - unit, and $t$ is the number of time - units. Given that the population doubles in 2 hours. Let $P_0$ be the initial population and $P(2) = 2P_0$. Substituting into the formula $P(t)=P_0(1 + r)^t$, we have $2P_0=P_0(1 + r)^2$. Divide both sides by $P_0$ (since $P_0
eq0$), we get $2=(1 + r)^2$. Then, take the square root of both sides: $1 + r=\sqrt{2}$ (we take the positive root since $r>0$ for growth), so $r=\sqrt{2}-1\approx0.414$ or $41.4\%$ per hour.

Step2: Write the population equation

The general population equation for exponential growth is $P(t)=P_0(1 + r)^t$. Since $P_0 = 120$ and $r=\sqrt{2}-1$, the equation for the number of cells $P$ at time $t$ (in hours) is $P(t)=120(\sqrt{2})^t$.

Step3: Calculate the population at $t = 3.5$ hours

First, convert 210 minutes to hours. Since 1 hour = 60 minutes, $t=\frac{210}{60}=3.5$ hours. Substitute $t = 3.5$ into the equation $P(t)=120(\sqrt{2})^t$. We know that $(\sqrt{2})^t=2^{\frac{t}{2}}$, so when $t = 3.5$, $(\sqrt{2})^{3.5}=2^{\frac{3.5}{2}}=2^{1.75}$. $P(3.5)=120\times2^{1.75}$. Using the property $a^b = e^{b\ln(a)}$, $2^{1.75}=e^{1.75\ln(2)}\approx e^{1.75\times0.693}=e^{1.21275}\approx3.36$. Then $P(3.5)=120\times3.36 = 403.2\approx403$ cells.

Answer:

a) The growth rate of the yeast cells is approximately $41.4\%$ per hour.
b) The equation for the number of cells $P$ at $t$ hours is $P(t)=120(\sqrt{2})^t$.
c) The population of yeast cells after 210 minutes (3.5 hours) is approximately 403 cells.