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 $P$ is the final population, $P_0$ is the initial population, $r$ is the growth rate per time - unit, and $t$ is the number of time - units. We know that when $t = 2$ hours, $P = 2P_0$. Let $P_0=120$, then $2P_0=P_0(1 + r)^2$. Divide both sides by $P_0$ (since $P_0
eq0$), we get $2=(1 + r)^2$.

Step2: Solve for $r$

Take the square root of both sides: $1 + r=\sqrt{2}$ (we take the positive root since $r$ represents a growth rate). Then $r=\sqrt{2}-1\approx1.414 - 1=0.414$. To convert to a percentage, we multiply by 100, so $r = 41.4\%$ per hour.

Step3: Write the population equation

Using the general form $P = P_0(1 + r)^t$, with $P_0 = 120$ and $r=0.414$, the equation is $P = 120(1.414)^t$.

Step4: Convert time to hours

Since $210$ minutes, and $1$ hour = $60$ minutes, then $t=\frac{210}{60}=3.5$ hours.

Step5: Calculate the population

Substitute $t = 3.5$ into the equation $P = 120(1.414)^t$. 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 $P=120\times3.365 = 403.8\approx404$ cells.

Answer:

a) $41.4\%$
b) $P = 120(1.414)^t$
c) $404$ cells