# Question 071e7

Feb 16, 2016

$6.1 \cdot {10}^{7} {\text{cells mL}}^{- 1}$

#### Explanation:

Your strategy here will be to calculate the average number of cells per box, then use the dimensions of a single box to find the number of cells per ${\text{mm}}^{3}$ for the diluted culture.

Once you know this values, use the know dilution factor to find the number of cells per ${\text{mm}}^{3}$ in the original culture.

Finally, convert the resulting density from cells per cubic millimeter to cells per milliliter.

So, you know that the five boxes contain

• $\text{Box no. 1: " "12 cells}$
• $\text{Box no. 2: " "17 cells}$
• $\text{Box no. 3: " "17 cells}$
• $\text{Box no. 4: " "14 cells}$
• $\text{Box no. 5: " "16 cells}$

The average number of cells per box will be equal to

$\text{avg. no. of cells" = ((12 + 17 + 17 + 14 + 16)color(white)(a)"cells")/"5 boxes}$

$\text{avg. no. of cells " = " 15.2 cells/box}$

Now, I'm not really sure how to interpret the dimensions of a box. You provided the area of a single box, which is said to be $\frac{1}{400} {\text{mm}}^{2}$, but you still need to know the depth of such a box in order to be able to calculate its volume.

To keep the calculations simple, I will assume that each box has a depth of $\text{1 mm}$. This will make the volume of each box equal to

$\textcolor{b l u e}{{V}_{\text{box" = "area" xx "depth}}}$

${V}_{\text{box" = 1/400 "mm"^2 * "1 mm" = "0.0025 mm}}^{3}$

Now, you know that you have an average of $\text{15.2 cells/box}$, which means that the number of cells per ${\text{mm}}^{3}$ will be equal to

$15.2 {\text{cells"/color(red)(cancel(color(black)("box"))) * overbrace((1 color(red)(cancel(color(black)("box"))))/("0.0025 mm"^3))^(color(brown)("volume of a box")) = "6080 cells mm}}^{- 3}$

So, your diluted culture contains an average of $6080$ cells per ${\text{mm}}^{3}$.

You know that you used a $1 : 10$ dilution to get this diluted culture. This should automatically tell you that the original sample contained more cells per ${\text{mm}}^{3}$.

When you diluted the original culture, you essentially kept the number of cells constant, but you increased the volume of the culture $\to$ you decreased the concentration of the original culture.

In your context, you can think of the concentration of cells as being equivalent to the number of cells per ${\text{mm}}^{3}$.

Mathematically, you can write this as

color(blue)(overbrace(c_1 xx V_1)^(color(purple)("no. of cells in original culture")) = overbrace(c_2 xx V_2)^(color(purple)("no. of cells in diluted culture"))" ", where

${c}_{1}$, ${V}_{1}$ - the concentration of cells and the volume of the original culture
${c}_{2}$, ${V}_{2}$ - the concentration of cells and the volume of the diluted culture

The dilution factor is equal to

$\textcolor{b l u e}{\text{D.F.} = {c}_{1} / {c}_{2} = {V}_{2} / {V}_{1}}$

In your case the dilution factor is equal to $10$. This means that the volume of the diluted culture is $10$ times bigger than the volume of the original culture.

Likewise, this is equivalent to saying that the initial culture was $10$ times more concentrated than the diluted culture.

Therefore, you can say that the initial culture contained $10$ times as many cells per ${\text{mm}}^{3}$ than the diluted sample

$\text{D.F." = c_1/c_2 implies c_1 = "D.F.} \times {c}_{2}$

c_1 = 10 * "6080 cells mm"^(-3) = color(green)("60800 cells mm"^(-3))
${\text{1 mL" = 10^3"mm}}^{3}$
60800 "cells"/color(red)(cancel(color(black)("mm"^3))) * (10^3color(red)(cancel(color(black)("mm"^3))))/"1 mL" = color(green)(6.1 * 10^7"cells mL"^(-1))#