# Question #071e7

##### 1 Answer

#### Answer:

#### 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 **diluted culture**.

Once you know this values, use the know **dilution factor** to find the number of cells per **original culture**.

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

So, you know that the five boxes contain

#"Box no. 1: " "12 cells"# #"Box no. 2: " "17 cells"# #"Box no. 3: " "17 cells"# #"Box no. 4: " "14 cells"# #"Box no. 5: " "16 cells"#

The **average number of cells per box** will be equal to

#"avg. no. of cells" = ((12 + 17 + 17 + 14 + 16)color(white)(a)"cells")/"5 boxes"#

#"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 **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

#color(blue)(V_"box" = "area" xx "depth")#

#V_"box" = 1/400 "mm"^2 * "1 mm" = "0.0025 mm"^3#

Now, you know that you have an average of

#15.2"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

You know that you used a **dilution** to get this diluted culture. This should automatically tell you that the original sample contained **more cells** per

When you diluted the original culture, you essentially kept the number of cells **constant**, but you **increased** the volume of the culture **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

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

The **dilution factor** is equal to

#color(blue)("D.F." = c_1/c_2 = V_2/V_1)#

In your case the dilution factor is equal to **times** bigger than the volume of the original culture.

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

Therefore, you can say that the initial culture contained **times** as many cells per

#"D.F." = c_1/c_2 implies c_1 = "D.F." xx c_2#

In your case, you have

#c_1 = 10 * "6080 cells mm"^(-3) = color(green)("60800 cells mm"^(-3))#

Finally, you know that

#"1 mL" = 10^3"mm"^3#

This means that the original culture had a density of

#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))#