Question #071e7

1 Answer
Feb 16, 2016

Answer:

#6.1 * 10^7"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 #"mm"^3# for the diluted culture.

Once you know this values, use the know dilution factor to find the number of cells per #"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

  • #"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 #1/400"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 #"1 mm"#. This will make the volume of each box equal to

#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/box"#, which means that the number of cells per #"mm"^3# will be equal to

#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 #6080# cells per #"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 #"mm"^3#.

When you diluted the original culture, you essentially kept the number of cells constant, but you increased the volume of the culture #-># 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 #"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

#color(blue)("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 #"mm"^3# than the diluted sample

#"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))#