Question #071e7
1 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
Once you know this values, use the know dilution factor to find the number of cells per
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
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
When you diluted the original culture, you essentially kept the number of cells constant, but you increased the volume of the 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
Likewise, this is equivalent to saying that the initial culture was
Therefore, you can say that the initial culture contained
#"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))#