2. You have a stock solution of 1 x 10\' spores/ml. You serially dilute the spor
ID: 195249 • Letter: 2
Question
2. You have a stock solution of 1 x 10' spores/ml. You serially dilute the spores as described in the table below. Dilution 1 |:10 Dilution 3 Dilution 4 Dilution 2 ID 100 l dilution 3 100 1 stock solution | 900 | water | 100 1 dilution 1 100 1 dilution 2 900 1 water | 900 1 water 900 1 water + + + 2A. (2 pts) You load 10 ul of the stock solution onto a hemocytometer. How many spores do you expect to see in a 4 nl volume? 000 sPvd te 100 1 of dilution 4 omPD plate. How many colonies do you expect to 2B (2 pts) You pla see after incubation? iniExplanation / Answer
Ans. #2A. Given, stock [spore] = 1 x 107 spores/ mL
To calculate number of spores in 4 nL volume.
# Note the following points-
1 L = 103 mL = 109 nL
So, 1 mL = 106 nL
Now, stock [Spore] = 1 x 107 spores/ mL = 1 x 107 spores/ (106 nL) = 10 spores / nL
And,
No. of spores in 4 nL stock = stock [spore] x Volume of stock solution
= (10 spores / nL) x 4.0 nL
= 40 spores
Therefore, 40 spores are expected to be observed in 4 nL volume of stock solution.
Note: There is no need to calculating values using 10.0 uL volume loaded to the haemocytometer because the loaded volume does not affect cell count.
#2B. [Spore] in dilution 4 =
Stock [spore] x total dilution factor from stock soln. to dilution 4
= (1 x 107 spores/ mL) x (1 : 105)
= (1 x 107 spores/ mL) x 10-5
= 100 spores/ mL
Therefore, spore concertation in dilution 4 tube = 100 spores /mL
# Given, volume of dilution 4 taken = 100 uL = 0.100 mL
Now,
No. of spores in 100 uL of dilution 4 = dilution 4 [spore] x Volume of dilution 4 taken
= (100 spores/ mL) x 0.100 mL
= 10 spores
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