You then use the LB and plates to do a serial dilution experiment as well as a p
ID: 65768 • Letter: Y
Question
You then use the LB and plates to do a serial dilution experiment as well as a plaque assay.
a. For the serial dilution experiment you take 10µL of an E. coli culture and dilute it in 990µL LB. You do a total of 4 dilutions and plate 100µL of each dilution on a separate LA plate. You see 29 colonies on the second to last plate. What is the cell concentration of the original bacterial culture?
b. How many colonies would you see on the last plate?
c. For the plaque assay, your T4 bacteriophage titer is 7.8x108 pfu/mL. You do six 1:10 dilutions of the virus, putting 10µL of each dilution in a tube of top agar with 100µL of E. coli. After pouring the contents of the top agar onto an LA plate, you incubate overnight. How many plaques would you expect to see on the final plate?
Explanation / Answer
1. the dilution factor could be calculated by the following formula :
volume of solution taken/ total volume of the tube
therefore here, the dilution factor would be- 10ul / 990 + 10 ul
= 10/1000 which gives 1:100 dilution factor.
now the question says that there are 29 colonies on the second to last plate which means that dilution factor is (1/100 )^3
because it was diluted 3 times till then. which means dilution factor is 10^6
the formula for calculating cfu/ml in the original solution is:
cfu/ml = (no. of colonies x dilution factor) / volume of culture plate
therefore, cfu/ml = 29 x 10^6 / 0.1
the volume of culutre plate is 0.1 ul because 100ul of culture is plated which means 100/1000 ul = 0.1 ul
therefore cfu/ml = 29 x 10^7
2. using the formula for the cfu/ml again we can find out the no of colonies in the last plate. here the dilution factor becomes 10^8 because it is diluted 4 times by then.
29 x 10^7 = no. of colonies x 10^8 / 0.1
29 x 10^7 = no. of colonies x 10^9
no. of colonies = 29 x 10^7 / 10^9
no of colonies = 0.29
3. here, initial pfu/ml = 7.8 x 10^8 pfu/ml
after dilution = 7.8 x 108 x 10^-1
similarly pfu ater 6 dilutions = 7.8 x 10^8 10^-6
= 7.8 x 10^2 pfu/ml
however only 10ul is used so pfu contained = 7.8 x 10^2 / 10/1000
= 7.8 pfu so in the final plate you would see around 7 to 8 plaques
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