Question

A grad student has a culture of E. coli with a total volume of 600 mL, and OD reading at 550 nm of 1.7. He makes serial dilutions of the culture, starting with a transfer of 1 mL culture into 9 mL tryptone broth. Next, he transfers 0.1 mL of this dilution into 0.9 mL tryptone broth, and repeats this two more times to make a total of 4 serial dilutions. He then pipets 200 uL from the fourth dilution on an agar plate, spreads it over the surface, and puts the plate in an incubator to grow overnight. The next day, he counts 152 cfus (colony forming units) on the plate.

How many viable cells are present in the 600 mL starting culture?

Answer is 4,560,000,000 but need to show work.

Answer 1

Ans:- First of all we will calculate all the dilution.

Dilution 1:- 1ml culture was taken and transferred to 9 ml broth.

So, the dilution here will be 10 fold.

Dilution 2:- 0.1 ml was taken from dilution 1 and diluted to 0.9 ml media.

Here, again the dilution is 10 fold.

Now, in question it is mentioned that this dilution was repeated two times.

So, the total dilution factor from dilution 1 + 3 * dilution 2.

Total dilution is 10000.

Now, see 200 ul of culture was plated means to know the total number of colonies in ml you have to multiply the colony obtained by 5.

Now, calculating the CFU/ ml in stock solution.

So, stock solution CFU/ ml = 152 * 5 * 10000

Total CFU / ml in stock solution is 7600000.

Now, this CFU / ml but if you want to know the total CFU in 600 ml of stock solution. You have to multiple CFU/ml *600

Total CFU = 7600000 CFU/ml * 600 ml

Total CFU in bottole is 4,560,000,000 in the your stock solution.

Dear student if you find the answer convincing please do give the feedback.