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.

What is the dilution of cells spread onto the agar plate?

Answer is 0.00002 but need to show work.

Answer 1

dilution= aliqout volume/final volume

first 1 mL culture is transferred into 9 mL tryptone broth.

dilution=`1/10 ( final volume=9+1=10 mL)

dilution at the second step is

dilution= aliqout volume/final volume

= 0.1/1 ( final volume=0.9+0.1=1 mL)

this was repeated twice, so that total four dilutions are made.

so total dilution= product of dilution at each step

= 1/101/101/101/10

= 10^-4

volume plated= 200 uL

volume plated in mL= 200/1000=0.2 mL

dilution of the plate= volume plated in mL dilution of the tube

= 10^-40.2

= 2`0^-5

so the answer is 0.00002 (0.00002=2`0^-5)