Question

You diluted a starting culture of E. coli 1:10 ten times. From these dilutions you plated 0.1mL of each dilution onto an agar plate. You then counted the number of bacterial colonies that grew after incubation (see below). Give the average estimated starting concentration of bacteria per milliliter from the plates shown below. Don't forget units! Show your work. 1 x 10^-9 dilution (1/1,000,000,000) 1 x 10^-10 dilution (1/10,000,000,000)

          You diluted a starting culture of E. coli 1:10 ten times. From these
dilutions you plated 0.1mL of each dilution onto an agar plate. You
then counted the number of bacterial colonies that grew after
incubation (see below). Give the average estimated starting
concentration of bacteria per milliliter from the plates shown below.
Don't forget units! Show your work.
1 x 10^-9 dilution
(1/1,000,000,000)
1 x 10^-10 dilution
(1/10,000,000,000)

You diluted a starting culture of E. coli 1:10 ten times. From these
dilutions you plated 0.1mL of each dilution onto an agar plate. You
then counted the number of bacterial colonies that grew after
incubation (see below). Give the average estimated starting
concentration of bacteria per milliliter from the plates shown below.
Don't forget units! Show your work.
1 x 10^-9 dilution
(1/1,000,000,000)
1 x 10^-10 dilution
(1/10,000,000,000)

Added by Bryan M.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

Instant Answer

Solved by Expert Shaiju T

04/18/2022

Step 1

- 1x10^-9 dilution plate: 12 colonies - 1x10^-10 dilution plate: 5 colonies Show more…

Show all steps

Thanks for your feedback!

You diluted a starting culture of E. coli 1:10 ten times. From these dilutions you plated 0.1mL of each dilution onto an agar plate. You then counted the number of bacterial colonies that grew after incubation (see below). Give the average estimated starting concentration of bacteria per milliliter from the plates shown below. Don't forget units! Show your work. 1 x 10^-9 dilution (1/1,000,000,000) 1 x 10^-10 dilution (1/10,000,000,000)