Added 100 microL of E. coli to top agar of two plates Have tubes 8 tubes 10 -1 t

Question

Added 100 microL of E. coli to top agar of two plates

Have tubes 8 tubes 10-1 thru 10-8, and add 900 microliters to each tube and perform a series of 1:10 dilutions of viral sample

Two plates divided in fourths put each one of the dilutions of 10 microliters on each fourths.

Resulted in 41 PFU in the 10-8.

A serial dilution is required to be calculated--then the original concentration of phage particles can be calculated by dividing the number of plaques by the volume plated on the plate (in mL) and the dilution factor. The resulting number is the plaque forming units (PFU) per milliliter.

HOW TO DO CALCALUTATIONS? MUST SHOW WORK FOR CREDIT

Explanation / Answer

For preparation of serial dilution we have been provided with eight tubes labeled 10-1 to 10-8, each containing 900µl of diluents.

So, in order to perform 1:10 dilution of viral samples, we will add 100 µl of viral sample in the first tube and mix thoroughly. The total volume of liquid in this tube becomes 1000 µl ( 900 µl + 100 µl ).

The dilution of the viral sample= volume of sample/ total volume= 100/1000= 1/10 or 0.1 or 10-1.

In the next step, 100 µl sample will be taken out from the 10-1 tube and added to the next tube containing 900 µl of diluent.

So the dilution in the second tube= (100x 0.1)/1000= 10/1000= 1/100 or 0.01 or 10-2

In this manner, the next dilution (10-3) can be prepared by transferring 100 µl sample from the 10-2 tube into the next tube containing 900 µl of diluents. Similarly the rest of dilutions can be prepared up to 10-8.

For calculation of PFU/ml

No. of plaques= 41

Vol. of viral sample plated= 10 µl= 10 x 10-3 ml= 10-2 ml

Dilution= 10-8, so dilution factor= 1/10-8= 108

PFU/ml= No. of plaques x Dilution factor/ volume of viral sample plated= 41 x 108/ 10-2= 41 x 108 x 102= 41 x 1010= 4.1 x 1011 PFU/ml

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