Question

You repeat the viral plaque assay adding 10 microliters of viral solution to 990 microliters of LB. You perform this serial dilution five times. You add 10 microliters from each dilution to the corresponding host bacteria molten top agar mixture which is then spread onto an LB agar plate. The next day you count 39 plaques on the fourth plate.

Why is it that on plate 3 the number of plaques will be 39 x 10^2 and on plate 5 there will be almost no plaques to do plaque assay. EXPLAIN THIS PLEASE.

Answer 1

In serial dilution concept we are decreasing the intial concentration of the cells gradually by transfering to the next tube.

In the above context the intial concentration of the viral solution is unknown but upon serial dilution the plaques formed on plate no: 3 is 39*10^2 , plate no: 4 is 39 and plate no: 5 is 0.

If we are doing back calcution.

Starting from 5th plate to 1st plate, then the concentrations are as follows.:

5th plate : 0
4th plate : 39
3rd plate : 39*10^2
2nd plate : 39*10^4
1st plate : 39*10^6

i.e, initial concentration of your 10microliters of viral solution to 990 microliters of LB consists of 39*10^6 viruses.

So, initial concentration was decreased gradually by 10^2 folds.

Note:

To perform any phase calculations minimum 30 - 300 plaques required.

Hope this explanation helped to understand better.

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