Answer:
61 for Plate 1 ,
59 for Plate 2
Explanation:
Given that the original stock has a concentration of 6 * 10^6 infectious virus particles /mL
and it was diluted by 1:10000
New concentration = ( 6 * 10^6 ) / 10000 = 600 infectious virus particles/ mL
Also given that the virologist plates 100 mL of each sample
therefore; 0.1 mL has a count of 60.
The experimental results that he is most likely to observe in the duplicate plates will be ;
61 for (Plate 1) ,
59 for (Plate 2)