Question

plete WT ril#2 ut of E. coli B E. coli B E. coli B E. coli B Lysate Lysate Lysate Lysate titer on titer on titer on titer on B KB KB KB K pfu/ml 109 109 10° 103 10° 102 109106 Approximately what percent of the 106/ml ril#1 x rll#2 progeny that can grow on E. coli Kare the result of reversion rather than recombination? Select one: a. 100% b. 10% c. 1% o 10%

Answer 1

ANSWER :-

Accordingly, the revertants are the recombinant phenotypes that have reverted back to the wild-type phenotype.

These are thus able to infect the K-strain of E.coli which in turn will lead to the formation of plaques indicating their growth.

When the Wild-type are grown on the K-strain, the lysate shows presence of 10⁹ PFU/ml. The double recombinants (rII#1 and rII#2) shows presence of 10⁶ PFU/ml.

So, for finding out the percent of revertants, we should consider the recombination frequency which accounts to (10⁶/10⁹ = 10^-3). So, percent of revertants equals to (10^-3 × 100 = 0.1).

So, accordingly, option 'D' seems to be the appropriate answer.

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