You are co infecting e coli klambda with two independent rii...

You are co-infecting E. coli K(lambda) with two independent rII mutants and find that there is no plaque formation. This means that: a) no recombination occurs between the two mutations b) the two rII mutants are different genes, one in rIIA and the other in rIIB c) the two rII mutants are in the same rII gene, either rIIA or rIIB d) one of the mutants is dominant

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Transcript

00:01 Hello students, in the given question we have e.

00:03 Coli strain k lambda which is having infection with two independent r2 mutants of vacuum utilities.

00:13 But plate is not formed.

00:16 What does this mean? this is asked in the given question.

00:20 Here we have been provided with a, b, c, d options.

00:23 Let us consider the explanation and move to the options.

00:27 E.

00:28 Coli.

00:29 K lambda, this strain.

00:32 Is a non permissive stream.

00:35 Whenever this is infected with t4 bacterial phase, here we find that the e.

00:42 Coli k lambda stream is a nonpermissive stream.

00:46 So, whenever t4 bacteria phase has mutation, mutation in the r2 region, that is r2a genes and r2b genes, they infect this e.

00:59 Coli that is co -infection occurs that to two independently r2 mutants they infect is equal like k we find flakes are not formed flakes are not formed there are no recombinants they contain mutation mutation is present in the same r2 gene and same defective product is also formed so whenever we come across light cycle in virus, lytic cycle results in formation of a plaque.

01:42 Plake will be formed but due to mutation, plaque is not formed here.

01:51 So considering this, let us observe the mutations here...

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