You decide to focus first on Mutation 1 and Mutation 2.
The F' plasmid appears to "rescue" mutant strain 1 by restoring wild-type inducible expression.
The F' plasmid does not change the phenotype of mutant strain 2; there is still no expression of the lac operon.
There are three possible mutations that could lead to lack of expression: a super-repressor (Is), a promoter mutation (P-) or a mutation in the lacZ gene (Z- ). To determine the effect of introducing a wild-type F' plasmid into a bacterial cell with each of these mutations, you draw out the three possible scenarios.
F' l+ cap+
Determine the expression of the lac operon for each possible partial diploid and for the "unknown" mutation partial diploids (Mutation 1 and Mutation 2).
1. F' lac+/ ts
2. F' lac+/ P
3. F' lac+/ Z
4. F' lac+/ Mutation 1
5. F lac+/ Mutation 2
A. Inducible (wild type)
B. No expression (mutant)

Respuesta :

Answer:

Super-repressor (Is) - Uninducible dominant

Binding region is modified by the protein called repressor which prevents the lactose to bind with it. Thus, even the lactose is present the repressor protein is bound to the promoter and can not be detach from it. Transcription is not inducible.

Promoter mutation (P)- - Uninducible recessive

Mutated promoter region inhibits the binding of RNA polymerase that leads to trranscription.

The lacZ gene ( Z-) is mutated - recessive while the normal LacZ gene (Z +) -  dominant .

                   Inducible (Wild type)        No expression (Mutant)

F' lac+/ Is             No                                Yes

F' lac + / P-            Yes                                 No

F' lac + / Z-            Yes                                 No

F' lac +/ Mutation1     Yes                                 No

F' lac+ / Mutation2      No                                 Yes

Mutation 1 and F' plasmid - Inducible wild type

Mutation 2 and F' plasmid - Mutant no expression