Question

. For each of the following E.coli merodiploid genotypes, Explain briefly (1-3 sentences) whether they would be able to metabolize lactose, and if so,would they synthesize lacZ inducibly or constitutively? (2 PTS EACH)

a) i⁺p⁺o⁺z⁺/i⁺ p⁺o^cz⁺

b) i⁺p⁺o⁺z⁺/ i⁺ p^-o^cz⁺

c) i^-p⁺o⁺z⁺/ i⁺ p^-o^cz^-

d) i^-p^-o⁺z⁺/ i⁺ p⁺o⁺z^-

e) i⁺p⁺o⁺z⁺/ i⁺ p⁺o^cz^-

f) i⁺p⁺o⁺z^-/ i⁺ p^-o^cz⁺

Answer 1

A) Yes, it will metabolise lactose in constitutive manner. Because of presence of O^c mutation. This mutation makes the operator to work in a constitutive manner. It can not allow repressor to bind to it. RNA polymerase binds to the promoter. It will proceed further to Z gene and there is found expression of Z gene.

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B) Yes,it will able to metabolize lactose. It will metabolize the lactose in inducible manner. When the inducer is present, then it will bind to the repressor and inactivates repressor. The repressor cannot bind to the operator. That's why the expression of Z gene occurs.

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C) Yes, it will able to metabolize lactose in inducible manner. The wild type and mutant type pressure is encoded by I+ and I- gene. It will combined to form while type repressor. When lactose is present, then lactose into the repressor and inactivate this repressor. This repressor cannot bind to the operator. That's why there is found expression of z gene in presence of lactose.

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D) No, it can not metabolise the lactose. Because of P- in one strand and z- on another strand.

5) Yes, it is able to metabolise the lactose in inducible manner. Because of presence of I+ gene. The O^c on other DNA has Z- in cis position to it. That's why the O^c expression does not occur.

6) No, it can not metabolise the lactose. Because Z- in one DNA and P- in another DNA.

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