Question

What will the expression pattern be for the following lac operon partial diploid?

I- Oc Z+ Y- / I+ O+ Z- Y+

A) both genes inducible

B) lacZ constitutive, lacY inducible

C) lacZ inducible, lacY constitutive

D) Both genes (lacZ and lacY) constitutive

Answer 1

The Oc mutation is a mutant form of operator which cannot bind the repressor. Hence, this operon will be constitutively active. I- indicates that there is no repressor present. I+ indicates that there is repressor expressed.

The + sign indicates expression of the gene while - sign indicates that there is non-functional enzyme/no enzyme. Inducible expression is differential expression of the genes in presence and absence of lactose. When the presence or absence of lactose doesn’t inhibit expression of the genes, it is constitutive expression.

The first chromosome in the partial diploid has an Oc operator that cannot bind the repressor. Hence, genes on this operon are always turned off. This operator only affects the lac genes on first chromosome (cis). Presence (from second chromosome) and absence of repressor (from first chromosome) will still cause expression of lac Z. LacY will be nonfunctional (Y-). Thus, there will be constitutive expression of lacZ. This expression is not inducible due to the Oc present.

In the second chromosome, the lac I gene will encode the repressor, which binds the operator in absence of lactose (inducer). This will inhibit the transcription of lac Y. However, in presence of inducer, the allolactose will bind the repressor. Hence, RNA polymerase can bind promoter to induce Lac Y expression. The expression of lacZ is not affected in this chromosome as the partial diploid is lacZ-.

Thus, there is constitutive lacZ expression from chromosome 1 and inducible lacY from the second chromosome.

Right choice: B) lacZ constitutive, lacY inducible