Question

The genotypes of the lac operon of several haploid and partial diploid E. coli strains are given. For each genotype, indicate the status of B-galactosidase synthesis as either Yes or No. Assume that no glucose is present and cAMP is rich in the growth medium.

• Oc is a mutant operator that cannot be bound by the wild type repressor protein.

• Is is a mutant of repressor gene and cannot bind to operator

Haploid Partial Diploid IOZY IOCZY ISOZY IOZYWIOCZY IOZYISOZY B-gal enzyme without lactose With lactose

Answer 1

In first case, where E .coli is haploid:

1. In first case every gene/operator is function. Therefore in withour lactose repressor (I gene product) is produced and it inhibit the synthesis of β- galactosidase. In second condition where lactose is present it inhibit repressor ( I-gene product) and it produce β- galactosidase. Answer is without lactose No and with lactose yes.
2. In second case, repressor cannot bind to operator (O^c) due to therefore β- galactosidase is synthesized in presence and absence of lactose.
3. In third case, where repressor (I_s) can not bind to operator, hence β- galactosidase is synthesized in presence and absence of lactose.

In second case, where E. coli is partially diploid:

1. First and second set of gene will produce repressor protein but repressor only bind with first operon where operator (O) is functional but in second gene operator is not bound to (O^c) repressor. In this case second gene produce β- galactosidase enzyme in presence and absence of lactose.
2. In second case repressor functional repressor is produced by first gene but second gene is not producing the functional repressor (I_s). In this case first gene functional repressor binds to operator and regulates the expression of β- galactosidase. So in absence of lactose β- galactosidase production is inhibited but in presence of lactose β- galactosidase is synthesized.