Question

knowledge about the lac operon system AND knowledge Problem set about the Lac operon (these problems require knowledge about the lac operon system AND about consequences of mutations) You are working with three strains of E.coli in the laboratory: A wild-type strain. A strain that contains a 15 base-pair deletion in the lac operon operator sequence. A strain that contains a nonsense mutation in the gene that encodes the CRP protein. This nonsense mutation occurs in the region that corresponds to the DNA binding domain of the CRP protein Strain #1 Strain #2 Strain #3 Unfortunately the labels have come off of your media plates and you don't know which strain is which! You decide to label them strain #1, #2, and #3 until you figure it out. You grow each E.coli strain in different media conditions and assay for Lacz gene product activity (that is measured in Miller Units). The following table has the results of your experiment. Strain Media Condition Lac Z gene product activity (recorded in Miller Units) 100 units 11 no lactose, high glucose 100 units high lactose, high glucose 10,000 units high lactose, no glucose 10,000 units Silent no lactose, no glucose 100 units no lactose, high glucose 100 units high lactose, high glucose 10,000 units high lactose, no glucose 100 units no lactose, no glucose 100 units no lactose, high glucose 100 units high lactose, high glucose 100 units high lactose, no glucose 100 units no lactose, no glucose

A. Which strain is the wild-type E.coli? Explain how you know this.

B. Which strain contains the nonsense mutation in the CRP protein gene? Explain how you know this.

C. Which strain contains the deletion mutation in the lac operon operator sequence? Explain how you know this.

Answer 1

Answer-A Strain 2 is wild type because in presence of lactose and in absence of glucose the lac Z gene activity is high as there will be high cAMP in presence of low glucose concentration which will bind to the catabolite activating protein and thus transcription will be above basal level.

Answer- B Strain 3 as in each condition the lac Z gene product activity is same thus suggesting that even when lactose and cAMP are high in concentration and glucose low in concentration the cell is not able to produce high lac z gene product because of the mutation in CRP protein due to which it is not able to bind DNA and regulate transcription positively.

Answer- C Strain 1 because even when lacose is absent , the lac Z gene product activity is high thus indicating the repressor is not able to recognise the mutated opreator sequence