Question 10 [13 marks 13 minutes] The hypothetical BSc operon of bacteria encode

Question

Question 10 [13 marks 13 minutes] The hypothetical BSc operon of bacteria encodes the HD enzyme and the P enzyme. Expression of this operon is regulated in response to substance X and substance Y. Five different mutants have been isolated that show abnormal regulation of expression of the BSc operon. The table below shows the level of enzyme activity for these different mutants. First the bacterial cells were incubated in minimum medium, then compound 1 was added, and then compound 2 was added after further incubation And then after substance Y added Minimal mediumm After substance X added HD enzyme Penzyme HD enzyme Penzyme HD enzyme Penzyme Cells wildtype mutant 1 mutant 2 mutant 3 mutant 4 mutant5 activit ?? act zero high low high zero low zero high high zero zero zero a) b) c) Which observations indicate that the BSc operon is inducible? Which observations indicate that the BSc operon is also repressible? Which mutant shows gene expression consistent with being a loss-of-function mutation in: [I mark] [I mark] i. the promoter of this operon ii. the DNA sequence bound by the positive regulator ii. the DNA sequence bound by the negative regulator iv. the gene encoding HD enzyme v. the gene encoding P enzyme 7 marks d) Predict the enzyme activity (using the same format as the table above) that would result from a loss-of-function mutation in i. the gene encoding the positive regulator ii. th e gene encoding the negative regulator iii. both (i) and () .e. a double mutant 4 marks]

Explanation / Answer

Answering any 3 of the questions based on CHEGG rules:

a.Mutant 2 shows that BSc operon is inducible as the mutation has resulted in no expression of HD and P enzyme in presence of compound X whereas wild type is induced by compund X

b. Mutant 4 behaves more or less like wild type and shows that upon addition of X, BSc operon gets activated, whereas compound Y leads to repression of the same.

c.

1.Mutant 5 as it does not show transcription in basal, after addition of substance X or Y due to loss in promotor activity.

II.Mutant 2 shows the loss of function due to consistent to DNA sequence bound by a positive regulator.

III.Mutant 4 shows the loss of function due to consistent to DNA sequence bound by a negative regulator as the expression is high like wild type and low when compund 2 is added

IV.Mutant 1 shows no activity for HD enzyme as the function is abrogated.

V.Mutant 3 shows no function for P enzyme as the mutation has caused the abrogation of P enzyme function.

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