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 minimunm medium, then compound 1 was added, and then compound 2 was added after further incubation And then after substance Y added Minimal medium After substance X added HD enzyme activit Penzyme activit HD enzyme activit P enzyme HD enzyme activit P enzyme activit Cells wildtype mutant 1 mutant 2 mutant 3 mutant4 mutant 5 activit high high low zero zero zero high low low zero zero zero high zero zero zero high zero low high high low low zero low zero low low low zero zero zero zero zero zero a) Which observations indicate that the BSc operon is inducible? b) Which observations indicate that the BSc operon is also repressible? c) Which mutant shows gene expression consistent with being a loss-of-function mutation in mark [1 mark] i. the promoter of this operon ii. the DNA sequence bound by the positive regulator iii. the DNA sequence bound by the negative regulator iv. the gene encoding HD enzyme v. the gene encoding P enzyme 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. the gene encoding the negative regulator iii. both (i) and (ii), i.e. a double mutant 4 marks]

Explanation / Answer

a) first four observations in the table includes wild-type, mutant 1, mutant 2, mutant 3 are inducible.

They are not expressed until the inducer (x substance) is added to the solution. Their expression rate is zero in resting stage hence they are in inducible condition.

b) Mutant 4 indicates that it is a repressible operon.

In this case, the gene expression is normally high but reduced under specific conditions.

c) i. mutant 5 in the table have promoter mutations hence it is not synthesised even after induction.

ii. mutant 2 have positive regulator mutation, due to this mutation positive regulator cannot bind to DNA sequence to increase its expression.

iii. Mutant 4 have negative regulator mutation, hence the gene expression is possible without induction.

iv. Mutant 1 has a mutation in HD enzyme-coding gene. hence it is unable to express this enzyme even after induction

v. Mutant 3 have mutation in P enzyme coding gene . hence it is not expressing this enzyme even after induction.

d) The enzyme activity in

i. Loss of function mutations in positive regulator causes the genes to express at a low level. here both Hd and P enzymes are expressed but at a low level because in this mutations there is no scope of induction of expression.

ii. Loss of function mutations in negative regulator enable the structural genes HD and P to express at high levels without induction. But externally added diffusable negative regulator added to the bacteria may again cause the reduction of gene expression (LIke mutant 4).

iii. The double mutant in which the regulator is unable to bind to reduce gene expression so genes are expressed, but because of the mutation in positive regulater the gene expression cannot be induced, so the gene expression will be

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