The Leu operon (prokaryotic gene regulation) There are two proteins that are inv
ID: 259912 • Letter: T
Question
The Leu operon (prokaryotic gene regulation) There are two proteins that are involved in the synthesis of the amino acid leucine. Leu1 is required for transport of the precursor oxopentanoate into the bacterial cell. Leu2 converts oxopentanoate into leucine. Their genes are both under the control of the same promoter. The operon is negatively regulated by a protein encoded by Leu3, under the control of a constitutive promoter. Transcription from the Leu operon is induced by the presence of oxopentanoate. Basal levels of transcription from this operon are very high. Repressed levels are undetectable
A. B. C. D. Is the gene product of Leu3 an ACTIVATOR or a REPRESSOR protein? In the absence of the inducer, will there be any detectable transcription of Leu1? In the presence of leucine, will there be any detectable transcription of Leu2? In the absence of the inducer, will there be any detectable transcription of Leu3? You create five mutations (A-E) that are leucine auxotrophs.Explanation / Answer
Ans.
A) gene product of Leu3 is an repressor protein as operon is negatively regulate by the the protein encoded by Leu3.
B) in the absence of inducer , repressor(leu3 protein) will repress the transcription of the operon hence there will not be any detectable transcription of leu1.
C) transcription of leu2 is independent of presence or absence of leucine and depends on the presence of inducer i..e., oxopentanoate.
D) as given in the question, leu3 is under the control of a constitutive promoter and it is not part of the operon hence in eithr the absence or the presence of the inducer , there will always be a detectable transcription of leu3.
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