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If lactose isn't present in the environment of an E. coli, what is the state of the lac operon? O a. It is "on" (high transcription) because will glucose will still be present, so the operon's products will be needed. O b. It is "on" (high transcription) because CAP will be bound to the promoter O c. It is "off" (no transcription) because RNA polymerase will be inactivated O d. It is "off" (no transcription) and the repressor will be bound to the operator

If lactose isn't present in the environment of an E. coli, what is the state of the lac operon? O a. It is "on" (high transcription) because will glucose will still be present, so the operon's products will be needed. O b. It is "on" (high transcription) because CAP will be bound to the promoter O c. It is "off" (no transcription) because RNA polymerase will be inactivated O d. It is "off" (no transcription) and the repressor will be bound to the operator

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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If lactose isn't present in the environment of an E. coli, what is the state of the lac operon? O a. It is "on" (high transcription) because will glucose will still be present, so the operon's products will be needed. O b. It is "on" (high transcription) because CAP will be bound to the promoter O c. It is "off" (no transcription) because RNA polymerase will be inactivated d. It is "off" (no transcription) and the repressor will be bound to the operator
Transcribed Image Text:If lactose isn't present in the environment of an E. coli, what is the state of the lac operon? O a. It is "on" (high transcription) because will glucose will still be present, so the operon's products will be needed. O b. It is "on" (high transcription) because CAP will be bound to the promoter O c. It is "off" (no transcription) because RNA polymerase will be inactivated d. It is "off" (no transcription) and the repressor will be bound to the operator
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