Most of what we know about the lac operon of E. coli has come from genetic analysis of various mutants. In the following list are the genotypes of seven strains of E. coli. Part A For each strain, indicate whether the lacZ gene product will be expressed ( ) or not (-) in the presence of lactose and in the absence of lactose. Explain your reasoning in each case. Unless noted otherwise, assume that glucose levels are low. Drag the appropriate labels to their respective targets.

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hihirasohail hihirasohail · Answer 1 · 2020-05-06T03:50:20+02:00

Answer:

Strain Presence of Lactose Absence of Lactose

[tex]I^{+} P^{+}O^{+} Z^{+}[/tex] (+) (-)

For the wild type case, the lacZ gene product will be expressed in the presence of lactose only.

[tex]I^{S} P^{+}O^{+} Z^{+}[/tex] (-) (-)

Because of the super repressor, transcription will not take place regardless of the presence of lactose.

[tex]I^{+} P^{+}O^{C} Z^{+}[/tex] (+) (+)

The constitutive operator will allow for transcription to occur, whether or not lactose is present.

[tex]I^{-} P^{+}O^{+} Z^{+}[/tex] (+) (+)

The repressor is non-functional, transcription will occur with or without presence of lactose.

[tex]I^{S} P^{+}O^{C} Z^{+}[/tex] (+) (+)

The super repressor will not be able to stop transcription as it cannot bind to the constitutive operator.

[tex]I^{+} P^{-}O^{+} Z^{+}[/tex] (-) (-)

The promoter is non-functional, transcription will not start and hence, not gene product formed.

[tex]I^{+} P^{+}O^{+} Z^{+}[/tex] (-) (with glucose present) (-)

E. coli will not metabolize lactose in the present of glucose, since glucose is the preferred source of nutrition.

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