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Genetics - Heather > Lecture 2, the lac operon > Flashcards

Lecture 2, the lac operon Flashcards

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Biology 101 flashcards
1
Q

Catabolic operons

A

Catabolism is breaking down molecules to make something that is needed. Catabolic operons control the expression of enzymes used in sugar and metabolism.

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2
Q

E.coli - what happens if there is not enough sugar?

A

Glucose is the primary carbon source, if there is not enough of this then E.coli will switch on the expression of enzymes that will break down other components and release sugars they can use.

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3
Q

What type of operon is the lac operon?

A

A catabolic operon

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4
Q

Describe lactose

A

Disaccharide sugar made up of glucose and galactose. E.coli can use the monosaccharides

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5
Q

What enzyme breaks down lactose

A

B-galactosidase can break down lactose into glucose and galactose by breaking the B-galactose-1,4-glucose link

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6
Q

What are the 3 steps for lactose to be used in E.coli cells?

A

Detection - detect that lactose is available outside the cell
Import lactose into the cell
Cleavage- break lactose into monosaccharides

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7
Q

What is lactose transported through?

A

The lactose permease channel

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8
Q

B galactose enzyme

A

Breaks down lactose, can introduce errors to create a lactose analogue that can be toxic. Transacetylase will de-toxify the lactose arrangement by adding acetyl groups.

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9
Q

What induces the lac-operon

A

The presence of lactose

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10
Q

What are the 3 genes in the operon and who found them?

A

LacZ
LacY
LacA
Jacob and Monod

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11
Q

What is co-ordinate regulation

A

All the genes in the operon are regulated in the same way. The three genes LacZ, LacY and LacA constitute the operon. Structural genes transcribed from a single promoter into polycistronic mRNA

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12
Q

Structural genes

A

Genes that code for enzymes

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13
Q

Regulatory genes

A

Control the operon

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14
Q

LacI

A

Produces the regulatory protein and has its own promoter and terminator.

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15
Q

Jacob & Monod LacI

A

Showed that LacI was a regulatory gene because if they disrupted its function by mutating it all of the other three genes were expressed all the time. When its not disrupted there is control of Lac Z, Lac Y and Lac A.

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16
Q

Function of LacI

A

Encodes a repressor (tetramer).
Regulatory gene that is not part of the promoter but produces a protein that regulates the expression of Lac Z, Lac Y and Lac A. It is expressed all the time and has its own promoter/terminator region.

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17
Q

Function of Lac Z

A

Encodes enzyme B- galactosidase (tetramer) which cleaves lactose into galactose and glucose. When all the glucose has been used up the galactose operon begins. B-galactosidase is a large enzyme and is very stable so will not easily break down.

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18
Q

Function of Lac Y

A

Lac Y encodes a permease (membrane bound) which transports lactose into bacteria.

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19
Q

Function of Lac A

A

Encodes an enzyme transacetylase (dimer)

Adds acetyl groups (CH3COO-) to other molecules and toxic sugars make them harmless.

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20
Q

How can Lac Z be identified?

A

By the addition of X-gal which forms a blue product in colonies that express LacZ. If LacZ is present then LacY and LacA is present.

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21
Q

Describe what happens to a cell when you add lactose to its growth media

A

Appearance of lac mRNA. Increasing concentration of B-galactosidase.

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22
Q

What did Jacob and Monod predict would happen if LacI or LacO were mutated?

A

The expression of LacZ would be affected.

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23
Q

Where is LacO positioned?

A

Next to LacZ and is the operator sequence

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24
Q

How does the repressor function in the absence of lactose?

A

LacI gene is transcribed and produces mRNA that gets translated into a protein. 4 couples of this protein aggregate to form a tetramer. The tetramer is the repressor which binds to the operator region in the lac operon’s promoter and stops transcription happening.

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25
Q

How does the repressor interfere with the binding of RNA polymerase?

A

Steric hindrance
Expression of operon prevented.
The RNA polymerase covers the -10 and -35 boxes and starts transcription at the promoter. The lac repressor is at the operator region. The RNA polymerase and repressor overlap so there is steric hindrance.

26
Q

How does lactose induce expression of the operon?

A

Inactivates the lac repressor so it cannot bind. RNA polymerase will then bind at the promoter and transcribes the operon.

27
Q

How does lactose initially get into the cell?

A

There is always some permease in the plasma membrane and some B-galactosidase so the permease imports a few molecules of lactose and this is enough to start the process.

28
Q

What are the first two steps that follow after a few molecules of lactose enter the cell?

A

The inducer binds to the repressor and causes it to undergo a conformational change so it can no longer recognise and bind to the base sequence of the operator.
RNA polymerase can now bind and express the operon

29
Q

Beta-galactosidase products & side products

A

Beta galactosidase can produce some side products and rearrange lactose rather than cleave it. Allolactose is one of the side products which is beta-galactose-1,6-glucose. It is produced by the isomerisation of lactose. As allolactose binds to the tetramer it acts as an inducer molecule so there is a conformational change in the tetramer inactivating it.

30
Q

How does RNA polymerase have access to the promoter region when lactose is present?

A

The repressor is deactivated by the binding of allolactose

31
Q

What mutations did Jacob and Monod isolate that meant the cells did not respond normally to lactose?

A

lacl-

lacOc

32
Q

LacI- mutation

A

The operon is never repressed.
LacZ is expressed all the time. This showed that LacI is involved in repression.
Recessive mutant as regulation can be restored by the addition of normal wild type genes.
As LacI is a protein coding gene it does not need to be attached to the chromosome that it acts on. LacI works in trans as it encodes a diffusible protein.

33
Q

LacOc mutation

A

A change in the DNA operator region so it is not recognised by a functional wild type repressor. Operon is continually switched on, Laz Z always expressed.
Dominant mutation

34
Q

Strains made by Jacob and Monod

A

Strain 1: mutant operator is linked to the normal wild type lacZ gene on the bacterial chromosome. The F’ plasmids adds a normal wild type lacO operator linked to a mutant lacZ gene.
Strain 2: entirely wild type chromosome and added to this strain is the F’ plasmid containing a mutant operator region and a mutant LacZ gene.

35
Q

What type of mutations are cis-dominant?

A

Cis-dominance is indicative of a site at which regulatory proteins act. LacO does not produce something that can diffuse across the cell and affect genes elsewhere.
It is not a gene that produces a protein that can move to where it is needed.

36
Q

Lacl^(s)

A

Super repressed mutations that are un-inducible. Expression of the operon genes is always repressed.
Repressor produced with no binding sites for allolactose so the repressor is always active and binding to the operator region.

37
Q

Describe strain 1 of LacOc

A

LacOc lacZ+ / plasmid: F’lacO+ lacZ-

Operon always expressed

38
Q

Describe strain 2 of LacOc

A

LacO+ lac Z+ / plasmid: F’ lacOc lac Z-

Repression of operon can occur. The 2 mutations on the plasmid cannot destroy normal functioning

39
Q

Strain 1 of LacI

A

LacI - Lac Z+
Plasmid: Lacl+ LacZ-
Showed normal induction and the operon worked

40
Q

Strain 2 of LacI

A

LacI+ LaxZ+
Plasmid: LacI- LacZ-
Strain shows normal induction

41
Q

Conclusions made from the strains of LacI

A

It does not matter if the functional LacI gene is on the chromosome/connected to the functional LacZ gene.
Shows that LacI works in trans as it can move to where it is needed so LacI makes a diffusible product (protein)

42
Q

Dominant mutation

A

Adding a normal wild type of the gene will not correct the mutation

43
Q

Recessive mutation

A

Adding a normal wild type of the gene corrects the mutation

44
Q

Conclusions made from the LacOc strains

A

LacOc does not encode a diffusible protein as it is a regulatory piece of DNA bound to LacI.
LacOc acts in Cis so needs to be connected to the genes it is influencing.

45
Q

What type of mutation is LacI^(S)

A

Dominant mutation- cannot be corrected by adding a wild type version of the gene

46
Q

Explain what a Diauxic growth curve shows

A

When glucose is available growth progresses exponentially.

As glucose levels decline the lac opeon is induced and B-galactosidase activity.

47
Q

Describe Adenyl cyclase

A

An enzyme that catalyses the conversion of ATP into cyclic AMP.
Growth on glucose inhibits adenyl cyclase.

48
Q

What reaction does Adenyl cyclase catalyse?

A

ATP -> Cyclic-5’,3’ AMP

49
Q

What is seen in E.coli CYA mutants?

A

They are unable to make cAMP so cannot induce the operon, even when glucose is absent.

50
Q

What transports glucose in the cell?

A

Transport of glucose is carried out by a complex of membrane proteins called phosphoenolpyruvate (PEP) : glucose phosphotransferase system. This also phosphorylates sugars. Phosphoenolpyruvate is the source of the phosphate group.

51
Q

What causes inhibition of adenyl cyclase?

A

IIA^Glc: a protein component that transports sugars into the bacterium, specific to glucose.

52
Q

What proteins are involved in the phohsphoenolpyruvate: glucose phosphotransferase system?

A

IIC, IIB, IIA, HPr

53
Q

Function of HPr, IIA, IIB

A

Transfer phosphate groups from one to another in a chain.

54
Q

What happens to IIA^Glc when glucose is present?

A

IIA^Glc becomes dephosphorylated. Adenyl cyclase activity is inhibited so no cAMP is produced and transcription of the operon does not occur.

55
Q

What stimulates adenyl cyclase to produce cAMP?

A

Phosphorylated IIA in the absence of glucose

56
Q

Does cAMP interfere with the operon?

A

No, cAMP binds to a CAP protein which will then bind to the lac DNA promoter so transcription can occur.

57
Q

What does CAP require to bind to the DNA?

A

cAMP to be bound to it

58
Q

Where does the CAP protein bind?

A

CAP is in contact with the -10 and -35 binding sites which facilitates binding of RNA polymerase and increases transcription.
Bends DNA to force it apart

59
Q

The lac Promoter

A

Slightly unusual as there is are two base differences in the -10 (6) base sequence.
A swapped to T
And
A swapped to G
No longer 6 weakly paired bases as G makes the interactions stronger so the sigma factor can not open this section.

60
Q

Where does the transcription bubble open up?

A

Pribnow box

Genetics - Heather (8 decks)

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