Goals of today’s lecture

1) Describe the basics of prokaryotic gene regulation -operons, negative and positive regulation

2) Illustrate the use of genetics in understandingcellular processes

3) Cover some aspects of DNA-binding proteins

Gene Expression• Regulating the amount of active gene

product (protein) by:– Control of gene expression at the level of

transcription– Control of at the level of translation– Post-translational control

• Constitutive expression = gene product made continuously

• Regulated expression = gene product made on demand; expression can be induced or repressed

• Replication = produce exact copy of DNA for mitosis (cell division) or reproduction (pass to the next generation)

• Transcription = transcribe DNA code into RNA (uses same ‘language’ of nucleic acids)

• Translation = translate nucleic acid code into a sequence of amino acids (the primary structure of polypeptides)

• Post-translational modification = chemical modification to activate a protein so it can function in the cell

DNA RNA Protein

Replication

Transcription Translation

Protein*

Post-translation

Gene Expression & Information Flow

Regulation of Gene Expression in Prokaryotes

Catabolic Metabolism

• E. coli use many sugars for metabolism

• Glucose is the preferred source of carbon for E. coli –Why?

• If glucose is not available, bacteria can break down lactose to generate glucose

Figure 17-2b-setup

Figure 17-2c-results

• Jacques Monod found that -galactosidase is not expressed in E. coli cells grown in medium containing glucose or glucose + lactose but only in medium containing lactose and no glucose.

• E. coli produces high levels of -galactosidase, the enzyme that cleaves lactose to glucose + galactose, only when lactose is present in the environment. Thus, lactose (actually it is a metabolite of lactose) acts as an inducer—a molecule that stimulates the expression of a specific gene.

Genetic screeningunmutagenized cells

mutagenized cells

Mutant that can’t utilize lactose as acarbon source

105 cells

mutagen

each cell hasa different mutation

How to find the needle in the genetic haystack?Which mutations effect lactose utilization?

Replica plating allows the identification of

genes that are essential to utilize lactose

~5000 colonies/plate

• Three classes of E. coli mutants defective in lactose metabolism were isolated (Table 17.1): lacZ, lacY, and lacI. lacI is a constitutive mutant—one that has lost the ability to regulate expression of a particular gene because it produces a product at all times, not just when an inducer is present.

E. coli

-GalactosidaseGalactosidepermease

LactosePlasma membrane

Glucose

Galactose

Two proteins are critical for E. coli to use lactoseand one is critical for regulation of their expression

Section of E. colichromosome

lacl product

lacl

-Galactosidase

lacZ product lacY product

lacZ lacY

Galactosidepermease

Far away on the chromosome

Model for Operons in Prokaryotes

• Portion of DNA including a set of genes involved in a specific metabolic pathway

• Single regulatory region (operator + promoter)

• Generates single polycistronic RNA

• RepressorRepressor binds the operator and blocks RNA polymerase

• RepressorRepressor is the product of a regulatory gene

lacZ lacY5’ 3’

AUG AUGUGA UAA

Figure 17-6a

Negative control: Regulatory protein shuts downtranscription

No negativecontrol…

With negativecontrol…

RNA polymerase

Regulatory protein

TRANSCRIPTION

Gene sequence

No transcription

Figure 17-6b

Positive control: Regulatory protein triggers transcription

No transcription

RNA polymerase

No positivecontrol…

Regulatory protein

With positivecontrol…

TRANSCRIPTION

Gene sequence

• Jacob and Monod proposed that the lacI gene produces a repressor (the LacI+ protein) that exerts negative control over the lacZ and lacY genes. The repressor was thought to bind directly to DNA near or on the promoter for the lacZ and lacY genes (Figure 17.7).

Repressor present, lactose absent:

Repressor present, lactose present:

No repressor present, lactose presentor absent:Transcription occurs.

Repressorsynthesized

DNA

lacl+

RNA polymerasebound to promoter

(blue DNA)

lacZ lacY

TRANSCRIPTION BEGINS

-Galactosidase Permease

mRNA

lacZ lacY

RNA polymerasebound to promoter

(blue DNA)

Lactose-repressorcomplex

Repressorsynthesized

No functionalrepressor synthesized

mRNATRANSCRIPTION BEGINS

-Galactosidase Permease

lacZ lacY

RNA polymerasebound to promoter

(blue DNA)

Lacl –

Repressor binds to DNA.No transcription occurs.

Lactose binds to repressor,causing it to release fromDNA. Transcription occurs(lactose acts as inducer).

Normallacl gene

Normallacl gene

lacl+

Mutantlacl gene

The repressor blocks transcription

When tryptophan is present, transcription is blocked.

Repressor Tryptophan

No transcription

Operator

RNA polymerasebound to promoter

When tryptophan is absent, transcription occurs.

RNA polymerasebound to promoter

TRANSCRIPTION

5 genes coding for enzymes involvedin tryptophan synthesis

The Trp operon is also under negative control, but with a twist

Figure 17-10

lac operon trp operon

Catabolism Anabolism(breakdown of lactose) (synthesis of tryptophan)

Repressor RepressorLactose Tryptophan

Lactose bindsto repressor

Tryptophanbinds torepressor

Lactose-repressorcomplexreleases fromoperator

Operator

Operator

Tryptophan-repressorcomplex bindsto operator

No moretranscriptionof trp operon

Transcriptionof lac operon

TRANSCRIPTION

Catabolite Repression

Why doesn’t beta-galactosidase get induced in media containing Glucose?

When cAMP is present:

When cAMP is absent:

RNA polymerase boundtightly to promoter (blue DNA)

RNA polymerase boundloosely to promoter (blue DNA)

FREQUENT TRANSCRIPTION

INFREQUENT TRANSCRIPTION

cAMP CAP

CAP site

CAP

CAP site

Operator

Operator

lacZ lacY

lacZ lacY

lacA

lacA

cAMP binds to CAP and thecAMP-CAP complex bindsto DNA at the CAP site.RNA polymerase bindsthe promoter efficiently.Transcription occurs frequently.

CAP does not bind to DNA.RNA polymerase bindsthe promoter inefficiently.Transcription occurs rarely.

CAP regulates lac operon positively and requires cAMP for DNA binding

Glucose inhibits the activity of the enzyme adenylyl cyclase, which catalyzes production of cAMP from ATP.

The amount of cAMP and the rate of transcription of the lac operon are inversely related to the concentrationof glucose.

ATP Adenylyl cyclase

Glucose inhibitsthis enzyme

cAMPTwo phosphate

groups

Infrequent transcriptionof lac operon

(Cell continues to useglucose as energy source.)

CAP does notbind to DNA

CAP

LOWcAMP

INACTIVEadenylyl cyclase

HIGHglucose

concentration

LOWglucose

concentration

ACTIVEadenylyl cyclase

HIGHcAMP

cAMP

CAP

CAP-cAMP complex binds to DNA

Frequent transcriptionof lac operon

(Cell uses lactoseif lactose is present.)

Cyclic AMP (cAMP) is synthesized when glucose levels are low

Dual Regulation of lac operon

• Negative control by lac repressor >> needs the inducer (lactose) to inactivate the lac repressor

• Positive control by CAP (activated by high [cAMP] resulting from low [glucose]) >> determines rate of transcription if the operator is NOT blocked by the repressor

Figure 17-15lac operon

Promoter Repressor

INFREQUENT TRANSCRIPTION

INFREQUENT TRANSCRIPTION

CAPsite

CAPsite

CAPsite

FREQUENT TRANSCRIPTION

Operator

Operator

Operator

RNA polymerase boundloosely to promoter

RNA polymerase boundloosely to promoter

RNA polymerase boundtightly to promoter

Glucose HIGH

Glucose HIGH

Glucose LOW

Lactose LOW

Lactose HIGH

Lactose HIGH

lacZ

lacZ

lacZ

lacY

lacY

lacY

lacA

lacA

lacA

Inducer-repressor complex

Figure 17-11-1

DNA FOOTPRINTING

Radioactiveatom

DNA Repressorprotein

No repressor

1. Generate fragments from the DNA region of interest, such as the lac operon of E. coli. Attach a label to end of fragments.

2. Divide fragmentsinto two samples.Add repressor proteinto one sample. Therepressor will bind to the operator.

3. Cut fragments with nucleaseto produce fragments of differentlengths. Repressor protectsoperator DNA from nucleasecleavage.

Figure 17-11-2

DNA FOOTPRINTING

“Footprint”No cuts occurred in the DNAregion protected by the repressor.This region must be the operator.

Largest fragments(cut far from label)

Smallest fragments(cut close to label)

4. Load fragmentsinto two lanes in agel. Sort by size viaelectrophoresis. (Thefragments with alabel will be visible.)

A DNA sequencingreaction can beused to determinethe sequence of the“footprint.”

You isolate an E. coli mutant that has high expression of thelacZ gene even in the absence of lactose (media has no glucose). You think you have a mutation in the lacI repressor, but this turns out not to be the case. Which of the following do you think best explains your mutant?

A) Mutation in the lacZ gene that increases its activity.B) Mutation in the O1 binding site that reduces lacI binding.C) Mutation in the O1 binding site that increases lacI binding.D) Mutation inactivating the lacY gene.

Summary of Prokaryotic Gene Regulation

• Prokaryotic genes that code for enzymes in a specific metabolic pathway are clustered in groups and regulated together = operon

• Lac operon discovered by Jacob & Monod • Key advantage: single ‘on-off’ switch to coordinate

gene expression • Switch = operator that controls access of RNA

polymerase to the promoter• Repressor protein binds to the operator• Repressor proteins are subject to regulation (positive or

negative) by the metabolic substrates and products of the pathway.