Prokaryotic Transcription

Question 1

What is the basic meaning of DNA transcription, and what are the main steps?

Answer 1

Transfer of genetic information from DNA to RNA.

Initiation - Elongation - Termination

Question 2

How many RNA polymerases do PROKARYOTES have?

Answer 2

ONLY ONE

Question 3

Discuss the features of prokaryotic RNA polymerases.

Answer 3

RNA polymerase synthesises RNA in a 5’ - 3’ direction

It does NOT need a primer

The RNA product does NOT remain base-paired to the template DNA strand

Question 4

CODING STRAND ===

TEMPLATE STRAND ===

Answer 4

SENSE STRAND

ANTISENSE STRAND

Question 5

How does RNA polymerase know where to start its function along the nucleotide sequence?

Answer 5

There are INITIATION SEQUENCES

e.g. start codons = AUG

Question 6

What is the unit of transcription in prokaryotes?

Answer 6

Polycistronic transcripts

Question 7

Outline the structure and function of the subunits of E. coli RNA polymerase holoenzyme

Answer 7

CORE POLYMERASE:
alpha(2)-beta-beta’-omega

beta’: core polymerase subunit (DNA binding)

beta: core polymerase subunit (polymerase active site)
alpha: scaffolding role
omega: unknown role

SIGMA FACTOR:
sigma: types, role in transcription initiation (promoter recognition), main type is sigma(70)

Question 8

Briefly outline the process of INITIATION in prokaryotic transcription.

Answer 8

The bacterial RNA polymerase positions itself on the promoter through the sigma factor

Question 9

Describe the structure of a bacterial promoter.

Answer 9

Sigma(70) factor;

-35 and -10 regions are instrumental in positioning the polymerase in the time leading up to initiation;

transcription start site is at +1

Question 10

What is meant by ‘consensus’?

Answer 10

Most commonly occurring sequence out of all the successful sequences recorded for that particular gene?

Question 11

The closer a promoter sequence is to the consensus…

Answer 11

… the stronger the promoter will be, i.e. more effective transcription; this depends on how well the polymerase is bound.

Question 12

Sigma and alpha subunits recruit RNA polymerase to the promotoer; elaborate.

Answer 12

Sigma(2) subunit binds to -10 region –> region that melts within the active site

Sigma(4) binds to -35 region –> recognition of promoter site

UP-element: further encourages good binding of polymerase in strong promotoers, e.g. rRNA genes

Question 13

Briefly outline the initiation stage of prokaryotic transcription.

Answer 13

INITIATION
Involves recognition and binding of RNA polymerase to the promoter (closed complex), promoter melting (open complex) and formation of a stable ternary complex (transcription bubble).

Question 14

Briefly outline the elongation stage of prokaryotic transcription.

Answer 14

ELONGATION

Polymerase advances 3’-5’ down template strand, adding rNTPs to growing RNA chain

Question 15

Briefly outline the termination stage of prokaryotic transcription.

Answer 15

TERMINATION
Transcription is terminated by signals within the RNA sequence; polymerase releases completed RNA and dissociates from DNA.

Question 16

Define ‘terminators’

Answer 16

Sequences which trigger the elongating polymerase to dissociate from the DNA and release the nascent RNA.

Question 17

State the two types of termination.

Answer 17

1. Rho-dependent

2. Rho-independent

Question 18

What are the features of the ‘Rho factor’ involved in Rho-dependent termination?

Answer 18

RHO FACTOR:

• six-ringed protein;
• binds RNA at termination recognition sites called ‘rut sites’;
• binds to single-stranded RNA;
• doesn’t bind RNA being translated;
• rho pulls the RNA out of the polymerase, or induces polymerase conformational change such that the polymerase terminates

Question 19

How does Rho induce termination of transcription?

Answer 19

Rho pursues polymerase by moving along the growing, newly transcribed RNA strand.

A ‘hairpin’ is formed, causing polymerase to pause and allowing Rho to catch up.

The combination of protein action and destabilizing effect of hairpin loop allows Rho to induce termination

Question 20

What is meant by Rho-independent termination?

Answer 20

Rho-independent termination requires intrinsic terminators which DO NOT involve other factors.

For example, such a terminator could include a short inverted repeat in a stretch of 8 A-T base pairs

Question 21

What is the benefit of the lactose operon (lac operon)?

Answer 21

The lac operon allows E. coli to use lactose rather than glucose, its typical food source. It does this by hydrolyzing (breaking down) lactose into glucose and galactose.

Question 22

State the name and function of each of the three genes on the lac operon.

Answer 22

lacZ, lacY, lacA

lacZ - produces beta-galactosidase, which hydrolyzes lactose into glucose and galacotse

lacY - produces beta-galactoside permease, which enables lactose to transport into the cell through the membrane

lacA - encodes beta-galactoside transacetylase (biological role uncertain)

Question 23

Summarise transcription by RNA polymerase

Answer 23

INITIATION
involves recognition and binding of RNA polymerase to the promoter (closed complex), promoter melting (open complex) and formation of stable ternary complex (transcription bubble)

ELONGATION
polymerase advances 3’-5’ down template strand, adding rNTPs to growing RNA chain

TERMINATION
transcription is terminated by signals within the RNA sequence, polymerase releases completed RNA and dissociates from DNA

Question 24

What is the lactose (lac) operon?

Answer 24

A cluster of inducible genes including:
  ==> lacZ, lacY, lacA (all transcribed as a single mRNA)
  ==> a promoter 
  ==> a regulator
  ==> an operator

Question 25

In what state is the lac operon normally in?

Answer 25

Normally, lac operon is turned off and a repressor protein binds the operator region upstream of the operon preventing RNA polymerase from reading the three lac genes.

Question 26

Discuss the regulation of the lac operon.

Answer 26

When lactose is in the cell, lactose binds to the repressor. This causes a structural change in the repressor, causing the repressor to lose its affinity for the operator.

Thus, RNA polymerase can then bind to the promoter and transcribe the structural genes, meaning lactose acts as an effector molecule.

Question 27

What are the main factors that regulate transcription in prokaryotes?

Answer 27

Prokaryotic transcription is mainly regulated through initiation of transcription.

Sigma factor acts as an initiation factor, and most bacterial genes are regulated by the sigma(70) factor; however, other sigma factors are available.

Question 28

What is the general mechanism by which prokaryotic transcription is regulated?

Answer 28

The general mechanism involves a repressor that binds to the operator region of a gene or operon and blocks transcription.

The repressor is regulated through the binding of a small molecule ligand(s) which control its DNA binding activity

Some bacterial genes also respond to activators (positive regulation, e.g. CAP in lac operon).

Activators also regulated by binding of small molecule ligands (e.g. cAMP)

Question 29

How many RNA polymerases are found in prokaryotes and eukaryotes?

Answer 29

PROKARYOTES: ONLY ONE
e.g. T. aquaticus RNA polymerase:
beta, beta’, alpha, a, psi

EUKARYOTES: FIVE
e.g. Yeast RNA polymerase II:
RPB1, RPB2, RPB3, RPB11, RPB6

Question 30

For each of the eukaryotic RNA polymerases, state the name, the gene transcribed, and the sensitivity to alpha-amanitin.

Answer 30

RNA Polymerase I
Ribosomal RNA (45S precursor of 28S, 18s and 5.8S rRNA)
Insensitive

RNA Polymerase II
All protein-coding genes, small nuclear RNAs, U1, U2, U3, etc.
Very sensitive

RNA polymerase III
tRNA, 5S rRNA, snRNA U6, some repeated sequences (Alu)
Moderately sensitive

Mitochondrial RNA Polymerase
All mitochondrial genes

Chloroplast RNA Polymerase
All chloroplast genes

Question 31

How is pre-rRNA processed?

Answer 31

The primary transcript is cut to yield the mature rRNAs found in ribosomes.

These are modified by methylation of the specific ribose 2’-OH.

Specific uridine (ribonucleoside of uracil) residues are then converted into pseudouridine, an isomeric form.