Transcription

Question 1

Transcription

Answer 1

Process where RNA is synthesized using the “template strand” of the DNA

Question 2

List the 4 steps in the flow of information in a eukaryotic cell

Answer 2

A gene is first transcribed to pre-mRNA in nucleus, using the complementary DNA strand as template.

mRNA use is to allow the cell to separate information storage and information utilization.

mRNA use is to greatly amplify the synthetic output of information. (1 DNA template can serve as a template for many mRNA molecules)

mRNA are mobile units that are processed after they which they can travel out of the nucleus into the cytoplasm.

Question 3

Structure of a double stranded DNA

Answer 3

2 bars bottom and top 5’ to 3’ and 3’ to 5’

Each bar consists of non template strand, regulatory sequence, promoter, RNA-coding region inclusive of the transcribed sequence and terminator from left to right.

Question 4

Promoter

Answer 4

Signals the beginning of transcription

Question 5

Terminator

Answer 5

Signals the end of transcription

Question 6

Regulatory sequence

Answer 6

Site for the binding of regulatory proteins.

It is before the promotor

Question 7

Regulatory proteins

Answer 7

Influence the rate of transcription

Question 8

RNA polymerase function

Answer 8

RNA polymerase incorporates nucleotides one at a time (5’ to 3’) to form a strand that is complementary to the DNA template

Question 9

The 1st step of mRNA synthesis

Answer 9

Association of the RNA polymerase to the DNA template at the promoter region, thus initiating the transcription process.

Question 10

The 3 steps of transcription in prokaryotes

Answer 10

1. Initiation
2. Elongation
3. Termination

Question 11

Initiation step

Answer 11

Promoter function as a recognition site for sigma factor.
RNA polymerase is bound to sigma factor, causing it to bind to the promoter.

DNA is unwound to form an open complex following binding of mRNA.

Question 12

Elongation

Answer 12

When 10-12 nucleotides have been successfully incorporated, the sigma factor dissociates from the transcription elongation complex.

The core-enzyme catalyzes the addition of ribonucleotides complementary to the DNA template and joins the ribonucleotides using phosphodiester bonds.

RNA polymerase slides along the template strand in the 3’ to 5’ direction, while it synthesizes RNA in the reverse direction, 5’ to 3’.

Question 13

Termination

Answer 13

When RNA reach the terminator, the terminator and RNA transcript dissociate from the DNA>

Question 14

RNA transcript

Answer 14

Note that the RNA is a copy of a coding region between the promotor and terminator

Question 15

E.coli RNA polymerase

Answer 15

Consists of 5 subunits (beta, beta prime, alpha 1, alpha 2 and omega), which form the core enzyme

There is loose association between DNA and core enzyme, hence RNA chains that are begun are not initiated at proper sites.

Core enzyme + sigma factor become complete RNA polymerase.

Question 16

Core enzyme and sigma factor

Answer 16

When the σ-factor is bound to
the core enzyme, the enzyme
has strong affinity for
promoter sites on the DNA
strand.

Sigma factor participates in recognition of promotor region.

Question 17

Promoter binding site

Answer 17

The promoter consists of -35 and -10 regions. and that is where the RNA Polymerase binds to.

Question 18

Closed promoter complex

Answer 18

Sigma unit and the other subunits together with the promoter sequence forms the closed promoter complex.

Sigma factor + core enzyme + promoter sequence

Question 19

Consensus sequences

Answer 19

They are conserved and apparent in the transcriptional

promoter region for prokaryotes

Question 20

Termination: Intrinsic termination

Answer 20

RNA polymerase recognize the terminator that indicates the end of a prokaryotic gene

Regions of RNA molecules can form secondary structures or hairpin loops that cause transcribing RNA polymerase to stall

RNA transcript then separates from the DNA template and RNA polymerase dissociates

Question 21

What do most terminators code for

Answer 21

• > a single stranded region that folds on itself due to hydrogen bonding between complementary base pairs
• > the final region containing several uracils (poly-U)

Question 22

2 classes of prokaryotic termination

Answer 22

Rho-dependent

Rho-independent (intrinsic termination)

Question 23

Rho protein independent termination

Answer 23

Physical modified RNA structure terminates transcription

Poly A tails signal the termination site

RNA molecule getting synthesized have a couple of GC rich regions.

GC rich region fuse with each other and form a hairpin loop.

The poly U is the weak point, allowing the RNA transcript to be removed.

Question 24

Rho protein dependent termination

Answer 24

Helicase protein Rho terminates transcription

Question 25

A rho independent terminator

Answer 25

Contains an inverted repeat followed by a string of approximately 6 A bases The inverted repeats are transcribed into RNA The inverted repeats in RNA fold into a hairpin loop, causing RNA polymerase to pause. Hydrogen bonds in the AU of the poly U tail break RNA transcript separates from template, terminating transcription. Conclusion : Transcription terminates when inverted repeats form a hairpin followed by a string of U bases.

Question 26

Rho dependent termination

Answer 26

- Some terminators require terminator proteins such as the Rho termination factor to be active. - Rho dependent terminators lack a poly-U region in the 3’ end of the terminator. - A hairpin loop is likely to be present after Rho factor binds to a terminator pause site on the RNA molecule which causes the transcribing RNA polymerase to stall. - Rho then interacts with the RNA polymerase stalled at the terminator site and catalyzes the dissociation of RNA polymerase from the DNA strand. It then dissociates as well.

Question 27

-35 and -10

Answer 27

-35 and -10 upstream from the initiation site.

Question 28

Ribo-nucleoside triphosphate

Answer 28

Used to synthesize RNA bases. | Examples are ATP, UTP, GTP, CTP

Question 29

Basic Elements of a Promoter Region in E. coli

Answer 29

-> The new RNA chain is transcribed from DNA template and +1 denotes the start of a gene (initiation site). -> RNA polymerase reads the DNA template in a 3’ to 5’ direction, giving rise to a mRNA strand synthesized from 5’ to 3’.

Question 30

List the 3 types of RNA polymerase

Answer 30

RNA polymerase 1 RNA polymerase 2 RNA polymerase 3

Question 31

RNA polymerase 1

Answer 31

Transcribes genes encoding large precursor rRNA, which is processed into 5.8S, 18S and 28S rRNAs Transcribes genes that code for most of the ribosomal RNA’s Note: Large precursor rRNA and small ribosomal RNA

Question 32

RNA polymerase 2

Answer 32

Transcribes all protein coding genes, it is key in transcribing mRNAs Note: mRNA

Question 33

RNA polymerase 3

Answer 33

Transcribes tRNA, 5 SrRNA and other small rRNAs genes Note: tRNA and small rRNA

Question 34

Eukaryotic gene structure

Answer 34

- 80 CAAT box - 30 TATA box Start codon to stop codon forms hnRNA (heterogenous nuclear RNA) transcript AAUAA poly A signal

Question 35

What are the differences between eukaryotes and prokaryotes transcription

Answer 35

1. Transcription in nuclear membrane in eukaryotes , mRNA must leave nucleus 2. Upstream promoter consists of TATA box at -30 and CAAT box -80 possibly present as well. 3. Transcription factors facilitate binding of RNA polymerase 2 to the promoter 4. Enhancers may be present upstream, downstream or inside the gene to enhance transcription 5. PTM can occur

Question 36

PTM post translational modification

Answer 36

From pre-mRNA transcribed, junk DNA is removed before mRNA is formed, pre-mRNA is called heterogenous nuclear RNA (hnRNA) Undergoes PTM to become mature transcript ready for translation Includes adding 7mG caps, adding poly A tails and removal of introns

Question 37

Why do we need PTM

Answer 37

hnRNA must undergo 3 basic PTM before it is ready to go to cytoplasm 1) Addition of 7mG cap to 5' end 2) Addition of poly A tail to 3' end 3) Splicing

Question 38

Addition of 5' 7mG cap

Answer 38

- > Aids in the export of mRNA from nucleus to cytoplasm - > Recognized by cap-binding proteins that enable mRNA to bind to ribosome - > Protects 5’ end of mRNA from attack by 5’ ribonucleases

Question 39

Addition of 3' poly A tail

Answer 39

- Protects 3’ end of mRNA from degradation by 3’ ribonuclease - Increases the lifespan of the mRNA

Question 40

Splicing - removal of introns

Answer 40

- Provides contiguous coding sequences for translation by ribosome - Plays a role in gene expression via alternate splicing

Question 41

3' Poly A tail 2 characteristics

Answer 41

- added enzymatically to the 3' end | - a string of 100-200 adenine nucleotide added after transcription

Question 42

Spliceosome mediated splicing

Answer 42

- Introns within the pre-mRNA contain conserved regions required for splicing. - The splicing reaction is catalyzed by a spliceosome consisting of small nuclear ribonucleoproteins (snRNPs). - The GU dinucleotide at the 5’ end of intron and the AG dinucleotide at the 3’ end of intron determines the intron boundaries to be spliced

Question 43

How spliceosome mediated splicing work

Answer 43

1. The snRNPs (designated U1, U2, U4, U5 and U6) are made up of small nuclear RNA (snRNA) and proteins. 2. U1 attaches to the 5’ end of the intron by base-pairing while U2 attaches to an A nucleotide at the branch point. 3. Others snRNPs (U4, U5 and U6) will then attach to the pre-mRNA to form the spliceosome. 4. U1 and U4 dissociate from the pre-mRNA while U2 attaches to U6. 5. A loop (or lariat) is formed which gets excised and the exons are ligated together.

Question 44

Video of spliceosome

Answer 44

Snurps are made of small RNA and proteins 1 type of Snurp bind to 5' end of the intron and another type of Snurp bind to the 3' end of the intron. Additional Snurps interact with the complex, bringing the 2 end of the introns together, forming a loop. These gather the snurps and form a spliceosome. Introns is then cut out, exons are joined or spliced together. Spliceosome then dissociates as the Snurps are released.

Question 45

Introns

Answer 45

Intervening sequences

Question 46

Exons

Answer 46

Expressed sequences

Question 47

Lifespan of mRNA in eukaryotes and prokaryotes

Answer 47

Different mRNAs within the same cell have distinct stabilities. * In bacterial cells, individual mRNAs can survive from seconds to more than an hour, shorter lifespan. * In eukaryotes (e.g. mammalian cells), mRNA lifespans range from several minutes to days, longer lifespan. * The longer the lifespan of an mRNA the more protein may be produced from that mRNA.