4. DNA Transcription and RNA processing

Question 1

Intergenic Regions:

Answer 1

Regulate Transcription
Each intergenic region contains a sequence that:
* Signals the end of transcription of the upstream gene
* Signals the start of transcription of the downstream gene

Question 2

Transcription purpose:

Answer 2

mRNA molecule is transcribed from DNA template strand.

Question 3

Where does Transcription occur?

Answer 3

Nucleus

Question 4

Where does mRNA transcript go after being transcribed?

Answer 4

It is transported to the cytoplasm where it will used to generate a protein through translation.

Question 5

Polypeptide Chain

Answer 5

resulting chain of translated amino acids

Question 6

Differences between Transcription and DNA Replication

Answer 6

1. Transcription bubble is much smaller (10-14bp)
2. Initiation of RNA synthesis does not require a primer.
3. No Ligase.
4. Only helicase needed
5. MAJOR DIFFERENCE: DNA unwinds at front of transcription bubble and then rewinds (glues back).

Question 7

What synthesizes RNA in transcription?

Answer 7

RNA Polymerase. 5’-3’ direction.
* Catalyzes phosphodiester bonds

Question 8

DNA Polymerase

Answer 8

1. Synthesize DNA using DNA template.
2. A, T, G, C
3. Polymerizes 5’-3’
4. Requires primer to start polymerization

Question 9

RNA Polymerase

Answer 9

1. Synthesizes RNA using DNA template
2. A, U, G, C
3. Polymerizes 5’-3’
4. Reads template strand 3’-5’
5. Does not require primer to start polymerization

SLIDE 14 DIAGRAM

Question 10

Which RNA polymerase is involved in gene transcription?

Answer 10

RNA Polymerase II

Question 11

Promoter Sequence:

Answer 11

Determines where template strand starts.
* RNA Pol II recognizes promoter at start of transcription.

Question 12

Promoter has two main elements:

Answer 12

1. Consensus Sequences: -35 and -10 (pribnow box)
2. +1 Transcription start site

Question 13

Consensus Sequences:

Answer 13

Attracts RNA Pol II
* -35 to -10 direction: determines orientation of RNA Pol II.

Question 14

Sigma Factor:

Answer 14

Recognizes Pribnow Box and brings RNA Pol II complex to the promoter.
* DNA IS UNWOUND LOCALLY and RNA Pol II starts to create mRNA transcript.

Question 15

What happens to sigma factor when transcription starts?

Answer 15

Sigma factor is released from holoenzyme and RNA Pol II continues chain elongation.

Question 16

Holoenzyme:

Answer 16

Binds the promoter tightly and unwinds the double-stranded DNA, creating an open complex.
* Made up of several polypeptides

Question 17

How does RNA Pol II start transcription?

Answer 17

RNA Transcription is initiated when core RNA Polymerase binds to the promoter with the help of sigma factor.

SLIDE 22 DIAGRAM

Question 18

Terminator Sequence:

Answer 18

Tells RNA Pol II where to stop.

Question 19

Three forms of RNA Polymerase:

Answer 19

1. I: Produces rRNA
2. II: Transcribes all protein-coding genes (makes mRNA)
3. III: Transcribes all tRNA genes and one component of rRNA

Question 20

Transcription in Eukaryotes: Differences with Prokaryotes:
1. The Presence of Nucleosomes. Chromatin must be relaxed.

Answer 20

1. Nucleosome Remodeling Complexes (NRC)
2. Histone Modification Complexes (HMC)
   Locally relax chromatin and make it accessible for transcription

Question 21

Difference 2: Presence of multiple transcription factors

Answer 21

Examples:
2. 1. TFIIA and TFIIB
2. p53
3. c-Myc
4. CREB
5. c-FOS and c-Jun

Question 22

RNA Pol II Recruitment:

Answer 22

1. TFIIA/B - G rich sequence: On -35
2. TBP (TATA Binding Protein): On -25

Question 23

Difference 4: Presence of other regulatory sequences beyond promoter regions

Answer 23

• Enhancers or silencers: Can be found after the gene and even in an intron in the middle of the gene. Usually upstream the promoter.

Question 24

Enhancer Sequences:

Answer 24

Promote transcription initiation. More common than Silencers.

Question 25

Silencer sequences:

Answer 25

Inhibit Transcription

Question 26

Activator Proteins:

Answer 26

Bind to enhancers and speed rate of transcription.

Question 27

Inhibitors / Repressor Proteins:

Answer 27

Bind to silencers and slow transcription.

Question 28

Coactivators:

Answer 28

“Adapter” molecules that integrate signals from activators and perhaps repressors.

Question 29

Basal Transcription Factors:

Answer 29

Responds to injunctions from activators:
* Factors position RNA polymerase at start of transcription and initiate transcription process

Question 30

Difference 5: Processing/maturing of RNA transcript

Answer 30

Pre-RNA molecule created by RNA Pol II processed in three steps:
1. 5’ Capping
2. Poly A tail formation
3. Splicing

Question 31

5’ Capping:

Answer 31

PROTECTS starting 5’ from exonuclease activity.
* 7’ methylguanosine (methyl group on 7’ C) added to 5’ end of pre-mRNA
* 5’-5’ tri-phosphodiester bond
* Pre-mRNA is 5’ capped even before transcription is finished.
* Exonuclease cannot bind to active site to chew up the RNA strand

Question 32

Poly A Tail Formation:

Answer 32

PROTECTS from 3’ exonuclease activity.
1. Series of Cleavage Factors recognize specific sequences around 3’ end.
2. Factors recruit PolyA Polymerase
3. PolyA Polymerase: CLEAVES later portion of 3’end. Starts adding multiple A to create a Poly A Tail.
4. PAB (polyA binding protein): recognize and bind to created polyA tail, further stabilizing 3’ end.

Question 33

What is a good place to put PolyA Signal?

Answer 33

Exon.
* If put on intron, gene strand will be cut.

Question 34

Splicing:

Answer 34

Removes Introns and keeps Exons.

Question 35

Spliceosome:

Answer 35

Group of proteins that perform splicing.

Question 36

Intron sequences:

Answer 36

• Start: GU sequence
• Finish: AG sequence
• Branch point

Question 37

Intermediate Lariat (loop)

Answer 37

Initial loop that forms after intron branch point and initial GU sequence is recognized by splicosome machinery.

Question 38

Formation of a Larait (detailed):

Answer 38

1. Intron is first cut at 5’ splice site (beginning of intron).
2. Lariat structure forms: 5’ end of intron is covalently linked to BRANCH POINT A residue through 2’ Oxygen of branch-point to 5’ end of intron. Phosphodiester bond forms between the 2’-5’.

Question 39

Resolution of a Lariat:

Answer 39

1. Spliceosome recognizes the AG terminal sequence.
2. 3’ end of intron is then cut and 5’ and 3’ ends of exons are joined together through transesterification reaction.

SLIDE 51 DIAGRAM

Question 40

Transesterification Reaction

Answer 40

OH group at 3’ end of exon ATTACKS phosphodiester bond at 3’ splice site. Results in covalent bonding of exons and release of lariat strucuture.

SLIDE 51 DIAGRAM

Question 41

What is the evolutionary advantage to having exons and introns?

Answer 41

It allows alternative splicing
* One gene can give rise to multiple types of proteins.
* Form new forms of proteins without having to increase the number of genes

Question 42

Alternative Splicing

Answer 42

The ability to create different types of mRNA transcripts from a single gene by changing how the pre-mRNA is spliced.
* Different combination of exons