Section 4 - Regulation of Gene Expression

Question 1

Describe the components needed for transcription in bacterial cells

Answer 1

Basically everything that’s in a bacterial Operon:
- Start Site: called +1. In order for RNA Pol to start, it must recognize the start of the gene.

• Promoter: the DNA sequence (Around 100 bp) required to initiate transcription of a gene of operon. The sigma subunit of RNA Pol recognizes promoter, and holoenzyme latches firmly onto this specific sequence of nucleotides. Lies immediately upstream (-1) of start site.
• has -10 and -35 sequences
• the polarity of the promoter orients the polymerase and determines which strand is transcribed (the one that is 3’to5’ so that it can transcribe 5’-3’)
• not transcribed into RNA
• Repressor/Terminator: the DNA sequence required to stop transcription. Transcribed into RNA.
• RNA Polymerase: The multi-subunit enzyme that makes specific RNA transcripts using DNA as a template and nucleoside triphosphates (NTPs) as substrates.
• Moves stepwise 3’-5’ along the DNA, unwinding it as it goes. Adds ribonucleotides one by one to the RNA chain. Catalyzes the formation of phosphodiester bonds that link the nucleotides + the backbone. The two strands of DNA come right back together. No primer needed.

Question 2

Describe mechanisms for transcriptional activation and repression in bacteria. (Trp Operon)

Answer 2

- increased transcription by enhancing activity of RNA Pol or enhancing recruitment to the site 
Trp operon (expression of Trp) negatively regulated by Trp repressor
   - Little/no Trp = transcription (of the 5 genes responsible for Trp biosynthesis, aka. Trp operon) can occur.  
   -When aa tryptophan is high in concentration, 2 Tryptophans bind trp repressor--> undergoes a conformational change now can bind to operator(btwn-10 and-35) impeding recruitment of RNA Pol --> no transcription.  Trp repressor binds DNA as a dimer --- helix 5 on each of the two monomers of repressor makes BASE SPECIFIC contacts in consecutive major grooves of the DNA.

Question 3

Explain Trp Operon and how it is regulated

Answer 3

When is it expressed?

What does the operon encode for?

Question 4

Explain Lac Operon and how it is regulated

Answer 4

When is it expressed?
What does the operon encode for?
What does Catabolite Repression have to do with it?
-you shut off the genes required to metabolize other carbon sources in the presence of glucose. so lac operon is subject to this.
(Glucose, CAP, cAMP)
What is the role of CAP?

Question 5

Two ways to regulate transcription

Answer 5

1. Stimulate/Repress recruitment of RNA Pol
2. Stimulate/Repress activity of RNA Pol
   then activation is through protein-protein interactions (CAP to RNA Pol)

Question 6

Identify the structural characteristics of some bacterial DNA-binding proteins
- Trp repressor

Answer 6

• 107 residues. a monomer
• 6 helices + some unstructured loops. Alpha helices 4 and 5 make up the helix-turn-helix motif.
• acts as a dimer – 2-fold symmetry. 2 monomers held together by protein-protein interactions. so actually needs TWO tryptophans to bind
• at either end (the heads), its helices are pointed inward (blocking interaction with operator. When 2 Trps binds to it, –> conformational change: the two recognition helix 5s on each of the two monomers tilt back out to make BASE SPECIFIC contact in consecutive major grooves of the DNA. highly specific recognition

Question 7

Compare and contrast prokaryotic and eukaryotic transcription

Answer 7

1. Euks have 3 diff RNA Pols, Proks have 1
   Pol 1 – rRNA (RNA)
   Pol 2 – mRNA
   Pol 3 – tRNA , snRNA (5s rRNA)
2. Euks no operon (genes transcribed as single units – monocistronic)
3. Promoter Structure
   
   • Euk: no sigma factor/-10-35 sequences. RNA needs Promoter recognition help from/by transcription factors to start transcription. Namely TFIID, which has TATA-binding protein (TBP) attached. TBP recognizes and binds TATA box (a DNA sequence, promoter element) which is ~ 20 bps up from start site +1. (this makes it bend uniquely which attracts other transcription factors here, like TFIIB)
4. Euk gene activation can occur thousands of bps away from +1 where RNA Pol is. The enhancer DNA site + activator protein (trans factor) are brought close to promoter +Pol by DNA looping. Held together by Mediator.
5. Nucleosomes and higher order chromatin structure can regulate transcription by blocking/uncovering transcription factors –> by recruiting chromatin remodelling complexes to reposition nucleosomes allowing for TATA box to be accessed)
   Or recruit histone-modifying enzymes (eg, histone acetyltransferases) to add acetyl groups to specific histones which then serve as binding sites for proteins
6. Combinational Control
   
   • groups of proteins work together to determine the expression of a single gene
   • whereas in prok, it’s either on or off. Euk gets a range of expression
     SEE FIG. 8-12

Question 8

Examples of protein-protein interactions (prok)

Answer 8

• CAP and RNA Pol

- holoenzyme: the sigma factor/subunit that helps RNA Pol find promoter

Question 9

Describe the fundamental features of chromatin and how it regulates
transcription.

Answer 9

Transcription factors, regulators, and RNA Pol must get to the DNA amidst its high-order packing. Nucleosomes can inhibit initiation of transcription by physically blocking the promoter from the general trans factors or RNA Pol.
-Chromatin remodelling complexes

Question 10

Describe the fundamental features of chromatin and how it regulates
transcription.

Answer 10

Transcription factors, regulators, and RNA Pol must get to the DNA amidst its high-order packing. Nucleosomes can inhibit initiation of transcription by physically blocking the promoter from the general trans factors or RNA Pol.
-Chromatin remodelling complexes: use ATP energy to loosen the DNA and push it along the histone octamer, exposing the DNA
-Reversible Chemical Covalent modifications of the histone proteins: eg. acetyl, methyl and phosphate groups can be added to/removed from the tails by enzymes that are in nucleus. e.g. recruitment of acetyl transferases promote attachment of acetyl groups to selected lysines in the tail of histones – alters/loosens chromatin structure, allowing accessibility + groups attract proteins that promote transcription (ie, trans factors)
Thus can also attract histone deacetylases, thereby reversing pos effects
- These modifications also act as docking sites for regulatory proteins

Question 11

Outline mechanisms that a eukaryotic cell could use to regulate gene expression

Answer 11

Using example of regulating genes responsible for galactose metabolism in yeast
Positive Reg: Yes Galactose = Yes transcription
-

Question 12

Outline mechanisms that a eukaryotic cell could use to regulate gene expression

Answer 12

Using example of regulating transcription at estrogen receptor:
□ Has 3 domains
a) Transcription-activating domain –Calls RNA polymerase to promoter
b) DNA-binding domain in middle –Zinc finger.
c) Ligand binding domain –Binds inhibitory proteins and Ligand = estrogen
No estrogen = No transcription (bc inhib proteins)– in cytoplasm
Yes estrogen = Estrogen binds to LBD, conf change of ER, bye inhib protein, translocation to nucleus (now ACTIVE). Now ER binds to ERE (DNA sequence) on promoter. Co-activator protein binds, attracting CRCs and histone modifiers —> TRANSCRIPTION
-Can be ligand/estrogen-INdependent: environment phosphorylates ER, mimics, releases inhib protein, etc. Results in transcription of specific genes with ERE. Application in Cancer.
-Can activate non-ERE genes (ER=promiscuous): Estrogen-ER complex (or phosphorylated ER) binds to transcription factors that bring it to promoters that don’t have ERE (eg, trans factor AP1 brings it to its promoter –> increased transcription)

Question 13

When do yeast express the genes required to metabolize galactose?

Answer 13

idk

Question 14

Provide examples of where protein-protein interactions regulate transcription. (euk)

Answer 14

Spliceosome holding the snRNPs in place for splicing

Question 15

Describe the steps that occur in the maturation of an RNA to form an mRNA and
allow it to be translated.

Answer 15

Pre-mRNA –> 5’ capping, 3’ polyadenylation, Splicing –> translatable mRNA
must go through some processing.
- 5’ Capping: cap 5’ end with an atypical nucleotide (7-methylguanosine) by 5’-5’ triphosphate bridge. Occurs long before RNA Pol has finished transcribing full gene.
- 3’ Polyadenylation: provides newly transcribed mRNA with a long 300bp polyA tail (Adenine nucs)
- Splicing: introns are removed as lariats from the coding sequence (exons) by snRNPs–complex of small nuclear RNA and protein via 2 transesterification rxns

Question 16

Outline why these steps of RNA processing occur

Answer 16

Why process the primary RNA Transcript?

• stability of mRNA
• export from nucleus to cytoplasm
• to mark it as completed
• translational signal
• prevents the cell from translating partial fragments of mRNA
• protection from exonucleases (degradation)
• Splicing increases the coding capacity of the genome

Question 17

What is a utility of introns?

Answer 17

They provide the opportunity for differential splicing –aka, the mRNA can be assembled in different ways generating functionally related but distinct gene products
especially for creating tissue-specific forms of a protein

Question 18

Describe alternative splicing and outline models for how it might occur.

Answer 18

Self-Splicing Introns

• an example of RNA having catalytic potential: pre mRNA in protozoa, mitochondria, chloroplasts catalyze their own conversion to RNA
• another: ribozymes – catalyze the cleavage of other RNA molecules (RNA polymerization, alkylation, DNA ligations, etc)

Question 19

Describe how problems in splicing can result in disease.

Answer 19

Hemoglobin deficiency caused by abnormal splicing of Betaglobulin RNA, a subunit of hemo
Problem 1: Defect in 3’ Splice Junction at end of 2nd intron
Problem 2: Mutation at 3’ Splice Junction at end of 1st intron

Question 20

3 main sequences in splicing

Answer 20

5’ splice junction
Branch point – absolutely essential
3’ splice junction

Question 21

Describe the properties of the genetic code and provide a rationale for each of
these properties.

Answer 21

1. Universal –all derived from one common ancestor; found in all organisms
2. Non-overlapping – if it did overlap, there would be restrictions on which codon could come next
3. No gaps –no introns, no blank space
4. Redundancy – some codons specify the same aa (usually happens in 3rd position = WOBBLE POSITION). (61 codons for 20 amino acids) (3 stop codons, 1 start codon)
5. But amino acids found less frequently in proteins have less codons
6. Functionally related aas have similar codons(because increases chance of single base mutation still being a functional protein)

Question 22

Describe the structure/function relationships of tRNAs.

Answer 22

Structure: - ~80 nucs in length, cloverleaf sec structure (called stem-loop, looks like L) bc of base pairing to itself causing double helical regions. Have some unusual bases caused by chemical modifications after its been synthesized.

Function: Molecular adaptors that link amino acids to codons (bring aas to growing peptide chain).
Each tRNA has its own anticodon, and thus matching aa.
Their anticodon region hybridizes with the codon, and the right aa is covalently linked to its 3’ end

Question 23

Which mutation is least deleterious to the function of a protein?

Answer 23

Insertion of 3 consecutive bases to a gene. because it doesn’t change the frame, just adds an aa.

Question 24

Define the different mutation types:
Missense
Nonsense
Frameshift

Answer 24

Missense: single base CHANGE –> one diff aa
Nonsense: ^^ but results in a stop codon
Frameshift: insertion or deletion of a base –> changes the whole sequence of aas from there on

Question 25

Reason for Wobble

Answer 25

tRNA is less specific in base pairing and will usually just account for the first two positions of the codon.
5’GAA3’ anticodon will work with 5’UUC3’ and 5’UUU3’ –>both coding for Phe

Question 26

Describe the role of aminoacyl-tRNA synthetases in translation.

Answer 26

Enzyme links the 3’ end of tRNA to its correct aa, using ATP, creating high-energy ester bond. Specific enzyme that recognizes only this aa and its tRNA.
Important for:
1. Providing a stored energy source for peptide bond formation later
2. Providing specificity by matching the correct amino acid to the specific tRNA (has proofreading function)

Question 27

Describe how the structure of the ribosome relates to its function.

Answer 27

Ribosomes (protein RNA complexes) are where protein synthesis happens.
Have 2 subunits: large and small – composed of protein and rRNA
- Small: matches tRNA to codon
- Large: catalyzes the formation of peptide bonds
3 sites on large for tRNAs: (occupy 2 at a time only)
- A site (binds aminoacyl-tRNA)
- P site (binds peptidyl-tRNA)
- E site (tRNA exits)

Question 28

Compare translational initiation in prokaryotes and eukaryotes,
and provide a rationale for any differences.

Answer 28

Euks:
-the initiator tRNA (carrying Met) + translation initiation factors/proteins binds small subunit. And this complex binds mRNA at 5’ Cap – all this (and below) REQUIRED
- small subunit SCANS for AUG codon and now initiation factors dissociate and large subunit binds, putting met initiator in P site. Now aminoacyl-tRNA comes into A site
Proks:
- can’t just scan for 5’ cap because there are multiple start sites (AUGs) (ORFs) along the mRNA, so it would always just translate the first one gene
- so instead the ribosome recognizes internal (AUGs)ribosome BINDING SITE found just upstream from each functional AUG, that each gene has
- Each ribosome binding site can have a different affinity allowing for differential translation

Question 29

Key points in translation (6)

Answer 29

i. Translation occurs in the cytoplasm
ii. Occurs in the 5’-3’ direction along the RNA making a protein from N to C terminus
iii. The mRNA is decoded one codon at a time (3 then 3 then 3)
iv. Energy for peptide bond synthesis comes from the high energy aa-tRNA ester bond – indirectly from ATP
v. Complex reaction involving:
1) Both RNA and protein molecules
2) Conformation shape changes in the ribosome
vi. Specificity comes from the:
1) Aminoacyl-tRNA synthetases (putting the right aa on the tRNA)
2) Requirement for base pairing in the A site of the ribosome

Question 30

What’s something that could decrease translation in eukaryotes?

Answer 30

If a hairpin loop was in the mRNA, the small subunit might not be able to pass it

Question 31

Compare translational termination in prokaryotes and eukaryotes,
and provide a rationale for any differences.

Answer 31

Euks: When A site reaches stop codon, release factor binds to the site. (No tRNAs for stop codon). Catalyzes addition of water to the chain which breaks bond–> releases peptide chain from ribosome (with COOH at end). Ribosome dissociates
Prok: no difference?

Question 32

Describe mechanisms for transcriptional activation and repression in bacteria. (Lac Operon + Catabolite Repression)

Answer 32

Expression of Lactose negative regulated by Lac repressor

• No lactose = no transcription. (Lac repressor binds promoter, inhibiting RNA Pol. )
• Yes lactose = Lactose binds Lac Repressor, resulting in conformational change in the protein–> repressor falls off lac operator sequence (DNA element btwn -10,-35) in promoter –> TRANSCRIPTION

Lac operon is subject to positive regulation by CAP. (when glucose not around)

• CAP encourages transcription. also a sensor for glucose
  - high glucose inhibits synthesis of cAMP
  - cAMP binds CAP, so now CAP can bind CAP site (5’ to the -35 sequence), interacts with RNA Pol- helping it do more transcription of Lac

Catabolite Repression: glucose is the main/preferred carbon source so,

• No glucose, yes lac: cAMP(is here)+CAP lots of transcription
• Yes glucose, yes lac: no cAMP, so little transcription

Question 33

What does a Closed Complex refer to?

Answer 33

RNA Polymerase holoenzyme attached to promoter sequence on DNA
this is a protein-DNA interaction