1
Q
What is the role of Shine-Dalgarno sequence?
A
- it is the conserved sequence 8-13 nucleotides upstream the start codon in prokaryotes, that is complementary to the 16S rRNA in the small 30S subunit of the ribosome
- also called the ribosomal binding site (RBS), this sequence enables the ribosome to position itself correctly with respect to the start codon (at P site of first methionine)
- 5’ – AGGAGGU – 3’
2
Q
What is the role of Kozak sequence?
A
- Kozak sequence is a consensus sequence in eukaryotes that has the AUG embedded in it as: 5’ – CC A/G CC AUG G – 3’
- this sequence is recognized by the 40S subunit (which already has initiator tRNA bound), which binds to the methyl cap and scans mRNA for the Kozak sequence
- while AUG is the start codon, it must be present within the right context for proper binding of the ribosome to the mRNA
3
Q
What are the major differences in initiation of translation between eukaryotes and prokaryotes (remember: first AA and the way mRNA and tRNA bind to ribosomes)?
A
- translation initiation in eukaryotes:
o AUG embedded in Kozak sequence
o 40 S subunit recognizes 5’cap and scans until AUG in a Kozak context is found
o tRNA comes to ribosome before addition of mRNA - translation initiation in prokaryotes
o AUG downstream from Shine-Dalgarno sequence
o ribosome’s 16S rRNA complementary to SD and base pairs
o tRNA with f-Met is added to complex after addition of mRNA
4
Q
Describe in detail events during initiation of the Prokaryotic translation.
A
- IF3 binds to free 30S subunit (prevention of binding of 50S subunit)
- IF1 binds (prevents potential binding of tRNA to A site)
- IF2 (GTPase) complexes with GTP and binds
- mRNA binds to 30S subunit through interaction of Shine-Dalgarno sequence with 16S rRNA
- Initiator tRNA binds (anticodon-codon base pairing) to P site
- IF3 is released (not needed anymore)
- This forms the 30S initiation complex
- 50S subunit binds
- This displaces IF1 and IF2, and GTP is hydrolyzed – energy consuming step
- This forms the 70S initiation complex
5
Q
Describe in detail events during initiation of the Eukaryotic translation.
A
- Free 40S subunit complexes with eIF3 (large protein) and eIF4C and eIF1A to keep it apart from 60S subunit)
- Ternary complex forms (i.e., initiator tRNA, eIF2, and GTP)
- This binds to 40S forming the 43 pre-initiation complex
- Different eIF4 factors are involved in recognition of 5’ methyl cap; they keep mRNA free of any secondary structures using energy from ATP
- mRNA will bind to 43S complex through the 5’ methyl cap
- mRNA is scanned by 43S complex to find the right AUG (anticodon from initiator tRNA has to base pair with the first Met codon) – role of the Kozak sequence; ATP hydrolysis is energy used for scanning
- eIF1A, eIF3 and eIF4 are released when AUG is found
- 40S initiation complex is formed
- eIF5 displaces eIF2 and eIF3 and GTP is hydrolyzed – energy for binding of 60S subunit
- eIF4C assists binding of 60S subunit and after that is released, the 80S initiation complex is formed
- eIF2-GDP complex is recycled by eIF2B (converted back into eIF2 and GTP); rate of recycling can be regulated
6
Q
Briefly explain two cases of RNA having enzymatic capability (how many ribozymes did we mention in class)?
A
- Ribozymes: RNA molecules that catalyze chemical reactions
- peptidyl transferase: a 23S rRNA from the large ribosomal subunit that affects transfer of peptide chain to amino acid
- snRNA are catalytically active components of snRNP (spliceosomal splicing)
7
Q
What is a peptidyl transferase? What is catalyzed by peptidyl transferase?
A
- enzyme that affects the rate of growth of the peptide chain; 23S rRNA on large subunit
- energy consumed by peptidyl transferase: 1 ATP and 2 GTP
8
Q
How is gene expression controlled at the level of translation (mRNA is in contact with ribosomes or their subunits/translation factors; three things mentioned in the class)?
A
1) phosphorylation of translation factors
2) multiple AUG codons
3) internal ribosome entry sites
9
Q
State an example of phosphorylation of translation factors in gene expression
A
if eIF2 (plays a role in making ternary complex that activates/completes pre-initiation complex) gets phosphorylated, translation initiation rate is slowed
10
Q
State an example of multiple AUG codons in gene expression
A
- since viral DNA (and some cellular mRNA) lack 5’ cap, instead have multiple AUGs in their 5’UTR
- two types of multiple AUGs:
o if Kozak sequence is very “weak” – the scanning of the ribosomal subunits for the AUG is “leaky scanning”, therefore the resulting proteins differ in N-terminus, which is typically responsible for cell destination
o AUG/UGA combo – short open reading frame between 5’ end of mRNA and beginning of actual gene; upstream open reading frame (uORF); results in dysfunctional polypeptide, thereby stalling/decreasing the translation of the actual downstream gene by “trapping” the scanning ribosome and causing it to drop down from the mRNA (before reaching the AUG of real sequence)
11
Q
State an example of multiple entry sites in gene expression
A
- allow for cap-independent translation to occur (no need for eIF4 family)
- around 450 nucleotide long region with complex secondary structure; 43S pre-initiation complex bind to IRES and scans from there
12
Q
Which two factors greatly influence efficiency of protein synthesis?
A
- simultaneous translation of a single mRNA by multiple ribosomes
o form polysomes
o after first ribosome has moved, 70-80 nucleotides from start, second can bind and start
o can be up to 50 ribosomes on one mRNA strand - rapid recycling of ribosomal subunits
o function of poly A-binding protein I, PABI causes circularization of eukaryotic mRNAs, so everything stays close to the start site for repeated synthesis (plays a role in helping prevention of degradation as well)
13
Q
What are the roles of PABI? What is the role of PABII?
A
- PABI:
o involved in circularizing mRNA so that once a ribosome finishes, start of the mRNA (and all needed factors) and units are nearby to just start translating again - PABII
o extends the Poly(A) tail of pre-mRNA within the host cell nucleus, allows mRNA to exit the nucleus
o adds around 200-250 nucleotides in the rapid phase
14
Q
Explain what happens with polypeptides after translation.
A
- the first step is posttranslational modification and transport
- formation of mature functional protein involves: folding and co-factor binding, cleaving, covalent modification, binding to other protein subunits
15
Q
How are protein modifications related to the control of gene expression?
A
- protein modifications are involved in processing proteins to functional molecules; their stability and modifications can be regulated to control protein stability
16
Q
List possible protein modifications
A
- modifications necessary for transport from cytoplasm
- proteins that are membrane bound or are destined for secretion are synthesized by ribosomes association with the membranes of the endoplasmic reticulum; all contain a N-terminus (signal sequence or signal peptide) that direct transport - proteolytic cleavage
- most undergo this; simplest form is the removal of the first methionine - acylation: acetyl group added to the N-terminal; Acetyl-CoA is the acetyl donor
- glycosilation: add sugars by covalently linking with Asn
- methylation: occurs at lysine residues from S-adenosylmethionine
17
Q
List some more possible protein modifications
A
- prenylation: addition of compounds derived from the cholesterol biosynthetic pathway
- phosphorylation: most important and common protein modification; regulates biological activity of protein; transient (added and later removed)
- kinases: transfer a phosphate group from a donor (usually ATP) to the acceptor (protein)
- phosphorylases: transfer a phosphate group from an inorganic phosphate
- phosphatases: remove phosphate2s - sulfation: tyrosine residues; added permanently, not regulatory modification
- vitamin c-dependant modifications
- vitamin k-dependant modifications
18
Q
Define cytoplasmic male sterility in plants.
A
- the failure of the plant to produce functional anthers or pollen
- it is under extra nuclear genetic control (reside outside of nucleus) and does not follow Mendelian inheritance
19
Q
Explain endosymbiotic theory.
A
- organelles require both their own and nuclear gene products for their construction
- theory exists that mitochondria and chloroplasts existed as separate prokaryotic organisms that were taken inside the cell as endosymbionts
- organelle’s genomes resemble prokaryotic genomes more than nuclear genomes (small, circular, no histones associated)
- organelle’s protein-synthesizing machinery resembles prokaryotic machinery more than eukaryotic nuclear machinery (i.e., prokaryotic-like rRNA, ribosomes; fMet is first amino acid; different effects of antibiotics and protein inhibitors; rifampicin inhibits bacterial and mitochondrial RNA polymerases, not nuclear)
20
Q
Which genome encodes for proteins found in mitochondria. Explain.
A
- the mitochondrial genome encodes for some proteins (mostly those involved in oxidative phosphorylation and protein synthesis)
- the nuclear genome encodes most of the proteins for organelle function and structural proteins; these are then imported from the cytoplasm
21
Q
What are the characteristics of mitochondrial transcripts?
A
- double stranded circular DNA molecule; 17000 bp long
- heavy strand (G rich) and light strand (C rich)
- 37 genes: 28 on heavy, 9 on light; 22 tRNA, 2 rRNA, and 13 polypeptide encoding genes
- no introns, compact (only 1 or 2 bases separate genes), two genes overlap
- incomplete termination codons; insert UAA at the transcriptional level instead
- polyadenylation at 3’ end, but no 5’ cap
22
Q
How is normal tRNA: rRNA ratio maintained during the transcription of mitochondrial DNA?
A
- H-strand transcription often stops at 3’ end of the 16S rRNA gene and starts again from PH, therefore 15-60 times more rRNA transcripts made than full length H-strand transcripts so the normal rRNA: tRNA ratio is ensured
23
Q
Define and explain the importance of heteroplasmy and homeoplasmy.
A
- heteroplasmy: a mutation in one mitochondrial DNA molecule within a cell leads to a mixture of mutant and normal mtDNA molecules
- this occurs because it is a matter or change which mtDNAs will be passed to each daughter cell during mitosis
- homeoplasmy: either pure mutant or normal mtDNA
- mutations in mtDNA mostly affects metabolically active tissues; can cause late onset disease (combinations of inherited and somatic mutations eventually push the cell into disease state)
24
Q
Is mutation rate in mitochondrial DNA high or low? Explain why.
A
- mutation rate is high due to:
o exposure to oxygen free-radicals during respiration
o lack of protective histones
o less efficient mtDNA repair system (e.g., thymine dimers cannot be repaired)
o high rate of replication of mtDNA
