Study Questions Set 10

Question 1

What is the role of Shine-Dalgarno sequence?

Answer 1

• 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’

Question 2

What is the role of Kozak sequence?

Answer 2

• 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

Question 3

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)?

Answer 3

• 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

Question 4

Describe in detail events during initiation of the Prokaryotic translation.

Answer 4

1. IF3 binds to free 30S subunit (prevention of binding of 50S subunit)
2. IF1 binds (prevents potential binding of tRNA to A site)
3. IF2 (GTPase) complexes with GTP and binds
4. mRNA binds to 30S subunit through interaction of Shine-Dalgarno sequence with 16S rRNA
5. Initiator tRNA binds (anticodon-codon base pairing) to P site
6. IF3 is released (not needed anymore)
7. This forms the 30S initiation complex
8. 50S subunit binds
9. This displaces IF1 and IF2, and GTP is hydrolyzed – energy consuming step
10. This forms the 70S initiation complex

Question 5

Describe in detail events during initiation of the Eukaryotic translation.

Answer 5

1. Free 40S subunit complexes with eIF3 (large protein) and eIF4C and eIF1A to keep it apart from 60S subunit)
2. Ternary complex forms (i.e., initiator tRNA, eIF2, and GTP)
3. This binds to 40S forming the 43 pre-initiation complex
4. Different eIF4 factors are involved in recognition of 5’ methyl cap; they keep mRNA free of any secondary structures using energy from ATP
5. mRNA will bind to 43S complex through the 5’ methyl cap
6. 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
7. eIF1A, eIF3 and eIF4 are released when AUG is found
8. 40S initiation complex is formed
9. eIF5 displaces eIF2 and eIF3 and GTP is hydrolyzed – energy for binding of 60S subunit
10. eIF4C assists binding of 60S subunit and after that is released, the 80S initiation complex is formed
11. eIF2-GDP complex is recycled by eIF2B (converted back into eIF2 and GTP); rate of recycling can be regulated

Question 6

Briefly explain two cases of RNA having enzymatic capability (how many ribozymes did we mention in class)?

Answer 6

• 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)

Question 7

What is a peptidyl transferase? What is catalyzed by peptidyl transferase?

Answer 7

• 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

Question 8

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)?

Answer 8

1) phosphorylation of translation factors
2) multiple AUG codons
3) internal ribosome entry sites

Question 9

State an example of phosphorylation of translation factors in gene expression

Answer 9

if eIF2 (plays a role in making ternary complex that activates/completes pre-initiation complex) gets phosphorylated, translation initiation rate is slowed

Question 10

State an example of multiple AUG codons in gene expression

Answer 10

• 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)

Question 11

State an example of multiple entry sites in gene expression

Answer 11

• 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

Question 12

Which two factors greatly influence efficiency of protein synthesis?

Answer 12

• 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)

Question 13

What are the roles of PABI? What is the role of PABII?

Answer 13

• 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

Question 14

Explain what happens with polypeptides after translation.

Answer 14

• 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

Question 15

How are protein modifications related to the control of gene expression?

Answer 15

• protein modifications are involved in processing proteins to functional molecules; their stability and modifications can be regulated to control protein stability

Question 16

List possible protein modifications

Answer 16

1. 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
2. proteolytic cleavage
   - most undergo this; simplest form is the removal of the first methionine
3. acylation: acetyl group added to the N-terminal; Acetyl-CoA is the acetyl donor
4. glycosilation: add sugars by covalently linking with Asn
5. methylation: occurs at lysine residues from S-adenosylmethionine

Question 17

List some more possible protein modifications

Answer 17

6. prenylation: addition of compounds derived from the cholesterol biosynthetic pathway
7. 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
8. sulfation: tyrosine residues; added permanently, not regulatory modification
9. vitamin c-dependant modifications
10. vitamin k-dependant modifications

Question 18

Define cytoplasmic male sterility in plants.

Answer 18

• 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

Question 19

Explain endosymbiotic theory.

Answer 19

• 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)

Question 20

Which genome encodes for proteins found in mitochondria. Explain.

Answer 20

• 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

Question 21

What are the characteristics of mitochondrial transcripts?

Answer 21

• 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

Question 22

How is normal tRNA: rRNA ratio maintained during the transcription of mitochondrial DNA?

Answer 22

• 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

Question 23

Define and explain the importance of heteroplasmy and homeoplasmy.

Answer 23

• 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)

Question 24

Is mutation rate in mitochondrial DNA high or low? Explain why.

Answer 24

• 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