Previous Exam Questions

Question 1

Draw the chemical structure of the second messenger cyclic-AMP and the Watson-Crick base pair adenine-uracil!

Answer 1

Question 2

Answer 2

• A) Replication
  
  • 2) DNA Polymerase
  • 4) dNTP
  • 8) Primer
• B) Transcription
  
  • 1) RNA polymerase
  • 6) NTP
  • 7) mRNA
  • 9) Promotor
• C) Translation
  
  • 3) Ribosome
  • 5) tRNA

Question 3

Answer 3

• A) T-State - 4) Less active state of an allosteric protein
• B) R-State - 3) More active state of an allosteric protein
• C) Preprotein - 1) Protein with signal sequence
• D) Kinase - 6) Protein phosphorylation
• E) Phosphatase - 5) Removes phosphates
• F) Zymogen - 2) Proenzyme

Question 4

Outline the principle regulatory components of the E.Coli Lac operon!

Answer 4

• Make sure that underlined ones are mentioned.
• P - Promotor: site that directs RNA polymerase to correct transcription inititiation site.
• i gene - encodes the lacl repressor
• o site - operator site - repressor attaches here to prevent transcription of the structural genes by RNA polymerase.
• Lacl repressor: binds to operator site and prevents transcription of Lacxya genes. Binds tightly in absence of lactose, releases in presence of lactose.
• Structural genes (Lac-xya) - Code for mRNA that creates 3 proteins (known as a polycistronic transcript)
  
  • LacZ - codes for beta-galactosidase
  • LacY - codes for galactoside permease
  • Laca - codes for thiogalactoside transacetylase
• Cyclical nature of the operon: In the presence of lactose, the repressor releases and allows beta-galactosidase gene (LacZ) to be expressed. This enzyme cleaves lactose, and then the lack of lactose causes repressor to bind again. This way, beta-galactosidase is only expressed when needed.

Question 5

Answer 5

Answer:

D) Polymer of Glucose

Question 6

Answer 6

Answer:

C) Lipids

Question 7

Answer 7

• Current 20 amino acids are encoded by triplet codons, coding for all 20 + 3 stop codons. With 4 bases to work with, this is 4³ = 64 triplet codons.
• If only 16 amino acids need to be coded for, this can be done with doublet codons since 4² = 16
• However, if at least 1 stop codon also needs to be coded for, this would make 17 codons needed. Thus, triplets would still be needed.

Question 8

Answer 8

• The main distinguishing characteristic of a retrovirus is that it has RNA as its genetic material.
• A retrovirus uses an enzyme called reverse transcriptase to synthesize complementary (cDNA) in the host cell.
• Thus, the retrovirus flow of genetic information is as follows:
  
  • Viral RNA –> DNA (in host) —> RNA (transcribed by host)
• The normal cellular flow of genetic information is:
  
  • DNA —-> RNA

Question 9

Answer 9

• A) Transcribes tRNA genes
  
  • RNA Polymerase III
• B) Is localized in nucleolus
  
  • RNA Polymerase I
• C) Transcribes most protein coding genes
  
  • RNA Polymerase II

Question 10

Answer 10

Answer to A)

1. Treat the complex with RNAse to digest RNA.
2. Rebind proteins to affinity matrix.
3. The proteins will only elute together if they are still associated with each other.
4. If they elute together, RNA does not hold them together, if they elute seperatly it does.

Answer to B)

1. Incubate the heterodimer in beta-mercaptoethanol to break disulfide bonds.
2. Rebind proteins to affinity matrix.
3. If they elute together, they are still associated with each other, if they elute seperatly, then they are not.
4. Eluting together indicates that disulfide bridges were not holding them together. Eluting seperatly indicates that they were in fact held together by disulfide bridges.

Question 11

Answer 11

1. A) NMR spectroscopy at near atomic resolution
   
   1. Membrane protein means its bad for x-ray
   2. Small, so good for NMR
   3. In large amounts, so good for NMR
2. B) Cryo-EM single particle analysis at near atomic resolution
   
   1. Native complex - not good for x-ray
   2. Not much sample - not good for NMR
3. C) X-ray Crystallography at atomic to near atomic resolution
   
   1. Large amounts - good for X-ray
   2. Is stable - good for crystallization
   3. Thermophilic makes it bad for cryo-EM

Question 12

Answer 12

Underlined parts are the necessary words.

1. The Shine-Dalgarno sequence is necessary for translation initiation in prokaryotes.
2. The 3’-end of the 16S rRNA base pairs with this sequence.
3. This ensures the correct positioning of the AUG start codon on the 30S ribosomal subunit in preparation for the pairing of the start codon with the initiator tRNA anti-codon.

Question 13

Answer 13

1. The two targeting factors required are the Signal Recognition Protein (SRP) and the SRP receptor.
2. The hydrophobic signal sequence on the nascent protein determines if this pathway will be used. The SRP binds to this signal sequence.
3. SecA ATPase provides energy for post translational protein translocation in prokaryotes.
4. The Hsp70 ATPases, Bip and Kar2p provide energy for post translational protein translocation in eukaryotes.

Question 14

Answer 14

1. mRNA Splicing - The Nucleus
2. Citric acid cycle (krebs cycle) - The mitochondrial matrix
3. Beta-oxidation of fatty acids - The Mitochondrial matrix
4. Photosynthesis - Thylakoid stacks in stroma of chloroplasts.
5. Electron transport (respiration) chain - Inner mitochondrial membrane.
6. Disulfide bond formation - ER

Question 15

Answer 15

1. Protein kinases catalyze:
   
   1. Unphosphorylated —-> Phosphorylated protein
2. GTPases catalyze:
   
   1. GDP-bound protein —> GTP-bound protein
3. These changes lead to Conformational change that leads to changes in activity. This can then lead to downstream signaling. Molecular switches like this are perfect, becuase pathways are off when there is no signal from them, and on when there is a signal from them.

Question 16

Answer 16

Cell Growth

1. Increase in protein mass
2. Occurs in G_1, then S phase and Mitosis

Cell Division

1. The replication of DNA in S phase and then mitosis.

Checkpoints

1. G₁ checkpoint/restriction point:
   
   1. based on growth (nutrients) in yeast
   2. Based on mitogens in metazoan cells.
2. G₂-M-Checkpoint:
   
   1. makes sure that DNA replication was successful
   2. checks for improperly copied DNA
3. M-Checkpoint:
   
   1. Checks that spindle has successfully been assembled.
   2. Checks that sister chromatids are properly attached to it.

Question 17

Draw the complete structure for the tripeptide Met-Phe-Ser, as it would appear at pH 7!

Answer 17

Question 18

Answer 18

• A) Fructose-2,6-bisphosphate activates phosphofructokinase (a key enzyme in glycolysis) and inhibits Fructose-1,6-bisphosphatase (a key enzyme in gluconeogenesis).
• B) Phosphofructokinase activation would stimulate glycolysis. Fructose-1,6-bisphosphatase inhibition would inhibit gluconeogenesis. This would mean that anaerobic energy use would increase.
• C) Blood vessels do not provide sufficient oxygen to tumors, meaning that they often grow in hypoxic conditions. Thus, an increase in anaerobic ATP production would be very beneficial to them.

Question 19

Answer 19

1. HMG-CoA reductase is the enzyme that catalyzes teh comitted step in cholesterol synthesis.
2. Inhibition of HMG-CoA reductase will lower cholesterol levels.
3. In heart disease, high cholesterol in plasma can lead to the formation of atherosclerotic plaques in blood vessels, potentially leading to heart attacks.
4. Lowering cholesterol can help prevent this.

Question 20

Answer 20

Part A

1. The Ribonucleotide reductase catalyzed reaction
   
   1. This is a critical enzyme in dNTP creation
2. The Thymidylate synthase reaction
   
   1. converts dUMP to dTMP

Part B

1. Ribonuclease reductase reaction
   
   1. can be inhibited by hydroxy-urea at the radical in the active site.
2. Thymidylate synthase
   
   1. Anticancer drug fluorouracil forms a suicide inhibitor in vivo for thymidylate synthase.
   2. Thymidylate synthase can also be inhibited by methotrexate which stops the regeneration of methylene tetrahydrofolate by inhibiting dihydrofolate reductase.