1. Questions 1-19 Proteins and Enzymes
MBLG1001 Past Paper
Questions 1-19
Proteins and Enzymes

2. 1. Carbon based life
Life evolved to be carbon based rather than silicon based because:
Silicon only allows 2 options for coding (1/0), whereas carbon allows 4 (A/T/G/C)
Silicon has a smaller atomic radius
Silicon reactions, unlike carbon, are controlled kinetically rather than thermodynamically
Si-O bonds are stronger than Si-Si bonds
Carbon is more abundant than silicon
silly! But nice to start with some humour!
No, carbon is above silicon on the periodic table.
Answer: D. Option A is more a science fiction option. Option B is wrong carbon has the smaller radius, although a small radius results in strong covalent bonds. Option C is again false. Carbon reactions are controlled kinetically rather than thermodynamically. This enables enzymes to regulate reactions. This is unlike the reactions of many other elements. Option D is the correct one, silicon is found mainly as silicates or glass. Option E is wrong, silicon is actually more abundant than carbon but as in option D found as silicates.
ummm.., see notes below
yes. example sand & glass
Opposite, there is heaps more silicon than carbon

3. 2. Alpha helicies
Which of the following statements about the alpha helix is FALSE?
The alpha helix is found in zinc finger motifs and leucine zippers
The twist in the alpha helix is the result of repulsive forces acting on the phosphates
An alpha helix from any source has approximately the same phi angle
Proline is rarely found as an internal residue in alpha helices
Alpha helices interact with the DNA bases through the major groove.
yes
no. hydrogen bonding… um, phosphates??
Answer: Option B, watch the distracter there the twist of DNA comes from the phosphates! Key concept, the alpha helix is a rigid secondary structure which polypeptide chains can assume. It fits brilliantly into the major groove of the DNA double helix and enables the side chains to interact with the bases in a sequence specific manner. So it is not surprising that zinc finger motifs and leucine zippers, which are protein structures found in many transcription factors, contain alpha helices which do the interacting with the DNA. Proline is that nasty little amino acid where the side chain wraps around in forms a ring with the amino group. This restricts the conformations it can assume.
yes, that’s what makes it a helix
yes, it’s a weird, inflexible shape peptide bond-wise
indeed, they do

4. 3. Anfinsen’s Experiment
Which of the following statements concerning Anfinsen’s experiment is FALSE?
The aim of the experiment was to show that hydrophobic interactions drive water soluble proteins to fold
The 6 M guanidine HCl was added to reversibly disrupt the hydrophobic interactions
Measuring the enzyme activity of RNase can be used to monitor the extent of folding
The fully denatured protein was inactive
Dialysis was used to slowly remove the denaturant
Intuitively the statement is correct but didn’t they think that before got results?
yes
Answer: Option A. This was assumed. The aim of the experiment was to determine whether the amino acid sequence alone dictates the fold. So while the statement is correct insofar as it describes the driving force behind the folding of water-soluble proteins it was not the aim of the experiment. The other options are all correct.
extent of re-folding
yes
yes

5. 4. More Anfinsen!
Caution must be applied when extrapolating Anfinsen’s conclusions to the folding of ALL proteins in vivo. Which of the following is NOT a valid reason for this caution?
The cytoplasm contains many proteins ([protein] ~300 mg/mL) which may interfere with folding
Some large proteins need chaperones to fold
Some proteins fold in the membrane
Ribonuclease is a small, unusually stable protein
The experiment was never done with proteins that need to form disulphide bonds
intuitively not. Confused by concn
sounds right
Answer: Option E. Options A to D are all reasons for caution when extrapolating the conclusions. Option A is certainly a factor for caution. The cytoplasm is often thought of as a low salt solution in which a few organelles float around. It actually has a very high protein concentration (lots of different ones of course). This gives our protein lots of binding partners during its folding which will interfere with the process. Option C is considering the proteins which are not water-soluble. These guys are actually translated into the membrane so the polpeptide chain is “extruded” from the ribosome into the membrane of the endoplasmic reticulum. See you can’t let these fellows fold in an aqueous environment, they will fold inside out. RNase, although very stable (as in option D) does have disulphide bonds and these formed correctly. One of the key findings of Anfinsen’s set of experiments was that disulphide bonds don’t form randomly, otherwise full activity would not have been restored once the denaturant was removed.
they don’t ‘become’ membranes
very stable! a bit erudite, any enzyme that can be boiled and it survives is stable!!
sounds reasonable and important

6. 5. Glycine Titration Curve
amino group protonated
carboxylic acid group 50:50
[H+] low but not swamping
H+ added
OH- added pH E D C B A
Answer: Option B.
In which region of the titration curve (A – E) would you have
Equal amounts of H3N+ -CH2-COOH and H3N+-CH2-COO-.

7. 6. Glycine Titration Curve
amino group protonated
carboxylic acid group not
H+ added
OH- added pH E D C B A
Answer: Option C, this is the pI or isoelectric point.
In which region of the titration curve (A – E) would you have
Predominantly H3N+-CH2-COO-.

8. 7. Glycine Titration Curve
Which region would make a good buffer?
Any region could be used as the solution can be made to any pH
No region would be suitable as the pH changes with the addition of H+
Region C, as it is in the middle of the pH range
Only regions B or D
Only regions A or E
no. Some very reactive
counter intuitive!
a disaster
Answer D. Both these regions resist a change in pH when there is an addition of either H+ or OH-. The plateau region tells me the pH is not changing over a certain range of acid or base additions. This is what a buffer does; resists a change in pH
H+ added
OH- added pH A B C D E
yes
a disaster
looking for a small change in pH after adding acid or base

9. 8. Peptide Linking C B E D
Answer: E, cysteine forms disulphide bonds with other cysteines under oxidizing conditions. Two molecules would be able to covalently dimerise through their cysteines. A
Peptide K is found as a covalent dimer under some conditions. Which of the circled features (A – E) would enable it to form covalent dimers?
disuphide bonds - E

10. 9. Absorbing Stuff C B E D
Answer: A, the aromatic ring will absorb at ~280 nm. A
Which of the circled features (A – E) would enable its detection at 280 nm?
ring - A

11. 10. Hydrogen Bonds C B E D
Answer: B, all the others will form H-bonds. A
Peptide K is often found associated with other neuropeptides by hydrogen bonding. Which of the circled features (A – E) could NOT participate in this hydrogen bonding?
hydrophobic methyl - B

12. 11. Phosphorylation C B E D
Answer: A, tyrosine is often phosphorylated during regulation, as are serines and threonines. All 3 amino acids have an OH in their side chain. A
As part of its neurotransmission regulatory role Peptide K is sometimes phosphorylated. Which of the circled features (A – E) would be most likely to be phosphorylated?
looking for serine.. but tyrosine OK - A

13. 12. Charges
Your team has isolated a second neuropeptide, Peptide J, which is slightly longer. Peptide J was found to be composed of equal amounts of the following amino acids
What is the overall charge of Peptide J at pH 13?
ends will have a single negative – amine unprotonated and neutral, carboxylic acid unprotonated and negative
Answer: -2
A unprotonated and neutral, B not affected, C unprotonated and negative, D not affected, E not affected
minus 2

14. 13. Charges
Your team has isolated a second neuropeptide, Peptide J, which is slightly longer. Peptide J was found to be composed of equal amounts of the following amino acids
How many charged groups will Peptide J have at pH 7?
four
Answer: 4 (the N and C terminals, the COO- of residue C and the NH3+ of residue A)
ends will both be charged – amine protonated and positive, carboxylic acid unprotonated and negative
A protonated and positive, B not affected, C unprotonated and negative, D not affected, E not affected

15. 14. Turns
Your team has isolated a second neuropeptide, Peptide J, which is slightly longer. Peptide J was found to be composed of equal amounts of the following amino acids
Peptide J was found to have a sharp bend in its backbone conformation, unlike peptide K. Which amino acid (A – E) would be responsible for this unusual backbone conformation
Answer: the proline, residue E. The side chain ring forces the backbone into a bent shape, often found in beta turns..
Proline - E

16. 15. Equilibria
One of the enzymes of glycolysis, Fructose Bisphosphate aldolase, is found in all cells. The reaction it catalyses has a Keq ([product]/[substrate] at equilibrium) of 6 X 10-5 at 25oC. What can NOT be concluded from this information?
The ΔGof of the product is greater than the ΔGof of the substrate
The reaction is endergonic
The reverse reaction (formation of substrate) will be favoured
The small Keq means the reaction rate will be slow
The ΔGo for this reaction will be positive
not sure how we infer this from the Keq
Answer: D. You cannot predict the rate of reaction from the thermodynamic data. At equilibrium there is ~ times more substrate than product so the reaction favours substrate formation (option C), the product has a higher energy than the substrate (option A), the reaction is endergonic with a DGo of kJ/mol (option B and E).
linked to E. delta G is positive
at equlib [product] <<< [substrate] – so reaction wants to give substrate!
can’t deduce speed from delta G
it will. Keq < 1

17. 16. Equilibria
Which of the following statements concerning equilibrium is CORRECT?
For a cellular reaction to reach equilibrium the product must be continually used by the cell elsewhere
At equilibrium the [products] always equals the [substrate] i.e. Keq = 0
Reactions with large negative ΔGo reach equilibrium quicker
At equilibrium the entropy of the system is zero i.e. ΔS = 0
The reaction will probably reach equilibrium quicker in the presence of an enzyme
opposite. if used then equilibrium never attained
no. Keq is 1 when P=S
: Option E.The enzyme can’t change the thermodynamics of the reaction, so the DGo and the Keq are unchanged. At equilibrium there is no net flow of energy in either direction. Option A stops the reaction reaching an equilibrium. The flow of matter and energy in and out of the cell allows the cell to maintain a steady state non-equilibrium position. The further the cell is away from equilibrium the more energy to do work. Equilibrium is when DG is 0 not Keq which would be 1 anyway if the [P] = [S] and the DS is 0 at absolute zero not equilibrium.
no relationship between delta G and speed
depends on the enthalpy
yes
Enzymes don’t change the equlibrium position, just the rate it gets there!!

18. 17. Weak Forces
Which of the following statements is CORRECT concerning weak forces?
Hydrogen bonding is only significant in macromolecules such as proteins and DNA.
Hydrophobic interactions result from hydrogen bonding between hydrophobic molecules.
Hydrogen bonding results from minimizing the loss of entropy rather than forming bonds.
The strength of ionic interactions decreases in an environment with higher [salt].
Unlike the other weak forces (hydrogen bonding, ionic and hydrophobic interactions) van der Waal’s forces cannot be induced.
what about water?
not H-bonding
Answer: Option D.The higher the ionic strength the more ions to shield the charge on the molecules or charged parts the molecules of interest
No, this is hydrophobic interactions
It does as the ions shield the charges of interest
they are induced and temporary

19. 18. Protein Structure
Which of the following statements best describes the primary structure of a protein?
The number of amino acid residues in the polypeptide chain.
The percent amino acid composition of the polypeptide chain.
The sequence of amino acids in the polypeptide chain.
The regular folding of a single polypeptide chain in repeated patterns - the local conformation of the polypeptide backbone.
The unique three-dimensional structure of a polypeptide chain
not bad.. but need identity
it’s 100!
that’s it!
Answer: Option C
conformation not described. Secondary.
Tertiary.

20. 19. Km Which of the following is INCORRECT concerning Km? Km is:
Used to predict the maximum attainable rate of the reaction.
The [substrate] that gives half the maximum attainable rate of reaction
The [substrate] where half the enzyme present in the reaction is bound to substrate i.e. ES = Efree
Used to determine the [substrate] when measuring the activity of an enzyme
A measure of the affinity the enzyme has for the substrate
possibly. If know the rate at Km but by itself you can’t.
yes
Answer: Option A. KM does not predict the Vmax or Kcat. An enzyme can have a low KM (high affinity) but still have a low Kcat. It is the [S] that gives half Vmax, the [S] that results in half the enzyme present in the reaction bound to substrate at any given time i.e. ES. It is used to work out what substrate concentration to use in an enzyme dependent assay (>10*KM) and it is a measure (in most cases) of the affinity the enzyme has for the substrate. A low KM means a high affinity and vice versa.
Yes, at the Km 50% of the enzyme exists as ES at any time.
yes, want [S] >> Km
yes, low Km is high affinity