MBLG1001 Past Paper

Questions 1-19

Proteins and Enzymes

1. Carbon based life

• Life evolved to be carbon based rather than silicon based because:

A. Silicon only allows 2 options for coding (1/0), whereas carbon allows 4 (A/T/G/C)

B. Silicon has a smaller atomic radius

C. Silicon reactions, unlike carbon, are controlled kinetically rather than thermodynamically

D. Si-O bonds are stronger than Si-Si bonds

E. Carbon is more abundant than silicon

silly! But nice to start with some humour!

No, carbon is above silicon

on the periodic table.

ummm.., see notes below

yes. example sand & glass

Opposite, there is heaps more silicon than carbon

2. Alpha helicies

• Which of the following statements about the alpha helix is FALSE?

A. The alpha helix is found in zinc finger motifs and leucine zippers

B. The twist in the alpha helix is the result of repulsive forces acting on the phosphates

C. An alpha helix from any source has approximately the same phi angle

D. Proline is rarely found as an internal residue in alpha helices

E. Alpha helices interact with the DNA bases through the major groove.

yes

no. hydrogen bonding… um, phosphates??

yes, that’s what makes it a helix

yes, it’s a weird, inflexible shape peptide bond-wise

indeed, they do

3. Anfinsen’s Experiment

• Which of the following statements concerning Anfinsen’s experiment is FALSE?

A. The aim of the experiment was to show that hydrophobic interactions drive water soluble proteins to fold

B. The 6 M guanidine HCl was added to reversibly disrupt the hydrophobic interactions

C. Measuring the enzyme activity of RNase can be used to monitor the extent of folding

D. The fully denatured protein was inactive

E. Dialysis was used to slowly remove the denaturant

Intuitively the statement is correct but didn’t they think that before got results?

yes

extent of re-folding

yes

yes

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?

A. The cytoplasm contains many proteins ([protein] ~300 mg/mL) which may interfere with folding

B. Some large proteins need chaperones to fold

C. Some proteins fold in the membrane

D. Ribonuclease is a small, unusually stable protein

E. The experiment was never done with proteins that need to form disulphide bonds

intuitively not. Confused by concn

sounds right

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

5. Glycine Titration Curve

H+ added OH- added

pH

A

B

C

DE

In which region of the titration curve (A – E) would you have

Equal amounts of H3N+ -CH2-COOH and H3N+-CH2-COO-.

amino group protonated

carboxylic acid group 50:50

[H+] low but not swamping

6. Glycine Titration Curve

H+ added OH- added

pH

A

B

C

DE

In which region of the titration curve (A – E) would you have

Predominantly H3N+-CH2-COO-.

amino group protonated

carboxylic acid group not

7. Glycine Titration Curve

H+ added

OH- added

pH

A

B

C

D E

• Which region would make a good buffer?

A. Any region could be used as the solution can be made to any pH

B. No region would be suitable as the pH changes with the addition of H+

C. Region C, as it is in the middle of the pH range

D. Only regions B or D

E. Only regions A or E

looking for a small change in pH after adding acid or base

a disaster

no. Some very reactive

counter intuitive!

a disaster

yes

8. Peptide Linking

H3N+ CH C

CH2

O

OH

HN CH C

CH3

O

HN CH C

CH2

O

C

NH2

O

HN CH C

CH2

O-

O

SH

A

B

C

D

E

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

9. Absorbing Stuff

H3N+ CH C

CH2

O

OH

HN CH C

CH3

O

HN CH C

CH2

O

C

NH2

O

HN CH C

CH2

O-

O

SH

A

B

C

D

E

Which of the circled features (A – E) would enable its detection at 280 nm?

ring - A

10. Hydrogen Bonds

H3N+ CH C

CH2

O

OH

HN CH C

CH3

O

HN CH C

CH2

O

C

NH2

O

HN CH C

CH2

O-

O

SH

A

B

C

D

E

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

11. Phosphorylation

H3N+ CH C

CH2

O

OH

HN CH C

CH3

O

HN CH C

CH2

O

C

NH2

O

HN CH C

CH2

O-

O

SH

A

B

C

D

E

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

12. ChargesH2N CH C

CH2

OH

O

CH2

CH2

CH2

NH2

H2N CH C

CH

OH

O

CH3

CH2

CH3

HN

C OH

O

H2N CH C

CH2

OH

O

CH2

C

OH

O

H2N CH C

CH2

OH

O

C

NH2

O

A B

C D

E

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

A unprotonated and neutral, B not affected, C unprotonated and negative, D not affected, E not affected

minus 2

13. ChargesH2N CH C

CH2

OH

O

CH2

CH2

CH2

NH2

H2N CH C

CH

OH

O

CH3

CH2

CH3

HN

C OH

O

H2N CH C

CH2

OH

O

CH2

C

OH

O

H2N CH C

CH2

OH

O

C

NH2

O

A B

C D

E

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?

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

four

14. TurnsH2N CH C

CH2

OH

O

CH2

CH2

CH2

NH2

H2N CH C

CH

OH

O

CH3

CH2

CH3

HN

C OH

O

H2N CH C

CH2

OH

O

CH2

C

OH

O

H2N CH C

CH2

OH

O

C

NH2

O

A B

C D

E

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

Proline - E

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?

A. The ΔGof of the product is greater than the ΔGo

f of the substrate

B. The reaction is endergonic

C. The reverse reaction (formation of substrate) will be favoured

D. The small Keq means the reaction rate will be slow

E. The ΔGo for this reaction will be positive

not sure how we infer this from the Keq

linked to E. delta G is positive

can’t deduce speed from delta G

it will. Keq < 1

at equlib [product] <<< [substrate] – so reaction wants to give substrate!

16. Equilibria

• Which of the following statements concerning equilibrium is CORRECT?

A. For a cellular reaction to reach equilibrium the product must be continually used by the cell elsewhere

B. At equilibrium the [products] always equals the [substrate] i.e. Keq = 0

C. Reactions with large negative ΔGo reach equilibrium quicker

D. At equilibrium the entropy of the system is zero i.e. ΔS = 0

E. 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

no relationship between delta G and speed

depends on the enthalpy

Enzymes don’t change the equlibrium position, just the rate it gets there!!

yes

17. Weak Forces

• Which of the following statements is CORRECT concerning weak forces?

A. Hydrogen bonding is only significant in macromolecules such as proteins and DNA.

B. Hydrophobic interactions result from hydrogen bonding between hydrophobic molecules.

C. Hydrogen bonding results from minimizing the loss of entropy rather than forming bonds.

D. The strength of ionic interactions decreases in an environment with higher [salt].

E. 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

No, this is hydrophobic interactions

It does as the ions shield the charges of interest

they are induced and temporary

18. Protein Structure• Which of the following statements best describes the

primary structure of a protein?

A. The number of amino acid residues in the polypeptide chain.

B. The percent amino acid composition of the polypeptide chain.

C. The sequence of amino acids in the polypeptide chain.

D. The regular folding of a single polypeptide chain in repeated patterns - the local conformation of the polypeptide backbone.

E. The unique three-dimensional structure of a polypeptide chain

not bad.. but need identity

it’s 100!

that’s it!

conformation not described. Secondary.

Tertiary.

19. Km

• Which of the following is INCORRECT concerning Km? Km is:

A. Used to predict the maximum attainable rate of the reaction.

B. The [substrate] that gives half the maximum attainable rate of reaction

C. The [substrate] where half the enzyme present in the reaction is bound to substrate i.e. ES = Efree

D. Used to determine the [substrate] when measuring the activity of an enzyme

E. 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

Yes, at the Km 50% of the enzyme exists as ES at any time.

yes, want [S] >> Km

yes, low Km is high affinity