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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) silly! But nice to start
with some humour!
B. Silicon has a smaller atomic radius
No, carbon is above silicon
on the periodic table.
C. Silicon reactions, unlike carbon, are controlled
kinetically rather than thermodynamically ummm.., see
D. Si-O bonds are stronger than Si-Si bonds
E. Carbon is more abundant than silicon
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
yes
B.
The twist in the alpha helix is the result of repulsive forces
acting on the phosphates
no. hydrogen bonding… um, phosphates??
C.
An alpha helix from any source has approximately the same
phi angle yes, that’s what makes it a helix
D.
Proline is rarely found as an internal residue in alpha helices
E.
Alpha helices interact with the DNA bases through the major
groove.
indeed, they do
yes, it’s a weird, inflexible shape peptide bond-wise
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
Intuitively the statement is correct but didn’t they think that before got results?
B.
The 6 M guanidine HCl was added to reversibly disrupt the
hydrophobic interactions
yes
C.
Measuring the enzyme activity of RNase can be used to
monitor the extent of folding
extent of re-folding
D.
The fully denatured protein was inactive
E.
Dialysis was used to slowly remove the denaturant
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 intuitively not. Confused by concn
B.
Some large proteins need chaperones to fold
sounds right
C.
Some proteins fold in the membrane
D.
Ribonuclease is a small, unusually stable protein
they don’t ‘become’ membranes
very stable! a bit erudite, any enzyme that can be boiled and it survives is stable!!
E.
The experiment was never done with proteins that need to
form disulphide bonds
sounds reasonable and important
5. Glycine Titration Curve
D
amino group
protonated
carboxylic acid
group 50:50
C
pH
E
[H+] low but not
swamping
B
A
H+ added
OH- added
In which region of the titration curve (A – E) would you have
Equal amounts of H3N+ -CH2-COOH and H3N+-CH2-COO-.
6. Glycine Titration Curve
D
amino group
protonated
carboxylic acid
group not
C
pH
E
B
A
H+ added
OH- added
In which region of the titration curve (A – E) would you have
Predominantly H3N+-CH2-COO-.
7. Glycine Titration Curve
•
Which region would make a good buffer?
A. Any region could be used as the solution can be
made to any pH
no. Some very reactive
B. No region would be suitable as the pH changes with
the addition of H+ counter intuitive!
C. Region C, as it is in the middle of the pH range
a disaster
D
D. Only regions B or D
yes
E. Only regions A or E
a disaster
looking for a small change in pH
after adding acid or base
C
p
H
B
A
H+
added
OHadded
E
8. Peptide Linking
C
O
H3N+
CH
CH2
C
O
HN
CH
C
O
HN
CH
CH3
CH2
B
C
NH2
C
O
HN
CH
C
O-
CH2
O
SH
E
D
OH
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
9. Absorbing Stuff
C
O
H3N+
CH
CH2
C
O
HN
CH
C
O
HN
CH
CH3
CH2
B
C
NH2
C
O
HN
CH
C
O-
CH2
O
SH
E
D
OH
A
Which of the circled features (A – E) would enable its detection at 280 nm?
ring - A
10. Hydrogen Bonds
C
O
H3N+
CH
CH2
C
O
HN
CH
C
O
HN
CH
CH3
CH2
B
C
C
O
HN
CH
C
O-
CH2
O
NH2
SH
E
D
OH
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
11. Phosphorylation
C
O
H3N+
CH
CH2
C
O
HN
CH
C
O
HN
CH
CH3
CH2
B
C
NH2
C
O
HN
CH
C
O-
CH2
O
SH
E
D
OH
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
O
A
H2N
O
B
CH
C
OH
H2N
CH
C
CH2
CH
CH3
CH2
CH2
CH2
CH3
OH
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
CH2
NH2
O
C
H2N
CH
C
O
D
OH
H2N
CH
CH2
CH2
CH2
C
C
O
NH2
OH
O
E
HN
C
OH
C
12. Charges
OH
What is the overall charge of
Peptide J at pH 13?
O
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
O
A
H2N
O
B
CH
C
OH
H2N
CH
C
CH2
CH
CH3
CH2
CH2
CH2
CH3
OH
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
CH2
NH2
O
C
H2N
CH
C
O
D
OH
H2N
CH
CH2
CH2
CH2
C
C
O
NH2
OH
O
E
HN
C
OH
C
O
13. Charges
OH
How many charged groups will
Peptide J have at pH 7?
four
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
O
A
H2N
O
B
CH
C
OH
H2N
CH
C
CH2
CH
CH3
CH2
CH2
CH2
CH3
OH
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
CH2
NH2
O
C
H2N
CH
C
O
D
OH
H2N
CH
CH2
CH2
CH2
C
C
O
NH2
OH
O
E
HN
C
OH
C
O
14. Turns
OH
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 ΔGof of the
substrate
not sure how we infer this from the Keq
B.
The reaction is endergonic linked to E. delta G is positive
C.
The reverse reaction (formation of substrate) will be favoured
at equlib [product] <<< [substrate] – so reaction wants to give substrate!
D.
The small Keq means the reaction rate will be slow
can’t deduce speed from delta G
E.
The ΔGo for this reaction will be positive
it will. Keq < 1
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
opposite. if used then equilibrium never attained
B.
At equilibrium the [products] always equals the [substrate] i.e.
Keq = 0
no. Keq is 1 when P=S
C.
Reactions with large negative ΔGo reach equilibrium quicker
no relationship between delta G and speed
D.
At equilibrium the entropy of the system is zero i.e. ΔS = 0
depends on the enthalpy
E.
The reaction will probably reach equilibrium quicker in the
presence of an enzyme yes
Enzymes don’t change the equlibrium position, just the rate it gets there!!
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.
what about water?
B.
Hydrophobic interactions result from hydrogen bonding
between hydrophobic molecules.
not H-bonding
C.
Hydrogen bonding results from minimizing the loss of entropy
rather than forming bonds. No, this is hydrophobic
interactions
D.
The strength of ionic interactions decreases in an environment
with higher [salt]. It does as the ions shield the charges of
interest
E.
Unlike the other weak forces (hydrogen bonding, ionic and
hydrophobic interactions) van der Waal’s forces cannot be
induced.
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.
not bad.. but need identity
B.
The percent amino acid composition of the polypeptide chain.
it’s 100!
C.
The sequence of amino acids in the polypeptide chain.
that’s it!
D.
The regular folding of a single polypeptide chain in repeated
patterns - the local conformation of the polypeptide backbone.
conformation not described. Secondary.
E.
The unique three-dimensional structure of a polypeptide chain
Tertiary.
19. Km
•
Which of the following is INCORRECT concerning Km?
Km is:
A.
Used to predict the maximum attainable rate of the reaction.
possibly. If know the rate at Km but by itself you can’t.
B.
The [substrate] that gives half the maximum attainable rate of
reaction
yes
C.
The [substrate] where half the enzyme present in the reaction
is bound to substrate i.e. ES = Efree
Yes, at the Km 50% of the enzyme exists as ES at any time.
D.
Used to determine the [substrate] when measuring the activity
of an enzyme yes, want [S] >> Km
E.
A measure of the affinity the enzyme has for the substrate
yes, low Km is high affinity