Biochemistry 2000
Sample Question
(1) Enzymes are potent catalysts because they:
a. are consumed in the reactions they catalyze.
b. are very specific and can prevent the conversion of products back to substrates.
c. drive reactions to completion while other catalysts drive reactions to equilibrium.
d. increase the equilibrium constants for the reactions they catalyze.
e. lower the activation energy for the reactions they catalyze.
(2) The role of an enzyme in an enzyme-catalyzed reaction is to:
a. bind a substrate analog that cannot be converted to product.
b. ensure that all of the substrate is converted to product.
c. ensure that the product is more stable than the substrate.
d. increase the rate at which substrate is converted into product.
e. make the free-energy change for the reaction more favorable.
(3) Identify each of the designated regions on the energy diagram for the conversion of a
substrate to product with and without an enzyme.
(4) Ingestion of methanol is a medical emergency. It is often treated by the administration
of ethanol, which prevents the dangerous effects of methanol metabolism. In the
body, methanol is oxidized to formaldehyde (methanal), a toxic molecule that cannot
be further oxidized and that damages proteins. Ethanol is oxidized by the same
enzyme (alcohol dehydrogenase), to acetaldehyde, a metabolite that can be further
processed. If ethanol and methanol are substrates for the same enzyme, how does
ethanol prevent the oxidation of methanol?
(5) The enyzme urease enhances the rate of urea hydrolysis at pH 8.0 and 20°C by a
factor of 1014. If a given quantity of urease can completely hydrolyze a given quantity
of urea in 5.0 min at 20°C and pH 8.0, how long would it take for this amount of urea
to be hydrolyzed under the same conditions in the absence of urease? (Assume that
both reactions take place in sterile systems so that bacteria cannot attack the urea.)
(6) What are the two major hypothesis that describe the interaction of enzymes and their
substrates? How do they differ?
(7) The affinity of Hemoglobin for dissolved O2 is regulated by multiple different effector
molecules.
(a) Dissolved O2 binding to Hemoglobin increases the protein's affinity for dissolved
O2. What type of effector molecule O2 and what is the type of regulation?
1
Proteins
Biochemistry 2000
Sample Question
(b) 2,3 bisphosphoglycerate (2,3-BPG) binding to Hemoglobin increases the protein's
affinity for dissolved O2. What type of effector molecule is 2,3-BPG and what is the
type of regulation?
(8) Draw a suitable Reaction Coordinate for a spontaneous overall reaction composed of
two elementary steps in which the first step is rate determining. Label all elements of
the plot.
(9) Write an overall reaction that has a second order reaction rate constant.
(10) Consider the following reaction mechanism.
k1
k2
k3
A -----> I1 -----> I2 -----> P
Draw a reaction coordinate consistent with the following information:
The overall reaction is spontaneous. The elementary reaction producing I2 is not
spontaneous. The rate constants for the individual steps have the following
relationship, k1 = k3 >> k2.
(11) Give brief definitions or unique descriptions of the following terms:
(a) prosthetic
(b) apoenzyme
(c) Melting temperature of DNA
(d) KM
(e) nucleotide
(12) Draw a stick diagram for each of the following:
(a) guanine
(b) cytidine
(c) deoxyadenosine in the syn conformation
(d) deoxythymidine-5-phosphate
(e) Watson-Crick G:C base pair (indicate H-bonds)
(13) Draw an initial rates plot for an enzyme that catalyzes a first order reaction and has a
Vmax of 0.2 mM s-1. Label all elements of the plot.
(14) How would you determine the KM of an enzyme catalyzed reaction from a
Lineweaver-Burk plot?
(15) Consider an enzyme with a Michealis-Menten mechanism, a KM of 0.1 mM and a
Vmax of 0.01 mM s-1. Answer the following:
(a) At what substrate concentration would the enzyme have an initial velocity of 1
M s-1?
(b) What fraction of all active sites are filled if the initial velocity is 8 M s-1?
(16) Give the long name for the following nucleic acid fragments?
(a) pAp
(b) d(TpC)
(c) pppU
(d) Gppp
(17) At what concentration will a Michaelis-Menten enzyme have an initial velocity of ¼
2
Proteins
Biochemistry 2000
Sample Question
vmax?
(18) Answer the following question regarding a 1000 base pair (1 kb) dsDNA fragment:
(a) If the fragment contains 22% G, what is the A content?
(b) If you slowly heat the dsDNA fragement while measuring the UV absorption,
what will happen?
(10) Many DNA binding proteins have an approximate overall shape that resembles a 'U'
or a horseshoe. The DNA binds inside the 'U' and makes extensive contacts with the
protein. Answer the following questions:
(a) What is the most likely inside diameter of 'U' shaped DNA binding proteins?
(b) Would you expect 'U' shaped DNA binding proteins to bind to dsRNA with an
equivalent sequence? Explain.
3
Proteins
Biochemistry 2000
Sample Question
Answers:
(1)
a. lower the activation energy for the reactions they catalyze.
(2)
d. increase the rate at which substrate is converted into product.
(3)
A free energy of substrate
B activation energy without enzyme
C activation energy with enzyme
D free energy of product
(4) Methanol and ethanol compete for the active site of alcohol dehydrogenase, they are
both substrates for this enzyme. If ethanol is present in excess over methanol, it
will “win this competition” thereby preventing methanol from being oxidized.
(Instead methanol will be secreted from the body.)
(5) 5 min * 1014 = 5 1014 min = 8.333 * 1012 h = 3.47 1011 days = 9.5 * 108 years
(6)
(i) Lock & Key Hypothesis – suggests enzymes have rigid, preformed active sites that
do not undergo changes upon substrate binding.
(ii) Induced Fit Hypothesis – suggests enzymes have flexible active sites that adapt
their shape to the substrate they bind.
The two models differ with respect to the shape of the active site prior to substrate
binding and whether or not the active site undergoes conformational changes upon
substrate binding. Experimental evidence suggests most enzyme active sites are
largely preformed and only undergo subtly conformational changes upon substrate
binding.
(7)
(a) O2 is a homotropic activator and is a cooperative regulator of Hemoglobin activity.
(b) 2,3-BPG is a heterotropic acitivator and is an allosteric regulator of Hemoglobin
activity.
4
Proteins
Biochemistry 2000
Sample Question
(8) The reaction is spontaneous as the relative energy
of 'P' is lower than that 'A'.
The 1st elementary reaction is rate limiting as its
activation energy barrier (G‡) is larger than that
of step two.
(9) A reaction will have a second order rate constant if the reaction rate depends upon the
concentration of two reactants.
A + B -----> P
or
2A -----> P
(33) The overall reaction is spontaneous as the relative energy of 'P' is lower than that of
'A'. The elementary reaction producing I2 is not spontaneous as the relative energy of 'I2'
is higher than that of 'I1'.
The reaction rate constants k1 and k3 are equal as their activation energy barriers are
equal. The reaction rate constant k2 is lower than that of k1 and k3 as its activation energy
barrier is much larger.
TS2
TS1
TS3
DG
DG‡
DG‡
A
I1
DG‡
I2
P
Reaction Coordinate
(18)
g. a cofactor that is permanently associated with its protein
h. an inactive protein or enzyme in the absence of its cofactor
(1)
a – The coding sequence that is retained in mature mRNA and directs eucaryotic
protein synthesis.
b – The concentration of substrate at which the enzyme has ½ its maximal initial
5
Proteins
Biochemistry 2000
Sample Question
Proteins
velocity.
c – The covalent bond between the C1' of D-ribose and the N1(or N9) atom of the
nucleic acid base.
d – The temperature at which 50% of dsDNA or dsRNA has separated into single
strands (denatured).
e – A nucleoside phosphate or the repeating unit of a nucleic acid polymers.
f – Nucleic acid base structures that differ in the location of single hydrogen atom.
These structures have altered covalent and hydrogen bonding patterns.
g – A stretch of 10s to 100s of adenylic acid nucleotide added to the 3' terminii of
eucaryotic mRNA.
h – The transcription initiation complex in which the DNA strands of the bacterial
promoter (-10 - +4) are single stranded.
(2)
a–
b–
d–
e–
6
c–
Biochemistry 2000
Sample Question
(3) The theoretical maximum efficiency of any catalyst occurs when the catalytic rate is
limited by diffusion of the substrate and enzyme. In this case, k2 >>> k-1 and every
“collision” between substrate and catalyst results in the formation of product.
Catalytic perfection is measured using kcat/KM, where kcat represents the “turnover
number” or the rate constant for product formation (typically equivalent to k2 for
Michaelis-Menten enzymes) and KM is the Michaelis constant indicating the
concentration of substrate at which the initial velocity is ½ maximal. This value is ~108 –
109 M-1 s-1 for catalytically perfect enzymes.
Note: Recall that enzyme efficiency depends upon both its maximal rate AND the
concentration of substrate.
Initial Velocity (mM s-1)
(4)
0.2
This question does not indicate magnitude
of KM so we cannot numerically quantify
the substrate concentrations along the xaxis.
Vmax
0.1
0
KM
3 KM
5 KM
Substrate Concentration (mM)
(5) – There are two solutions.
a – Measure the x-intercept which is (-1/KM)
b – Determine Vmax from the y-intercept and then determine KM from the slope
(KM/Vmax)
(6)
a – Starting with the Michaelis-Menten equation, vo = (Vmax [S]) / (KM + [S])
We substitute the given values for vo, Vmax and KM, which yields:
1 M s-1 = (0.01 mmol s-1 [S]) / (0.1 mM + [S])
Converting M to mM and solving for [S] produces:
0.001 mM s-1 (0.1 mM + [S]) = (0.01 mM s-1 [S])
0.0001 mM2 s-1 + 0.001 mM s-1 [S] = 0.01 mM s-1 [S]
0.0001 mM2 s-1 = 0.009 mM s-1 [S]
[S] = 0.011 mM
7
Proteins
Biochemistry 2000
Sample Question
b – Vmax is 0.01 mM s-1 or 10 M s-1. At Vmax, ET = ES (all enzyme has substrate
bound).
If the initial velocity of the enzyme is 8 M s-1, the fraction of sites filled is
simply 8 M s-1/10 M s-1 or 0.80 (80%).
Note: Part b only holds for enzymes with a Michaelis-Menten mechanism.
(7)
a – adenosine – 3', 5' – diphosphate
b – deoxythymidyl – 3',5' – deoxycytidine
c – uridine – 5' – triphosphate
d – guanosine – 3' – triphosphate
(8) Starting with the Michaelis-Menten equation, vo = (Vmax [S]) / (KM + [S]).
Substitute vo = ¼ Vmax, and solve for [S]
¼ Vmax = (Vmax [S]) / (KM + [S])
KM + [S] = (Vmax [S]) / ¼ Vmax
KM + [S] = 4 [S]
[S] = KM / 3
(9)
a – Using Chargaff's rule, G = C and G + C = 44%
A + T = 100% - 44% = 56% and A = 28%
b – The UV absorption will remain constant and relatively low until the melting
temperature is
reached. The UV absorption will rapidly rise over a narrow
temperature range as the dsDNA
cooperatively denatures and will then remain
constant and relatively high.
Note: Denature and Unwind are not equivalent or interchangable.
(10)
a – Virtual all dsDNA in cells is B-DNA and has a diameter of 2.0 nm. The 'U'
shape would require a closely similar diameter to maintain geometric complementarity.
b – The dsRNA is unlikely to bind as it adopts an A-DNA like conformation. The
overall size and shape of dsRNA is significantly different from B-DNA and generally
disrupts the geometric
complimentarity required for protein-ligand interactions. In
particular, the double helix is wider (2.6 vs 2.0 nm) and there are significant difference
in the size of the major and minor grooves. The chemical differences (2'-OH and U)
between RNA and DNA may also adversely affect specific binding interactions.
Note: The difference in helix diameter and groove size are intimately related to
the
differences in other structural parameters (eg. helix pitch, rise, base pair twist,
ribose conformation, etc.).
8
Proteins