CH 339K
Dr. Ready
Biochemistry I
EXAM 2
Fall 2011
10/25/2011
There are 104 total points on this test (including extra credit). Unless otherwise stated, assume all
temps are 37o C. Graph paper is attached; if you need more, it's available. Good luck.
Theme for the Test: The year is 2035. Texas is an independent republic. U.S. President-for-life Obama
has unleashed a biological attack using a militarized version of a little known West Indies virus called
Zombivirus A. Infected individuals die within 24 hours, but are maintained in an animated state by the
virus. The animated corpses of the walking dead attack the living, surviving on human brains and
transmitting the virus through their bites. As a medical officer in the Army of the Republic, you are part
of the team working to control the rapidly spreading epidemic of ravenous zombies, who even now seek
to break into the laboratory and devour you alive. No pressure...
1) Zombivirus A produces an enzyme, zombie cerebral corticase, which converts material from the
decaying brain proteins of devoured victims into the novel amino acid zombine, which acts as a
stimulant causing the contraction and relaxation of dead muscle tissue.
cadaverine + putrescine ⇌ zombine
(Cadaverine and putrescine are, by the way, real substances formed in decaying tissues)
You are searching for inhibitors of the corticase enzyme, which will deny the evil dead the nutrient
necessary for their continued ravenous march. Prozac turns out to be an effective agent. You are
testing Prozac as an inhibitor and measure the velocity of the cerebral corticase reaction with and
without inhibitor, obtaining the following results (concentration is in M, velocity in M/second):
[cadaverine] v (no Prozac)
v (10 uM Prozac)
1
750
250
3
1125
563
10
1364
1000
30
1452
1286
100
1485
1429
a) Using the graph paper provided, draw a Lineweaver-Burke plot for the experiment, with
and without the inhibitor. (6 pts)
Page 1 of 7
CH 339K
Dr. Ready
Biochemistry I
Fall 2011
10/25/2011
EXAM 2
0.0045
0.004
y = 0.0033x + 0.0007
0.0035
0.003
0.0025
0.002
0.0015
y = 0.0007x + 0.0007
0.001
0.0005
0
-1
-0.5
0
0.5
1
b) What kind of an inhibitor is Prozac? (6 pts)
Competitive
c) What is Km for the enzyme with the inhibitor? (4 pts)
5uM
d) The enzyme concentration in the experiment is 1 nM (1 * 10-9 M). What is kcat for the
uninhibited enzyme? (4 pts)
Vmax = 1.,43*10-3 Msec-1
Kcat = 1.43*10-3/1 * 10-9 M = 1.43*106 sec-1
e) Fear and terror result in the secretion of several unique compounds in human sweat. One
of these is a powerful allosteric activator of zombie cerebral corticase. What the heck is an
allosteric activator? (4 pts)
An allosteric activator binds to the enzyme at a ste separate from the active site and
increases enzyme activity.
f)
Zombie cerebral corticase lowers the activation energy for the reaction:
cadaverine + putrescine ⇌ zombine
from +40 to +10 kJ/mol. What is the change in the equilibrium constant? (Don’t forget the
kilo in kilojoules) (4 pts)
0 – zip – zilch –nada. Activation energy controls rate, not Keq!!!!!!!!!
Page 2 of 7
CH 339K
Dr. Ready
Biochemistry I
EXAM 2
2) Provide short (but clear) definitions to the following. (4 pts each)
a) Restriction Endonuclease
b) Supercoiling
c) Ribozyme
d) Sugar Pucker
e) Molecular Chaperone
3) Match up the following: (3 pts each)
Epimers
__d___
Enantiomers
__e___
Anomers
__a___
Diastereomers
__c___
Tautomers
__b___
Page 3 of 7
Fall 2011
10/25/2011
CH 339K
Dr. Ready
Biochemistry I
EXAM 2
(a)
(b)
(c)
(d)
Fall 2011
10/25/2011
(e)
4) Since zombies don't breathe, there are serious issues with oxygen transport to keep the bloodthirsty
corpse supplied with sufficient O2 to keep moving. The weaponized zombivirus A also produces an
enzyme which converts a large fraction of the body's hemoglobin to the new form, zombiglobin.
Zombiglobin pools immediately beneath the skin (accounting for the bruised appearance of
zombies) , where it absorbs atmospheric O2 and passes it to tissues by diffusion. (Low internal O2
concentrations are why zombies can only shuffle along slowly.) Below is a table containing O2binding data for zombiglobin as a function of the partial pressure of oxygen:
pO2
θ
0
15
30
60
100
130
150
0.000
0.021
0.100
0.369
0.666
0.789
0.841
a) Using the graph paper provided, draw a Hill plot for oxygen binding by zombiglobin. (NOTE:
this data does not contain the characteristic tails at each extreme of pO2, only the central
region. (It's hard enough catching a zombie to draw blood from, and they don't last very
long.) (6 pts)
Page 4 of 7
CH 339K
Dr. Ready
Biochemistry I
Fall 2011
10/25/2011
EXAM 2
0.75
y = 2.3928x - 4.4857
0.25
-0.25
0
0.5
1
1.5
2
-0.75
-1.25
-1.75
b) What is the minimum number of subunits in the intact zombiglobin molecule? (5 pts)
nH = 2.4, so there are at least 3 subunits
c) What is p50? (5pts)
1.875
10
= 75 mm
d) You also determine, in a separate experiment, that zombiglobin does not exhibit a Bohr
effect. Your terrified but beautiful/handsome (as appropriate) assistant turns to you and
asks, "What the heck is a Bohr effect?" (5pts)
The Bohr effect is the property of hemoglobin by which decreased pH shifts the binding curve to the
right – i.e. lowers affinity – i.e. stabilizes the transition state.
e) From the data shown, would moving to very high altitude help you survive the zombie
apocalypse? (1 pt) Why or why not? (4 pts)
At 10,000 feet (pO2 = 55mM, like I said), their globin would only be about 35% saturated, little
different than what is required by their tissues. No O2 exchange. For a human, it would still be about
85% saturated.
5) For 4 points, why does DNA not contain uracils?
Cytidine spontaneously deaminates to uracil at a low rate. This mutation will be propogated
when the DNA replicates. A DNA repair system constantly scans DNA for the presence of Uracils,
clips them out, and repairs them to match the adjoining strand.
6) Identify the following by name: (3 pts each)
Page 5 of 7
CH 339K
Dr. Ready
Biochemistry I
Fall 2011
10/25/2011
EXAM 2
__Thymine______________
__Galactose______________
_ribose_______________
__adenine______________
Physical Constants and Parameters
Constant
Avagadro’s Number
Boltzmann’s Constant
Charge on electron
Gas Constant (R)
k = 1/(4)
ln x
Value
23
6.02 * 10
1.3807 * 10-23 JK-1
-1.602 x 10-19 coulomb
8.314 JK-1mol-1
8.99 * 109 Nm2coulomb-2
2.303 log x
Page 6 of 7
CH 339K
Dr. Ready
Biochemistry I
EXAM 2
Dielectric constants:
Vacuum 1
Benzene 2.3
Water 78.54
Page 7 of 7
Fall 2011
10/25/2011