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1
BIOCHEMISTRY I (CHEM-UA 881)
Fall 2023
MIDTERM EXAM 2, Nov 3, 2023
One point extra credit for completing the 3 lines below properly
NAME
:
_______
KEY
____________________
N-NUMBER
:
__________________________
TA & RECITATION TIME
:
__________________________
PLEASE WRITE LEGIBLY & BE CONCISE. Your answers must be your own! No
calculators, no use of electronic devices, no communication between students. Partial
credit will be given.
(There are a total of
16
pages in the exam, including this page).
An Appendix with useful
information is on the last page, which can be removed.
QUESTION
PTS
YOUR SCORE
1
20
2
25
3
10
4
10
5
10
6
15
7
10
TOTAL
100
2
Part 1.
Multiple Choice/Fill in the Blanks. For multiple choice questions, circle the correct
answer for each. Responses with more than one answer will be marked as incorrect. (2
pts each, 20 total)
1.
Which of the following amino acid residues is most likely present within the region of
an
a
-helix that spans the cell membrane?
A. Leu
B. Lys
C. Arg
D. Asp
E. Glu
2. How do 2,3-bisphoglycerate (BPG) levels help regulate hemoglobin binding to O
2
?
A. BPG levels are higher in tissues versus lungs, which leads to the release of
O
2
in the tissues.
B. BPG protonates hemoglobin in tissues, which favors the T state.
C. BPG acts as a negative effector by making ion-ion contacts that stabilize the
T-state of hemoglobin.
D. BPG competes directly for the O
2
binding site in hemoglobin, which remains
oxygenated under high [O
2
] in the lungs.
E. None of the above.
3.
In the structure below, circle the letter of the arrow that points to the site where the
disaccharide lactose could be oxidized.
Answer: D
O
HO
HO
OH
OH
O
O
HO
OH
OH
OH
A
B
C
D
3
4. Which statement or statements best describe the relationship between
b
-
D
-Glucose
and
b
-
D
-Mannose?
A. They are epimers.
B. They are anomers.
C. They have the same molecular formula.
D. They are tetroses.
E. Both A and C.
5. What is bioorthogonal chemistry?
A. Chemical reactions performed in flasks that cannot be performed in biological
systems.
B. Chemical reactions that typically occur in biological systems and can be used
for modification of cells.
C. Chemical reactions that do not typically occur in biological systems and can
be used for modification of cells.
D. Chemical reactions that do not typically occur in biological systems and
cannot be used for modification of cells.
E. None of the above
6. How does aquaporin achieve specificity in the transportation of water across
membranes?
A. By dehydrating water molecules as they pass through the membrane.
B. Using a specificity pore with a constriction point that forms H-bonds with
water.
C. Using metal ions that coordinate to water to provide the energy of transport.
D. Using the power of ATP hydrolysis, aquaporin undergoes conformational
changes that lead to selective binding of water molecules.
E. All of the above
7. The oxyanion hole of chymotrypsin is formed by which of the following?
A. Asp side chains
B. Glu side chains
C. Lys side chains
D. Arg side chains
E. Peptide backbone atoms
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8. The structure below depicts the
__tetrahedral___
intermediate in the chymotrypsin-
catalyzed cleavage of peptide bonds. (R
1
and R
2
represent the rest of the peptide
chain). [Please fill in the blank]
[1 pt for oxyanion or two answers]
9. The general structure of glycerophospholipids is shown below.
Which head-group (represented by
X
) would lead to a net charge of -1 at pH = 7
resulting in a net negative charge in membranes enriched in the resulting
glycerophospholipids. (R
1
and R
2
represent fatty acid chains.)
A. Asp
B. Lys
C. Ser
D. Ethanolamine
E. Arg
10. The function of the flippase enzymes is to:
A. Transfer peripheral membrane proteins from the outer leaf of the plasma
membrane bilayer to the inner leaf.
B. Transfer peripheral membrane proteins from the inner leaf of the plasma
membrane bilayer to the outer leaf.
C. Transfer phospholipids from the outer leaf of the plasma membrane bilayer
to the inner leaf.
D. Transfer phospholipids laterally in the plasma membrane bilayer.
E. Generate lipid rafts by transferring cholesterol to the outer leaf of the
plasma membrane bilayer.
R
2
N
H
R
1
O
-
O
Ser
CH
2
OR
1
CHOR
2
CH
2
O
P
O
O
O
-
X
5
Part 2.
Short response. Please respond to each of the following questions or
statements in 1-2 sentences. (5 pts each, 25 total)
A.
Explain why hemoglobin binding to O
2
produces a sigmoidal curve while
myoglobin binding to O
2
does not.
Hemoglobin is a tetrameric protein, and binding to one ligand molecule (O
2
) causes a
change in affinity for the next binding event in another subunit.
This cooperativity of
binding between subunits produces the hallmark sigmoidal binding curve. Myoglobin is
a monomer, so cooperativity in binding is not observed because the protein only binds
one molecule of oxygen per myoglobin protein (and so a hyperbolic curve results).
Additional possible response: hemoglobin needs to more sensitive to the amount of [O
2
]
because it is a transport protein that needs to bind O
2
at high [O
2
] and release it at low
[O
2
] while myoglobin is a storage protein that primarily binds O
2
. (3 pts awarded)
B.
The Michaelis-Menten equation relies on a “steady state assumption”. What is
assumed to be at a steady state during the course of an enzymatic reaction?
The Michaelis-Menten analysis assumes that the concentration of the intermediate
Enzyme-Substrate [ES] complex remains constant for an enzymatic reaction, at least for
the period of time under evaluation.
It is also reasonable to state the rate of the formation of the ES complex and the rate of
its breakdown are equal.
C.
Under what circumstances does a
K
M
value begin to resemble a
K
D
(which would
represent the dissociation constant of a substrate binding to a protein)?
This is the case for a special condition – when product formation is slow relative to the
rate of substrate binding. For the typical Michaelis-Menten situation:
When k
-1
>> k
2
, then Km = k
-1
/k
1
= K
d
6
D.
Chymotrypsin is a protease that cleaves after aromatic residues. Elastase
cleaves after glycine and alanine residues. What mutations would you make to
engineer chymotrypsin to have substrate recognition that resembles elastase?
Since the specificity pocket is responsible for the recognition of substrates, one could
mutate the specificity pocket to introduce bulky/branched amino acids (such as Thr or
Val) instead of the smaller side chains (Gly) found in the specificity pocket of
chymotrypsin.
E.
Explain why passive transport channels and passive transport carriers are
different. Why do they display different trends in translocation rates in response
to increasing concentration of the solute being transported (as shown in the
graphs below)?
Both passive channels and transporters move molecules in the same direction as the
concentration gradient across a membrane (they do not require an input of energy).
Channels serve as semi-selective pores such that transport occurs at high rates as
shown in the graph on the left (can approach the diffusion controlled limit).
Carriers are
typically more selective and show a rate of transport that is high initially, but then levels
off at higher concentrations) consistent with the saturation of the carrier observed in the
graph on the right (as seen in Michaelis-Menten kinetic analysis).
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7
Part 3. Free response/Data Analysis.
Please respond to each question in the space
provided.
3.
(10 points) The graph below shows the rates of chemical reactions catalyzed by
several enzymes (their names are abbreviated as ADC, ODC,
etc
.). The distribution of
catalyzed versus uncatalyzed rates is indicated on the right side of the graph.
A.
Estimate the “rate enhancement” for the enzyme Ketosteroid Isomerase (KSI).
Note that the scale is logarithmic on the y-axis on the left.
(4 pts)
Rate enhancement
k
cat
/
k
uncat
. For KSI,
k
uncat
= 10
-7
and
k
cat
= 10
5
; hence the rate
enhancement is 10
5
/10
-7
= 10
12
.
8
B.
Why does there seem to be a maximum to the catalyzed rates (
k
cat
)? In your
response, explain what parameter determines the “speed limit” at which an
enzyme can operate, which puts an upper limit on catalytic efficiency.
(3 pts)
In the formation of the ES complex: E + S
↔
ES, the rate in the forward direction (
k
1
)
is limited by the rate of diffusion (how fast molecules can diffuse in water = 10
9
M
-1
s
-
1
). This explains why enzyme-catalyzed reactions all exhibit a maximum in their
catalyzed rates.
C.
Could you determine the catalytic efficiency of the enzyme FUM from this graph?
Why or why not?
(3 pts)
No. Need both
k
cat
and
K
M
to determine catalytic efficiency. This graph lacks data
on the
K
M
.
9
4.
(10 points) The enzyme Ketosteroid Isomerase (KSI) catalyzes the
transformation below:
The active sites of two different KSI enzymes are illustrated below. The wild-type
enzyme is shown in (i) and a mutant form is shown in (ii). The relevant residues are
labeled in the wild-type enzyme.
A.
Note Asp1 is an aspartate. Is Asp1 acting as general acid, general base, specific
acid, or specific base? Explain your choice.
(2 pts)
General base, as it is a not a water molecule and abstracts a proton in the reaction
mechanism.
Specific acid/base catalysis is mediated by water.
(i)
(ii)
k
cat
= 10 s
-1
k
cat
= 5 x 10
-4
s
-1
Asp1
Asp2
Tyr
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10
B.
What is the V
max
of the wild-type reaction if the total enzyme concentration is 1 mM?
(2 pts)
V
max
= k
cat
[E
t
]
= 10 s-
1
(1 mM)
= 10 mM/s
C.
What is the probable role of Asp2 (aspartic acid) and Tyr?
(2 pts)
These residues stabilize the growing negative charge in the carbonyl transition state,
acting as an oxyanion hole.
D.
What residues are Asp2 and Tyr mutated to in part (ii)? Do the mutations shown
speed the enzymatic rate or slow it down? Based on your knowledge of enzyme
mechanisms, explain why the mutations have the observed effect.
(4 pts)
Asp2 is mutated to Leu and Tyr is mutated to Phe. The mutations slow down the
enzymatic rate (the
k
cat
of the mutant is lower than that of the wild-type enzyme).
Since
there are no longer H-bond donors in the mutant’s side chains, the TS of the reaction is
no longer stabilized by an oxyanion hole as seen in the wild-type.
Herschlag D, Natarajan A. Fundamental challenges in mechanistic enzymology:
progress toward understanding the rate enhancements of enzymes.
Biochemistry
2013
;
52
, 2050-67
11
5.
(10 points) In the Lineweaver-Burk plot below, one line represents a plot of an
enzyme’s activity in the presence of an inhibitor and one line represents a plot of
the enzyme activity in the absence of an inhibitor. The equations of the lines are
shown next to the relevant line. The x-intercept of each line is -0.5 mM
-1
.
Remember to provide units in your answers.
A.
Which line in the plot indicates inhibition [top (grey) or bottom (black) line]?
Provide a brief explanation for your choice.
(3 pts)
The top line. All inhibitors that we discussed lower the V
max
and the top line has a
lower V
max
than the bottom line, as indicated by comparison of the 1/V
max
value of
each (the y-intercept of each line).
y = 0.4x + 0.2
y = x + 0.5
-0.5
0
0.5
1
1.5
2
-0.55
-0.45
-0.35
-0.25
-0.15
-0.05
0.05
0.15
0.25
0.35
0.45
1/V (min/mM)
1/[S] (mM^-1)
12
B.
Estimate the
K
M
for the uninhibited enzyme. (Please show calculations
performed.)
(2 pts)
The x-intercept = (-1/
K
M
)
x-intercept = -0.5 mM
-1
K
M
= -1/x-intercept = -1/-.5 mM
-1
K
M
= 2 mM
C.
Estimate the V
max
for the uninhibited enzyme. (Please show calculations
performed.)
(2 pts)
The y-intercept = (1/V
max
), which can be determined from the equation of the line
(y = mx + b).
y = 0.4x + .2
1/V
max
= 0.2 min/mM
V
max
= 5 mM/min
D.
What type of inhibition do these data represent? Can the inhibitor bind to the free
enzyme or does the inhibitor only bind to the enzyme-substrate complex?
(3 pts)
Noncompetitive
or
mixed
inhibition (both accepted). Yes, it can bind the free enzyme
(
or the ES,
has equal affinity for both
).
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13
6.
(15 points) Professor Kirshenbaum is interested in regaining some cognitive
function, so he has been eating meals rich in fish oils, containing omega-3 and
omega-6 unsaturated fatty acids. Professor Lupoli prefers a diet of pizza
enriched in saturated fatty acids.
A.
In the space below, please draw the chemical structure of the omega-3 fatty
acid called:
a
-linolenic acid or
cis
18:3(
D
9,12,15
). Briefly explain why this lipid is
called an omega-3 fatty acid and not an omega-2 or omega-5 fatty acid
.
(5
pts)
(Either charge state of the fatty acid is accepted since the pH wasn’t specified.)
This is an omega-3 fatty acid because the first double bond occurs at carbon 3
when counted from the last carbon (the omega carbon).
B.
Would increasing the polyunsaturated fatty acid content of a cell membrane
increase or decrease the Tm of that membrane? Briefly explain your choice.
(5 pts)
Increasing the polyunsaturated fatty acid content would decrease the Tm (meting
temperature) because the introduction of “kinks” would lower the efficiency of
packing seen with saturated fatty acids, which are stabilized via van der Waals
and hydrophobic interactions due to their long (flexible) carbon tails.
OH
O
14
C.
Below is a graph of a representative Fluorescence Recovery After Photobleaching
(FRAP) experiment. This experiment can evaluate the rate of diffusion of
fluorescently labeled proteins within a membrane.
In the empty graph below, sketch a reasonable FRAP profile for membranes composed
exclusively of C
18
saturated fatty acids. In the same graph, also sketch the data for
membranes enriched in C
18
omega-3 fatty acids. Clearly label EACH line that you draw
as “omega-3” or “saturated”.
(5 pts)
In the graph, the FRAP profile for saturated fatty acids is in black. The FRAP profile for
unsaturated fatty acids is in red (top). Any graph where the slope up after F
0
is faster for
unsaturated versus saturated is correct.
Fluorescence Intensity
Time
15
7.
(10 points) Human blood groups can be distinguished by the sequence of sugars
found in glycoconjugates on red blood cells.
A)
Blood group determinants are
O
-linked glycans. What residues in glycoproteins
would you expect these glycans to be attached to?
(2 pts)
Ser / Thr (1/2 for only one or more than these two)
B)
An extra galactose is present in the glycoconjugates of B-type red blood cells that is
not found in other blood group types.
Draw the structure of
a
-
D
-galactose in chair
form in the space below AND draw an arrow that points to the anomeric carbon.
(4 pts)
C)
Mutations in the blood group B glycosyltransferase (GT-B) have been identified that
lead to changes in antigenic glycoconjugate structures. Kinetic parameters of wild-type
and mutant GT-B with indicated donors are shown below (Gal = galactose; GalNAc =
N
-
acetylgalactosamine).
Enzyme
K
M
(UDP-Gal)
μ
M
k
cat
(UDP-Gal)
s
-1
K
M
(UDP-GalNAc)
μ
M
k
cat
(UDP-GalNAc)
s
-1
Wild-type
88
5.1
180
0.42
P234S
4500
0.24
3740
14.4
J. Biol. Chem
.
2003,
278
, 12403.
Individuals that carry the P234S mutation in GT-B produce A-type blood glycoconjugates,
even though they do not carry the GT-A enzyme found in most A-type individuals. Based
on the data above and your knowledge of glycosyltransferases, provide an explanation
for this observation.
(4 pts)
P234S shows a lowered
k
cat
with the native donor UDP-Gal (it is reduced from 5.1 s
-1
with the wild-type enzyme to 0.24 s
-1
with the mutant enzyme) and a higher
k
cat
with the
non-native donor UDP-GalNAc than seen for the wild-type (it is increased from 0.42 s
-1
with the wild-type enzyme to 14.4 s
-1
with the mutant enzyme).
The mutant also has a
O
OH
HO
OH
OH
OH
anomeric
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16
slightly lower
K
M
for UDP-GalNAc than UDP-Gal, which demonstrates that the reaction
of the mutant enzyme with UDP-GalNAc is more catalytically efficient (
k
cat
/
K
M
) than that
with UDP-Gal. Since A-type individuals contain GalNAc on their antigenic
glycoconjugates, these data explain why individuals carrying this mutation can produce
A-type glycoconjugates without the presence of GT-A. [
Recall from lecture that a single
amino acid change in GT-A and GT-B leads to different donor specificities
].
Appendix
Useful pKa values:
α-carboxyl of free amino acid: 2
α-amino of free amino acid: 9
C-terminal carboxyl of peptide: 3
N-terminal amino of peptide: 8
Enzyme Kinetics Equations:
17
V
o
= V
max
[S] / (K
m
+ [S])
V
o
= k
cat
[ES]
V
max
= k
cat
[E
t
]
K
m
= (k
-1
+ k
2
)/k
1
D
G = − RTlnK
E
+
S
k
1
k
-
1
ES
k
2
-⇥
E
+
P
ES
]
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AIV-20: Getting Connected (Writing lonic Compounds)
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anion
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valence
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F-
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02-
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Interconverting compound SI units
Convert the following measurement.
2
kg•m
7 g.cm
х10
8.3 x 10
2
2
S
Ar
Explanation
Check
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Engaging ActivitiesACTIVITY 1Directions: Identify the element described in each statement.___________1. An element with 26 protons and 30 neutrons.___________2. Has a mass number of 11 and an atomic number of 5.___________3. Has 117 neutrons and an atomic number of 78.___________4. Has 69 neutrons and 50 protons.___________5. Has 125 neutrons and an atomic number of 85.
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Plot your values of In(Ksp) VS. 1/Tand find the slope and y-intercept of the best fit line. Use the equation for the best fit
line and the following equation
In(K) = - AH°
RT
AS°
R
to calculate AH and AS for dissolving Borax
What is the slope of your best fit line in the plot?
What is AH (kJ/mol)?
What is the y-intercept of your best fit line in the plot?
What is AS (J/mol)?
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Based on results can you help me answer these two questions being asked of me
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Question 35
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Only 100% sure experts solve it correct complete solutions each okk take 24hrs but solve all okkk must accurate students own words okk don't use Ai okk.
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All one question
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Radiation can be used to__________.
provide electricity
sterilize the surface of meats, fruits and vegetables
diagnose certain diseases
kill cancer
all of the above
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I need help completing this I just need the formula and names of each box, there are some that I did.
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36Ar18 + 4He2 --> __________ + 0?γ0 (?γ = gamma ray)
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