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BIOCHEMISTRY (CHEM 360)
EXAM 3
November 18, 2011
There are seven pages, TEN questions, and a total of 114 points in this exam.
Please read each question carefully and possibly more than once. Good luck…
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1.
The first decarboxylation reaction in Kreb’s cycle is initially an oxidation. Write the
missing structure of the “not so stable” oxidation product below.
The half reaction for this process is conventionally written as follows (in reduction
potential tables)
1.1.
Based on your knowledge of Kreb’s cycle write the overall net reaction for the
decarboxylation process; except instead of giving specific oxidizing/reducing
agents simply write “electron donor” and/or “electron acceptor” at the
appropriate part(s) of the equation .
1.2.
Give a (generic) half reaction for your electron acceptor and electron donor (as it
would be conventionally written). Make sure that your half reaction is balanced
and consistent with the other half reaction given above and with the net equation
you have written.
1.3.
Assuming that the overall net equation you have written (for question 1.1) is
spontaneous, propose a reduction potential, Eo’ for the half reaction you have
written for question 1.2 (in comparison with the one given numerically above).
a less negative value than -0.38 eV
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2.
Glucose, 14C labeled at C-1, is fed to various living organisms.
2.1. Identify the position of 14C in pyruvate at the end of glycolysis.
Show your work (no points for blind guessing).
2.2. Identify the position of 14C in -ketogluterate in the first turn of the citric acid cycle.
(Draw the structure of -ketogluterate and circle the labeled carbon)
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3.
Complete the following reactions by showing the full structures of the missing products
Please do not use abbreviations.
3.1.
3.2.
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4.
Assume that ribulose-1-phosphate, shown below, can undergo a reaction catalyzed by a
lyase enzyme, similar to that of fructose-1,6-bisphosphate (when catalyzed by aldolase).
Write down the structures for the products of ribulose-1-phosphate.
Count your carbons!..
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5.
Circle the best correct answer for the multiple choice questions given below:
5.1.
Which of the following is (are) considered “energy rich”?
(1) only A
(2) only B
(3) only C
(5) they are all considered “energy rich”
5.2.
(4) both A and C
Valinomycin is an anti-bacterial that kills bacteria by surrounding K+ ions and
shuttling them down their concentration gradient and across membranes. Which
of the following might be the cause of cell death?
(1) Disruption of secondary transport processes that depend on K+ ion
concentration gradient.
(2) Change in the pH of the bacterial cytosol.
(3) Blocking of bacterial pores with K+ ions.
(4) Massive denaturation of bacterial proteins upon change of the K+ ion
concentration.
5.3.
5.4.
Which of the reactions below are utilized in the following conversion?
(A) Reduction
(B) Phosphate addition
(1) Only (A)
(4) Both (A) and (B)
(2) Only (B)
(3) Only (C)
(5) Both (B) and (C)
Once inside the cell, glucose is rapidly phosphorylated to glucose-6-phosphate.
What is the main purpose of this phosphorylation?
(1) To keep glucose inside the cell
(3) To form a high-energy compound
5.5.
(C) Oxidation
The citric acid cycle is
(1) anabolic (2) catabolic
(2) To prevent mutarotation
(4) To activate PFK-1
(3) both anabolic and catabolic
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(4) linear
5.6.
Why should it not be surprising that for many cells water requires a protein for its
transport across a membrane?
(1) All molecules require transport proteins to cross a membrane.
(2) Water is very polar, which inhibits its free diffusion across the
membrane.
(3) There is never a concentration gradient for water across the membrane to
drive its transport.
(4) The transport protein is needed to prevent the hydrolysis of the phospholipid
chains as water crosses the membrane.
5.7.
How many ATP molecules are consumed in the hexose stage of glycolysis for
every one molecule of glucose?
(1) 1
(2) 2
(3) 3
(4) 4
(5) ATP is produced, not consumed.
5.8.
Which of the following is/are (an) oxidizing agent(s)?
(1) only A
(2) only B
(5) both A and C
(3) only C
(4) both A and B


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The only reaction where a C-C bond is formed in the citric acid cycle is given below:
In a mutant organism the cycle is fueled by propionyl CoA instead of acetyl CoA.
Assuming that the mutant organism possesses the necessary enzyme to catalyze the
reaction, identify the structure of the metabolite formed instead of citrate
(show your work).
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submitted (mostly) by Cheryl
7.
Iodoacetate inhibits glyceraldehyde-3-phosphate dehydrogenase by interacting with an
active site cycsteine residue. Show the reaction which occurs between iodoacetate and
the enzyme.
(Draw an illustration of the enzyme with its relevant key features)
7.1.
Draw the “tetrahedral intermediate” formed in the absence of iodoacetate.
7.2.
Outline a curly arrow mechanism of a reaction between the “tetrahedral
intermediate” and a common metabolite whose key features are given
below.
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8.
Give brief specific answers for each of the questions below.
8.2.
Although one turn of the citric acid cycle only generates one GTP, it is assumed
that the cycle in fact has a net yield of 10 ATP. Explain.
Citric acid cycle also generates 3 NADH (2.5 ATP equivalents each) and 1 QH2
(1.5 ATP equivalents)
8.4.
Why is it generally accepted that the lability (reactivity) of 1-arseno-3phosphoglycerate, shown below, effectively uncouples the oxidation and
phosphorylation events.
glyceraldehyde-3-phosphate is oxidized into glycerate-3phosphate and at the same time phosphorylated at C-1 into
a metabolite with a strong phosphorylating potential.
Replacement of the phosphate at C-1 with an arsenate
prevents glycerate from being able to phosphorylate ADP
and thus “uncouples” oxidation from phorsphorylation.
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9.
Glucose enters some cells by simple diffusion through channels and pores, but glucose
enters red blood cells by passive transport. On the plot below, indicate which line
represents diffusion through a channel or pore, and which line represents passive
transport. Explain why the rates of the two processes (and the shapes of the two plots) are
different.
“A” shows passive transport
A hyperbolic curve such as
“A indicates the
involvement of a protein
which can be saturated at
high glucose concentration
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10.
Phosphatidylinositol-4,5-bisphosphate, PIP2, undergoes hydrolysis catalyzes by PLC to
produce two “second messengers.”
10.1.
Draw the structures of the two “second messengers”.
(The hydrolysis site of PIP2 is indicated by an arrow in boldface)
10.2.
One of the second messengers diffuses to the intercellular organelles, where the
release of Ca2+ is activated. On the other hand, the other second messenger
remains in the plasma membrane, where it activates a Ca2+-dependent protein
kinase.
Identify which messenger does what.
IP3 diffuses to the intercellular organelles; DAG remains in the plasma
membrane
10.3.
On the basis of the physical properties of each of the secondary messengers
explain your answer for 10.2 (why each messenger goes where it does).
IP3 is water soluble and is capable of reaching the organelles in the cytosol
DAG is hydrophobic and stays in the lipophilic regions in the plasma membrane
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