Download Answer Key - Department of Chemistry ::: CALTECH

Fall 2016
Due November 22
Bi/Ch110 Problem Set 4
Problem 1: Overview of metabolism (33 points)
a. (12 points) Metabolic pathways often reuse reaction motifs to perform the
necessary transformations required. List six major reaction types and the
generalized outcome achieved by each reaction/their reactive purpose.
- Oxidation-reduction reactions: Derives energy from the oxidation of carbon
compounds.
- Ligation reactions: form bonds by using free energy from ATP cleavage;
necessary for anabolism.
- Isomerization reactions: rearrange atoms necessary for subsequent
reactions such as oxidation-reduction.
- Group transfer reactions: activate compounds for subsequent reactions or
modify compounds for regulatory purposes (trap glucose in the cell for
further catabolism)
- Hydrolytic reactions: Hydrolysis for the degradation of large molecules for
metabolism or to reuse some of the components for biosynthetic purposes
- Carbon-carbon bond cleavage: release of CO2 or H2O during metabolic
pathways that help drive reactions forward.
b. (6 points) To ensure proper homeostasis, describe three principle ways that
cells control metabolic processes.
- Control the amounts of enzymes: Occurs at the level of transcription
- Control catalytic activity: Allosteric control; feedback loops; covalent
modifications; energy levels of the cell
- Control the accessibility of substrates: Compartmentalization of substrates;
substrate sequestration
c. (5 points) Sometimes, the substrate (or sites on the substrate) for a
metabolic pathway are structurally similar: A sample of deuterated reduced
NAD was prepared by incubating CH3CD2OH and NAD+ with alcohol
dehydrogenase. This reduced coenzyme was added to a solution of 1,3-DPG
and glyceraldehyde 3-phosphate dehydrogenase. The NAD+ formed by this
second reaction contained one atom of deuterium, whereas the
glyceraldehyde 3-phosphate, the other product, contained none. What does
this experiment reveal about the nature of the two enzymes?
Fall 2016
Due November 22
The Alcohol Dehydrogenase and glyceraldehyde dehydrogenase have
opposite selectivities as to which H is taken from/replaced to NADH.
d. (5 points) Suppose you discovered a mutant yeast whose glycolytic pathway
was shorter because of the presence of a new enzyme catalyzing the reaction
glyceraldehyde-3-phosphate + NAD+  3-phosphoglycerate + NADH + H+.
Would shortening the glycolytic pathway in this way benefit the cell? Explain.
No. There would be no anaerobic productions of ATP; aerobic ATP
production would be diminished only slightly.
e. (5 points) By choosing glycolysis over the citric acid cycle, what tradeoffs do
cancer cells choose?
Cancer cells will produce less ATP but will produce many more important
metabolites such as nucleic acids, amino acids, and fatty acids for their
proliferation. Glycolysis also provides energy at a much quicker rate.
Problem 2: The Citric Acid Cycle (40 points)
a. (5 points) The output of glycolysis is pyruvate, but the citric acid cycle uses
acetyl CoA. Briefly explain how glycolysis and the citric acid cycle are linked.
The pyruvate dehydrogenase complex converts pyruvate into acetyl CoA by
the reaction given below:
Pyruvate + CoA + NAD+  acetyl CoA + CO2 + NADH + H+
b. (5 points) The citric acid cycle produces only one molecule of ATP per cycle,
yet it is key to producing the majority of energy used by the cell. Explain how
this is possible.
While the citric acid cycle itself does not produce much ATP, it also produces
NADH and FADH2 by reducing NAD+ and FAD. It does so by transferring 8
high-energy electrons from acetyl CoA to three molecules of NAD+ (6 e-) and
one molecule of FAD (2 e-). When NADH and FADH2 are then oxidized during
oxidative phosphorylation, those electrons then are used to produce nine
ATP.
c. (5 points) Explain how the citric acid cycle is a cycle.
The 4-carbon molecule oxaloacetate is a reactant and then a product of the
cycle such that there is no net gain or loss of oxaloacetate over one cycle. In
the first step of the citric acid cycle, oxaloacetate and the acetyl group of
Fall 2016
Due November 22
acetyl CoA are condensed to form a 6-carbon molecule. Over the citric acid
cycle, that 6-carbon molecule loses two carbons in the form of CO2 such that
the last reaction in the citric acid cycle, catalyzed by malate dehydrogenase,
regenerates oxaloacetate.
d. (15 points) The citric acid cycle is tightly regulated. Explain how each of the
following molecules affect the activity of isocitrate dehydrogenase and why
each of these outcomes are logical for the regulation of the citric acid cycle.
i. ADP
ADP is an allosteric activator of the enzyme. A high [ADP] in the cell
indicates the cell is energy-poor; therefore the activation of the citric
acid cycle to produce more energy is in the best interest of the cell.
ii.
ATP
ATP is an allosteric inhibitor of the enzyme. A high [ATP] in the cell
indicates that the cell is energy-rich; therefore, it is in the best interest
of the cell to shut down the citric acid cycle since no more energy is
required.
iii.
NADH
NADH is a product/competitive inhibitor of the enzyme. A high
[NADH] in the cell indicates that the cell is energy-rich; therefore, it is
in the best interest of the cell to shut down the citric acid cycle since
no more energy is required.
e.
(10 points) Pyruvate, labeled with 14C as shown below in red, was added to
cell extracts that contain functioning metabolic enzymes. Where does the 14C
label end up after one turn of the citric acid cycle? Show your work by tracing
the atom through the relevant metabolic reactions.
Fall 2016
Due November 22
After one turn of the citric acid cycle, the label ends up on the internal carbon atoms
of oxaloacetate. Because succinate is a symmetric molecule, you cannot trace the
label specifically after that step. Instead, 50% will be on one internal carbon
(carbonyl) while the other 50% of the label will be on the other internal carbon
(aliphatic).
Problem 3: Gluconeogenesis (27 points)
Fall 2016
Due November 22
a. (7 points) As you have learned, glycolysis has several irreversible steps. Yet,
gluconeogenesis is essentially a reversal of the process of glycolysis. How is
that possible, given that some of the reactions of glycolysis are so
energetically favorable?
While most of the reactions in gluconeogenesis are the reverse of those found
in glycolysis, the three irreversible steps, which are very energetically
favorable in the glycolysis direction, are replaced by more kinetically
favorable reactions. Hexokinase, phosphofructokinase, and pyruvate kinase
enzymes of glycolysis are replaced with glucose-6-phosphatase, fructose-1,6bisphosphatase, and phosphoenolpyruvate carboxykinase. These enzymes
are able to perform the same function as the original enzymes in a way that is
favorable during gluconeogenesis.
b. Consider the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which is
involved in gluconeogenesis, but not glycolysis.
i. (10 points) What is its role, metabolically speaking? Which hormones
regulate its expression and why?
PEPCK is an essential enzyme in gluconeogenesis which catalyzes the
conversion of oxaloacetate into phosphoenolpyruvate. It is one of the
enzymes that circumvents the irreversible steps of glycolysis. Its gene
expression is regulated by cAMP, insulin, cortisol and glucagon in
order to maintain glucose homeostasis.
ii.
(5 points) What would happen if PEPCK was overexpressed? Name a
disease that could occur as a result of this overexpression.
Increasing gene levels of PEPCK would result in more conversion of
oxaloacetate into phosphoenolpyruvate, which ultimately results in
more production of glucose. The increased level of glucose could
result in the development of Type II diabetes mellitus from the
persistent high circulating blood glucose levels.
iii.
(5 points) What would happen if there were a deficiency in PEPCK?
A lower level of PEPCK would result in less conversion of oxaloacetate
into phosphoenolpyruvate, which means that less production of
glucose would occur. This would result in hypoglycemia. Lactic
acidemia can result from the increased acid in the bloodstream, the
liver can become enlarged and impaired, and normal growth could be
inhibited.