FCH 530 Homework 7
1. Glycolysis has the overall stoichiometry:
Glucose + 2ADP + 2Pi + 2NAD+  2 pyruvate + 2ATP + 2NADH + 2H+ + 2H2O
wheras that of gluconeogenesis (glucose synthesis) is:
2 pyruvate + 4ATP +2NADH + 2H+ + 4H2O  glucose + 4ADP + 4Pi + 2NAD+
What is the overall stoicheimetry of the glycolytic breakdown of 1 mol of glucose
followed by its gluconeogenic synthesis?
To balance the equations:
Glucose + 2ADP + 2Pi + 2NAD+  2 pyruvate + 2ATP + 2NADH + 2H+ + 2H2O
2 pyruvate + 2ATP 4ATP +2NADH + 2H+ + 2H2O 4H2O  glucose + 2ADP 4ADP
+ 2 Pi 4Pi + 2NAD+
2ATP+ 2H2O2ADP + 2 Pi
Explain why it is necessary that the pathways of these two processes be
independently controlled and why they must differ by at least one reaction.
These pathways must be independently controlled or the ATP formed in glycolysis
would be automatically used in gluconeogenesis thereby uselessly dissipating the
ATP formed. Conversely, any glucose formed by gluconeogenesis would be rapidly
broken down again by glycolysis. These pathways cannot be separately controlled
unless they differ in at least one reaction. This is because altering the activity of an
enzyme changes the rate of the forward reaction by the same proportion as that of
the back reaction. Moreover, since glycolysis is exergonic, at least one reaction of
gluconeogenesis must differ from that of glycolysis so that free energy can be
supplied to make gluconeogenesis also exergonic.
2. Consider the reaction catalyzed by hexokinase:
ATP + glucose  ADP + glucose-6-phosphate
A mixture containing 40 mM ATP and 20 mM glucose was incubated with hexokinase at
pH 7 and 25 °C. Calculate the equilibrium concentrations of the reactants and products
(see Table 16-3).
ATP + H2O  ADP + Pi : G0’ = -30.5 kJ mol-1
glucose-6-phosphate+ H2O  glucose + Pi : G0’ = -13.8 kJ mol-1
Subtracting the two reactions,
ATP + glucose  ADP + glucose-6-phosphate : G0’ = -30.5 kJ mol-1+13.8 kJ mol-1= 16.7 kJ mol-1
K’eq = [G6P][ADP] = e-Gº’/RT= e-x298 = 846
[G][ATP]
[ATP] + [ADP] = 40 mM = 0.040M
[G6P] +[G] = 20 mM = 0.020M
Stoichiometry
[G6P] = [ADP] = x
x2
(0.020-x)(0.040-x)
= 846
x2 = 846 x2 -50.8x +0.677
845 x2 -50.8x +0.677 = 0
Using the quadratic equation,
x = 50.8 + √50.82- 4  845  0.677
2  845
=
50.8+ 17.1
1690
x = 0.04018M or 0.01994M
The former root is physically unrealistic since it is larger than the maximum possible,
0.040M. Hence,
[ADP] = [G6P] = 0.01994M
[G]=0.0200-0.01994 = 0.00006M
[ATP}=0.04000-0.01994=0.02006M
3. Calculate ∆Go′ for the following pairs of half-reactions at pH 7 and 25 °C. Write a
balanced equation for the overall reaction and indicate the direction in which it
occurs spontaneously under standard conditions.
a. (H+/1/2 H2) and (1/2 O2 + 2H+/H2O)
1/2 O2 + 2H+ + 2 e-  H2O Eº’ =0.815V
2(1/2 H2H+ + e-)
Eº’ =0.421V
H2 +1/2 O2  H2O
Eº’ =1.235V
Gº’ = -nFEº’ = -2 X 96,494 J V-1 mol-1 X 1.235 V X 10-3 kJ J-1 =-238.3 kJ mol-1
The reaction as written proceeds to the right since Gº’ is negative.
b. (pyruvate- +2H+/lactate-) and (NAD+ + H+/NADH).
NAD+ + H+ + 2e-  NADH
Lactate  Pyruvate + 2 H+ + 2e-
Eº’ = -0.315V
Eº’ =+0.185V
NAD+ + Lactate  Pyruvate + NADH+ H+ Eº’ = -0.130V
Gº’ = -nFEº’ = -2 X 96,494 J V-1 mol-1 X (-0.130V) X 10-3 kJ J-1 =-25.1 kJ mol-1
The reaction as written proceeds to the right since Gº’ is negative.
4. Define the following terms as they relate to biochemistry.
a. Autotroph-organism that can use CO2 as their starting carbon source.
b. Heterotroph-organism that use preformed complex organic molecules as carbon
sources
c. Organotroph-organism that obtains H or e- from organic compounds.
d. Lithotroph-organism that uses an inorganic substrate to obtain reducing
equivalents.
e. Anabolism-biosynthetic pathways; metabolism that builds molecules, energy
requiring.
f. Catabolism-degradative pathways, usually energy yielding.
g. Amphibolic pathway-metabolic pathways that serve both catabolism and
anabolism.
h. Primary metabolism-metabolism used for the construction of essential building
blocks and energy metabolism.
i. Secondary metabolism-nonessential metabolism
j. Feedback inhibition-product of pathway down regulates activity of an early step
in the pathway.
k. Feedforward activation-metabolite produced early in the pathway activates
downstream enzyme activity.
5. What are 5 essential features for metabolism?
Metabolic pathways are irreversible. (large negative free energy change)
Catabolic and anabolic pathways must differ.
Every metabolic pathway has a first committed step.
All metabolic pathways are regulated.
Metabolic pathways in eukaryotic cells occur in specific cellular locations.
6. Write out a mechanism for the isomerization of G6P to F6P. Show all chemical
structures, electron flow and amino acids involved. (see below) What type of
reaction is this? Isomerization of aldose (G6P) to a ketose (F6P). Does it use a
general acid or base? Uses both. What are they? Lys acts as general acid and
imidazole of His acts as a base to remove the acidic proton from C2 of G6P.
7. Write out a mechanism for phosphofructokinase. Show all chemical structures,
electron flow and amino acids involved. See below. What type of reaction is this?
This reaction is a phosphorylation reaction, similar to hexokinase.
Does it use a general acid or base? No. What are they? Not applicable.
Mg2+
N
O
N
H2N
N
O - O- OP
P
P O
O
O
O
O
O
O
O
P
O
OH
OH
O
HO
HO
OH
HO
OH
N HO
adenosine triphosphate
N
N
H2N
N
N HO
O - OP
P O
O
O
O
O
O
-2O
HO
3PO
HO
O
P
O - OO
O
OH
OH
adenosine diphosphate
8. Write out a mechanism for aldolase. Show all chemical structures, electron flow
and amino acids involved. See below. What type of reaction is this? Aldol
cleavage (retro aldol condensation). Does it use a general acid or base? Uses a
Schiff base mechanism (covalent catalysis) and acid (aspartate) What are they?
Active site lysine and aspartic acid.
9. Write out a mechanism for TIM. Show all chemical structures, electron flow and
amino acids involved. What type of reaction is this? Isomerization. Does it use a
general acid or base? Yes. What are they? Glutamatic acid and Histidine Why
does this favor GAP formation over DHAP? GAP is rapidly removed for latter
steps in the pathway so that rapidly removed from this equilibrium.