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Answers to Problem Set 2
Lecture 10 (More kinetics)
1. a. In the absence of inhibitor, Vmax is about 44.6 µmol/minute and KM is 10 mM (1.0
x 10-5 M). In the presence of inhibitor, Vmax is about the same and the apparent KM is 20
mM (3.0 x 10-5 M).
b. Competitive. KI = KM[I]/(KM,apparent – KM) = 1.0 x 10-5[2 x 10-3]/(3.0 x 10-5 –1.0 x 10-5)
= 1.0 x 10-3 M.
2. a. Vmax is about 44.6 µmol/minute; Vmax,apparent is 8.9 µmol/minute. KM is 1.0 x 10-5, the
same as without inhibitor.
b. Non-competitive. KI = Vmax,apparent [I]/(Vmax – Vmax,apparent) = 8.9[100 x 10-6]/(44.6 –
8.9) = 2.5 x 10-5 M.
3. a. When [S+] is much greater than the value of KM, pH will have a negligible effect on
the enzyme because S+ will interact with E- as soon as the enzyme becomes available.
b. When [S+] is much less than the value of KM, the plot of Vo versus pH becomes
essentially a titration curve for the ionizable groups, with enzyme activity being the
titration marker. At low pH, the high concentration of H+ will keep the enzyme in the
EH form and inactive. As the pH rises, more and more of the enzyme will be in the Eform and active. At high pH (low H+), all of the enzyme is E-.
c. The midpoint on this curve will be the pKa of the ionizable group, which is stated to
be pH 6.
Lecture 11 (Mb, Hb, allostery, and motors)
1. a. In humans, 1.44 x 10-2 g (4.49 x 10-4 moles) of O2 per kilogram of muscle. In
sperm whale, 0.144 g (4.49 x 10-3 moles) of O2 per kilogram.
2. a. 2; b. 4; c. 2; d. 1
GG15-5. For n = 2.8, Ylungs = 0.98 and Ycapillaries = 0.77.
For n = 1.0, Ylungs = 0.79 and Ycapillaries = 0.61.
Thus, with an n of 2.8 and a P50 of 26 torr, hemoglobin becomes almost fully saturated
with O2 in the lungs and drops to 77% saturation in resting tissue, a change of 21%. If n
is 1.0 and P50 is 26 torr, hemoglobin would become only 79% saturated with O2 in the
lungs and would drop to 61% in resting tissue, a change of 18%. The difference in
hemoglobin O2 saturation condition between the values for n = 2.8 and 1.0, 3%
saturation, seems small, but note that the potential for O2 delivery (98% versus 79%) is
large and becomes crucial when pO2 in actively metabolizing tissue falls below 40 torr.
GG16-8. At 2000 nm/s and 8 nm/step, kinesin takes 250 steps/s. To cover 0.1 m would
require (0.1)/(8 x 10-9) = 12.5 x 106 steps, and this would take 5 x 104 s (833 min). At 1
ATP/step, the amount of ATP hydrolyzed is 12.5 x 106/6 x 1023 = 2 x 10-17 = 200 fmol.
The rate of hydrolysis is 250/6 x 1023 = 41 x 10-23 mol/s.
GG16-13. Energy (or work) = force x distance; rearranging, force = energy/distance. If
one ATP is hydrolyzed per step and the available energy of hydrolysis, ΔG’ is assumed to
be -50 kJ/mol (= 50,000/6 x 1023 = 8.3 x 10-20 J/molecule), then available force = 8.3 x
10-20 J/step size. Note that 1 J = 1 newton-meter.
For the kinesin-1 motor, force = 8.3 x 10-20 J/(8 x 10-9) m = 1 x 10-11 newtons
For the myosin-V motor, force = 8.3 x 10-20 J/(36 x 10-9) m = 2.2 x 10-12 newtons
For the dynein motor, force = 8.3 x 10-20 J/(28 x 10-9) m = 2.9 x 10-12 newtons
(The answer in the book for the kinesin-1 motor is 6.25 x 1012 newtons. What did they
do wrong?)
Lecture 12 (Carbohydrates)
1. Three amino acids can be linked by peptide bonds in only six different ways.
However, three different monosaccharides can be linked in a plethora of ways. The
monosaccharides can be linked in a linear or branched manner, with α or ß linkages, with
bonds between C-1 and C-3, between C-1 and C-4, between C-1 and C-6, and so forth.
There are 12,288 possible trisaccharides but only 6 different tripeptides.
2. a. ß-D-mannose. b. ß-D-galactose. c. ß-D-fructose. d. ß-D-glucosamine.
3. a. Not a reducing sugar; no open-chain forms are possible.
b. D-galactose, D-glucose, D-fructose.
c. D-galactose and sucrose (glucose + fructose)
GG7-4. a. The systematic name for trehalose is α-D-glucopyranosyl-(1-1)- α-Dglucopyranoside.
b. Not a reducing sugar. Both anomeric carbons are occupied in the disaccharide
linkage.
GG22-7. Glycogen molecules do not have any free reducing ends, regardless of the size
of the molecule. If branching occurs every 8 residues and each arm of the branch has 8
residues (or 16 per branch point), a glycogen molecule with 8000 residues would have
about 500 ends. If branching occurs every 12 residues, a glycogen molecule with 8000
residues would have about 334 ends.
Lecture 13 (Lipids)
1. a. n-dodecanate = laurate
b. cis-Δ9-hexadecenoate = palmitoleate
c. cis,cis- Δ9, Δ12-octadecadienoate = linoleate
2. Triacylglycerols from plants may have many cis double bonds or have shorter fatty
acid chains than those from animals.
3. The C16 fatty acid is attached by an ether linkage. The C-2 carbon atom of glycerol
has only an acetyl group attached by an ester linkage instead of a fatty acid, as is the case
with most phospholipids.
4. Instead of forming a soluble micelle that will be washed down the drain, the
magnesium or calcium salts will precipitate, forming a scum-like bathtub ring (or, if you
shower, scum on you).
GG8-1. a. Because the question specifically asks for triacylglycerols that contain stearic
acid and arachidonic acid, we can discount the triacylglycerols that contain only stearic or
only arachidonic acid. In this case, there are six possibilities:
S_S_A
S _A_S
A_S_S
S_A_A
A_S_A
A_A_S
(TMM Note: it is not clear how the first and third structures can be distinguished; same
for the fourth and sixth. If they cannot be, then there are four possibilities.)
b. Using the same logic, there are two phosphatidylserine isomers that can be formed
from palmitic and linolenic acids: P L PS and L P PS
GG8-3. a. Phosphatidylethanolamine and phosphatidylserine have a net positive charge
at low pH.
b. Phosphatidic acid, phosphatidylglycerol, phosphatidylinositol, phosphatidylserine, and
diphosphatidylglycerol normally carry a net negative charge.
c. Phosphatidylethanolamine and phosphatidylcholine carry a net zero charge at neutral
pH.
Lecture 14 (Membrane proteins)
1. Databases could be searched for proteins with stretches of 20 hydrophobic amino
acids. This is analogous to using a hydropathy plot (see Fig. 9.14 in GG 4th Edn).
2. An ion channel must transport ions in either direction at the same rate. The net flow
of ions is determined only by the composition of the solutions on either side of the
membrane.
GG9-4. r = (4Dt)1/2. According to this equation, a phospholipid with D = 1 x 10-8 cm2
will move approximately 200 nm in 10 msec.
GG9-5. a. Fibronectin: for t = 10 msec, r = 1.67 x 10-7 cm = 1.67 nm.
Rhodopsin: r = 110 nm
b. All else being equal, the value of D is roughly proportional to (Mr)-1/3. Molecular
weights of rhodopsin and fibronectin are 40,000 and 460,000, respectively. The ratio of
diffusion coefficients in thus expected to be (40,000)-1/3 / (460,000)-1/3 = 2.3. On the other
hand, the values given for rhodopsin and fibronectin give an actual ratio of 4286. The
explanation is that fibronectin is anchored in the membrane via interactions with
cytoskeletal proteins, and its diffusion is severely restricted compared with that of
rhodopsin.
GG9-9. Plot [S]/v for each value of [S]. This Haynes-Woolf plot makes it clear that
[S]/v should be constant for all [S] for the case of passive diffusion. In the present case,
[S]/v is a constant value of 0.0588 (L/min)-1.
Lecture 15 (Introduction to metabolism)
1. a. ATP + creatine <==> creatine-phosphate + ADP (to the left)
b. ATP + glycerol <==> glycerol-3-phosphate + ADP (to the right)
c. ATP + pyruvate <==> phosphoenolpyruvate + ADP (to the left)
d. ATP + glucose <==> glucose-6-phosphate + ADP (to the right)
2. a. ΔGo’ = 31.4 kJ/mol; and Keq’ = 3.06 x 10-6.
b. 3.28 x 104.
3. Under standard conditions, ΔGo’ = -RT ln [products]/[reactants]. Substituting 23.8
kJ/mol for ΔGo’ and solving for [products]/[reactants] yields 7 x 10-5. In other words the
forward reactions does not take place to a significant extent. Under intracellular
conditions, ΔG is -1.3 kJ/mol. Using the equation ΔG = ΔGo’ + RT ln
[products]/[reactants] and solving for [products]/[reactants] gives a ratio of 3.7 x 10-5.
Thus, a reaction that is endergonic under standard conditions can be converted into an
exergonic reaction by maintaining the [products]/[reactants] ratio below the equilibrium
value. This conversion is usually attained by using the products in another coupled
reaction as soon as they are formed.
4. Liver: -45.2 kJ/mol; muscle: -48.1 kJ/mol; brain: -48.5 kJ/mol. The free energy of
hydrolysis is most negative in the brain.
GG17-7 (modified). See the chapter. ATP, NAD+, and NADPH all can promote
reactions for which they are substrates. ATP is a phosphate-transfer agent; NAD+ is an
oxidizing agent; NADH and NADPH are reducing agents. In general, in eukaryotic
cytoplasm, NAD+/NADH is maintained in a mostly oxidized state, while
NADP+/NADPH is maintained in a reduced state.
Lecture 16 (Glycolysis)
1. a. 3 ATP; b. 2 ATP; c. 2 ATP; d. 2 ATP; e. 4 ATP.
GG18-1. a. Phosphoglucoisomerase, fructose bisphosphate aldolase, triose phosphate
isomerase, glyceraldehyde-3-P dehydrogenase, phosphoglycerate mutase, and enolase.
b. Hexokinase/glucokinase, phsophofructokinase, phosphoglycerate kinase, pyruvate
kinase, and lactate dehydrogenase..
c. Hexokinase and phosphofructokinase.
d. Phosphoglycerate kinase and pyruvate kinase.
e. Reactions in which the number of reactant molecules differs from the number of
product molecules exhibit a strong dependence on concentration. Using this criterion, we
can predict that the free energy changes of the fructose bisphosphate aldolase and
glyceraldehyde-3-P dehydrogenase reactions will be strongly influenced by changes in
concentration.
f. Reactions that occur with ΔG near zero operate at or near equilibrium.
GG18-11. a. 7.5.
b. 0.0266 mM.
GG18-14. a. -13.8 kJ/mol
b. Keq = 211.
c. [FBP]/[fructose-6-P] = 528.
Lecture 17 (More glycolysis)
1. Without triose isomerase, only one of the two three-carbon molecules generated by
aldolase could be used to generate ATP. Only two molecules of ATP would results from
the metabolism of each glucose. But two molecules of ATP would still be required to
form fructose-1,6-bisphosphate, the substrate for aldolase. The net yield of ATP would
be zero, a yield incompatible with life.
2. a. The 14C is in the methyl carbon atom of pyruvate.
b. 5 mCi/mmol. The specific activity is halved because the number of moles of product
(pyruvate) is twice that of the labeled substrate (glucose).
GG18-3. Increased [ATP], [citrate], or [glucose-6-phosphate] inhibits glycolysis.
Increased [AMP], [fructose-1,6-bisphosphate], and [fructose-2,6-bisphosphate]
stimulates glycolysis.
GG18-8. Ignoring the possibility that 32P might be incorporated into ATP, the only
reaction of glycolysis that utilizes Pi is glyceraldehyde-3-P dehydrogenase, which
converts glyceraldehyde-3-P to 1,3-bisphosphoglycerate. The label will be lost in the
next reaction, and no other glycolytic intermediates will be directly labeled. (Once the
label is incorporated into ATP, it will also show up in glucose-6-P, fructose-6-P, and
fructose-1,6-bisphosphate.)
GG18-15 a. -19.1kJ/mol
b. Keq = 1651
c. [ATP]/[ADP] = 13.8.