Download Book Problems Chapter 2

Book Problems Chapter 2 1.
Identify the potential hydrogen bond donors and acceptors in the following molecules:
(a)
(b)
(c)
2.
Occasionally, a C−H group can form a hydrogen bond. Why would such a group be more likely to be a hydrogen
bond donor group when the C is next to N?
8.
Draw the structures of the conjugate bases of the following acids:
a.
b.
c.
d.
8.
Draw the structures of the conjugate bases of the following acids:
a.
b.
c.
d.
10.
2 mL solutiion of pure water to whichh has been addded 50 mL off 1 mM HCl.
Calculate the pH of a 200
14.
w
acid HA
A if a solutionn containing 0.1
0 M HA andd 0.2 M A− haas a pH of 6.55?
What is thhe pK of the weak
15.
How manny grams of so
odium succinaate (formula weight
w
140 g · mol−1) and disodium succcinate (formuula weight
162 g · mool−1) must be added to 1 L of water to produce
p
a soluution with pH
H 6.0 and a tottal solute concentration of
50 mM?
16.
t volume off a solution of 5 M NaOH that must be added to adjuust the pH froom 4 to 9 in 1000 mL of a
Estimate the
100 mM solution
s
of ph
hosphoric acidd.
17.
(a) Wouldd phosphoric acid or succinnic acid be a better
b
buffer at
a pH 5? (b) Would
W
ammoonia or piperiddine be a
better bufffer at pH 9? (c)
( Would HE
EPES or Tris be
b a better buuffer at pH 7.55?
Chaapter 2 Answers:
A
:
1.
(a)
D
Donors:
NH1, NH2 at C2, NH9;
N
acceptorrs: N3, O at C6,
C N7.
(b)
D
Donors:
NHl, NH
N 2 at C4; accceptors: O att C2, N3.
(c)
D
Donors:
group, OH group;
g
acceptoors: COO− grroup, OH grouup.
2.
A protonaated (and therrefore positiveely charged) nitrogen
n
would promote thhe separation of charge in the
t adjacent
C−H bondd so that the C would havee a partial neggative charge and the H woould have a paartial positivee charge. Thiss
would maake the H morre likely to bee donated to a hydrogen boond acceptor group.
g
8.
(a)
(b)
(c)
(d)
10.
ue to the addiition of HCl iss (50 mL) (l mM)/(250
m
mL
L) = 0.2 mM = 2 × 10−4 M. Because thee
The increaase in [H+] du
− M, is relativ
− ) or 3.7.
[H+] of puure water, 10−7
vely insignificant, the pH of
o the solutionn is equal to −log
−
(2 × 10−4
14.
Use the Henderson–Ha
H
asselbalch equuation (Eq. 2--9) and solve for pK:
15.
Let HA = sodium succinate and A− = disodium succinate.
HA] = 0.05 M,, so [A−] = 0.005 M − [HA]
[A−] + [H
From Eq. 2-9 and Tablle 2-4,
H − pK = 6.0 − 5.64 = 0.366
log ([A−]/[HA]) = pH
[A−]/[H
HA] = antilog 0.36 = 2.29
(0.05 M − [HA])/[HA
A] = 2.29
[HA] = 0.015 M
0
M − 0.01
15 M = 0.0355 M
[A−] = 0.05
grams of
o sodium succcinate = (0.015 mol ⋅ L−1)((140 g ⋅ mol−1) × (1 L) = 2.1
2 g
− )(162 g ⋅ mo
grams of
o disodium su
uccinate = (0.035 mol ⋅ L−1
ol−1) × (1 L) = 5.7 g
16.
At pH 4, essentially
e
alll the phosphooric acid is in the
form, and at pH 9, essentiially all is in the
t
−
form (Fig. 2-18). Thereefore, the conncentration off OH required is equivalennt to the conccentration of the
t acid: (0.1
mol ⋅ L−1 phosphoric accid)(0.1 L) = 0.01 mol NaO
OH required = (0.01 mol)((1 L/5 mol NaaOH) = 0.0022 L = 2 mL
17.
(a)
Suuccinic acid; (b)
mmonia;
am
(c)
PES.
HEP
Chaapter 4
1.
Identify the
t amino accids that difffer from eachh other by a single methyyl or methyllene group.
3.
Identify the
t hydrogen
n bond donoor and accepttor groups inn asparagine.
4.
Draw thee dipeptide Asp-His
A
at pH
H 7.0.
6.
Determinne the net ch
harge of the predominant
p
t form of Aspp at (a) pH 1.0,
1 (b) pH 3.0,
3 (c) pH 6.0, and (d)
pH 11.0.
7.
H and (c) Glu.
G
Calculatee the pI of (aa) Ala, (b) His,
9.
Circle thee chiral carb
bons in the foollowing com
mpounds:
13.
Some am
mino acids arre synthesizeed by replaciing the keto group (C=O
O) of an orgaanic acid knoown as an αketo acidd with an am
mino group (
from the following α-keto
α
acids:
). Identify thee amino acidds that can bee produced this way
14.
t amino accid residue from
fr
which the
t followingg compoundds are synthesized:
Identify the
15.
Draw thee peptide AT
TLDAK. (a) Calculate itss approximaate pI. (b) Whhat is its net charge at pH
H 7.0?
16.
The proteein insulin consists of tw
wo polypeptiides termed the
t A and B chains. Insuulins from diifferent
organism
ms have been
n isolated andd sequencedd. Human andd duck insulins have thee same aminoo acid
sequencee with the ex
xception of six amino aciid residues, as
a shown below. Is the pI
p of human insulin
lower thaan or higher than that of duck insulinn?
Amino acid
d residue A8
A9
A10
B1
B2
B
B27
Human
Thhr
Ser
Ile
Phe
Val
T
Thr
Duck
Gluu
Asn
Pro
Ala
Ala
S
Ser
Chapter 4 Answers:
1.
Ser and Thr; Val, Leu, and Ile; Asn and Gln; Asp and Glu.
3.
Hydrogen bond donors: α-amino group, amide nitrogen. Hydrogen bond acceptors: α-carboxylate group,
amide carbonyl.
4.
6.
(a)
+1;
(b)
0;
(c)
−1;
7.
(a)
pI = (2.35 + 9.87)/2 = 6.11
(b)
pI = (6.04 + 9.33)/2 = 7.68
(c)
pI = (2.10 + 4.07)/2 = 3.08
13.
(a)
Glutamate;
14.
(a)
Serine (N-acetylserine);
(c)
methionine (N-formylmethionine).
(d)
−2.
9.
(b)
aspartate
(b)
lysine (5-hydroxylysine);
15.
(a)
The pK’s of the ionizable side chains (Table 4-1) are 3.90 (Asp) and 10.54 (Lys); assume that the
terminal Lys carboxyl group has a pK of 3.5 and the terminal Ala amino group has a pK of 8.0 (Section
4-1D). Thhe pI is apprroximately midway
m
betw
ween the pK’’s of the twoo ionizations involving thhe neutral
species (tthe pK of Assp and the N-terminal
N
pK
K):
(b)
16.
T net charg
The
ge at pH 7.0 is 0 (as draw
wn above).
At positioon A8, duck
k insulin has a Glu residuue, whereas human
h
insullin has a Thrr residue. Sinnce Glu is
negativelly charged att physiological pH and Thr
T is neutraal, human inssulin has a higher
h
pI thann duck
insulin. (The
(
other am
mino acids thhat differ beetween the prroteins do noot affect the pI because they
t
are
unchargeed.)
Chaapter 5
1.
Which peeptide has grreater absorbbance at 2800 nm?
A.
G
Gln–Leu–Glu
u–Phe–Thr–L
Leu–Asp–G
Gly–Tyr
B.
Ser–Val–Trp–
–Asp–Phe–G
Gly–Tyr–Trpp–Ala
2.
(a) In whhat order wou
uld the aminno acids Arg, His, and Leeu be eluted from a carbboxymethyl column
c
at
pH 6? (b)) In what ord
der would Glu,
G Lys, andd Val be eluted from a diiethylaminoeethyl columnn at pH 8?
4.
Determinne the subun
nit compositiion of a proteein from the following innformation:
Molecuular mass by
y gel filtration: 200 kD
Molecuular mass by
y SDS-PAGE
E: 100 kD
Molecuular mass by
y SDS-PAGE
E with 2-merrcaptoethanool:
40 kD and
a 60 kD
6.
What fractionation prrocedure couuld be used to
t purify prootein 1 from a mixture off three proteeins whose
f
amino accid composittions are as follows?
1.
25% Ala, 20%
% Gly, 20% Ser, 10% Ilee, 10% Val, 5% Asn, 5%
% Gln, 5% Prro
2.
30% Gln, 25%
% Glu, 20% Lys, 15% Ser,
S 10% Cyss
3.
% Gly, 20% Asp, 20% Ser,
S 10% Lys, 5% Tyr
25% Asn, 20%
All three proteins aree similar in size
s and pI, and
a there is no
n antibody available foor protein 1.
9.
You must cleave
c
the fo
ollowing pepptide into sm
maller fragmeents. Which of the proteaases listed inn Table 5-3
would be liikely to yield the most fragments?
fr
T fewest?
The
NMTQGRCKPVNTFVHEPLVDVQNVCFKE
11.
Separate cleavage reactions of a polypeptide by CNBr and chymotrypsin yield fragments with the
following amino acid sequences. What is the sequence of the intact polypeptide?
CNBr treatment
1.
Arg–Ala–Tyr–Gly–Asn
2.
Leu–Phe–Met
3.
Asp–Met
6.
Gly–Asn
Chymotrypsin
4.
13.
Met–Arg–Ala–Tyr
5.
Asp–Met–Leu–Phe
Treatment of a polypeptide with 2-mercaptoethanol yields two polypeptides:
1.
Ala-Val-Cys-Arg-Thr-Gly-Cys-Lys-Asn-Phe-Leu
2.
Tyr-Lys-Cys-Phe-Arg-His-Thr-Lys-Cys-Ser
Treatment of the intact polypeptide with trypsin yields fragments with the following amino acid
compositions:
3.
(Ala, Arg, Cys2, Ser, Val)
4.
(Arg, Cys2, Gly, Lys, Thr, Phe)
5.
(Asn, Leu, Phe)
6.
(His, Lys, Thr)
7.
(Lys, Tyr)
Indicate the positions of the disulfide bonds in the intact polypeptide
Chapter 5 Answers
1.
Peptide B, because it contains more Trp and other aromatic residues.
2.
(a) Leu, His, Arg. (b) Lys, Val, Glu.
4.
The protein contains two 60-kD polypeptides and two 40-kD polypeptides. Each 40-kD chain is
disulfide bonded to a 60-kD chain. The 100-kD units associate noncovalently to form a protein with a
molecular mass of 200 kD.
6.
Because protein 1 has a greater proportion of hydrophobic residues (Ala, Ile, Pro, Val) than do proteins
2 and 3, hydrophobic interaction chromatography could be used to isolate it.
9.
Thermolysin would yield the most fragments (9) and endopeptidase V8 would yield the fewest (2).
11.
Asp–Met–Leu–Phe–Met–Arg–Ala–Tyr–Gly–Asn
13.
Chapter 6
5.
Globular proteins are typically constructed from several layers of secondary structure, with a
hydrophobic core and a hydrophilic surface. Is this true for a fibrous protein such as α keratin?
9.
Is it possible for a native protein to be entirely irregular, that is, without α helices, β sheets, or other
repetitive secondary structure?
12.
You are performing site-directed mutagenesis to test predictions about which residues are essential for a
protein’s function. Which of each pair of amino acid substitutions listed below would you expect to
disrupt protein structure the most?
Explain.
(a)
Val replaced by Ala or Phe.
(b)
Lys replaced by Asp or Arg.
(c)
Gln replaced by Glu or Asn.
(d)
Pro replaced by His or Gly.
15.
Describe the intra- and intermolecular bonds/interactions that are broken or retained when collagen is
heated to produce gelatin.
16.
Under physiological conditions, polylysine assumes a random coil conformation. Under what conditions
might it form an α helix?
18.
Which of the following polypeptides is most likely to form an α helix? Which is least likely to form a β
strand?
(a)
CRAGNRKIVLETY
(c)
QKASVEMAVRNSG
(b)
SEDNFGAPKSILW
Chapter 6 Answers
5.
A fibrous protein such as α keratin does not have a discrete globular core. Most of the residues in its
coiled coil structure are exposed to the solvent. The exception is the strip of nonpolar side chains at the
interface of the two coils.
9.
Yes, although such irregularity should not be construed as random.
12.
(a)
Phe. Ala and Phe are both hydrophobic, but Phe is much larger and might not fit as well in Val’s
place.
(b) Asp. Replacing a positively charged Lys residue with an oppositely charged Asp residue would
be more disruptive.
(c)
Glu. The amide-containing Asn would be a better substitute for Gln than the acidic Glu.
(d) His. Pro’s constrained geometry is best approximated by Gly, which lacks a side chain, rather
than a residue with a bulkier side chain such as His.
15.
Hydrophobic effects, van der Waals interactions, and hydrogen bonds are destroyed during denaturation.
Covalent cross-links are retained.
16.
At physiological pH, the positively charged Lys side chains repel each other. Increasing the pH above
the pK (>10.5) would neutralize the side chains and allow an α helix to form.
18.
Peptide c is most likely to form an α helix with its three charged residues (Lys, Glu, and Arg) aligned on
one face of the helix. Peptide a has adjacent basic residues (Arg and Lys), which would destabilize a
helix. Peptide b contains Gly and Pro, both of which are helix-breaking (Table 6-1). The presence of Gly
and Pro would also inhibit the formation of β strands, so peptide b is least likely to form a β strand.
Chapter 9
4.
Draw the structure of a glycerophospholipid that has a saturated C16 fatty acyl group at position 1, a
monounsaturated C18 fatty acyl group at position 2, and an ethanolamine head group.
13.
When bacteria growing at 20°C are warmed to 30°C, are they more likely to synthesize membrane lipids
with (a) saturated or unsaturated fatty acids, and (b) short-chain or long-chain fatty acids? Explain.
14.
(a) How many turns of an α helix are required to span a lipid bilayer (∼30 Å across)? (b) What is the
minimum number of residues required? (c) Why do most transmembrane helices contain more than the
minimum number of residues?
Chapter 9 Answers
4.
13.
(a) Saturated; (b) long-chain. By increasing the proportion of saturated and long-chain fatty acids, which
have higher melting points, the bacteria can maintain constant membrane fluidity at the higher
temperature.
14.
(a)
(1 turn/5.4 Å)(30 Å) = 5.6 turns
(b)
(3.6 residues/turn)(5.6 turns) = 20 residues
(c)
The additional residues form a helix, which partially satisfies backbone hydrogen bonding
requirements, where the lipid head groups do not offer hydrogen bonding partners.
Chapter 10
2.
Rank the rate of transmembrane diffusion of the following compounds:
3.
Calculate the free energy change for glucose entry into cells when the extracellular concentration is 5
mM and the intracellular concentration is 3 mM.
4.
(a) Calculate the chemical potential difference when intracellular [Na+] = 10 mM and extracellular [Na+]
= 150 mM at 37°C. (b) What would the electrochemical potential be if the membrane potential were −60
mV (inside negative)?
8.
The rate of movement (flux) of a substance X into cells was measured at different concentrations of X to
construct the graph below.
(a) Does this information suggest that the movement of X into the cells is mediated by a protein
transporter? Explain.
(b) What additional experiment could you perform to verify that a transport protein is or is not
involved?
10.
Endothelial cells and pericytes in the retina of the eye have different mechanisms for glucose uptake.
The figure shows the rate of glucose uptake for each type of cell in the presence of increasing amounts
of sodium. What do these results reveal about the glucose transporter in each cell type?
11.
T compoun
The
nd shown below is the anntiparasitic drug
d
miltefosine.
(a)
Iss this compo
ound a glycerrophospholippid?
(b)
H does miiltefosine likkely cross thee parasite cell membranee?
How
(c)
Inn what part of
o the cell would the druug tend to acccumulate? Explain.
E
(d) Miltefosine
M
binds
b
to a prootein that alsso binds som
me sphingolippids and som
me
glyceroophospholipiids. What feature common to all thesse compounds is recognnized by the protein?
p
Thee
proteinn does not bin
nd triacylglyycerols.
15.
Proteins known as multidrug
m
resiistance (MD
DR) transportters use the free
f energy of
o ATP hydrrolysis to
pump a variety
v
of hy
ydrophobic suubstances ouut of cells. (aa) Write the reaction for ATP hydrollysis and
explain, in
i structural terms, how the transporrter might takke advantage of this proocess to drivee transport.
There is no
n phosphorrylated proteein intermediate, as in thhe (Na+–K+)––ATPase. (b) Why would
overexprression of an
n MDR transpporter in a cancer
c
cell make
m
the canccer more diffficult to treaat?
Chaapter 10 Answerss
2.
3.
The less polar a subsstance, the faaster it can diffuse
d
througgh the lipid bilayer.
b
From
m slowest too fastest: C,
A, B.
4.
(a)
(b)
8.
T data do not
The
n indicate the
t involvem
ment of a trannsport proteiin, since the rate of transsport does
(a)
not approoach a maxim
mum as [X] increases.
(b)
T verify thatt a transport protein is innvolved, incrrease [X] to demonstratee saturation of
To
o the
transportter at high [X
X], or add a structural
s
annalog of X too compete with X for binnding to the transporter,
t
resulting in a lower flux
f
of X.
10.
The hypeerbolic curvee for glucosee transport innto pericytess indicates a protein-meddiated sodium
m-dependennt
+
process. The transporrt protein haas binding sittes for sodiuum ions. At low
l [Na ], glucose transpport is
+
+
+
directly proportional
p
to [Na ]. Hoowever, at high
h
[Na ], alll Na bindinng sites on thhe transport protein are
occupiedd, and thus gllucose transpport reaches a maximum
m velocity. Glucose
G
transsport into endothelial
+
cells is not sodium-d
dependent annd occurs at a high rate whether
w
or noot Na is preesent. There is not
enough innformation in
i the figure to determinne whether gllucose transpport into enddothelial cellls is protein-mediatedd.
11.
(a)
N there is no
No;
n glycerol backbone.
b
(b)
M
Miltefosine
iss amphipathiic and thereffore cannot cross
c
the parrasite cell meembrane by diffusion.
Since it is
i not a norm
mal cell compponent, it proobably does not have a dedicated
d
acctive transporrter. It most
likely entters the cell via a passivee transport protein.
p
(c)
T amphipaathic molecuule most likely accumulaates in membbranes, with its hydrophoobic tail
This
buried inn the bilayer and its polarr head groupp exposed to the solvent.
(d)
T protein reecognizes thhe phosphochholine head group, which also occurrs in some spphingolipids
The
and somee glyceropho
ospholipids. Since the prrotein does not
n bind all phospholipid
p
ds or triacylgglycerols, it
does not recognize th
he hydrocarbbon tail.
15.
(a) ATP + H2O → ADP + Pi
The transporter must include a cytosolic nucleotide binding site that changes its conformation when its
bound ATP is hydrolyzed to ADP. This conformational change must be communicated to the
membrane-spanning portion of the protein, where the transported substrate binds. (b) Overexpression of
an MDR transporter would increase the ability of the cancer cell to excrete anticancer drugs. Higher
concentrations of the drugs or different drugs would then be required to kill the drug-resistant cells.
Chapter 11
7.
Studies at different pH’s show that an enzyme has two catalytically important residues whose pK’s are
∼4 and ∼10. Chemical modification experiments indicate that a Glu and a Lys residue are essential for
activity. Match the residues to their pK’s and explain whether they are likely to act as acid or base
catalysts.
18.
Predict the effect of mutating Asp 102 of trypsin to Asn (a) on substrate binding and (b) on catalysis.
20.
Why is the broad substrate specificity of chymotrypsin advantageous in vivo? Why would this be a
disadvantage for some other proteases?
Chapter 11 Answers
7.
Glu has a pK of ∼4 and, in its ionized form, acts as a base catalyst. Lys has a pK of ∼10 and, in its
protonated form, acts as an acid catalyst.
18.
(a)
Little or no effect;
(b)
catalysis would be much slower because the mutation disrupts the function of the catalytic triad.
20.
As a digestive enzyme, chymotrypsin’s function is to indiscriminately degrade a wide variety of
ingested proteins, so that their component amino acids can be recovered. Broad substrate specificity
would be dangerous for a protease that functions outside of the digestive system, since it might degrade
proteins other than its intended target.
Chapter 12
6.
Explain why it is usually easier to calculate an enzyme’s reaction velocity from the rate of appearance of
product rather than the rate of disappearance of a substrate.
7.
At what concentration of S (expressed as a multiple of KM) will vo = 0.95 Vmax?
10.
Calculate KM and Vmax from the following data:
[S] (μM)
vo (mM · s−1)
0.1
0.34
0.2
0.53
0.4
0.74
0.8
0.91
1.6
1.04
13.
You are constructing a velocity versus [substrate] curve for an enzyme whose KM is believed to be about
2 μM. The enzyme concentration is 200 nM and the substrate concentrations range from 0.1 μM to 10
μM. What is wrong with this experimental setup and how could you fix it?
15.
Is it necessary to know [E]T in order to determine (a) KM, (b) Vmax, or (c) kcat?
16.
The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for
the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has
the higher apparent affinity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?
19.
Determine the type of inhibition of an enzymatic reaction from the following data collected in the
presence and absence of the inhibitor.
[S] (mM)
vo (mM · min−1)
vo with I present (mM · min−1)
1
1.3
0.8
2
2.0
1.2
4
2.8
1.7
8
3.6
2.2
12
4.0
2.4
25.
Sphingosine-1-phosphate (SPP) is important for cell survival. The synthesis of SPP from sphingosine
and ATP is catalyzed by the enzyme sphingosine kinase. An understanding of the kinetics of the sphingosine
kinase reaction may be important in the development of drugs to treat cancer. The velocity of the sphingosine
kinase reaction was measured in the presence and absence of threo-sphingosine, a stereoisomer of sphingosine
that inhibits the enzyme. The results are shown below.
[Sphingosine] (μM)
vo (mg · min−1) (no inhibitor)
vo (mg · min−1) (with threo-sphingosine)
2.5
32.3
8.5
3.5
40
11.5
5
50.8
14.6
10 72
25.4
20
87.7
43.9
50
115.4
70.8
Construct a Lineweaver–Burk plot to answer the following questions:
(a)
What are the apparent KM and Vmax values in the presence and absence of the inhibitor?
(b)
What kind of an inhibitor is threo-sphingosine? Explain.
Chapter 12 Answers
6.
Enzyme activity is measured as an initial reaction velocity, the velocity before much substrate has been
depleted and before much product has been generated. It is easier to measure the appearance of a small
amount of product from a baseline of zero product than to measure the disappearance of a small amount
of substrate against a background of a high concentration of substrate.
7.
vo = Vmax[S]/(KM + [S])
vo/Vmax = [S]/(KM + [S])
0.95 = [S]/(KM + [S])
[S] = 0.95KM + 0.95[S]
0.05[S] = 0.95 KM
[S] = (0.95/0.05)KM = 19KM
10.
Construct a Lineweaver–Burk plot.
KM = −1/x-intercept = −1/(−4 μM−1) = 0.25 μM Vmax = 1/y-intercept = 1/(0.8 mM−1 ⋅ s) = 1.25 mM ⋅ s−1
13.
The enzyme concentration is comparable to the lowest substrate concentration and therefore does not
meet the requirement that [E] ≪ [S]. You could fix this problem by decreasing the amount of enzyme
used for each measurement.
15.
(a, b) It is not necessary to know [E]T. The only variables required to determine KM and Vmax (for
example, by constructing a Lineweaver–Burk plot) are [S] and vo. (c) The value of [E]T is required to
calculate kcat since kcat = Vmax/[E]T.
16.
(a) N-Acetyltyrosine ethyl ester, with the lower value of KM, has greater apparent affinity for
chymotrypsin. (b) The value of Vmax is not related to the value of KM, so no conclusion can be drawn.
19.
The lines of the double-reciprocal plots intersect to the left of the 1/vo axis (on the 1/[S] axis). Hence,
inhibition is mixed (with α = α′).
[S]
1/[S]
1/vo
1/vo with I
1
1.00
0.7692
1.2500
2
0.50
0.5000
0.8333
4
0.25
0.3571
0.5882
8
0.125
0.2778
0.4545
12
0.083
0.2500
0.4167
25.
(a)
KM is determiined from thhe x-interceptt (= −1/KM). In the absennce of inhibiitor, KM = 1//0.14 μM−1 =
7 μM. In the presence of inhibitoor,
. Vmax is
i determinedd from the yy−1
intercept (= 1/Vmax). In
I the absennce of inhibittor, Vmax = 1//0.008 mg ⋅ min = 1255 mg ⋅ min−1. In the
presence of inhibitor,
.
(b)
T lines in th
The
he double-reeciprocal ploots intersect very
v
close too the 1/vo axxis. Hence, thhreosphingossine is most likely
l
a com
mpetitive inhiibitor. Comppetitive inhibbition is likelly also becauuse of the
structural similarity between
b
the inhibitor and the substraate, which alllows them to
t compete for
f binding
to the enzzyme active site.
Chaapter 13
2.
A certainn metabolic reaction
r
takees the form A → B. Its sttandard free energy channge is 7.5 kJ · mol−1. (a)
Calculatee the equilibrium constannt for the reaaction at 25°°C. (b) Calcuulate ∆G at 37°C
3
when the
t
concentraation of A iss 0.5 mM and the concenntration of B is 0.1 mM. Is the reactiion spontaneeous under
these connditions? (c) How might the reactionn proceed in the cell?
3.
Choose the
t best defin
nition for a near-equilibr
n
rium reactionn:
7.
a.
allways operattes with a faavorable freee energy channge.
b.
has a free eneergy change near zero.
c.
iss usually a co
ontrol point in a metabolic pathway..
d.
opperates very
y slowly in viivo.
Predict whether
w
creattine kinase will
w operate in
i the directiion of ATP synthesis
s
or phosphocreatine
synthesiss at 25°C wh
hen [ATP] = 4 mM, [AD
DP] = 0.15 mM,
m [phosphoocreatine] = 2.5 mM, annd [creatine]
= 1 mM.
8.
If intraceellular [ATP]] = 5 mM, [A
ADP] = 0.5 mM,
m and [Pi] = 1.0 mM,, calculate thhe concentraation of AMP
P
at pH 7 and
a 25°C und
der the conddition that thee adenylate kinase
k
reactiion is at equuilibrium.
13.
A hypothhetical three--step metaboolic pathwayy consists of intermediatees W, X, Y, and Z and enzymes
e
A,
B, and C. Deduce thee order of the enzymatic steps in the pathway froom the follow
wing inform
mation:
14.
1.
C
Compound
Q,
Q a metaboliic inhibitor of
o enzyme B, causes Z too build up.
2.
A mutant in enzyme
e
C requires Y forr growth.
3.
A inhibitor of
An
o enzyme A causes W, Y, and Z to accumulate..
4.
C
Compound
P, a metabolicc inhibitor of enzyme C,, causes W and
a Z to build up.
A certainn metabolic pathway
p
cann be diagram
med as
where A,, B, C, and D are the inteermediates, and
a X, Y, annd Z are the enzymes thaat catalyze thhe reactions..
The physsiological freee energy chhanges for thhe reactions are
a
(a) Which reaction iss likely to bee a major reggulatory poinnt for the patthway? (b) Iff your answeer in Part a
was in faact the case, in the presennce of an inhhibitor that blocks
b
the acctivity of enzzyme Z, wouuld the
concentraations of A, B, C, and D increase, deecrease, or not
n be affecteed?
Chaapter 13 Answerss
2.
(a)
Since ∆G°′ = −RT ln K,
(b)
The reaaction is not spontaneous since ∆G > 0.
(c) The
T reaction can proceedd in the cell if the producct B is the substrate for a second reacction such
that thee second reacction continuually draws off B, causinng the first reaction
r
to coontinually prroduce moree
B from
m A.
3.
b
7.
Calculatiing ∆G for th
he reaction ATP
A + creatiine ⇌ phosphhocreatine + ADP, usingg Eq. 13-1:
Since ∆G
G > 0, the reaaction will proceed
p
in thhe opposite direction
d
as written
w
above, that is, in the directionn
of ATP synthesis.
s
8.
Using thee data in Tab
ble 13-2, we calculate ∆G
G°′ for the adenylate
a
kinnase reactionn.
Since ∆G
G for a reactiion at equilibbrium is zeroo, Eq. 13-1 becomes
b
∆G
G°′ = −RT ln Keq so that Keq =
−∆G°′/RT
e
.
13.
14.
s catalyzed by enzymee Y is likely to be the maajor flux-conntrol point, since
s
this steep operates
(a) The step
farthest from
f
equilibrrium (it is ann irreversiblee step). (b) Inhibition off enzyme Z would
w
cause the
concentraation of D, th
he reaction’s product, too decrease, and
a it would cause C, thee reaction’s substrate,
s
to
accumulaate. The concentrations of
o A and B would
w
not chhange becausse the steps catalyzed byy enzymes X
and Y woould not be affected.
a
Thee accumulateed C would not be transfformed backk to B since the
t step
catalyzedd by enzymee Y is irreverrsible.
Chaapter 14
1.
Which off the 10 reacctions of glyccolysis are (aa) phosphoryylations, (b) isomerizatioons, (c) oxiddation–
reductionns, (d) dehyd
drations, andd (e) carbon––carbon bondd cleavages??
4.
Arsenatee (
), a structural analog of phhosphate, caan act as a suubstrate for any
a reaction in which
phosphatte is a substrrate. Arsenatte esters, unlike phosphaate esters, aree kinetically as well as
thermodyynamically unstable
u
and hydrolyze almost
a
instanntaneously. Write
W
a balannced overalll equation
for the coonversion off glucose to pyruvate
p
in the
t presencee of ATP, AD
DP, NAD+, and
a either (aa) phosphate
or (b) arssenate. (c) Why
W is arsenaate a poison??
7.
∆G°′ for the aldolasee reaction is 22.8 kJ · mool−1. In the ceell at 37°C, [DHAP]/[GA
[
AP] = 5.5. Calculate
C
the
−4
equilibriuum ratio of [FBP]/[GAP
[
P] when [GA
AP] = 10 M.
M
8.
The half--reactions in
nvolved in thhe lactate dehhydrogenasee (LDH) reacction and theeir standard reduction
r
potentials are
Calculatee ∆G at pH 7.0
7 for the LDH-catalyzeed reductionn of pyruvatee under the following
fo
connditions:
(a)
[llactate]/[pyru
uvate] = 1 annd [NAD+]/[[NADH] = 1
(b)
[llactate]/[pyru
uvate] = 1600 and [NAD+]/[NADH] = 160
(c)
[llactate]/[pyru
uvate] = 10000 and [NAD
D+]/[NADH]] = 1000
(d)
D
Discuss
the effect of the concentratio
c
n ratios in parts a–c on the
t directionn of the reacttion.
9.
Althoughh it is not thee primary fluux-control pooint for glyccolysis, pyruvvate kinase is
i subject to allosteric
regulation. (a) What is the metabbolic importaance of regullating flux thhrough the pyruvate
p
kinaase reaction??
(b) Whatt is the advan
ntage of activvating pyruvvate kinase with
w fructosee-1,6-bisphoosphate?
10.
Comparee the ATP yield of three glucose mollecules that enter
e
glycolyysis and are converted too pyruvate
with thatt of three glu
ucose molecuules that proceed throughh the pentose phosphate pathway such that their
carbon skkeletons (as two F6P andd one GAP) re-enter glycolysis and are
a metaboliized to pyruvvate.
Chaapter 14 Answerss
1.
(a) Reacttions 1, 3, 7, and 10; (b) Reactions 2,
2 5, and 8; (cc) Reaction 6;
6 (d) Reactiion 9; (e) Reeaction 4.
4.
(a)
G
Glucose
+ 2 NAD
N + + 2 ADP
A
+ 2 Pi → 2 pyruvate + 2 NADH + 2 ATP + 2 H2O
(b)
(c) Arsenate
A
is a poison becaause it uncouuples ATP geeneration froom glycolysiis. Consequeently,
glycolyytic energy generation
g
caannot occur.
7.
− M. Accord
When [G
GAP] = 10−4 M, [DHAP] = 5.5 × 10−4
ding to Eq. 1-17,
1
K = e−∆∆G°′/RT
8.
For the coupled
c
reacttion
Pyruvatee + NADH + H + → lacttate + NAD
D
+
∆ℰ°′ = (−
−0.185 V) − (−0.315 V) = 0.130 V.
Accordinng to Eq. 13--8
and ∆G = −nℱ∆ℰ (Eq. 13-7). Sinnce two electtrons are trannsferred in the
t above reaaction, n = 2.
2
(a)
(b)
(c)
(d) At
A the concen
ntration ratioos of Part a, ∆G
∆ is negatiive and the reaction
r
procceeds as writtten. As
[lactatee]/[pyruvate]] increases, the
t reaction ∆G increasees even thouggh [NAD+]/[[NADH] also increases
so that in Part b thee reaction is at equilibriuum (∆G = 0) and in Part c ∆G is posiitive and thee reaction
proceedds spontaneo
ously in the opposite direection.
9.
(a)
Pyruvate kinaase regulatioon is importaant for controolling the fluux of metaboolites, such as
a fructose
(in liver), which enteer glycolysis after the PF
FK step.
(b)
F is the pro
FBP
oduct of the third reactioon of glycolyysis, so it acts as a feed-forward actiivator of the
enzyme that
t catalyzees Step 10. This
T regulatoory mechanissm helps enssure that oncce metabolitees pass the
PFK stepp of glycolyssis, they willl continue thhrough the paathway.
10.
The three glucose molecules that proceed through glycolysis yield 6 ATP. The bypass through the
pentose phosphate pathway results in a yield of 5 ATP.
Chapter 15
1.
Indicate the energy yield or cost, in ATP equivalents, for the following processes:
(a)
glycogen (3 residues) → 6 pyruvate
(b)
3 glucose → 6 pyruvate
(c)
6 pyruvate → 3 glucose
2.
Write the balanced equation for (a) the sequential conversion of glucose to pyruvate and of pyruvate to
glucose and (b) the catabolism of six molecules of G6P by the pentose phosphate pathway followed by
conversion of ribulose-5-phosphate back to G6P by gluconeogenesis.
10.
Individuals with McArdle’s disease often experience a “second wind” resulting from cardiovascular
adjustments that allow glucose mobilized from liver glycogen to fuel muscle contraction. Explain why
the amount of ATP derived in the muscle from circulating glucose is less than the amount of ATP that
would be obtained by mobilizing the same amount of glucose from muscle glycogen.
Chapter 15 Answers
1.
(a)
+9 ATP,
(b)
+6 ATP,
2.
(a)
The equation for glycolysis is
(c)
−18 ATP.
Glucose +2 NAD + + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 4 H + + 2 ATP + 2 H2 O
The equation for gluconeogenesis is
2 Pyruvate + 2 NADH + 4 H + + 4 ATP + 2 GTP + 6 H2 O → glucose + 2 NAD + + 4 ADP + 2 GDP + 6 Pi
For the two processes operating sequentially,
2 ATP + 2 GTP + 4 H2 O → 2 ADP + 2 GDP + 4 Pi
(b)
The equation for catabolism of 6 G6P by the pentose phosphate pathway is
6 G 6 P + 12 NADP + + 6 H2 O → 6 Ru 5 P + 12 NADPH + 12 H + + 6 CO2
Ru5P can be converted to G6P by transaldolase, transketolase, and gluconeogenesis:
6 Ru 5 P + H2 O → 5 G 6 P + Pi
The net equation is therefore
G 6 P + 12 NAD
DP + + 7 H2 O → 12 NAD
DPH + 12 H
H + + 6 CO2 + Pi
10.
The convversion of cirrculating gluucose to lacttate in the muuscle generaates 2 ATP. If
I muscle glyycogen
could be mobilized, the
t energy yield
y
would be
b 3 ATP, since phosphoorolysis of glycogen
g
byppasses the
hexokinaase-catalyzed
d step that coonsumes AT
TP in the firsst stage of glyycolysis.
Chaapter 16
12.
Given thee following information, calculate thhe physiologgical ∆G of the
t isocitratee dehydrogennase reactionn
+
at 25°C and
a pH 7.0: [NAD ]/[NA
ADH] = 8, [α
α-ketoglutarate] = 0.1 mM;
m and [isoccitrate] = 0.002 mM.
Assume standard
s
con
nditions for CO
C 2 (∆G°′ iss given in Taable 16-2). Is
I this reactioon a likely siite for
metabolic control?
13.
Althoughh animals cannot synthessize glucose from acetyl-CoA, if a raat is fed 14C--labeled acettate, some off
the label appears in glycogen
g
exttracted from its muscles.. Explain.
Chaapter 16 Answerss
12.
For the reeaction isociitrate + NAD
D+ ⇌ α-ketogglutarate + NADH
N
+ CO
O2 + H+, we assume
a
[H+] and [CO2]
= 1. Accoording to Eq
q. 13-1,
With succh a large neg
gative free energy
e
of reaaction under physiologiccal conditionns, isocitrate
dehydroggenase is likeely to be a metabolic
m
conntrol point.
13.
Animals cannot carry
y out the nett synthesis off glucose froom acetyl-CooA (to whichh acetate is converted).
c
14
Howeverr, C-labeled
d acetyl-CoA
A enters the citric acid cycle and is converted
c
to oxaloacetatte. Some of
this oxalooacetate may
y exchange with
w the celllular pool off oxaloacetate to be convverted to gluccose throughh
gluconeoogenesis and
d subsequentlly taken up by
b muscle annd incorporaated into glycogen.
Chaapter 17
2.
How manny ATPs aree synthesizedd for every cytoplasmic
c
N
NADH
that participates in the glyceerophosphatee
shuttle inn insect fligh
ht muscle? How
H does this compare too the ATP yiield when NADH
N
reduciing
equivalennts are transfferred into thhe matrix viaa the malate–aspartate shhuttle?
3.
Calculatee ∆G°′ for th
he oxidation of free FAD
DH2 by O2. What
W is the maximum
m
nuumber of AT
TPs that can
be syntheesized, assum
ming standarrd conditions and 100% conservationn of energy??
5.
Why is itt possible for electrons too flow from a redox cennter with a more
m
positive ℰ°′ to one with
w a more
negative ℰ°′ within an
a electron-trransfer compplex?
7.
The diffeerence in pH
H between thee internal annd external suurfaces of thhe inner mitoochondrial membrane
m
is
1.4 pH unnits (externaal side acidicc). If the mem
mbrane poteential is 0.06 V (inside neegative) whaat is the free
energy reeleased on trransporting 1 mol of prottons back accross the mem
mbrane? Hoow many prootons must
be transpported to provide enoughh free energyy for the syntthesis of 1 mol
m of ATP (assuming
(
sttandard
biochemiical conditio
ons)?
11.
Explain why
w compou
unds such ass DNP increaase metaboliic rates.
Chaapter 17 Answerss
2.
When NA
ADH particiipates in the glycerophossphate shuttlle, the electroons of NAD
DH flow to FA
AD and thenn
to CoQ, bypassing
b
Complex
C
I. Thhus, about 2 ATP are syynthesized peer NADH. About
A
three ATP
A are
producedd when NAD
DH participattes in the maalate–aspartaate shuttle.
3.
The relevvant half-reaactions (Tablle 13-3) are
Since thee O2/H2O hallf-reaction has
h the more positive ∆ℰ
ℰ°′, the FAD half-reaction is reversedd and the
overall reeaction is
Since ∆G
G°′ = −nℱ∆ℰ
ℰ°′,
∆ G °′ = − (2)(96,4
485 J ⋅ V − 1 ⋅ mol − 1)(1.0034 V ) = − 200 kJ ⋅ mol − 1
The maximum numb
ber of ATP thhat could be synthesizedd under standdard conditioons is therefo
fore 200 kJ ⋅
mol−1/30.5 kJ ⋅ mol−11 = 6.6 mol ATP/mol
A
FA
ADH2 oxidizzed by O2.
5.
ℰ may diiffer from ℰ°°′, dependingg on the redoox center’s microenviron
m
nment and thhe concentraations of
reactantss and productts. In additioon, the tight coupling bettween succeessive electroon transfers within
w
a
complex may “pull” electrons soo that the oveerall process is spontaneous.
7.
For the trransport of a proton from
m outside to inside (Eq. 17-1),
1
∆ G = 2..3 RT[ pH ( in
i ) − pH ( out
o )] + Zℱ ∆ Ψ
The diffeerence in pH
H is −1.4. Sinnce an ion is transported from the positive to the negative sidde of the
membranne, ∆Ψ is neg
gative.
Since ∆G
G°′ for ATP synthesis is 30.5 kJ ⋅ mool−1 and 30.55/13.8 = 2.2, between tw
wo and three moles
m
of
protons must
m be transsported to prrovide the frree energy too synthesize one mole off ATP under standard
biochemiical conditio
ons.
11.
DNP andd related com
mpounds disssipate the prroton gradiennt required for
fo ATP syntthesis. The dissipation
d
off
this gradiient decreasees the rate of synthesis of
o ATP, decrreasing the ATP
A mass acction ratio. Decreasing
D
this ratio relieves thee inhibition of
o the electroon transport chain, causinng an increaase in metaboolic rate.
Chaapter 19
1.
Explain why
w individu
uals with a hereditary
h
deeficiency of carnitine
c
pallmitoyl transsferase II havve muscle
weaknesss. Why are th
hese symptooms more sevvere during fasting?
3.
The first three steps of
o β oxidatioon (Fig. 19-112) chemicallly resemble three successive steps of
o the citric
acid cyclle. Which steeps are thesee?
5.
Why are unsaturated fats preferabble to saturaated fats for an
a individuaal whose calooric intake must
m be
limited?
10.
1 CO used to
On what carbon atom
ms does the 14
t synthesizee malonyl-CooA from aceetyl-CoA apppear in
2
palmitatee?
11.
Explain why
w adipocy
ytes need gluucose as well as fatty aciids in order to
t synthesizee triacylglyccerols.
12.
Is the fattty acid show
wn below likkely to be synnthesized in animals? Exxplain.
13.
Comparee the energy cost, in ATP
P equivalents, of synthessizing stearate from mitoochondrial accetyl-CoA
to the eneergy recoverred by degraading stearatee (a) to acetyyl-CoA and (b) to CO2.
Chaapter 19 Answerss
1.
A defect in carnitine palmitoyl trransferase II prevents noormal transpoort of activatted fatty acidds into the
xidation. Tisssues such as muscle thaat use fatty acids
a
as metaabolic fuels therefore
t
mitochonndria for β ox
cannot geenerate ATP
P as needed. The problem
m is more sevvere during a fast because other fuells, such as
dietary glucose, are not
n readily avvailable.
3.
The first three steps of
o β oxidatioon resemble the reactions that convert succinate to oxaloacettate
(Sectionss 16-3F–16-3
3H).
5.
There are not as many usable nutritional calories per gram in unsaturated fatty acids as there are in
saturated fatty acids. This is because oxidation of fatty acids containing double bonds yields fewer
reduced coenzymes whose oxidation drives the synthesis of ATP. In the oxidation of fatty acids with a
double bond at an odd-numbered carbon, the enoyl-CoA isomerase reaction bypasses the acyl-CoA
dehydrogenase reaction and therefore does not generate FADH2 (equivalent to 2 ATP). A double bond at
an even-numbered carbon must be reduced by NADPH (equivalent to the loss of 3 ATP).
10.
The label does not appear in palmitate because 14CO2 is released in Reaction 2b of fatty acid synthesis
(Fig. 19-26).
11.
The breakdown of glucose by glycolysis generates the dihydroxyacetone phosphate that becomes the
glycerol backbone of triacylglycerols (Fig. 19-30).
12.
This fatty acid (linolenate) cannot be synthesized by animals because it contains a double bond closer
than 6 carbons from its noncarboxylate end.
13.
The synthesis of stearate (18:0) from mitochondrial acetyl-CoA requires 9 ATP to transport 9 acetylCoA from the mitochondria to the cytosol. Seven rounds of fatty acid synthesis consume 7 ATP (in the
acetyl-CoA carboxylase reaction) and 14 NADPH (equivalent to 42 ATP). Elongation of palmitate to
stearate requires 1 NADH and 1 NADPH (equivalent to 6 ATP). The energy cost is therefore 9 + 7 + 42
+ 6 = 64 ATP.
(a) The degradation of stearate to 9 acetyl-CoA consumes 2 ATP (in the acyl-CoA synthetase
reaction) but generates, in eight rounds of β oxidation, 8 FADH2 (equivalent to 16 ATP) and 8 NADH
(equivalent to 24 ATP). Thus, the energy yield is 16 + 24 − 2 = 38 ATP. This represents only about
half of the energy consumed in synthesizing stearate (38 ATP versus 64 ATP).
(b) The complete oxidation of the 9 acetyl-CoA to CO2 by the citric acid cycle yields an additional 9
GTP (equivalent to 9 ATP), 27 NADH (equivalent to 81 ATP), and 9 FADH2 (equivalent to 18 ATP)
for a total of 38 + 9 + 81 + 18 = 146 ATP. Thus, more than twice the energy investment of
synthesizing stearate is recovered (146 ATP versus 64 ATP).
Chapter 20
2.
Explain why the symptoms of a partial deficiency in a urea cycle enzyme can be attenuated by a lowprotein diet.
3.
Production of the enzymes that catalyze the reactions of the urea cycle can increase or decrease
according to the metabolic needs of the organism. High levels of these enzymes are associated with
high-protein diets as well as starvation. Explain this apparent paradox.
8.
Which of the 20 “standard” amino acids are (a) purely glucogenic, (b) purely ketogenic, and (c) both
glucogenic and ketogenic?
9.
Alanine, cysteine, glycine, serine, and threonine are amino acids whose breakdown yields pyruvate.
Which, if any, of the remaining 15 amino acids also do so?
Chapter 20 Answers
2.
The urea cycle transforms excess nitrogen from protein breakdown to an excretable form, urea. In a
deficiency of a urea cycle enzyme, the preceding urea cycle intermediates may build up to a toxic level.
A low-protein diet minimizes the amount of nitrogen that enters the urea cycle and therefore reduces the
concentrations of the toxic intermediates.
3.
An individual consuming a high-protein diet uses amino acids as metabolic fuels. As the amino acid
skeletons are converted to glucogenic or ketogenic compounds, the amino groups are disposed of as
urea, leading to increased flux through the urea cycle. During starvation, proteins (primarily from
muscle) are degraded to provide precursors for gluconeogenesis. Nitrogen from these protein-derived
amino acids must be eliminated, which demands a high level of urea cycle activity.
8.
(a)
Ala, Arg, Asn, Asp, Cys, Gin, Glu, Gly, His, Met, Pro, Ser, and Val
(b)
Leu and Lys
(c)
Ile, Phe, Thr, Trp, and Tyr
9.
Tryptophan can be considered a member of this group since one of its degradation products is alanine,
which is converted to pyruvate by deamination.