book problems chapter 2
TRANSCRIPT
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.
10.
14.
15.
16.
17.
Cha
1.
2.
8.
10.
d.
Calculate
What is th
How man162 g · mo50 mM?
Estimate t100 mM s
(a) Wouldbetter buf
apter 2 A
(a) D
(b) D
(c) D
A protonaC−H bondwould ma
(a)
(b)
(c)
(d)
The increa[H+] of pu
the pH of a 2
he pK of the w
ny grams of sool−1) must be
the volume ofsolution of ph
d phosphoric ffer at pH 9? (
Answers:
Donors: NH1,
Donors: NHl, N
Donors:
ated (and therd so that the Cake the H mor
ase in [H+] duure water, 10−
200 mL soluti
weak acid HA
odium succinaadded to 1 L
f a solution ohosphoric acid
acid or succin(c) Would HE
:
NH2 at C2, N
NH2 at C4; ac
group, OH g
refore positiveC would havere likely to be
ue to the addi−7 M, is relativ
ion of pure w
A if a solution
ate (formula w of water to p
f 5 M NaOH d.
nic acid be a bEPES or Tris b
NH9; acceptor
cceptors: O at
group; accepto
ely charged) ne a partial nege donated to a
ition of HCl isvely insignifi
ater to which
n containing 0
weight 140 g produce a solu
that must be
better buffer abe a better bu
rs: N3, O at C
t C2, N3.
ors: COO− gr
nitrogen wougative charge a hydrogen bo
s (50 mL) (l mcant, the pH o
h has been add
0.1 M HA and
· mol−1) and ution with pH
added to adju
at pH 5? (b) Wuffer at pH 7.5
C6, N7.
roup, OH grou
ld promote thand the H wo
ond acceptor g
mM)/(250 mLof the solution
ded 50 mL of
d 0.2 M A− ha
disodium sucH 6.0 and a tot
ust the pH fro
Would ammo5?
up.
he separation ould have a pagroup.
L) = 0.2 mM =n is equal to −
f 1 mM HCl.
as a pH of 6.5
ccinate (formutal solute con
om 4 to 9 in 10
onia or piperid
of charge in tartial positive
= 2 × 10−4 M−log (2 × 10−
5?
ula weight centration of
00 mL of a
dine be a
the adjacent e charge. This
. Because the−4) or 3.7.
s
e
14.
15.
16.
17.
Cha
1.
3.
4.
6.
7.
Use the H
Let HA =
[A−] + [H
From Eq.
log ([A−
[A−]/[H
(0.05 M
[HA] =
[A−] = 0
grams o
grams o
At pH 4, eform (Figmol ⋅ L−1
(a) Su
apter 4
Identify t
Identify t
Draw the
DeterminpH 11.0.
Calculate
Henderson–Ha
sodium succ
HA] = 0.05 M,
2-9 and Tabl
−]/[HA]) = pH
HA] = antilog
M − [HA])/[HA
0.015 M
0.05 M − 0.01
of sodium suc
of disodium su
essentially all. 2-18). Therephosphoric ac
uccinic acid;
the amino ac
the hydrogen
e dipeptide A
ne the net ch
e the pI of (a
asselbalch equ
inate and A−
, so [A−] = 0.0
le 2-4,
H − pK = 6.0
0.36 = 2.29
A] = 2.29
15 M = 0.035
ccinate = (0.0
uccinate = (0
l the phosphoefore, the concid)(0.1 L) =
(b) am
cids that diff
n bond dono
Asp-His at pH
harge of the p
a) Ala, (b) H
uation (Eq. 2-
= disodium s
05 M − [HA]
− 5.64 = 0.36
5 M
15 mol ⋅ L−1)(
.035 mol ⋅ L−
oric acid is in ncentration of
0.01 mol NaO
mmonia;
fer from each
or and accept
H 7.0.
predominant
His, and (c) G
-9) and solve
uccinate.
6
(140 g ⋅ mol−
−1)(162 g ⋅ mo
the f OH− requireOH required
(c) HEP
h other by a
tor groups in
t form of Asp
Glu.
for pK:
1) × (1 L) = 2
ol−1) × (1 L) =
form, and at d is equivalen= (0.01 mol)(
PES.
single methy
n asparagine
p at (a) pH 1
2.1 g
= 5.7 g
pH 9, essentint to the conc(1 L/5 mol Na
yl or methyl
.
1.0, (b) pH 3
ially all is in tcentration of taOH) = 0.002
lene group.
3.0, (c) pH 6
the the acid: (0.1 2 L = 2 mL
.0, and (d)
9.
13.
14.
15.
16.
Circle the
Some am
keto acidfrom the
Identify t
Draw the
The proteorganismsequencelower tha
e chiral carb
mino acids ar
d with an amfollowing α
the amino ac
e peptide AT
ein insulin cms have beene with the exan or higher
Amino acid
Human
Duck
bons in the fo
re synthesize
mino group (α-keto acids:
cid residue fr
TLDAK. (a)
onsists of twn isolated andxception of s
than that of
d residue A8
Th
Glu
ollowing com
ed by replaci
).
from which t
Calculate its
wo polypeptid sequencedix amino aciduck insulin
A9
hr Ser
u Asn
mpounds:
ing the keto
Identify the
the following
s approxima
ides termed td. Human andid residues, an?
A10 B1
Ile Phe
Pro Ala
group (C=O
e amino acid
g compound
ate pI. (b) Wh
the A and B d duck insulas shown be
B2 B
Val T
Ala S
O) of an orga
ds that can be
ds are synthe
hat is its net
chains. Insuins have thelow. Is the p
B27
Thr
Ser
anic acid kno
e produced t
sized:
charge at pH
ulins from die same aminopI of human
own as an α-
his way
H 7.0?
ifferent o acid insulin
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; (d) −2.
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
9.
13. (a) Glutamate; (b) aspartate
14. (a) Serine (N-acetylserine); (b) lysine (5-hydroxylysine);
(c) methionine (N-formylmethionine).
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
16.
Cha
1.
2.
4.
6.
9.
4-1D). Thspecies (t
(b) T
At positionegativelinsulin. (uncharge
apter 5
Which pe
A. G
B. S
(a) In whpH 6? (b)
Determin
Molecu
Molecu
Molecu
40 kD a
What fraamino ac
1. 2
2. 3
3. 2
All three
You must cwould be li
he pI is apprthe pK of As
The net charg
on A8, duckly charged at(The other amed.)
eptide has gr
Gln–Leu–Glu
er–Val–Trp–
hat order wou) In what ord
ne the subun
ular mass by
ular mass by
ular mass by
and 60 kD
ctionation prcid composit
5% Ala, 20%
0% Gln, 25%
5% Asn, 20%
proteins are
cleave the foikely to yiel
roximately msp and the N
ge at pH 7.0
k insulin has t physiologimino acids th
reater absorb
u–Phe–Thr–L
–Asp–Phe–G
uld the aminder would G
nit compositi
y gel filtratio
y SDS-PAGE
y SDS-PAGE
rocedure coutions are as f
% Gly, 20%
% Glu, 20%
% Gly, 20%
e similar in s
ollowing pepd the most fr
midway betwN-terminal pK
is 0 (as draw
a Glu residucal pH and That differ be
bance at 280
Leu–Asp–G
Gly–Tyr–Trp
no acids ArgGlu, Lys, and
ion of a prote
n: 200 kD
E: 100 kD
E with 2-mer
uld be used tfollows?
Ser, 10% Ile
Lys, 15% S
Asp, 20% S
size and pI, a
ptide into smfragments? T
ween the pK’K):
wn above).
ue, whereas hThr is neutraetween the pr
0 nm?
Gly–Tyr
p–Ala
, His, and Led Val be elut
ein from the
rcaptoethano
to purify pro
e, 10% Val,
Ser, 10% Cys
Ser, 10% Ly
and there is n
maller fragmeThe fewest?
’s of the two
human insulal, human insroteins do no
eu be eluteded from a di
following in
ol:
otein 1 from
5% Asn, 5%
s
s, 5% Tyr
no antibody
ents. Which
o ionizations
lin has a Thrsulin has a hot affect the
from a carbiethylaminoe
nformation:
a mixture of
% Gln, 5% Pr
available fo
of the protea
involving th
r residue. Sinhigher pI than
pI because t
boxymethyl cethyl column
f three prote
ro
or protein 1.
ases listed in
he neutral
nce Glu is n duck they are
column at n at pH 8?
eins whose
n Table 5-3
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
Chymotrypsin
4. Met–Arg–Ala–Tyr 5. Asp–Met–Leu–Phe 6. Gly–Asn
13. 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 (b) SEDNFGAPKSILW
(c) QKASVEMAVRNSG
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?
15.
Cha
2.
3.
11. T
(a) Is
(b) H
(c) In
(d) Mglyceroprotein
Proteins pump a vexplain, iThere is noverexpr
apter 10
The less A, B.
The compoun
s this compo
How does mi
n what part o
Miltefosine bophospholipin does not bin
known as mvariety of hyin structural no phosphor
ression of an
Answers
polar a subs
nd shown be
ound a glycer
iltefosine lik
of the cell w
binds to a proids. What fend triacylgly
multidrug resiydrophobic su
terms, how rylated prote
n MDR transp
s
stance, the fa
low is the an
rophospholip
kely cross the
ould the dru
otein that alsature commycerols.
istance (MDubstances outhe transpor
ein intermedporter in a c
aster it can d
ntiparasitic d
pid?
e parasite ce
ug tend to acc
so binds somon to all thes
DR) transportut of cells. (arter might takiate, as in th
cancer cell m
diffuse throug
drug miltefo
ll membrane
cumulate? E
me sphingolipse compoun
ters use the fa) Write the ke advantag
he (Na+–K+)–make the canc
gh the lipid b
sine.
e?
Explain.
pids and somds is recogn
free energy oreaction for e of this pro–ATPase. (bcer more dif
bilayer. From
me nized by the p
of ATP hydrATP hydrol
ocess to drive) Why woul
fficult to trea
m slowest to
protein? The
rolysis to lysis and e transport. d
at?
o fastest: C,
e
4.
8.
10.
11.
(a)
(b)
(a) Tnot appro
(b) Ttransportresulting
The hypeprocess. directly poccupiedcells is nenough inmediated
(a) N
(b) MSince it ilikely ent
(c) Tburied in
(d) Tand somedoes not
The data do noach a maxim
To verify thatter at high [Xin a lower f
erbolic curveThe transporproportional d, and thus glot sodium-dnformation i
d.
No; there is n
Miltefosine isis not a normters the cell
This amphipan the bilayer
The protein ree glycerophorecognize th
not indicate tmum as [X]
t a transport X], or add a sflux of X.
e for glucosert protein hato [Na+]. Ho
lucose transpdependent anin the figure
no glycerol b
s amphipathimal cell comp
via a passive
athic molecuand its polar
ecognizes thospholipids. he hydrocarb
the involvemincreases.
protein is instructural an
e transport inas binding sitowever, at hport reaches
nd occurs at ato determin
backbone.
ic and therefponent, it proe transport p
ule most liker head group
he phosphochSince the pr
bon tail.
ment of a tran
nvolved, incrnalog of X to
nto pericytestes for sodiu
high [Na+], al a maximuma high rate w
ne whether gl
fore cannot cobably does
protein.
ly accumulap exposed to
holine head rotein does n
nsport protei
rease [X] to o compete w
s indicates a um ions. At lll Na+ bindin
m velocity. Gwhether or nolucose transp
cross the parnot have a d
ates in membthe solvent.
group, whicnot bind all p
in, since the
demonstrateith X for bin
protein-medlow [Na+], gng sites on th
Glucose transot Na+ is preport into end
rasite cell mededicated ac
branes, with
h also occurphospholipid
rate of trans
e saturation onding to the t
diated sodiumlucose transphe transport sport into enesent. There dothelial cell
embrane by ctive transpor
its hydropho
rs in some spds or triacylg
sport does
of the transporter,
m-dependenport is protein are dothelial is not ls is protein-
diffusion. rter. It most
obic tail
phingolipids glycerols, it
nt
-
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.
Cha
2.
3.
7.
(a) K
7 μM. In intercept
presence
(b) Tsphingosstructurato the enz
apter 13
A certainCalculateconcentrathese con
Choose t
a. al
b. h
c. is
d. op
Predict wsynthesis= 1 mM.
KM is determi
the presenc(= 1/Vmax). I
of inhibitor
The lines in thsine is most ll similarity bzyme active
n metabolic re the equilibation of A isnditions? (c)
the best defin
lways operat
as a free ene
s usually a co
perates very
whether creats at 25°C wh
ined from th
e of inhibitoIn the absen
,
he double-relikely a combetween the site.
reaction takerium constan
s 0.5 mM anHow might
nition for a n
tes with a fa
ergy change
ontrol point
y slowly in vi
tine kinase when [ATP] =
he x-intercept
or, nce of inhibit
eciprocal plompetitive inhi
inhibitor an
es the form Ant for the read the concenthe reaction
near-equilibr
avorable free
near zero.
in a metabo
ivo.
will operate i4 mM, [AD
t (= −1/KM).
tor, Vmax = 1/
ots intersect vibitor. Compd the substra
A → B. Its staction at 25°ntration of Bn proceed in
rium reaction
e energy chan
lic pathway.
in the directiDP] = 0.15 m
In the absen
/0.008 mg−1
very close topetitive inhibate, which al
tandard free °C. (b) Calcu
B is 0.1 mM. the cell?
n:
nge.
.
ion of ATP smM, [phospho
nce of inhibi
. Vmax i ⋅ min = 125
.
o the 1/vo axbition is likelllows them t
energy chanulate ∆G at 3Is the reacti
synthesis or ocreatine] =
itor, KM = 1/
is determined5 mg ⋅ min−1.
xis. Hence, thly also becauto compete f
nge is 7.5 kJ 37°C when tion spontane
phosphocre2.5 mM, an
/0.14 μM−1 =
d from the y. In the
hreo-use of the for binding
· mol−1. (a) the eous under
atine nd [creatine]
=
y-
8.
13.
14.
Cha
2.
If intraceat pH 7 a
A hypothB, and C
1. C
2. A
3. A
4. C
A certain
where A,The phys
(a) Whicwas in faconcentra
apter 13
(a) S
(b)
The rea
(c) Tthat theB from
ellular [ATP]and 25°C und
hetical three-. Deduce the
Compound Q
A mutant in e
An inhibitor o
Compound P
n metabolic p
, B, C, and Dsiological fre
h reaction isact the case, ations of A,
Answers
ince ∆G°′ =
action is not
The reaction e second reac
m A.
] = 5 mM, [Ader the cond
-step metaboe order of th
Q, a metaboli
enzyme C re
of enzyme A
, a metabolic
pathway can
D are the inteee energy ch
s likely to bein the presenB, C, and D
s
−RT ln K,
spontaneou
can proceedction continu
ADP] = 0.5 mdition that the
olic pathwaye enzymatic
ic inhibitor o
quires Y for
A causes W,
c inhibitor o
n be diagram
ermediates, ahanges for th
e a major regnce of an inh
D increase, de
s since ∆G >
d in the cell iually draws
mM, and [Pi
e adenylate k
y consists of steps in the
of enzyme B
r growth.
Y, and Z to
f enzyme C,
med as
and X, Y, anhe reactions a
gulatory poinhibitor that becrease, or n
> 0.
f the producoff B, causin
i] = 1.0 mM,kinase reacti
intermediatepathway fro
, causes Z to
accumulate.
, causes W a
nd Z are the are
nt for the patblocks the acnot be affecte
ct B is the sung the first r
, calculate thion is at equ
es W, X, Y, om the follow
o build up.
.
and Z to buil
enzymes tha
thway? (b) Ifctivity of enzed?
bstrate for areaction to co
he concentrauilibrium.
and Z and ewing inform
d up.
at catalyze th
f your answezyme Z, wou
a second reacontinually pr
ation of AMP
enzymes A, mation:
he reactions.
er in Part a uld the
ction such roduce more
P
.
e
3.
7.
8.
13.
14.
Cha
1.
4.
b
Calculati
Since ∆Gof ATP s
Using the
Since ∆Ge−∆G°′/RT.
(a) The sfarthest fconcentraaccumulaand Y wocatalyzed
apter 14
Which ofreduction
Arsenatephosphat
ing ∆G for th
G > 0, the reasynthesis.
e data in Tab
G for a reacti
step catalyzefrom equilibration of D, thate. The conould not be ad by enzyme
f the 10 reacns, (d) dehyd
e ( ), te is a substr
he reaction A
action will p
ble 13-2, we
ion at equilib
d by enzymerium (it is anhe reaction’centrations oaffected. Thee Y is irrever
ctions of glycdrations, and
a structural rate. Arsenat
ATP + creati
proceed in th
calculate ∆G
brium is zero
e Y is likelyn irreversibles product, toof A and B we accumulatersible.
colysis are (ad (e) carbon–
analog of phte esters, unl
ine ⇌ phosph
he opposite d
G°′ for the a
o, Eq. 13-1 b
to be the mae step). (b) Io decrease, awould not ched C would
a) phosphory–carbon bond
hosphate, caike phospha
hocreatine +
direction as w
adenylate kin
becomes ∆G
ajor flux-connhibition of
and it would hange becausnot be transf
ylations, (b)d cleavages?
an act as a suate esters, are
+ ADP, using
written abov
nase reaction
G°′ = −RT ln
ntrol point, sf enzyme Z wcause C, these the steps formed back
isomerizatio?
ubstrate for ae kinetically
g Eq. 13-1:
e, that is, in
n.
Keq so that K
since this stewould cause e reaction’s scatalyzed by
k to B since t
ons, (c) oxid
any reaction as well as
the direction
Keq =
ep operates the
substrate, to y enzymes Xthe step
dation–
in which
n
X
7.
8.
9.
10.
Cha
1.
4.
thermodyfor the coor (b) ars
∆G°′ for equilibriu
The half-potential
Calculate
(a) [l
(b) [l
(c) [l
(d) D
Althoughregulatio(b) What
Comparewith thatcarbon sk
apter 14
(a) React
(a) G
(b)
(c) Aglycoly
ynamically uonversion ofsenate. (c) W
the aldolaseum ratio of [
-reactions ins are
e ∆G at pH 7
lactate]/[pyru
lactate]/[pyru
lactate]/[pyru
Discuss the e
h it is not then. (a) What
t is the advan
e the ATP yit of three glukeletons (as
Answers
tions 1, 3, 7,
Glucose + 2 N
Arsenate is a ytic energy g
unstable and f glucose to p
Why is arsena
e reaction is [FBP]/[GAP
nvolved in th
7.0 for the L
uvate] = 1 an
uvate] = 160
uvate] = 100
ffect of the c
e primary fluis the metabntage of activ
eld of three ucose molecutwo F6P and
s
and 10; (b)
NAD+ + 2 A
poison becageneration ca
hydrolyze apyruvate in tate a poison?
22.8 kJ · moP] when [GA
he lactate deh
DH-catalyze
nd [NAD+]/[
0 and [NAD+
00 and [NAD
concentratio
ux-control pobolic importavating pyruv
glucose molules that prod one GAP)
Reactions 2
ADP + 2 Pi →
ause it uncouannot occur.
almost instanthe presence?
ol−1. In the ceAP] = 10−4 M
hydrogenase
ed reduction
[NADH] = 1
+]/[NADH] =
D+]/[NADH]
n ratios in p
oint for glycance of regulvate kinase w
lecules that eceed throughre-enter gly
2, 5, and 8; (c
→ 2 pyruvate
uples ATP ge
ntaneously. We of ATP, AD
ell at 37°C, [M.
e (LDH) reac
n of pyruvate
1
= 160
] = 1000
arts a–c on t
colysis, pyruvlating flux thwith fructose
enter glycolyh the pentoscolysis and a
c) Reaction 6
+ 2 NADH
eneration fro
Write a balanDP, NAD+, a
[DHAP]/[GA
ction and the
e under the fo
the direction
vate kinase ihrough the pe-1,6-bispho
ysis and are e phosphate are metaboli
6; (d) Reacti
+ 2 ATP + 2
om glycolysi
nced overalland either (a
AP] = 5.5. C
eir standard r
following con
n of the react
is subject to pyruvate kinaosphate?
converted topathway su
ized to pyruv
ion 9; (e) Re
2 H2O
is. Conseque
l equation a) phosphate
Calculate the
reduction
nditions:
tion.
allosteric ase reaction?
o pyruvate ch that their vate.
eaction 4.
ently,
?
7.
8.
9.
When [G
For the c
Pyruvate
∆ℰ°′ = (−
Accordin
and ∆G =
(a)
(b)
(c)
(d) A[lactateso that proceed
(a) P(in liver)
(b) Fenzyme tPFK step
GAP] = 10−4
coupled react
e + NADH +
−0.185 V) −
ng to Eq. 13-
= −nℱ∆ℰ (E
At the concene]/[pyruvate]in Part b theds spontaneo
yruvate kina, which ente
FBP is the prothat catalyzep of glycolys
M, [DHAP]
tion
+ H + → lact
(−0.315 V)
-8
q. 13-7). Sin
ntration ratio] increases, te reaction is ously in the
ase regulatioer glycolysis
oduct of the es Step 10. Tsis, they will
= 5.5 × 10−
tate + NAD
= 0.130 V.
nce two elect
os of Part a, ∆the reaction at equilibriuopposite dire
on is importaafter the PF
third reactioThis regulatol continue th
−4 M. Accord
D +
trons are tran
∆G is negati∆G increase
um (∆G = 0) ection.
ant for controFK step.
on of glycolyory mechanishrough the pa
ding to Eq. 1
nsferred in t
ive and the res even thoug
and in Part
olling the flu
ysis, so it acsm helps ensathway.
1-17, K = e−∆
the above rea
reaction procgh [NAD+]/[c ∆G is posi
ux of metabo
ts as a feed-sure that onc
∆G°′/RT
action, n = 2
ceeds as writ[NADH] alsitive and the
olites, such a
forward actice metabolite
2.
tten. As o increases
e reaction
as fructose
ivator of the es pass the
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, (c) −18 ATP.
2. (a) The equation for glycolysis is
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
10.
Cha
12.
13.
Cha
12.
13.
Cha
2.
P + 12 NAD
The convcould be hexokina
apter 16
Given theat 25°C aAssume smetaboli
Althoughthe label
apter 16
For the re= 1. Acco
With sucdehydrog
Animals Howeverthis oxalogluconeo
apter 17
How manshuttle inequivalen
DP + + 7 H2 O
version of cirmobilized, t
ase-catalyzed
e following and pH 7.0: standard conc control?
h animals caappears in g
Answers
eaction isociording to Eq
ch a large neggenase is like
cannot carryr, 14C-labeledoacetate may
ogenesis and
ny ATPs aren insect flighnts are transf
O → 12 NAD
rculating gluthe energy yd step that co
information[NAD+]/[NAnditions for C
nnot synthesglycogen ext
s
itrate + NADq. 13-1,
gative free eely to be a m
y out the netd acetyl-CoAy exchange w
d subsequentl
e synthesizedht muscle? Hferred into th
DPH + 12 H
ucose to lactyield would bonsumes AT
, calculate thADH] = 8, [αCO2 (∆G°′ is
size glucose tracted from
D+ ⇌ α-ketog
energy of reametabolic con
t synthesis ofA enters the with the cellly taken up b
d for every cHow does thihe matrix via
H + + 6 CO2 +
tate in the mube 3 ATP, siTP in the firs
he physiologα-ketoglutars given in Ta
from acetylits muscles.
glutarate + N
action under ntrol point.
f glucose frocitric acid c
lular pool of by muscle an
cytoplasmic Ns compare toa the malate
+ Pi
uscle generance phospho
st stage of gly
gical ∆G of tate] = 0.1 mable 16-2). I
-CoA, if a ra. Explain.
NADH + CO
physiologic
om acetyl-Coycle and is c
f oxaloacetatnd incorpora
NADH that o the ATP yi–aspartate sh
ates 2 ATP. Iorolysis of gycolysis.
the isocitratemM; and [isocIs this reactio
at is fed 14C-
O2 + H+, we a
cal condition
oA (to whichconverted to e to be convated into gly
participates ield when Nhuttle?
If muscle glyglycogen byp
e dehydrogencitrate] = 0.0on a likely si
-labeled acet
assume [H+]
ns, isocitrate
h acetate is coxaloacetat
verted to gluccogen.
in the glyceNADH reduci
ycogen passes the
nase reaction02 mM. ite for
tate, some of
and [CO2]
converted). te. Some of cose through
erophosphateing
n
f
h
e
3.
5.
7.
11.
Cha
2.
3.
5.
7.
Calculatebe synthe
Why is itnegative
The diffe1.4 pH unenergy rebe transpbiochemi
Explain w
apter 17
When NAto CoQ, bproduced
The relev
Since theoverall re
Since ∆G
∆ G °′ =
The maxmol−1/30
ℰ may direactantscomplex
For the tr
∆ G = 2.
The diffemembran
e ∆G°′ for thesized, assum
t possible foℰ°′ within a
erence in pHnits (externaeleased on trported to proical conditio
why compou
Answers
ADH particibypassing Cd when NAD
vant half-rea
e O2/H2O haleaction is
G°′ = −nℱ∆ℰ
= − (2)(96,4
imum numb.5 kJ ⋅ mol−1
iffer from ℰ°s and product
may “pull”
ransport of a
.3 RT[ pH ( i
erence in pHne, ∆Ψ is neg
he oxidation ming standar
r electrons toan electron-tr
H between theal side acidicransporting 1vide enough
ons)?
unds such as
s
ipates in the Complex I. ThDH participat
actions (Tabl
lf-reaction h
ℰ°′,
485 J ⋅ V − 1 ⋅
ber of ATP th1 = 6.6 mol A
°′, dependingts. In additioelectrons so
a proton from
in ) − pH ( o
H is −1.4. Singative.
of free FADrd condition
o flow from ransfer comp
e internal anc). If the mem1 mol of proth free energy
s DNP increa
glycerophoshus, about 2tes in the ma
le 13-3) are
has the more
mol − 1)(1.0
hat could be ATP/mol FA
g on the redoon, the tight o that the ove
m outside to
out )] + Zℱ ∆
nce an ion is
DH2 by O2. Ws and 100%
a redox cenplex?
nd external sumbrane potetons back ac
y for the synt
ase metaboli
sphate shuttl2 ATP are syalate–asparta
positive ∆ℰ
034 V ) = −
synthesizedADH2 oxidiz
ox center’s mcoupling bet
erall process
inside (Eq. 1
∆ Ψ
transported
What is the mconservation
nter with a m
urfaces of thential is 0.06 cross the memthesis of 1 m
ic rates.
le, the electroynthesized peate shuttle.
ℰ°′, the FAD
200 kJ ⋅ m
d under standzed by O2.
microenvirontween succe is spontane
17-1),
from the po
maximum nun of energy?
more positive
he inner mitoV (inside ne
mbrane? Homol of ATP (
ons of NADer NADH. A
half-reactio
ol − 1
dard conditio
nment and thessive electroous.
sitive to the
umber of AT?
ℰ°′ to one w
ochondrial megative) wha
ow many pro(assuming st
DH flow to FAAbout three A
n is reversed
ons is therefo
he concentraon transfers w
negative sid
TPs that can
with a more
membrane is at is the free otons must tandard
AD and thenATP are
d and the
fore 200 kJ ⋅
ations of within a
de of the
n
11.
Cha
1.
3.
5.
10.
11.
12.
13.
Cha
1.
3.
Since ∆Gprotons mbiochemi
DNP andthis gradithis ratio
apter 19
Explain wweakness
The first acid cycl
Why are limited?
On what palmitate
Explain w
Is the fatt
Compareto the ene
apter 19
A defect mitochoncannot gedietary g
The first (Sections
G°′ for ATP must be transical conditio
d related comient decrease relieves the
why individus. Why are th
three steps ole. Which ste
unsaturated
carbon atome?
why adipocy
ty acid show
e the energy ergy recover
Answers
in carnitine ndria for β oxenerate ATPlucose, are n
three steps os 16-3F–16-3
synthesis is sported to pr
ons.
mpounds disses the rate o
e inhibition o
uals with a hhese sympto
of β oxidatioeps are these
fats preferab
ms does the 1
ytes need glu
wn below lik
cost, in ATPred by degra
s
palmitoyl trxidation. Tis
P as needed. not readily av
of β oxidatio3H).
30.5 kJ ⋅ morovide the fr
sipate the prf synthesis o
of the electro
hereditary deoms more sev
on (Fig. 19-1e?
ble to satura
14CO2 used t
ucose as wel
kely to be syn
P equivalentading stearate
ransferase II ssues such aThe problemvailable.
on resemble
ol−1 and 30.5ree energy to
roton gradienof ATP, decron transport
eficiency of cvere during
12) chemical
ated fats for a
to synthesize
l as fatty aci
nthesized in
s, of synthese (a) to acety
prevents nos muscle tha
m is more sev
the reaction
5/13.8 = 2.2, o synthesize
nt required foreasing the Achain, causin
carnitine palfasting?
lly resemble
an individua
e malonyl-Co
ids in order t
animals? Ex
sizing stearayl-CoA and
ormal transpoat use fatty avere during
s that conve
between twone mole of
for ATP syntATP mass acng an increa
lmitoyl trans
three succe
al whose calo
oA from ace
to synthesize
xplain.
te from mito(b) to CO2.
ort of activatacids as metaa fast becau
rt succinate
wo and three mf ATP under
thesis. The dction ratio. Dase in metabo
sferase II hav
ssive steps o
oric intake m
etyl-CoA app
e triacylglyc
ochondrial ac
ted fatty acidabolic fuels tse other fuel
to oxaloacet
moles of standard
dissipation ofDecreasing olic rate.
ve muscle
of the citric
must be
pear in
cerols.
cetyl-CoA
ds into the therefore ls, such as
tate
f
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 acetyl-CoA 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 low-protein 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.