Chapter 4 Problems
Page 1 of 7
11/1/2007
4.2 Frog muscle contains ATP, ADP and phosphate at respective concentrations of 1.25
mM, 0.5 mM and 2.5 mM. (a) Compute ΔG’ for ATP hydrolysis (assume T = 298 K and
pH = 7); (b) What is the maximum amount of mechanical work the muscle can do per
mol of ATP? (c) Creatine kinase catalyzes the phosphorylation of ADP to ATP by
converting phosphocreatine to creatine. The hydrolysis of phosphocreatine to creatine
plus phosphate has a ΔG0’ = -43.1 kJ/mol. Calculate K of the creatine kinase catalyzed
reaction.
a)
ATP(aq) + H2O(l) ↔ ADP(aq) + Pi(aq)
We need to know ΔG0’ for the hydrolysis of ATP. I found it in Voet & Voet’s
Biochemistry. It is ΔG0’ = -30.5 kJ/mol. The ΔG’ for different (non-standard-state)
concentrations of reactants and products for this reaction is
ΔG ' = ΔG 0 '+ RT ln
= −30.5
[ ADP][ Pi ]
kJ
J
(2.5 × 10 −3 )(0.5 × 10 −3 )
= −30.5
+ (8.314
)(298 K ) ln
[ ATP ]
mol
K • mol
(1.25 × 10 − 3 )
kJ
kJ
kJ
kJ
kJ
) ln(1.0 ×10 − 3 ) = −30.5
+ (2.48
− 17.1
= −47.6
mol
mol
mol
mol
mol
b) The maximum amount of non-PV work that can be extracted from a reaction is ΔG’, so
wmax, by the system = -wmax, on the system = -ΔG’ = 47.6 kJ/mol.
c) Here we need to add the two reactions we know about to get the final reaction. The
ΔG0’s add just like the reactions:
ATP(aq) + H2O(l) ↔ ADP(aq) + Pi(aq)
ΔG0’ = -30.5 kJ/mol
so
add
ADP(aq) + Pi(aq) ↔ ATP(aq) + H2O(l)
phos-Cr(aq) + H2O(l) ↔ Cr(aq) + Pi(aq)
ΔG0’ = 30.5 kJ/mol
ΔG0’ = -43.1 kJ/mol
net:
phos-Cr(aq) + ADP(aq) ↔ Cr(aq) + ATP(aq)
ΔG0’ = (30.5 - 43.1) kJ/mol
= -12.6 kJ/mol
From this we know K:
0
ΔG ' = − RT ln K or
−12,600kJ / mol
ΔG 0 '
−
K = e RT = e (8.314 J / K • mol )( 298 K ) = e 5.086 = 162
−
Chapter 4 Problems
Page 2 of 7
11/1/2007
4.4. In the red cell glucose transport is powered by ATP hydrolysis (ΔG0’ = -31.0 kJ/mol
is given here). Assume that transport is 100% efficient and that it is given by
ATP(aq) + H2O(l) + 2 glucose(out) ↔ ADP(aq) + Pi(aq) + 2 glucose(in)
(a) If [ATP] = [ADP] = [Pi] = 10 mM is held constant by cellular metabolism, what is the
maximum value of [gluc(in)]/[gluc(out)]?; (b) If the glucose transport stoichiometry
changed to 1, how would this affect the answer in (a)?; (c) In a real cell, the activity
coefficient for glucose is much less than 1. Would this increase or decrease the answer to
(a) if all other activity coefficients were assumed to equal 1?
a) We know the free energy available from ATP hydrolysis:
ΔG ' = ΔG 0 '+ RT ln
= −31.0
[ ADP][ Pi ]
kJ
J
(1×10 −2 )(1×10 −2 )
= −31.0
+ (8.314
)(298 K ) ln
[ ATP]
mol
K • mol
(1× 10 − 2 )
kJ
kJ
kJ
kJ
kJ
.
) ln(1.0 ×10 − 2 ) = −31.0
+ (2.48
− 11.4
= −42.4
mol
mol
mol
mol
mol
This is available for the transport of glucose. The free energy available out of ATP
hydrolysis equals the amount into glucose transport, i.e. ΔG’ATP = -ΔG’glucose. That is, ΔG’glucose = +42.4 Kj/mol. When the glucose transport is at equilibrium with no energy
input, [gluc(in)] = [gluc(out)] (by practical experience), so that means that K = 1 for this
reaction and thus ΔG0’ = 0. Therefore, for the glucose transport we have that
ΔG ' = ΔG 0 '+ RT ln
[ gluc(in)]2
[ gluc(out )]2
2
or ΔG ' = +42.4 kJ = 0 + RT ln [ gluc(in)] which says that
2
mol
[ gluc(out )]
42 , 400 kJ / mol
[ gluc(in)]
= e (8.314 J / K •mol )( 298 K ) = e17.1 = 5200 .
[ gluc(out )]
b) The answer here is done similarly to that for (a) except the concentrations of glucose are
not squared, and thus the final answer does not have the square root:
42, 400 J / mol
[ gluc(in)]
= e (8.314 J / K • mol )( 298 K ) = e17.1 = 2.67 ×10 7 .
[ gluc(out )]
c) Because the activity coefficient enters only in the glucose(out) ↔ glucose(in) reaction, it
will incorporate into the equation for the energetics of the glucose transport reaction:
2
[ gluc(in)]2 γ in
kJ
so that
ΔG ' = +42.4
= 0 + RT ln
mol
[ gluc(out )]2
42,400 J / mol
[ gluc(in)]γ in
[ gluc(in)] 5200
.
= e (8.314 J / K • mol )(298 K ) = e17.1 = 5200 or
=
[ gluc(out )]
[ gluc(out )] γ in
So if γ < 1, the in/out ratio will be bigger, i.e the concentration gradient is larger.
Chapter 4 Problems
Page 3 of 7
11/1/2007
4.6. (a) Equilibrium constants are given for the reaction of 3-PG to 2-PG at pH 7. Calculate
ΔH0’ from a lnK vs. 1/T plot. (b) What is ΔG0’? (c) What are the concentrations of 3-PG
and 2-PG when 0.15 M PG comes to equilibrium?
a) Convert K to lnK and T to 1/T (don’t forget to change temperature units to Kelvin). This
gives a straight line.
-1.84
-1.845
-1.85
lnK
-1.855
-1.86
-1.865
-1.87
-1.875
-1.88
0.0031
0.0032
0.0033
0.0034
0.0035
0.0036
0.0037
1/T (1/K)
The slope of the line is -ΔH0’/R. It can be gotten from least-squares fitting or by eye.
The least squares value is -60.6 K. So that says that ΔH0’ = -R(-60.6 K) = (8.314
J/mol·K)(60.6 K) = 503.8 J/mol.
b) We will use ΔG 0 ' = − RT ln K , so we need to estimate K at 25 ºC. We could extrapolate
from the 20 and 30 ºC values in the table, but at better way is to use our least squares fit
from (a). There the slope was -60.6 and the intercept was -1.65. So for 298 K, the value
of lnK is -1.65 -60.6(1/298) = -1.85, so K = e-1.85 = 0.157. This says that
ΔG 0 ' = − RT ln K = −(8.314 J / K • mol )(298 K ) ln(0.157) = 4.59kJ / mol
c) For the reaction [3-PG] ↔ [2-PG] we know the ratio of [2-PG]/[3-PG] = 0.157 where the
Keq came from the answer to (b). We also know that [3-PG] + [2-PG] = 0.15 M. So
plugging the second equation solved for [2-PG] into the first gives
0.15 − [3 − PG ]
= 0.157 . Solving for [3-PG] gives (1.157)[3-PG] = 0.15 M, or [3-PG] =
[3 − PG ]
0.130 M. That means that [2-PG] = 0.02 M. Check to see if this is correct:
(0.02)/(0.130) = 0.154, which is close to the expected Keq, so we must be close to correct.
Chapter 4 Problems
Page 4 of 7
11/1/2007
4.12 (a) From the ionization constant calculate ΔG0 for the ionization of acetic acid in water
at 25 ºC. (b) What is ΔG when the reaction is at equilibrium? (c) What is ΔG when [H+]
= 0.1 mM, [OAc-] = 10 mM and [HOAc] = 1 M for the reverse reaction? (d) What is ΔG
when the concentrations and reaction are as in (c) except [HOAc] = 10 μM? (e) What is
ΔG for transferring 1 mol of acetic acid from a 1 M solution to a 10 μM solution?
a) The ionization constant for the reaction [HOAc](aq) ↔ [H+](aq) + [OAC-] is given in
Ex. 4.6 as 1.80 × 10-5. So
ΔG 0 ' = − RT ln K = −(8.314 J / K • mol )(298 K ) ln(1.80 ×10 −5 ) = 27.1kJ / mol
b) ΔG = 0 at equilibrium.
c) The reaction is the protonation of acetate, just the opposite of the one in (a). So to
calculate ΔG for the ionization, we need ΔG0 for the protonation reaction, which from a)
is -27.1 kJ/mol (we’ve reversed the reaction). Then
ΔG = ΔG 0 + RT ln
[ HOAc]
+
−
= −27.1
kJ
J
(1)
+ (8.314
)(298 K ) ln
−
4
mol
K • mol
(1× 10 )(1×10 − 2 )
[ H ][OAc ]
kJ
kJ
kJ
kJ
kJ
for the protonation
= −27.1
+ (2.48
+ 34.3
= 7 .2
) ln(1.0 × 10 6 ) = −27.1
mol
mol
mol
mol
mol
reaction.
d) This is mostly repetitious of part c:
ΔG = ΔG 0 + RT ln
[ HOAc]
= −27.1
kJ
J
(1×10 −5 )
+ (8.314
)(298 K ) ln
mol
K • mol
(1× 10 − 4 )(1×10 − 2 )
[ H + ][OAc − ]
kJ
kJ
kJ
kJ
kJ
= −27.1
+ ( 2.48
) ln(10) = −27.1
+ 5 .7
= −21.4
mol
mol
mol
mol
mol
e) This is the free energy change between (d) and (c):
ΔGtransfer = ΔG10μM – ΔG1M = (-21.4 kJ/mol) – (7.2 kJ/mol) = -28.6 kJ/mol. For 1 mol, the
free energy change is -28.6 kJ.
Chapter 4 Problems
Page 5 of 7
11/1/2007
4.17 A single-strand DNA forms a self-complementary loop. (a) The K1 for this reaction is
0.86. If we start with 1 mM DNA, what will be the concentrations of open and looped
DNA at equilibrium? Will increasing the initial DNA concentration change the looped
fraction? (b) At 37 ºC K1 = 0.51. Calculate ΔH0, ΔS0 and ΔG0 at 37 ºC. (c) As the DNA
concentration increases, another reaction occurs: double stranded DNA forms. At 25 ºC
K2 = 1×10-2. Calculate the concentrations of all three species present for a total DNA
concentration of 0.1 M.
a) K1 = [loop]/[ss] and [loop] + [ss] = 1×10-3 M. So
1×10 −3 − [ ss ]
= 0.86 . This gives [ss] =
[ ss ]
5.38×10-4 M and [loop] = 1×10-3 M - 5.38×10-4 M = 4.62×10-4 M. There will be no effect
of changed concentration because the ratio of [loop] to [ss] is fixed.
b) ΔG 0 = − RT ln K = −(8.314 J / K • mol )(310 K ) ln(0.51) = 1.74kJ / mol
We can get the enthalpy change from a van’t Hoff plot, or by direct calculation:
ΔH 0 = − R
(ln K 2 − ln K1 )
(−0.673) − (−0.151)
Δ(ln K )
= −R
= −(8.314 J / K • mol )
= −33.4kJ / mol
−3
−3
1
1
1
Δ( )
(
−
)
(
3
.
23
10
3
.
36
10
)
×
−
×
T
T2
T1
Finally, ΔS0 = (ΔH0 - ΔG0)/T = (-33.4 kJ/mol – 1.74 kJ/mol)/310 K = -113 J/K·mol.
c) Mass conservation: [ss] + [loop] + 2 [ds] = 0.1 M. From the looping reaction we have
that [loop] = 0.86[ss]. From the double-strand reaction we have [ds]/[ss]2 = 1×10-2 or
[ds] = 1×10-2[ss]2. Plug these concentrations into the mass conservation equation:
[ss] + 0.86[ss] + 2×10-2[ss]2 = 0.1. This is a quadratic, the solutions for which are
[ ss ] =
− 1.86 ± (1.86) 2 − 4(2 × 10 − 2 )(−0.1)
2(2 × 10
−2
)
= 5.37 ×10 − 2 M where the negative square root
is impossible (negative concentration would result). Plug this in to the equilibrium
relations to get the other two concentrations: [loop] = 0.86(5.37×10-2 M) = 4.62×10-2 M.
Also, [ds] = 1×10-2(5.37×10-2)2 = 2.88×10-5 M.
Chapter 4 Problems
Page 6 of 7
11/1/2007
4.22. Conversion of β-hydroxybutyrate, bHB, to acetoacetate, AAc, uses O2 as the ultimate
oxidizing agent. (a) Using standard reduction potentials, calculate ΔG0’ and K at pH 7
and 25 ºC. (b) What is the ratio of [AAc] to [bHB] at equilibrium if P02 = 0.2 atm?
a) We need to combine the bHB/AAc reaction with the O2/H2O reaction to form a net
reaction that we want:
AAc + 2 H+ + 2 e- → bHB
So bHB → AAc + 2 H+ + 2 e
O2 + 4 H+ + 4 e- → 2 H2O
E0’ = -0.346 V
E0’ = 0.346 V
E0’ = 0.816 V
We add the electrode potentials (they are intensive), but need to balance the electron
flow:
bHB + 1/2 O2 → AAc + H2O
E0’ = 1.162 V
We know that ΔG0’ = -nFE = -(2) (96.485 kJ/V·mol) (1.162 V) = -224 kJ/mol.
− 224kJ / mol
ΔG 0 '
−
(8.314 J / K • mol )( 298 K )
RT
K =e
=e
= e 90.4 = 1.84 ×1039
−
b) The equilibrium constant is K =
[ AAc]
[bHB]( PO 2 ) 0.5
so
[ AAc]
= K ( PO 2 ) = 1.84 × 1039 (0.2) = 8.23 ×1038
[bHB]
Chapter 4 Problems
Page 7 of 7
11/1/2007
4.28 The half-cell reactions for formation of water and Cys-Cys are
2H+ + ½ O2 + 2e- → H20
E0’ = 0.816 V
2H+ + Cys-Cys + 2e- → 2Cys
E0’ = -0.34 V
(a) What is the ratio [Cys-Cys]/[Cys] at 298 K for a 0.01 M solution of Cys in
equilibrium with the O2 in the air (PO2 = 0.2 atm)? (b) What is ΔG for the reaction when
the activities are at equilibrium values?
a) We need to get the net reaction and the standard potential for the net reaction:
2H+ + ½ O2 + 2e- → H20
2Cys → 2H+ + Cys-Cys + 2e-
E0’ = 0.816 V
E0’ = 0.34 V
E0’ = 1.156 V
net:2Cys + ½ O2 → Cys-Cys
We get K from the relation ln K =
nF 0'
( 2e)(96.485kJ / eV )
E =
1.156V = 90.4 so
(
8
.
314
J / K • mol )( 298 K )
RT
K = 1.78×1039. This is extremely favorable, so the reaction at equilibrium is nearly
complete, i.e. [Cys-Cys] = 0.005 M. We can compute [Cys], then:
K=
[Cys − Cys]
2
[Cys] [ PO 2 ]
0.5
or [Cys] = [Cys − Cys] =
0.5
K [ PO 2 ]
0.005
39
(1.78 ×10 ) 0.2
Thus, [Cys-Cys]/[Cys] = (0.005 M)/(2.51×10-21 M) = 2×1018.
b) As always, ΔG = 0 at equilibrium.
= 2.51× 10 − 21 M .