CHAPTER 4
1.
a)
⎡G-1-P ⎤ ⎡ ADP ⎤ (770)(10−4 )
⎦⎣
⎦=
= 77.0
⎡G ⎤ ⎡ ATP ⎤
10−3
⎣ ⎦⎣
⎦
K=⎣
°′ = −RT ln K = − 8.314 298 ln77 = −10.76kJ mol−1
Δ rG298
( )( )
Δ rG°′ ⎛⎜⎝ kJ mol-1⎞⎟⎠
b)
G + ATP → G −1− P + ADP ; −10.76
ADP + Pi → ATP + H 2O
; +31.0
G + Pi
→ G −1− P + H 2O
;
20.2
ln K' = −Δ rG °′ RT = − ( 20,200) (8.314)( 298) = − 8.17
K' = 2.9×10−4
2.
a)
ATP Hydrolysis: ATP→ADP+ Pi
Following Eq. 4.15:
⎛⎡
ADP ⎤⎦ ⎡⎣ Pi ⎤⎦ ⎞⎟
⎜⎝ [ATP] ⎟⎠
Δ rG′ = Δ rG°′ + RT ln ⎜ ⎣
⎡⎛
−3⎞ ⎛
−3⎞ ⎤
⎢ ⎜⎝ 0.50×10 ⎟⎠ ⎜⎝ 2.5x10 ⎟⎠ ⎥
⎛
⎞
−3
⎥
=− 31.0+ ⎜⎝ 8.314×10 ⎟⎠ 298 ln ⎢
⎛ 1.25×10−3⎞
⎢
⎥
⎜
⎟
⎢⎣
⎥⎦
⎝
⎠
(
)
=−48.1kJ mol–1
b)
−w max,rev = Δ rG′ =48.1 kJ mol-1
(Eq. 4.55)
ΔG °′ (kJ mol–1)
c)
PC + H 2O → C+ Pi
; − 43.1
Pi + ADP → ATP+ H 2O
;
PC+ ADP → C+ ATP
; −12.1
31.0
ln K' = −Δ rG °′ RT =12,000 (8.314)( 298) = 4.88
K' =132
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
3.
ΔG °′ (kJ mol–1)
a)
b)
2
G − 6 − P + ADP → ATP + G
ATP + H 2O → ADP + Pi
; 16.7
; − 31.0
G − 6 − P+ H 2O→ G + Pi
; −14.3
ln K' =−Δ rG°′ RT = (16,700) (8.314)( 298) = 6.74
K' =846= ( ⎡⎣G − 6 − P ⎤⎦ ⎡⎣ ADP ⎤⎦) ( ⎡⎣G ⎤⎦ ⎡⎣ ATP ⎤⎦)
At equilibrium ⎡⎣G − 6 − P ⎤⎦ ⎡⎣G ⎤⎦ =846 when[ADP]=[ATP]
c)
G + Pi → G − 6 − P+ H 2O
(activity of H2O is 1)
ln K' =−Δ rG°′ RT =−14,300 ( 298)(8.314) =−5.77
K' =3.11×10−3 = ⎡⎣G-6-P ⎤⎦ ( ⎡⎣G ⎤⎦ ⎡⎣ Pi ⎤⎦)
⎡G-6-P ⎤ ⎡G ⎤ =3.11×10−3 ⎡ P ⎤ =3.11×10−5
⎣
⎦ ⎣ ⎦
⎣ i⎦
4.
a)
ATP+2G(out) → 2G(in)+ADP+ Pi ; Δ rG°′ =− 31.0kJ mol–1
ln K =−ΔG°′ RT =31,000 (8.314)( 298) =12.5
K =2.72×105 =[G(in)]2 [ADP][Pi ] [G(out)]2 [ATP]
1/2
⎡G(in) ⎤ ⎡G(out) ⎤ = ⎡(2.72×105 )(10−2 ) (10−2 )(10−2 ) ⎤
⎥⎦
⎣
⎦ ⎣
⎦ ⎢⎣
b)
⎡G(in) ⎤ ⎡G(out) ⎤ =2.72×107
⎣
⎦ ⎣
⎦
c)
If γ G(in) <1, then [G(in)] > 5200[G(out)]
=5,200
This would increase the concentration gradient.
5.
a)
Glu + Pyr ←⎯→ Ket + Ala ;
Ket + Asp ←⎯→ Glu + Ox;
Pyr + Asp ←⎯→ Ala + Ox;
°′ (J mol−1)
Δ rG303
−1004
4812
3808
K = [Ala][Ox] [Pyr][Asp]
°′ RT = −3808 (8.314)(303) = −1.51; K = 0.215
ln K ′ = − ΔG303
′
b)
°′ + RT ln[(10−4 )(10−5 ) (10−2 )(10−2 )]
Δ rG303
′ = Δ rG303
= 3808+ (8.314)(303)(−11.5) = −25,200J mol−1
The reaction proceeds in the forward direction under cytoplasmic conditions.
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
6.
a)
Least squares fit of lnK vs. 1/T gives: (Eq. 4.53)
ln K = −1.652 − 60.6(1 T )
ΔH ° = −(slope)(R) = 60.6 R = 504 J mol−1
b)
At 25˚C, ln K = −1.856, K = 0.1564 ; ΔG° = −RT ln K = 4598 J
c)
Notice it is possible to write two equations in two unknowns:
[2PG]
, [2PG] +[3PG] = 0.150 , then:
[3PG]
[2PG] = [3PG]K = (0.150 −[2PG])K
0.150 K (0.150)(0.1564)
[2PG] =
=
= 0.0203
1+ K
1.1564
[3PG] = 0.130
K=
7.
a)
All data in appendix A.5.
° = Δ H ° (SO ) − Δ H ° (SO ) − (1 2)Δ H ° (O )
Δ r H 298
3
2
f 298
f 298
f 298 2
= −395.72 + 296.83= −98.89kJ mol-1
° = S°(SO ) − S°(SO ) − (1 2)S°(O )
Δ r S298
3
2
2
= 256.87 − 248.33− (1 2)(205.25)
= −94.09 J mol−1 K −1
° = Δ H°−T Δ S° = −98.89 kJ mol−1 − (298 K)( − 0.09409 kJ mol−1 K −1)
Δ rG298
r
r
= −70.9 kJ mol−1
b)
c)
−1
−1 −1
°
K = e−ΔG / RT = e(70.9 kJ mol /(.008314 kJ mol K × 298 K)
[SO3]
K = 2.63×1012 =
[O2 ]1 2[SO2 ]
[SO3]
= K[O2 ]1/2 = 2.63×1012 (0.21)1/2 =1.20 ×1012
[SO2 ]
° = Δ H°(H SO ) − Δ H°(H O) − Δ H°(SO )
Δ r H 298
2
4
2
3
f
f
f
= −735.13+ 241.82 + 395.72
= −97.6 kJ mol−1
ΔS ° = 298.8−188.93− 256.87 = −147.0 J mol−1 K −1
ΔG ° = ΔH°−T ΔS ° = −53.79 kJ mol−1
-1
-1 -1
K = e(−ΔG°/ RT ) = e53.79kJ mol /(0.008314kJ mol K x 298K) = 2.66 ×109
[H 2SO4 ]
=
[H 2O][SO3]
[H 2SO4 ]
= K[H 2O] = (2.66 ×109 )(0.031) = 8.24 ×107
[SO3]
3
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
8.
a)
Δ rG° = −RT ln K = −(8.314 ×10−3)(298) ln(4) = −3.43 kJ mol−1
b)
Δ rG = 0 at equilibrium
c)
Δ rG = −3.43 kJ mol −1
d)
Δ rG = (2mol)(−3.43 kJ mol-1) = −6.86 kJ
e)
Δ r H° = (RTiT2 / ΔT )ln(K2 Ki ) (Eq. 4.53)
4
= (8.314 ×10−3kJ mol-1 K -1)(308K)(298 K 10K)ln(2) = 52.9 kJ mol−1
Δ r S° = (Δ r H°− Δ rG°) T
f)
= (52.9 + 3.4) ×103 298 =189 J K –1 mol−1
9.
Δ rG = Δ rG°+ RT ln([ADP][Pi ] [ATP])
a)
(Eq. 4.15)
= −31,000 + (8.314)(298)ln[(10−4 )(10−1) (10−2 )]= −48.1 kJ mol−1
b)
−wmax, rev = −Δ rG = 48.1 kJ mol−1
c)
Δ rG = −31,000 + (8.314)(298)ln[(10−1)(0.25) (10−7 )]= −206 J mol−1
−wmax, rev = −Δ rG = 206 J mol−1
10.
° = −394.36 +137.01= −257.35kJ mol-1
CO(g) + 12 O2 (g)→ CO2 (g); Δ rG298
ln K = − Δ rG° RT = 257,350 (8.314)(298) =103.87
K =1.29 ×1045 = PCO
PCO
1
2
PCO PO2
2
= (3×10−4 ) [(0.2)1 2 (1.29 ×1045 )]= 5.2 ×10−49 bar
No need to worry about the equilibrium concentration.
⎛
11.
a)
⎞
⎛ ° ⎞
⎛ ° ⎞ d ⎜ S° ⎟
°
°
°
H
H m,i
dH m,i
dSm,i
m,i ⎠
d ⎜ µm,i
⎟ = d ⎜ m,i ⎟ − ⎝
=
−
+
−
dT ⎜⎜ T ⎟⎟ dT ⎜⎜ T ⎟⎟
dT
TdT
dT
T2
⎝
Since
⎠
⎝
⎠
⎛ ° ⎞
°
°
°
dH m,i
dSm,
H°
d ⎜ µm,i
i Cm, p
⎟ = − m, i
°
= Cm,
and
=
,
only
the
first
term
survives:
p
T
dT
dT
dT ⎜⎜ T ⎟⎟
T2
⎝
b)
d ln K ⎛ d ln K ⎞ ⎛ dT ⎞ ⎛ Δ r H° ⎞ ⎛ 1 ⎞ −Δ r H°
=
=
=
(see eq. 4.51)
d(1 T ) ⎜⎝ dT ⎟⎠ ⎜⎝ d(1 T ) ⎟⎠ ⎜⎝ RT 2 ⎟⎠ ⎜⎝ −T −2 ⎟⎠
R
⎠
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
12.
a)
From Table 4.2, pK=4.76, so K =10−4.76 and
ln(K ) = ln(10−4.76 ) ≅ 2.303(−4.76) = −10.96
Δ rG° = −RT ln K = −(8.314)(298)(−10.96) = 27.15 kJ mol−1
b)
Δ rG = 0 at equilibrium
c)
Note: text should indicate 1 M for [HOAc]
Δ rG = Δ rG°+ RT ln[aHOAc aH+ aOAc− ]
5
= −27,155+ (8.314)(298)ln[(1) (10−2 )(10−4 )]= 7.07 kJ mol−1
13.
d)
Δ rG = −27,155+ (8.314)(298)ln10 = −21.45 kJ mol−1
e)
− 7.07 − 21.45= −28.5 kJ mol−1 (alt :ΔG = RT ln(10−5 ) = −28.5)
a)
Total of positive charges must equal total of negative charges.
(example: for 1 M MgCl2 , [Cl-]=2 M and [Mg2+]=1 M, so [Cl-]=2[Mg2+])
3−
[Na + ]+[H + ]= [OH − ]+[H 2PO−4 ]+ 2[HPO2−
4 ]+ 3[PO4 ]
b)
Look for approximations to simplify this and similar problems. A common strategy is to
identify species that are abundant or rare. Using K2 and pH=7:
−
+
−8
−7
[HPO2−
4 ]/[H 2 PO4 ]= K2 /[H ]= (6.2 ×10 ) / (10 ) = 0.62 .
So HPO42- and H2PO4- are present in similar amounts, and based on K1 and K3:
2−
−
[H 3PO4 ] and [PO3−
4 ]<<[HPO4 ] or [H 2 PO4 ] at pH 7 . Therefore:
−
−
−
[HPO2−
4 ]+[H 2 PO4 ]≅ 0.100M = 0.62 [H 2 PO4 ]+[H 2 PO4 ]
[H 2PO−4 ]= 0.100M 1.62 = 0.0617M
−
[HPO2−
4 ]= 0.62[H 2 PO4 ]= 0.0383 M
[H 3PO4 ]= (10−7[H 2PO−4 ]/ (7.1×10−3) = 8.7 ×10−7 M
2−
+
−13) / (10−7 ) =1.72 ×10−7 M
[PO3−
4 ]= [HPO4 ]K3 [H ]=(0.0383)(4.5×10
[Na + ] [H 2PO−4 ]+ 2[HPO2−
4 ]= 0.138 M from charge balance
Note that the small concentrations of H3PO4, PO43- justify the initial approximation.
c)
ln K(T2 ) = ln K(T1) − Δ r H°(T2−1 −T1−1) R
=−16.60 − (4,150)(−1.299 ×10−4 ) / 8.314 = −16.53 ( Table 4.2)
K(310K ) = 6.62 ×10−8
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
14.
a)
6
pK1 =1.82; pK2 = 6.00; pK3 = 9.16
At pH = 9.0, K3 dominates and HisH and His− are majority species (Fig. 4.12); thus
[His− ] = K3 = 10−9.16 = 0.6918
[HisH] [H + ] 10−9.00
[His− ]+[HisH]≅ 0.200 M
0.6918[HisH]+[HisH]= 0.200
[HisH]= 0.200 1.6918 = 0.118M
[His− ]= 0.082M
Solve for rest with remaining equilibria and charge balance:
+
+
+]
−6.00 , K = [His ][H ] =10−1.82
K2 = [HisH][H
=10
1
[His+ ]
[His2+ ]
(0.118)(10−9 )
[His+ ]=
=1.18×10−4 M
(10−6 )
(1.18×10−4 )(10−9 )
[His2+ ]=
= 7.81×10−12 M
(10−1.82 )
[OH − ]= K w 10−9 =10−5 M
[Na + ]= [His− ]+[OH − ]−[H + ]−[His+ ]− 2[His2+ ]= 0.082M
b)
Recalculate all K’s at the new temperature; K3 still dominates the chemistry near this pH:
Δ r H1° = 0 kJ mol−1; Δ r H 2° = 29.9 kJ mol−1; Δ r H3° = 46.6 kJ mol−1
⎡ K(40° C) ⎤ −Δ H° ⎛
1 − 1 ⎞ = (0.01934)(Δ H°)
r
⎥=
⎜
r
⎢ K(25° C) ⎥
R ⎝ 313 298 ⎟⎠
⎣
⎦
ln ⎢
K1(313 K) =10−1.82 ,
K2 (313 K) =1.783×10−6 ,
Kw (313 K) = 2.944 ×10−14 (ΔH w = 55.82 kJ mol−1 )
K3(313 K) = 2.463×10−9.16
Since [His-],[His] are the only significant histidine species:
[Na + ]≅ [His− ]since other charged species small.
[Na + ]= 0.082 M (same as part a) so [His- ]= 0.082 M
[His− ]+[HisH]≅ 0.200 M , [HisH]= 0.118 M
[H + ]= (0.118 0.082) K3 = 2.45×10−9 M ; pH = 8.61
c)
[His+ ]=
(0.118)(2.45×10−9 )
=1.62 ×10−4 M
(1.783×10−6 )
(1.62 ×10−4 )(2.45×10−9 )
= 2.63×10−11 M
(10−1.82 )
[OH − ]= (2.944 ×10−14 ) / (2.45×10−9 ) =1.20 ×10−5 M
[His2+ ]=
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
15.
a)
7
°' + RT ln([GDP][P ]/[GTP])
Δ rG'298 = Δ rG298
i
= −RT ln K'298 + RT ln[(5×10−3)(15×10−3) / (50 ×10−3)]
= (8.314)(298)ln[(1.5×10−3) (1.9 ×105 )]= −46.2 kJ mol−1
b)
Δ(Δ rG'298 ) = RT ln(1 2) = −1.7 kJ mol−1
c)
To attain equilibrium, the reaction mixture in (a) is spontaneous towards the right hand
side. Thus:
[GTP]= 0.050 − y, [GDP]= 0.005+ y, [Pi ]= 0.015+ y
A large value of K’ means that the final [GTP] will be very small, so recast the algebra
with x=0.050-y and then:
[GTP]= x, [GDP]= 0.055− x, [Pi ]= 0.065− x where x is small
K298
′ =1.9 ×105 = (0.055− x)(0.065− x) x ≅ (0.055)(0.065) x
[GTP]= x ≅ (0.105)(0.115) (1.9 ×105 ) =1.88×10−8 M
[GDP]= 0.055 M
[Pi ]= 0.065 M
K323 = (2.57 ×10−6 ) (9.97 ×10−4 ) = 2.58×10−3
16.
K373 = (1.4 ×10−4 ) (8.6 ×10−4 ) = 0.1628
Δ r H° = −R(T2−1 −T1−1)−1 ln(K373 / K323)
=− (8.314)(−2410)(4.146) = 83.05 kJ mol−1
17
a)
Set: [SS]=1.0 ×10−3 − x; [LOOP]= x;
K1 = [LOOP] [SS]= 0.86 = x (1×10−3 − x)
x = 8.6 ×10−4 − 0.86x
[LOOP]= x = 8.6 ×10−4 1.86 = 4.62 ×10−4 M
[SS]=10−3 − x = 5.38×10−4 M
If the solution is ideal, increasing the concentration will have no effect on the fraction of
hairpin loop.
b)
° = −RT ln K
−1
Δ rG310
310 = −(8.314)(310)ln0.51=1.74 kJ mol
Δ r H310 = −R(T2−1 −T1−1)−1 ln(K310 K298 )
= −(8.314)(−7,698)ln(0.51/ 0.86) = −33.4 kJ mol−1
° = (ΔH°− ΔG°) / T = (−33.4 +1.74) / 310 = −0.102 kJ K −1 mol−1
Δ r S310
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
18.
8
c)
Recognize three equations in three unknowns:
[SS]+[H]+ 2[DS]= 0.100M
[DS]= K2 ([SS])2
[H]= K1[SS]
Solve a quadratic (use the positive root) for [SS]:
[SS]+ 0.86[SS]+ 2(1.00 ×10−2)([SS])2 = 0.100 M
1.86[SS]+ 2.00 ×10−2[SS]2 = 0.100 M
[SS]= 0.0537 M
[H]= 0.0462 M
[DS]= 2.9 ×10-5 M
a)
The chemical potential of the H2O (l) decreases upon addition of some salt since now
χ H O(l) <1.0 ; at the instant the salt dissolves, the liquid water has a lower chemical
2
potential than the pure solid ice. Next, some of the ice spontaneously melts, which is
endothermic and therefore cools the entire vessel. As the solid ice cools its chemical
potential decreases and can approach that of the water in the saline solution. As the solid
ice melts, it also dilutes the saline solution and raises the chemical potential of the liquid
water. The new equilibrium is that temperature for which the pure solid ice has the same
chemical potential as the water in the saline solution.
b)
The persistent vacuum pump operation means that the partial pressure of water vapor in
the headspace of the vial will always be extremely low and thus so will the chemical
potential of the water vapor; the system will tend towards the low chemical potential of
the water vapor throughout the experiment. Initially, fast evaporation (endothermic) will
freeze the water faster than heat can re-enter the glass vial from the surroundings; then
sublimation takes over, further cooling the solid to temperatures below 273 K, until all of
the water is gone. Even though the ice will cool considerably and its chemical potential
will decrease accordingly, the vacuum pump ensures that: µH O(s),T <273 > µH O(g),T <273 .
2
19.
⎛
⎞
Set aH O(l) = [H 2O]/1 M = ⎜ 1000 g * 1 mol ⎟ /1 M = 55.5 M /1 M = 55.5 ;
18.02 g ⎠
2
⎝ L
Use molarities for activities:
a + a - ([H + ]/1 M)([OH - ]/1 M)
K = H OH =
aH O(l)
([H 2O]/1 M)
2
=
(1.0 ×10−7 )(1.0 ×10−7 )
=1.80 ×10−16
55.5
K333 can be estimated by assuming constant enthalpy of reaction:
−Δ r H −1 −1
(T2 − T1 ) = ln(K333 K298 )
R
−(55.82 kJ mol-1)(−3.53×10−4 K -1)
= ln(K333 /1.80 ×10−16 )
(8.314 ×10−3 kJ mol-1 K -1)
K333 = 1.92 ×10−15
2
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
20.
If temperature and volume are held constant for the system (the elastomer), then these are the
natural variables for defining the chemical potential based on the Helmholtz energy, A. Thus
Δ r A represents the maximum available non-pV work that could be extracted from this taut
elastomer.
21.
K obs = K [ Na + ] − χψ
log Kobs = − χψ log[Na + ]+ constant
9
Each set of data is well fit by a straight line
n = 4, slope = −3.68;
χ 4 = 4.18
n = 5, slope = −4.79;
χ5 = 5.45
assuming ψ = 0.88
The ratio χ5 / χ 4 =1.30 which is close to 5/4.
This indicates that the length of the binding site is proportional to the number of lysines in the
oligomers. Furthermore, each lysine appears to neutralize approximately 1 phosphate group.
22.
Equilibrium: K =
[ D]
;
[S] 2
Definitions to change variables:
f = 2[D] = fraction of single strands in double stranded form, so
c
[D]= fc 2 .
2[D]+[S]= c (total concentration of single strands) , so
[S]= C − 2[D]= c − fc = (1− f )c .
Express K in terms of f and c:
fc
2(1− f )2 c2
f
K=
2c(1− f )2
K=
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
23.
a)
GGGCCC 3' :
For 5'
3' CCCGGG 5'
(table 4.4)
ΔG° = ΔG°(initiation) + 4ΔG°(GG/CC) + ΔG°(GC/CG)
= +8.1+ 4(−7.7) − 9.1= −31.8 kJ mol−1
ΔH° = ΔH°(initiation) + 4ΔH°(GG/CC) + ΔH°(GC/CG)
= 0.8+ 4(−33.5) − 41.0 = −174.2 kJ mol−1
ΔS° = (ΔH°− ΔG°) T = −174.2 + 31.8 = −0.46 kJ K −1 mol−1
310
For:
5' GGTTCC 3'
3' CCAAGG 5'
ΔG° = ΔG°(initiation) + 2ΔG°(GG/CC) + ΔG°(GT/CA)
+ ΔG°(TT/AA) + ΔG°(TC/AG)
= 8.1+ 2(−7.7) − 6.0 − 4.2 −5.4= −22.9 kJ mol−1
ΔH° = 0.8+ 2(−33.5) − 35.2 − 33.1− 34.3= −168.8 kJ mol−1
ΔS° = −168.8+ 22.9 = −0.47 kJ K −1 mol−1
310
b)
For self complementary GGGCCC
K=
⎛ ⎞
f
at Tm , f = ⎜ 1 ⎟ K = 1
2
c
⎝ 2⎠
2c(1− f )
ΔG° = −RTmln K = ΔH°−TmΔS° ;
Tm (ΔS°− R ln K ) = ΔH°
ΔH°
ΔH°
=
ΔS°− R ln K ΔS°+ R ln c
−174.2 ×103 J mol−1
=
= 325 K = 52°C
−460 J K −1 mol−1 + (8.314 J K −1 mol−1 ln (1.0 ×10−4 ))
Tm =
For non-self-complementary GGTTCC + GGAACC
K = [D]
, where [D]= f c 2 , c = total concentration of single strands
[S1][S2 ]
[S1]= [S2 ]= c (1− f )
2
f
c
K=
= 2 f 2 ; K = 4 at f = 1 at Tm
c
c
2
2( (1− f )) c(1− f )
2
ΔH°
−167.6 ×103J mol−1
Tm =
=
⎛
ΔS°+ R ln (c 4)
−4 ⎞
−470 J K −1 mol−1 +8.314 J K −1 mol−1 ln ⎜ 2 ×10 ⎟
4 ⎟⎠
⎜⎝
= 303 K = 30°C
10
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
24.
a)
b)
c)
[D]
[P][T]
Cp = [P]+[D] ; CT = [D]+[T] ; f = [D] CT
Cp >>[D] so [P]= Cp
[D]= f CT
[T ]= CT −[D]= (1− f )CT
f CT
f
K=
=
Cp (1− f )CT (1− f )Cp
K=
ΔG° = ΔG°(initiation) + 3ΔG°(GG/CC) + 2ΔG°(GA/CT)
+ ΔG°(AA/TT) + ΔG°(AT/TA) + ΔG°(CA/GT)
= 8.1+ 3(−7.7) + 2(−5.4) − 4.2 − 3.7 − 6.0
ΔG° = −39.7 kJ mol−1
K = e−ΔG° RT = 9.0 ×106
(Table 4.4)
°
ΔH = 0.8+ 3(−33.5) + 2(−34.3) − 33.1− 30.2 − 32.7 − 35.6
ΔH ° = −299.9 kJ mol−1
ΔS ° = ΔH°− ΔG° = −267.2 + 39.7 = −0.839 kJ K −1 mol−1
T
310
°
ΔH°
Tm =
= ° ΔH
at f = 1
2
ΔS°− R ln K ΔS + R ln Cp
=
−299.9 ×103 J mol−1
−734 J K −1 mol−1 + (8.314 J mol−1) ln 10−4
Tm = 327 K = 54° C
d)
12log(0.5) = −3.61, 12log(.05) = −15.61, 12log(.005) = −27.61 ; then:
Tm, 0.5 M NaCl = 50.7°C
Tm, 0.05 M NaCl = 38.7°C
Tm, 0.005 M NaCl = 26.7°C
Higher NaCl stabilizes duplex formation by interaction of sodium cations with the
negative phosphate groups.
e)
As Cprobe increases, Tm increases (see equation above)
As Ctarget increases, Tm is unchanged: K is independent of target for large excess of
probe.
A single base mismatch decreases ΔG° and ΔH ° and decreases Tm
Lower salt concentration decreases complex stability and decreases Tm
11
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
25.
12
First using the quick ‘back of the envelope’ approach
For ATAT TAAT TTAA TATA : Tm 16 × 2°C = 32 °C
For GCCG CGCG GGGG CCCC : Tm 16 × 4°C = 64 °C
For GCTA AGCG GTAT CAGT : Tm  (8× 2°C)+(8× 4°C) = 48 °C
For GCTC AGCG GTAG CAGT : Tm  (6 × 2°C)+(10 × 4°C) = 52 °C
Next use neighbor method to estimate enthalpy and entropy changes:
For ATAT TAAT TTAA TATA count 5(AT)+5(TA)+5(AA,TT). Then
ΔH° = −467.5 kJ mol-1; ΔS° = −1.338 kJ mol-1K -1
For GCCG CGCG GGGG CCCC count 4(GC)+3(CG)+8(GG,CC). Then
ΔH° = −565.2 kJ mol-1; ΔS° = −1.417 kJ mol-1 K -1
For GCTA AGCG GTAT CAGT, count
2(GC)+(CT)+2(TA)+(AA)+2(AG)+(CG)+(GG)+2(GT)+(AT)+(TC)+(CA)
ΔH° = −522.0 kJ mol-1; ΔS° = −1.398 kJ mol-1 K -1
For GCTC AGCG GTAG CAGT,
ΔH° = −537.8 kJ mol-1; ΔS° = −1.415 kJ mol-1 K -1
Calculate Tm similar to 4.24; note C p = 0.25×10−6 M ;
(note: additional solutions shown here for 10-4 M probe can be an interesting contrast and point
of discussion)
ATAT TAAT
TTAA TATA
GCCG CGCG
GGGG CCCC
GCTA AGCG
GTAT CAGT
GCTC AGCG
GTAG CAGT
Quick Tm
Tm (0.05 M NaCl)
(0.25x10-6 M probe)
Tm (0.05 M NaCl)
(10-4 M probe)
32
30.5
41.7
64
77.5
48
53.7
65.2
52
60.1
71.8
89.7
Poor agreement of ‘back of envelope’ to calculated Tm’s when probe concentration is high (10-4 M). For
typical probe concentrations (~10-6 M), the quick Tm tends to be best when GC content is ca. 50%.
Chapter 4 solutions manual to accompany 5’th Edition of Physical Chemistry,
Tinoco, Sauer, Wang, Puglisi, Harbison, Rovnyak
26.
Find moles of macromolecule: (0.0020 L)(1.0 ×10−5 M) = 2.0 ×10−8 mol
Then the first injection bound fully to 2.0 ×10−9 mol , and so the enthalpy of binding is
−35×10−6 J = −17.5 kJ mol-1 .
2.0 ×10−9 mol
27.
In curve (a) the initial injections essentially bind completely.
13
In curve (a) a 1:1 binding stoichiometry is unambiguous, while it is also clear to see in (b). In (c)
the binding ratio is not well defined, however since an integer ratio is expected, (c) could be
consistent with 1:1 binding.
Strong is (a) and weakest binding is (c).
28.
a)
Modify Eqn. 4.56 first: [L]Tot = [L]+ nf M [ M ]Tot
Now continue with Eqn. 4.55 to obtain the analogue to Eqn. 4.65A
⎛
⎞
[L]Tot
[L]Tot
1
2 − f ⎜
fM
+
+1⎟ +
=0
M ⎜ n[ M ]
⎟⎠ n[ M ]Tot
⎝
Tot K n[ M ]Tot
We need to include n in Eqn. 4.57 also since it is needed to count the total number of
binding sites:
Q = nf M [ M ]TotV0Δ r H°
b)
By the inflection positions of the respective sigmoids, conclude that:
(a) 3 sites, (b) 2 sites, and (c) 1 site.