Solution Manual 69
CHAPTER 13
13.8 Using Equation 7.17, the table can be completed as follows:
Stepwise macroscopic Stepwise microscopic
constant constant
K1 = 100 M−1 β1 = K1 = 102 M−1
K2 = 100 M−1 β2 = K1K2 = 104 M−2
K3 = 100 M−1 β3 = K1K2K3 = 106 M−3
K4 = 100 M−1 β4 = K1K2K3K4 = 108 M−4
13.9 The easiest way to do this is to recognize that from Equation 7.17, Ki can be
obtained as a ratio of β’s:
βi
Ki =
βi−1
This formula is used in the table below:
Stepwise macroscopic Stepwise microscopic
constant constant
β1 = 104 M−1 K1 = β1 = 104 M−1
β2 106 M−2
β2 = 106 M−2 K2 = = = 102 M−1
β1 104 M−1
β3 = 107 M−3 β3 107 M−3
K3 = = = 101 M−1
β2 106 M−2
13.10 The constants in Problem 13.8 give rise to (modest) negative cooperativity.
Although it may seem surprising given that all four stepwise macroscopic
constants are identical, there are statistical factors that favor binding increase
the numerical value of the first macroscopic constant, and decrease the fourth.
The constants in Problem 13.9 are consistent with (fairly strong) negative
cooperativity. Although this may seem surprising given increasing values of
the overall constants, these constants cannot directly be compared to one
another, given their different units. Although statistical factors also affect
the numerical values of the overall constants (though to a lesser extent than
for the stepwise constants), for neutral cooperativity the values of overall
constants should increase by a constant factor (in this case, 104).
13.20 To keep things straight, call the first pair of constants (K1 < K2) “A,” and the
second set of constants (K1 > K2) “B.” With this nomenclature, the binding
polynomials are
PA = 1 + K1 x + K1K2 x 2
= 1 + 0.1x + 0.1×10 x 2
= 1 + 0.1x + x 2
PA = 1 + K1 x + K1K2 x 2
= 1 + 10 x + 10 × 0.1x 2
= 1 + 10 x + x 2
The populations of each macrostate are given by the Boltzmann factor divided
by P:
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Macrostate pi Case A Case B
(K1 < K2) (K1 > K2)
M0 1 1 1
p0 = p0, A = p0, B =
P PA PB
Mx K1 x 0.1x 10 x
p1 = p1, A = p1, B =
P PA PB
Mx2 K1K2 x 2 x2 x2
p2 = p2, A = p2, B =
P PA PB
Here are the species plots, with case A as solid lines, case B
as dashed lines, in black, red, and gray for M0, Mx, and Mx2,
respectively.
(A) (B)
1.0 1.0
0.8 0.8
0.6
Population
0.6
Population
0.4 0.4
0.2 0.2
0.0 0.0
0 5 10 15 20 25 0.01 0.1 1 10 100
x (m) log10x
13.29 This can be seen by recognizing that for the competitive scheme, there are
three possible states of binding; x bound (fx), y bound (fy), and nothing bound
(1 − fx − fy) and these states are mutually exclusive. For each and every fixed
fraction unbound f0, we can write fy in terms of fx: There was an equation
here that died. Treating f0 as a constant, differentiate the linkage coefficient
expression above with respect to fx and set to zero:
Λ x , y = −fx fy
= −fx (1− f0 − fx )
= fx 2 + (f0 − 1)fx
Substituting back in to get the relationship between fx and fy, using the
relationship 1 − f0 = fx + fy gives
1− f0 f + fy
fx = = x , or
2 2
f x = fy
This is true for every value of f0. So the only remaining question is, at what
value of f0 is the product fxfy maximal, subject to the constraint fx = fy? Clearly,
this occurs at f0 = 0.
y y y
13.31 If K 0,1 = K1,1, the K1,1 term can be eliminated from P:
P = 1 + K1x,0 [ x ] + K 0y,1[ y ] + K1x,0 [ x ]K1y,1[ y ]
= 1 + K1x,0 [ x ] + K 0y,1[ y ] + K1x,0 [ x ]K 0y,1[ y ]
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, Solution Manual 71
In this form, P depends on only two constants, one multiplying x and another
multiplying y. These terms can be factored into two separate pieces:
P = 1 + K1x,0 [ x ] + K 0y,1[ y ] (1 + K1x,0 [ x ])
= (1 + K 0y,1[ y ])(1 + K1x,0 [ x ])
= Px Py
Multiplication is the correct way to combine Px and Py, because if K 0y,1 = K1y,1,
these binding of y is independent of the binding of x (and vice versa).This
is equivalent to the product of two separate single-site polynomials.
Multiplication is the appropriate way to combine these two.
CHAPTER 14
14.1 ith these macroscopic binding constants, the binding polynomial of glycine
W
for hydrogen ions is
P = 1 + K1[ H + ] + K1K2 [ H + ]2
The fraction of hydrogen ions bound can be found by differentiation (see
Chapter 13 for the exact solution to the two-site model):
[ x ] dP K1[ H + ] + 2K1K2 [ H + ]2
fbound = =
2P d[ x ] 2(1 + K1[ H + ] + K1K2 [ H + ]2 )
To include the binding constants K1 and K2 in the fbound equation above, we can
invert the pK relationships, that is,
K1 = 10 pK1 = 109.60 M−1
K2 = 10 pK2 = 102.34 M−1
Substituting these values (it is convenient to leave the result in the
exponential form above) into fbound gives
109.60 [ H + ] + 2 ×109.60102.34 [ H + ]2
fbound =
2(1 + 109.60 [ H + ] + 109.60102.34 K2 [ H + ]2 )
109.60 [ H + ] + 2 ×1011.94 [ H + ]2
=
2(1 + 109.60 [ H + ] + 1011.94 [ H + ]2 )
This is plotted below:
1.0 1.0
0.8 0.8
0.6 0.6
fbound
fbound
0.4 0.4
0.2 0.2
0 2. × 10–9 4. × 10–9 6. × 10–9 8. × 10–9 1. × 10–8 0.00 0.02 0.04 0.06 0.08 0.10
[H+] (M) [H+] (M)
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On the scale on the left, only the high-affinity binding reaction is seen,
whereas on the scale on the right, only the low affinity binding reaction is
seen. A log [H+] scale is needed to see both binding reactions:
1.0
0.8
0.6
fbound
0.4
0.2
10–10 10–7 10–4 10–1
[H+] (M)
Plotting as a function of pH rather than [H+] concentration is equivalent
to using a log scale, only the scale is reversed. We simply need to
rearrange the relationship above to give fbound as a function of pH, using the
substitution
[ H + ] = 10− pH
This substitution gives
109.6010− pH + 2 ×1011.9410−2 pH
fbound =
2(1 + 109.6010− pH + 1011.9410−2 pH )
109.60− pH + 2 ×1011.94−2 pH
=
2(1 + 109.60− pH + 1011.94−2 pH )
A plot of this expression shows the two hydrogen ion binding events, each
centered at one of the two pKa values:
2.0
1.5
〈H+bound〉
1.0
0.5
2 4 6 8 10 12
pH
14.2 The stepwise scheme is
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CHAPTER 13
13.8 Using Equation 7.17, the table can be completed as follows:
Stepwise macroscopic Stepwise microscopic
constant constant
K1 = 100 M−1 β1 = K1 = 102 M−1
K2 = 100 M−1 β2 = K1K2 = 104 M−2
K3 = 100 M−1 β3 = K1K2K3 = 106 M−3
K4 = 100 M−1 β4 = K1K2K3K4 = 108 M−4
13.9 The easiest way to do this is to recognize that from Equation 7.17, Ki can be
obtained as a ratio of β’s:
βi
Ki =
βi−1
This formula is used in the table below:
Stepwise macroscopic Stepwise microscopic
constant constant
β1 = 104 M−1 K1 = β1 = 104 M−1
β2 106 M−2
β2 = 106 M−2 K2 = = = 102 M−1
β1 104 M−1
β3 = 107 M−3 β3 107 M−3
K3 = = = 101 M−1
β2 106 M−2
13.10 The constants in Problem 13.8 give rise to (modest) negative cooperativity.
Although it may seem surprising given that all four stepwise macroscopic
constants are identical, there are statistical factors that favor binding increase
the numerical value of the first macroscopic constant, and decrease the fourth.
The constants in Problem 13.9 are consistent with (fairly strong) negative
cooperativity. Although this may seem surprising given increasing values of
the overall constants, these constants cannot directly be compared to one
another, given their different units. Although statistical factors also affect
the numerical values of the overall constants (though to a lesser extent than
for the stepwise constants), for neutral cooperativity the values of overall
constants should increase by a constant factor (in this case, 104).
13.20 To keep things straight, call the first pair of constants (K1 < K2) “A,” and the
second set of constants (K1 > K2) “B.” With this nomenclature, the binding
polynomials are
PA = 1 + K1 x + K1K2 x 2
= 1 + 0.1x + 0.1×10 x 2
= 1 + 0.1x + x 2
PA = 1 + K1 x + K1K2 x 2
= 1 + 10 x + 10 × 0.1x 2
= 1 + 10 x + x 2
The populations of each macrostate are given by the Boltzmann factor divided
by P:
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Macrostate pi Case A Case B
(K1 < K2) (K1 > K2)
M0 1 1 1
p0 = p0, A = p0, B =
P PA PB
Mx K1 x 0.1x 10 x
p1 = p1, A = p1, B =
P PA PB
Mx2 K1K2 x 2 x2 x2
p2 = p2, A = p2, B =
P PA PB
Here are the species plots, with case A as solid lines, case B
as dashed lines, in black, red, and gray for M0, Mx, and Mx2,
respectively.
(A) (B)
1.0 1.0
0.8 0.8
0.6
Population
0.6
Population
0.4 0.4
0.2 0.2
0.0 0.0
0 5 10 15 20 25 0.01 0.1 1 10 100
x (m) log10x
13.29 This can be seen by recognizing that for the competitive scheme, there are
three possible states of binding; x bound (fx), y bound (fy), and nothing bound
(1 − fx − fy) and these states are mutually exclusive. For each and every fixed
fraction unbound f0, we can write fy in terms of fx: There was an equation
here that died. Treating f0 as a constant, differentiate the linkage coefficient
expression above with respect to fx and set to zero:
Λ x , y = −fx fy
= −fx (1− f0 − fx )
= fx 2 + (f0 − 1)fx
Substituting back in to get the relationship between fx and fy, using the
relationship 1 − f0 = fx + fy gives
1− f0 f + fy
fx = = x , or
2 2
f x = fy
This is true for every value of f0. So the only remaining question is, at what
value of f0 is the product fxfy maximal, subject to the constraint fx = fy? Clearly,
this occurs at f0 = 0.
y y y
13.31 If K 0,1 = K1,1, the K1,1 term can be eliminated from P:
P = 1 + K1x,0 [ x ] + K 0y,1[ y ] + K1x,0 [ x ]K1y,1[ y ]
= 1 + K1x,0 [ x ] + K 0y,1[ y ] + K1x,0 [ x ]K 0y,1[ y ]
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In this form, P depends on only two constants, one multiplying x and another
multiplying y. These terms can be factored into two separate pieces:
P = 1 + K1x,0 [ x ] + K 0y,1[ y ] (1 + K1x,0 [ x ])
= (1 + K 0y,1[ y ])(1 + K1x,0 [ x ])
= Px Py
Multiplication is the correct way to combine Px and Py, because if K 0y,1 = K1y,1,
these binding of y is independent of the binding of x (and vice versa).This
is equivalent to the product of two separate single-site polynomials.
Multiplication is the appropriate way to combine these two.
CHAPTER 14
14.1 ith these macroscopic binding constants, the binding polynomial of glycine
W
for hydrogen ions is
P = 1 + K1[ H + ] + K1K2 [ H + ]2
The fraction of hydrogen ions bound can be found by differentiation (see
Chapter 13 for the exact solution to the two-site model):
[ x ] dP K1[ H + ] + 2K1K2 [ H + ]2
fbound = =
2P d[ x ] 2(1 + K1[ H + ] + K1K2 [ H + ]2 )
To include the binding constants K1 and K2 in the fbound equation above, we can
invert the pK relationships, that is,
K1 = 10 pK1 = 109.60 M−1
K2 = 10 pK2 = 102.34 M−1
Substituting these values (it is convenient to leave the result in the
exponential form above) into fbound gives
109.60 [ H + ] + 2 ×109.60102.34 [ H + ]2
fbound =
2(1 + 109.60 [ H + ] + 109.60102.34 K2 [ H + ]2 )
109.60 [ H + ] + 2 ×1011.94 [ H + ]2
=
2(1 + 109.60 [ H + ] + 1011.94 [ H + ]2 )
This is plotted below:
1.0 1.0
0.8 0.8
0.6 0.6
fbound
fbound
0.4 0.4
0.2 0.2
0 2. × 10–9 4. × 10–9 6. × 10–9 8. × 10–9 1. × 10–8 0.00 0.02 0.04 0.06 0.08 0.10
[H+] (M) [H+] (M)
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On the scale on the left, only the high-affinity binding reaction is seen,
whereas on the scale on the right, only the low affinity binding reaction is
seen. A log [H+] scale is needed to see both binding reactions:
1.0
0.8
0.6
fbound
0.4
0.2
10–10 10–7 10–4 10–1
[H+] (M)
Plotting as a function of pH rather than [H+] concentration is equivalent
to using a log scale, only the scale is reversed. We simply need to
rearrange the relationship above to give fbound as a function of pH, using the
substitution
[ H + ] = 10− pH
This substitution gives
109.6010− pH + 2 ×1011.9410−2 pH
fbound =
2(1 + 109.6010− pH + 1011.9410−2 pH )
109.60− pH + 2 ×1011.94−2 pH
=
2(1 + 109.60− pH + 1011.94−2 pH )
A plot of this expression shows the two hydrogen ion binding events, each
centered at one of the two pKa values:
2.0
1.5
〈H+bound〉
1.0
0.5
2 4 6 8 10 12
pH
14.2 The stepwise scheme is
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