# Solutions Manual for Biomolecular Thermodynamics From Theory to Application Foundations of Biochemistry and Biophysics 1st Edition Barrick

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#### BIOMOLECULAR THERMODYNAMICS FROM THEORY TO APPLICATION FOUNDATIONS OF BIOCHEMISTRY AND BIOPHYSICS 1/E BY BARRICK SOLUTIONS MANUAL

Authors: Barrick
ISBN: 9781138068841 | 9781439800195 | 1439800197 | 9781439800201 | 1439800200 | 9781439800195 | 1439800197

# Solutions Manual for Biomolecular Thermodynamics From Theory to Application Foundations of Biochemistry and Biophysics 1st Edition Barrick

SAMPLE CHAPTER

 Solution Manual 13

e = 1000 it is ∼2 × 10−47. Evaluating the energy sum above to e = 100 gives <e> = 9.9971. Evaluating to e = 200, Mathematica returns <e> = 9.9999999. It seems pretty clear that the average is converging to 10.

1.19 The first    moment of the  Gaussian distribution can be calculated using Equation

1.43, namely

+∞

x 〉= ∫G ( x )dx

−∞

+∞

= ∫ x σ 12π e−( x x )22σ2 dx

−∞

+∞

• σ 12πxe−( x x )22σ2 dx

−∞

This integral can be calculated by a substitution of variables:

 ( x − ) u = x (or x = σu  2 + ) x σ 2

du = σdx2

This substitution simplifies the exponent, and it transforms the distribution to a new variable on which it is centered and symmetrical. Such transformations often allow integrals to be eliminated based on simple symmetry arguments. And although the transformation makes the leading term more complicated by introducing a second term, these two terms can be integrated separately:

 +∞ 1 +∞ ∫ x ρG ( x)dx = ∫ σ 2 ( u σ 2 + )e−u2 du x σ 2π −∞ −∞ 1 +∞ = ∫ σ 2 ( u σ 2 )e−u2 du σ 2π −∞ 1 +∞ + ∫ σ 2 xe−u2 du σ 2π −∞ σ 2 +∞ +∞ = ∫ ue−u 2 du + x ∫ e−u2 du π π −∞ −∞

+∞

= σ π2ueu2 du + xπ × π

−∞

= 0 + x = x

With respect to the new, centered variable u, the first integral is the product of an odd symmetrical function and an even symmetrical function. This product has odd symmetry, and its integral over all space is zero. Thus, the leading integral is zero, no matter what the value of sigma.

CHAPTER 2

2.14 As the above function is well-behaved (continuous and differentiable), its differential can be calculated using the exact differential formula given above (Equation 2.22).

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 ∂f ∂f dz = dx + dy ∂x y ∂y x { 2 2 { 2 2 ∂ x + y } ∂ x + y } dy = dx + ∂x ∂y y x

= 2xdx +2ydy

2.15 We can compare the differential calculated in Question 1.1 to the general differential relationship (Equation 2.38) we used to state the Euler criterion for exactness:

df = S ( x, y )dx +T( x , y)dy = 2xdx +2ydy

Because x and y are independent variables, and can each be incremented by any amount (including dx=0, dy≠0, and vice versa), the above equation implies that two separate equalities hold:

S ( x, y )dx = 2xdx, thus S ( x, y ) = 2x

and

T ( x, y )dy = 2ydy, thus T( x , y) = 2y

Using these expressions in the Euler criterion (Equation 2.41) gives

 ∂ ∂ = 2x = 0, and S ( x, y) ∂y ∂y x x ∂ ∂ ∂ 2y T ( x, y) = = 0 = S ( x, y) y y ∂y ∂x ∂x x

The differential satisfies the Euler criterion and is thus exact.

2.16 Starting with the differential of the paraboloid dz = 2xdx+2ydy, the first step integrates as follows:

x = 2 ,y =1

z1 =     ∫ {2xdx +2ydy}

 x =1, y =1 x=2 2x2 x=2 = ∫ 2xdx = = 4−1= 3 2 x=1 x=1

Similarly, the second step integrates as

x = 2 ,y =2

z 2 =    ∫ {2xdx +2ydy}

 x = 2 ,y =1 y =2 2y2 y =2 = ∫ 2ydy = = 4−1= 3 2 y =1 y =1

Because z = x2+y2 is a well-behaved state function, we can calculate the overall change in z by adding up any set of changes in z that connect up from the initial and final state. The Dz1 and Dz2 values from steps 1 and 2, calculated above, connect from the initial state (1,1) to the final state (2,2) when added, thus,

z =     z1 +   z2 = 3 +3 = 6

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 Solution Manual 15

This can be verified directly, since we have an analytic for z in terms of x and y:

z =           z(2,2) +  z(1,1) = (22 +22 ) − (12 +12 ) = 8 −2 −6

2.17 Expressed as a single variable, the differential of z has a particularly simple form analogous to Equation 2.7:

dz = dz dr = dr2 dr = 2rdr

dr           dr

Note that for completeness, you might be tempted to include θ, the other variable in a two-dimensional polar coordinate system. The exact differential would, in general, look like this:

 ∂z ∂z dθ dz = dr + ∂r θ ∂θ r

However, for the paraboloid, the second term is zero because

 ∂z ∂r 2 2 ∂1 = = r = 0 ∂θ ∂θ ∂θ r r r

2.18 In this problem, the path is along an increasing radial line, with no change in the polar angle q. Thus, the integral can be written quite simply:

 r2 r2 2r2 r2 z = ∫ dz = ∫ 2rdr = 2 2 = r2 −r1 2 r1 r1 r1

The problem is that we have initial and final position in the xy plane and we need to represent them as changes in r. This is done with the Pythagorean theorem:

 r = x2 + y2 r1 = 12 +12 = 2 r2 = 22 +22 = 8

 Using these values of r1 and r2 in the integral for z above gives 2r2 8 z = = 82− 22 =8−2 = 6 2 2

This is the same value as was calculated above using Cartesian coordinates.

2.28 The F matrix for this problem is 31 × 2 (n = 31 data points, m = 2

parameters)

 ∂f ( x1)  ∂f ( x1) ∂p1 ∂p2 ∂f ( x2 ) ∂f ( x2 ) ∂p1 ∂p2 F = ∂f ( x31) ∂f ( x31) ∂p ∂p 1 2

Left-multiplying F by FT by its transpose gives the 31 × 2 A matrix:

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 n 2 ∂f ( xi ) ∂p ∑ i=1 1 FTF = n ∂f ( xi ) ∂f ( xi ) ∂p ∂p ∑ i=1 1 2

 n ∂f ( xi ) ∂f ( xi ∑ ∂p1 ∂p2 i=1 n 2 ∂f ( xi ) ∑ ∂p2 i=1

)

The determinant of this matrix (which is also the determinant of the

transpose) is

F T F =n             f ( xi

i=1     p1

 2 n 2 n ∂f ( xi ) ∂f ( xi ) ∑ − ∑ ∂p1 i=1 ∂p2 i=1

2

)   f ( xi )

p2

Each of the cofactors (there are only three unique ones because FTF is

symmetric is

cof (FT F)  =n  f ( xi ) 2

1,1                                                                         i=1                p2

cof (FT F)   =n  f ( xi ) 2

2 ,2                                                                        i=1                p1

 T T n ∂f ( xi ) ∂f ( xi ) cof (F F)1,2 = cof (F F)2,1 = ∑ i=1 ∂p1 ∂p2

Together, these three cofactor equations can be combined with the

determinant equation to calculate the four elements of the covariance

matrix V. For the variance term for parameter p1,

 n ∝ cof1,1(F T F) = ∑(∂f ( xi )/∂p2 )2 v1,1 i=1 F T F n n ∑(∂f ( xi )/∂p1 )2 ∑(∂f ( xi )/∂p2 )2 i=1 i=1 n 2 ∑ (∂f ( x )/∂p )(∂f ( x )/∂p − ) i 1 i 2 i=1

A similar formula (with differentiation by p1 in the numerator instead of p2) is obtained for v2,2. The two covariance elements are given by

 v1,2 ∝ cof1,2 (FT F) F T F

n

• (∂f ( xi )/∂p1 )(∂ f ( x i )/∂p2 )
 = n n i=1 ∑(∂ f ( xi )/∂p1 )2 ∑(∂f ( x i )/∂p2 )2 i=1 i=1 n 2 ∑ (∂f ( x )/∂p )(∂f ( x − )/∂p ) i 1 i 2 i=1

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 Solution Manual 17

2.32 The mode of the distribution (that is, the peak) can be found by differentiating with respect to χ2 and setting the result to zero:

 dP 2 ν / 2 2 (ν−2)/2 2 /2 χ d −χ = 2 Γ(ν/2) (χ ) e dχ 2 2 dχ 2 ) (ν−4)/2  −χ2 /2 (χ 2 ) (ν−2)/2 −χ2 /2 = 2 ν 2 Γ(ν/2) (ν−2)(χ e − e 2 2

= 2ν2 Γ(ν/2) eχ 2 / 2 {(ν −2)(χ2 )(ν−4 )/2 −(χ2 )( ν−2)/2 }= 0

2

The zero in the equation above has to come from the term in the curly braces, since all of the multiplying terms outside the braces are everywhere positive. This means

(ν−2)(χ2 )(ν−4)/2 −(χ2 )(ν−2)/2 = 0

Rearranging gives

(ν−2)     (ν−4)

ν−2=(χ2)    2   2

• ν
• (χ2 )2−1−2+2

(ν2)      (ν4)

ν−2=(χ2)    2   2

• ν
• (χ2 )2−1−2+2

Cancelling the exponents gives the value of χ2 at the critical point,

χ2 =ν−2

Based on the shape of the probability distribution, it is clearly a

maximum. This can be confirmed by a second derivative test, but

the exponential from the third line above needs to be included in

the differentiation.

2.33 The normalization condition means that

• Pχ 22 = 1

0

The integral on the left-hand-side of the distribution can be broken into two parts, or “complements”:

χobs2                ∞

• Pχ 22 + ∫ Pχ 22 = 1
• χobs2

This is a result of the fundamental theorem of calculus, but it may be easier to see by thinking about the integrals as area under the curve. Rearranging the equation above gives the desired equality.

2.34 To carry out this test, we need to start with fits using both models. The fit for the exponential decay model (we will call it model 2) is shown in Figure 2.17. For the cubic model, the fit is shown in below.

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 1.0 0.8 y 0.6 0.4 0.2 20 40 60 80 100 x 0.03 0.02 0.01 ri 20 40 60 80 100 –0.01 –0.02 –0.03 x

As can be seen from the residuals, the quartic model does not fit as well as the exponential (this may not be a surprise, since we used an exponential model to generate the data). Using the SSR values for the two fits, we can calculate the fobs ratio as

 fobs (ν quartic , νexp ) = SSRquartic /νquartic = 0.0165/(101−5) =1.848 0.00902/(101−3) SSRexp /νexp

The f-ratio distribution, given the degrees of freedom ν1=96, ν2=98, is shown below:

 1.0 0.8 pf >1.84 (96, 98) 0.6 fobs (96, 98) = 1.84 0.4 0.2 0.5 1.0 1.5 2.0 2.5 3.0 f (96, 98)

As indicated by the arrow, there is very little of getting an f-ratio value

of 1.84 or higher from statistically equivalent models with 96 and

98 degrees of freedom. The exact probability can be calculated by

solving the equation

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 Solution Manual 19

pf >1.848 = ∫ pf df = 0.00137

1.848

(easiest done with the CDF[FRatioDistribution…] command in Mathematica). In other words, it is pretty unlikely (about one in a thousand) that you would get this f-value from two statistically equivalent models. Compared to the exponential model, SSR is unexpectedly high for the quartic model. We can certainly reject the model at the 95% confidence level.

CHAPTER 3

3.4    Starting with equation

dV = −κTVdp + αVdT +Vdn

Substituting the ideal gas results κT = p−1 , α = T−1, and the single-component molar volume definition V = V /n gives

dV = Vdpp + VdTT + Vdnn

Dividing by V separates variables:

dVV = dpp + dTT + dnn

We need to integrate this expression from starting values Vi, Ti, pi, and ni, to general values p, V, T, and n. Using the ideas developed in Chapter 2, we can integrate in three steps, first changing V at constant T and n, next changing T at constant V and n, and finally changing n at constant V and T. Each of these will lead to a change in p, which can be added to get the total change in pressure.

3.5    Assuming the ideal gas law,

 pVi 2.27× 105 Pa×0.0005 m3 Ti = = =273K nR − 1 −1 0.05 mol× 8.315 J⋅mol  K Tf  = pVf = 2.27× 105 Pa×0.002 m3 =1092 K nR − 1 −1 0.05 mol× 8.315 J⋅mol  K

As described above, to maintain the pressure throughout the expansion, the temperature increases.

3.9    To calculate the work done on the polymer (you can think of it as the system), integrate Equation 3.17. A nice way to do this is with a single limit of x = 0 end-to-end separation, giving work as an analytical function of of x:

w = −∫ Fdx

 0 kBT 1 1 x = − + dx Lp 4(1− x / Lc )2 4 Lc ∫ 0 kBT Lc x x2 = Lp 4(1− x / Lc ) − 4 + 2Lc 0 2 2 kBT Lc 2x = − x + 4Lp − x Lc Lc 0

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