Physical chemistry principles and applications in biological sciences 5th edition tinoco solutions m

Physical Chemistry Principles and Applications in Biological Sciences 5th Edition Tinoco Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/physical-chemistry-principles-and-applications-in-biological-sciences-5th-edition -tinoco-solutions-manual/

CHAPTER 10

1.

2.

A linear least-squares fit to a Lineweaver-Burk plot (left) gives vmax = 0.34 µmol m–1 and KM = 0.42 M. In contrast, a direct least-squares fit (plotted on the right) to the Michaelis-Menten equation v = vmax[S]/(KM + [S]) (which is statistically preferable) gives vmax = 0.35 µmol m–1 and KM = 0.45 M. a)

For the hydration reaction, a direct least-squares fit to the Michaelis-Menten equation v = vmax[S]/(KM + [S]) gives vmax = 2.5 × 10–4 M s–1 and KM = 10 mM. k2 = vmax/[E]0 = 2.5 × 10–4 M s–1/2.3 × 10–9 M = 1.1 × 105 s–1.

b)

For the dehydration reaction, a direct least-squares fit gives vmax = 7.5 × 10–5 M s–1 and KM = 12 mM. k2 = 7.5 × 10–5 M s–1/2.3 × 10–9 M = 3.3 × 104 s–1.

c)

The equilibrium constant is given by Eq. 10.24; K =

R vmax,F K M

vmax,R K F

= 4.0

M

3.

4.

5.

From Eq. 10.24 K =

R vmax,F K M

vmax,R K

= F

M

1.3× 103 × 1.0 × 10−5 800 × 4.0 × 10−6

= 4.1

a)

2700 × 10–5/(15 × 60) = 3 × 10–5 mol s–1 mg–1.

b)

10–3 g mg–1/20000 g mol–1 = 5 × 10–8 mol mg–1.

c)

3 × 10–5 mol s–1 mg–1/5 × 10–8 mol mg–1 = 6 × 102 s–1.

d)

The enzyme is at half maximum velocity at [PPi] = 5 × 10–6 M, which is therefore KM.

a)

vmax = 100 s–1 × 1.0 × 10–5 M = 1.0 × 10–3 M s–1. v = vmax[S]/(KM + [S]) = 10–3 × 10–1/(10–4 +10–1) = 0.001 M s–1.

b)

Ea = – R ln 2/[ (1/300 –1/280)] = 24 kJ mol–1.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

2

c)

KM = (k–1 + k2)/k1. If k–1 >> k2, then KM ~ 1/(k1/k–1) = 1/K = 104.

d)

The approximation used in the previous section can be applied at both temperatures K280 ~ 104; K300 ~ 6.7 × 103. Using the van’t Hoff equation ΔrH° = – R ln 0.67/[ (1/300 –1/280)] = –14 kJ mol–1

6.

In general a catalyst lowers Ea, and so it makes the rate coefficient proportionately less sensitive to temperature.

7.

a)

Assuming k2 is the rate determining step, ΔH‡ ~ Ea = – R ln 2.2/[ (1/298.15 –1/283.15)] = 37 kJ mol–1.

b)

ln A = ln k + Ea/RT = log[560] + 14.9 = 21.2. ΔS‡ = R[ln A + ln(h/kBT)] = –68.6 J mol–1 K–1. There is a comparatively low barrier, but the transition state is significantly more ordered than the bound ES complex.

c)

ln(1014) = –(105000 J mol–1/R)(1/T – 1/298.15 K); solving for T, we obtain 1248 K.

8.

Let KI = k3/k–3 = [EI]/[E][I]. We also have E0 = [E] + [ES] + [EI]. From Eq. 10 k [E][S] k [E][S] [E][I] ; substituting, we obtain E0 = [E] + 1 . So [ES] = 1 + k−1 + k2 k−1 + k2 KI ⎛ k [S] [I] ⎞ [E] = E0 ⎜ 1+ 1 + ⎟ ⎝ k−1 + k2 K I ⎠

−1

The forward velocity v = k2 [ES] =

k k [S] ⎛ k [S] [I] ⎞ Substituting for [E]: v = 1 2 E0 ⎜ 1+ 1 + k−1 + k2 ⎝ k−1 + k2 K I ⎟⎠

(

)

k k [S] ⎛ k−1 + k2 1+ [I] / K I + k1[S] ⎞ = 1 2 E ⎜ ⎟ k−1 + k2 0 ⎝ k−1 + k2 ⎠

−1

−1

= E0

(

k1k2 [S]

9.

a)

k1; k2; k3.

b)

k1 and k–1; k3 and k–3.

c)

Measure all six rate coefficients.

)

k−1 + k2 1+ [I] / K I + k1[S]

k2 [S] k−1 + k2 ⎛ [I] ⎞ 1+ ⎟ + [S] k1 ⎜⎝ KI ⎠ For the uninhibited enzyme KM = (k–1 + k2)/k1, so this becomes ⎛ k [S] [I] ⎞ v = E0 2 with K M ′ = K M ⎜ 1+ KM K I ⎟⎠ ′ + [S] ⎝ Dividing above and below by k1 v = E0

k1k2 [E][S] k−1 + k2

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 10.

11.

3

a)

Multiplying Eq. 10.9 above and below by [S], we get [ES] = E0[S]/([S] + KM). E0 = 0.04 × 10–6 g/(30000 g mol–1 × 10–3 L) = 1.33 × 10–9 M. [S] = 5 mM. So [ES] = 1.33 × 10–9 M × 5 mM/(5 mM + 0.05 mM) = 1.32 × 10–9 M. [ES]/E0 = 0.99.

b) c)

Since [S]0 >> KM, we can assume the rate is unchanged when [S]/[S]0 = 0.5. Therefore t = ([S]0 – [S])/v = ([S]0 – [S])/(k2E0) = 2.5 × 10–3 M/(2000 s–1 × 1.33 × 10–9 M) = 937.5 s.

[S] = KM = 5.0 × 10–5 M.

d)

It will decrease the answer to b) by 2/3, but will leave [S] in c) unaffected.

e)

At low substrate concentrations, v ~ E0k2[S]/K′M, with K′M = KM(1 + [I]/KI). To reduce v by a factor of 5, K′M = 5KM, so [I]/KI = 4. If KI = KM, then [I] = 4KM = 2 × 10–4 M.

a)

A Lineweaver-Burk plot (see below) gives a value of KM = 0.044 M. A more statistically robust least squares fit to the plot of v vs. [S] gives KM = 0.040 M.

b) A Lineweaver-Burk plot for both sets of data is shown above. It is clear the y-intercept is considerably altered by the inhibitor, indicating it is non-competitive (the x-intercept is also affected, but probably not within statistical significance).

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

4

12.

13.

a)

K1K2 = K3K4, where K1, K2 are the equilibrium constants for the E + S → ES → H+ES path, and K3, K4 the equilibrium constants for the E + S → H+E + S → H+ES path.

b)

d[P]/dt = k[H+ES]; E0 = [E] + [H+E] + [ES] + [H+ES] = [E] + K3[E][H+] + K1[E][S] + K3K4[E][H+][S] = [E](1 + K3[H+] + K1[S] + K3K4[H+][S]). Therefore E0 [E] = 1+ K3[H + ] + K1[S] + K3 K 4 [H + ][S] +

and [H ES] =

K3 K 4 [H + ][S]E0

1+ K3[H + ] + K1[S] + K3 K 4 [H + ][S]

. Thus

kK3 K 4 [H + ][S]E0 d[P] = k[H + ES] = dt 1+ K [H + ] + K [S] + K K [H + ][S] 3

3

4

( )} = V [S] (1+ K [H ]) ( K + K K [H ]) + [S] K + [S] with V = kE { K K [H ] ( K + K K [H ])} and K = (1+ K [H ]) ( K + K K [H ]) =

{

1

k K3 K 4 [H + ] K1 + K3 K 4 [H + ] [S]E0 3

max

M

c)

+

1

0

3

3

+

3

4

4

+

M

1

1

max

+

3

3

4

4

+

+

{

At high substrate concentrations the rate is v = kE0 K3 K 4 [H + ]

( K + K K [H ])} .

1

This is zero order in [H+] at low pH, and first order in [H+] at high pH.

3 4

+

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

14.

5

A linear least-squares fit to a Lineweaver-Burk plot (above) gives Vmax = 8.2 µM (100 s)–1 and KM = 14.5 µM. In contrast, a direct least-squares fit to the Michaelis-Menten equation v = Vmax[S]/(KM + [S]) (which is statistically preferable) gives Vmax = 8.9 µM (100 s)–1 and KM = 18.5 µM. The graph shows some indication of increased slope at high values of 1/[S], which could be because initial rates are no longer observed when [Pi]100 is a substantial fraction of [ATP]0. A more rigorous theoretical treatment could account for this.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

15.

16.

6

a) b)

From the plot Vmax = 1.5 M s–1 and KM = 3.0 × 10–3 M

c)

The Lineweaver-Burk plot is linear, but nearly crosses the origin. This means that both Vmax and KM are very large. This can happen if k2 is much greater than k1. With E0 = 4 × 10–7 M, k2 = 3.75 × 107 s–1 (although this value is very imprecise).

d)

It is first order with respect to O2–, because KM is much larger than any feasible concentration of O2–.

e)

–d[O2–]/dt = k[E]0[O2–]. At the highest velocity, 0.1 M s–1 = k × 4 × 10–7 M × 2 ×10–4 M, so k = 1.2 × 109 M–1 s–1. This is in the diffusion-controlled regime.

f)

At steady state, k1[E][O2–] = k2[E–][O2–] or [E–] = (k1/k2)[E]. Therefore v = k2[E–][O2–] = k1[E][O2–], consistent with the rate law.

g)

[E–]/[E] = k1/k2

h)

k1 = k = 1.2 × 109 M–1 s–1. k2 = 2k = 2.5 × 109 M–1 s–1.

a)

K1 = 0.01 × 10–9/(0.6 × 10–9)2 = 2.8 × 107. K2 = 0.39 × 10–9/(0.01 × 10–9) = 39. d[C] = k1[A][B] + k–2 [D] − k–1 + k2 [C] dt d[D] = k2 [C] − k–2 [D] dt

(

b)

)

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

7

c)

d)

17.

On the left are shown the plotted concentrations obtained by an exact solution of the coupled differential equations. On the right is the plot of the intermediate [C] expanded 20 fold on the y axis and four-fold on the x axis. k–2 = .05 s–1/39 = 0.0013 s–1.

a)

b)

The Lineweaver-Burk (left) plot looks linear, indicating a Michaelis-Menten model is compatible with the data. The direct plot (right) also looks like a good match between data and fit.

Vmax is the reciprocal of the y intercept. KM can either be obtained from the negative of the inverse of the x intercept or by dividing the slope by the value of the y intercept. A linear least-squares fit to a Lineweaver-Burk plot gives Vmax = 1.2 µM s–1 and KM = 0.47 mM. In contrast, a direct least-squares fit to the Michaelis-Menten equation v = Vmax[S]/(KM + [S]) (which is statistically preferable) gives Vmax = 1.2 µM s–1 and KM = 0.50 mM.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

8

c) The effect is shown both on a Lineweaver-Burk plot and a direct plot.

18.

a) The graph for initial alcohol levels of 60 mL (lighter line) and 120 mL (heavier line) is shown above. Because the alcohol content of the stomach and the amount converted to acetaldehyde can be solved by direct integration, we can obtain the alcohol content of

(

body fluid in closed form as [B] = [A]0 1− e min) and k2 = 1/6 mL min–1 above. b)

−k1t

) − k t . This is plotted for k = ln 2/(4 2

Differentiating [B] and setting its value to zero gives tmax =

1

ln A0 + ln k1 − ln k2 . k1

Inserting this in the expression for [B] gives [B]max, which in this case is 55 mL ethanol at tmax = 24 min. Since we have 60 mL ethanol degraded at 10 mL per hour, regardless of concentration, the time to get to zero is 6 hours after the start of alcohol absorption, or 5 hours and 36 minutes after maximum intoxication.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak c)

The graph is given in part a) above. Legal intoxication is reached after 3.6 min and continues until 390 min. The hangover period lasts from 390 min to 720 min, a total of 330 min.

d)

Legal intoxication from 2.1 min to 750 min; hungover from 750 min to 1080 min.

Reminds me of my college days!

e)

10 mL (the amount metabolized in 1 hour!)

f) The ‘social drinker’ (heavier line) takes longer to become intoxicated, and reaches a somewhat lower level of intoxication, but after 120 min the curve is indistinguishable from the drinker who downs 120 mL of alcohol in one gulp (lighter line)! k

+ 1  LADH + NAD+    LADH i NAD k−1

k

+ 2  LADH i NAD+ + C2 H 5OH    LADH i NAD i C2 H 5OH k−2

19.

a)

k

+ 3  LADH i NAD+ i C2 H 5OH    LADH i NADH i H i CH 3CHO k−3

k

+ 4  LADH i NADH i H + i CH 3CHO    LADH i NADH + H + CH 3CHO k−4

k

5  LADH i NADH    LADH+NADH

k−5

b)

The Michaelis constant KM for ethanol is low, so that LADH is running at maximum velocity (zero order kinetics) until ethanol levels drop below some very low value.

9

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak c)

The maximum velocity Vmax of the enzyme is 10 mL ethanol hr–1. The density of ethanol at 25°C is 0.789 g mL–1 and its molar mass M is 46.07 g mol–1, so this corresponds to 10 mL × 0.789 g mL–1/46.07 g mol–1 = 0.171 mol hr–1 or 4.76 × 10–5 mol s–1. Since Vmax = k2E0, we have E0 = 4.76 × 10–2 mmol s–1/3.1 s–1 = 1.5 × 10–2 mmol.

d)

From Table 9.4, τ =

10

1 . Reciprocating the values of τ, we have two k−1 + k1([A]eq + [B]eq )

simultaneous equations: 606 s–1 = k–1 + k1(10–3 M) and 127 s–1 = k–1 + k1(10–4 M).

Solving in the usual way, we obtain k1 = 5.3 × 105 M–1 s–1 and k–1 = 73 s–1. e)

20.

a)

Ethanol is an inhibitor for both methanol and ethylene glycol. In the presence of large amounts of ethanol, the rate of metabolism of both is reduced, leading to a lower rate of production of toxic products. (Either that, or if you're going to die, better to die drunk!)

The two plots against ethanol and NAD+ as substrates are shown above left and above right, respectively.

b)

0.44 mM for ethanol; 0.017 mM for NAD+.

c)

Competitive inhibition (Vmax is unaffected, but KM is changed).

d)

Noncompetitive inhibition (KM is unaffected, but Vmax is changed).

e)

It is likely pyrazole binds to the LADH.NAD+ complex, competing with ethanol binding to the same complex.

f)

Probably not. The inhibitor is competitive with ethanol, which already has a very low KM, much lower than ethanol concentration during intoxication. Even a substantial increase in the effective KM for ethanol will not affect enzyme turnover substantially.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak 21.

22.

11

a)

Vmax = kcatE0 v0 = Vmax[S]/(KM + [S]) = 150 s–1 × 10–8 M × 5 × 10–6 M/(2 × 10–6 + 5 × 10–6)M = 1.1 × 10–6 M s–1.

b)

v0 = Vmax[S]/{KM(1+[I]/KI) + [S]} = (1.1 × 10–6 M s–1))/2 = 5.4 × 10–7 M s–1. Solving for KI , we get 1.4 × 10–3 M.

c)

If we assume prior equilibrium for both reactions, then in both cases KM for one substrate is its inhibitory constant for the other substrate. So we have v0A = kcat,AE0[A]/{KMA(1+[B]/KMB) + [A]}and v0B = kcat,BE0[B]/{KMB(1+[A]/KMA) + [B]}. Simplifying the ratio, we get v0A/v0B = [A]kcat,A KMB/{[B]kcat,B KMA} = 7.5.

a)

Fitting both sets of temperature data directly to the Michaelis Menten equation, we obtain Vmax,20 = 1.2 × 10–6 M s–1, KM,20 = 5.0 × 10–4 M, and Vmax,40 = 5.9 × 10–6 M s–1, KM,40 = 5.0 × 10–4 M. Temperature only significantly affects the maximum velocity; KM values are the same.

b) c)

kcat,20 = Vmax,20/E0 = 0.12 and kcat,40 = 0.59.

Solving ln(kcat,40 /kcat,20) = –(Ea/R)(1/313.15 K – 1/293.15 K) for Ea, we obtain Ea = 60 kJ mol–1.

Enzymes usually lower the energy of activation.

d) 23.

In reversible non-competitive inhibition Vmax is reduced by a factor (1 + [I]/KI). Setting Vmax/(1 + [I]/KI) = 0.1Vmax, we obtain 1 + [I]/KI = 10, or [I] = 9KI. Therefore [I] = 2.6 × 10–3 M.

24.

a)

The key insight in solving this problem is that there are two pairs of enzyme or enzyme complexes; internally each pair is in fast equilibrium, and so we can tie the concentrations together, since [GE] = [E][G]/KG and [GES] = [ES][G]/KG Therefore [E]+[GE] = [E](1+[G]/KG) and [ES] + [GES] = [ES](1+[G]/KG). The total enzyme concentration is therefore E0 = [E](1+[G]/KG) + [ES](1+[G]/KG). This is very reminiscent of Eq. 10.8, so let’s set [E*]=[E]+[EG] = [E](1+[G]/KG) and [ES*] = [ES] + [GES] = [ES](1+[G]/KG). And as with Eq. 10.8, we must assume that the amount of G and S bound to enzyme is insignificant. The rate of the irreversible step giving [EP] is v = kc[GES]. [GES] = [ES][G]/KG = [ES*]([G]/KG) ]/(1+[G]/KG) = [ES*]/(1 + KG/[G]). Therefore v = k2[ES*] with k2 = kc/(1 + KG/[G]).

We now have finished the transformation to Michaelis Menten, and so [ES*] is given E0 [S]

by Eq. 10.9: [ES*] = Finally, [GES] = [ES*]/(1 + KG/[G]), so K M + [S] . ⎛ K ⎞ [GES] = ⎜ 1+ G ⎟ ⎝ [G] ⎠

−1

(

)

k + k / 1+ KG / [G] k +k E0 [S] with K M = −1 2 = −1 c . k1 k1 K M + [S]

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

12

b)

25.

26.

27.

c)

k1 and k–1 must be measured in the absence of G, where the E + S ⇋ ES equilibrium is unproductive. They could be measured either by rapid mixing or temperature jump, at S concentrations in the region of 10–4 – 10–5 M E0 and [S].

a)

From the (rather noisy) plot of ln 2/τ vs. [ATP]2, we obtain k = 3.6 × 108 M–1 s–1.

b)

A brief pulse of light is needed because if the generation of ATP is incomplete before the muscle starts relaxing, the second-order kinetics will be masked.

a)

The time needed to reduce substrate to 10–6 M is 2.31 ms.

b)

The time needed to reduce substrate from 1 M to 10–1 M is 902 ms.

c)

The respective times in the presence of the inhibitor are 4.61 ms and 904 ms. At low substrate concentrations, a competitive inhibitor will substantially reduce the reaction rate. At high substrate concentrations, it has very little effect.

a)

v = Vmax/2, so KM = [S] = 3 mM

b)

v = Vmax/2, so KʹM = [S] = 6 mM. Since KʹM = KM(1+[I]/KI), KI = [I] = 10–4 M.

Chapter 10 solutions manual to accompany 5th edition of Physical Chemistry Tinoco, Sauer, Wang, Puglisi Harbison, Rovnyak

13

c)

28.

d)

k3, which would corresponding to the Michaelis-Menten rate coefficient Vmax/E0, since this is the rate limiting step. From the diagram, the ΔH‡ for the reversed first reaction is 21 kJ mol–1.

e)

The last step, which has the highest barrier.

a) From the plot, the inhibition looks competitive. b)

(KʹM/KM) – 1 = [I]/KI = 15.8, so KI = [I]/15.8 = 1.3 µM.