All the answers refer to the 2nd Edition of  CBM

1. Assume the energy of A>B and  A = ΔG1, B = ΔG2. Draw an energy diagram that shows the progress of A→Β. Label both axes. Show ΔG and ΔG* when

a. There are no unstable intermediate. (3 points)

ANS: Fig 6.3, p.129

b. When there is an unstable intermediate. (3 points)

ANS: Fig. 6.7, p. 131

2. How does a transition state differ from an unstable intermediate? (3 points)


A transition state occurs at an energy maximum. An unstable intermediate occurs at an energy minimum that is higher than the final energy level.  (Fig. 6.7, p.131)

3. Define the Principle of Microscopic Reversibility. (3 points)

ANS: Rule 6.2, p. 132

4. What does the following equation tell us? (By convention, the energy level of the reactants is subtracted from the energy level of the products).

ΔG = ΔGº + RT ln[P]/[R]

where R = 1.986 cal/mol/ºK  (8.314 joules/mol/ºK, T = ºK,  [P] = molar concentration of the product(s), [R] = molar concentration of the reactants


a.  ΔG is negative for a spontaneous reactions (by convention). (Section 6.3, p 132) (2 points)

b. The reaction is as equilibrium when ΔG = 0 (Rule 6.4) (2 points)

c. At equilibrium,  ΔGº = – RT ln Keq  (Equation 2, p. 133) (2 points)

This equation  tells us that if we measure the equilibrium concentrations of reactants and products at a specified temperature and pressure, we can evaluate  ΔGº (R = 1.987 calories/mol/degree K). Indeed, this is the way many of the ΔGº values we find in the literature were measured.

d. When a reaction is at equilibrium, the ratio of concentrations of reactants to those of the products is a constant that is characteristic of the reaction in question at a specified temperature and pressure. (Rule 6.5, p.133) (2 points)

5. What is the difference between ΔG, ΔGº, ΔG’ and ΔGº’?


ΔG = the free energy under the actual reaction conditions (2 points)

ΔGº = the free energy under standard conditions (2 points)

ΔG’ = the free energy of the actual reaction concentrations  when the pH = 7.0 (2 points)

ΔGº’ = the free energy under standard conditions except pH = 7.0. (2 points)

6. Define standard thermodynamic conditions. (4 points)


T = 273º K (0º C), [reactants] = [products] = 1 M.

7. How many cal/mol are available when [H+] is diluted from 1 M to 10-7 M at 25º?  (10 points)


RTln {[10-7]/[1]} = (1.986)(273º + 25º) (-1.612) = -9,539 cal/mol.

8. Explain why ATP has such a high ΔGº’ (- 7,l300 cal/mol) of hydrolysis. (5 points)


This is primarily an entropy effect due to dilution of protons. (section 6.10, p.143)

9. Explain why phosphoenol pyruvate has such a high ΔGº’ (- 13,800 cal/mol) of hydrolysis. (5 points)


This is primarily an entropy effect involving keto and enol tautomers.(section 6.8, p. 141)

10. Write a mechanism for the formation of glucose 6-phosphate from glucose and ATP. (Phosphates usually react by substitution) (4 points)

ANS: Fig. 6.25, p. 150

11. Define a coupled reaction and give an example. (4 points)


Reactions with a common intermediate are said to be coupled. (Rule 6.8, p. 152)

12. What is the difference between a “closed system” and an “open system”?  (5 points)


A closed system exchanges heat but not mass with the environment. An open system exchanges both heat and mass with the environment. (section 6.16)

13. Assume the pathway A→ B→C→D occurs in an open system at steady state at 37º C where


a. From these data what can you conclude about the pathway under standard conditions?


The standard free energy of B→C is positive so this reaction cannot proceed under standard conditions. (4 points)

b. Calculate ΔG’ for each reaction at 37ºC if the steady state concentrations of [A] = 6mM and [D] = 0. 1 mM.  [Remember all concentrations must be M, T is in ºK and R = 1.986 cal/molºK]


The velocity of an enzymatic reaction is given by: v= Vmax[S]/{KM + [S]} We are given that [A] = 0.006 M. Therefore, the initial velocity of A→B is v = 90(0.006 M/(5 x 10-3 M + 0.006 M) = 49 arbitrary units. (4 points)

At steady state all reactions run at the same velocity. Therefore, 49 = 80[B]/(0.01M + [B] and [B] =0.016 M (4 points)

also 49 = 100[C]/(10-3 + [C]) and [C] ≈ 0.001M (4 points)

Using these concentrations, we can calculate ΔG’ for each reaction. (T able 3, p..157) ( 3 points for each reaction = 9 points)

The discussion also shows how to determine the concentrations graphically.

14. The Nomenclature Committee of the International Union of Biochemistry has defined ligases as “—enzymes catalyzing the joining of two molecules coupled with the hydrolysis of a pyrophosphate bond in ATP or a similar triphosphate. The bonds formed are often “high-energy bonds”. (p. 153)

What is wrong with this definition? (10 points)


Ligases catalyze coupled reactions between two substrates and ATP or a similar triphosphate. The common intermediate is usually a mixed  anhydride derived from the reaction of ATP and one of the starting materials. The mixed anhydride then reacts with the second starting material to form the final product. Hydrolysis of ATP is never involved. Furthermore, the term “high energy bond” is unacceptable in this context since most of the energy in a compound of this sort is due to entropy and is not concentrated in any bond p. 153).

Grade Yourself:  90-100 = A, 75-89 = B, 50-74 = C, 25-49 = D, <25 = F


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