Question

The reaction A + B → C has a ∆G^0’ of -20 kJ/mol at 25 ⁰C. Starting under

       Standard Conditions, one can predict that:

A. at equilibrium, the concentration of B will exceed the concentration of A.

B. at equilibrium, the concentration of C will be less than the concentration of A.

C. at equilibrium, the concentration of C will be much greater than the

     concentration of A or B.

D. C will rapidly break down to A + B.

E. when A and B are mixed, the reaction will proceed rapidly toward formation

     of C.

29. Estimate the value of K_m for an enzyme-catalyzed reaction for which the

      following data points were obtained.

[S] (M)                       V₀ (mM/min)

2.5 x 10^-6                                28

4.0 x 10^-6                                40

1.0 x 10^-5                                70

2.0 x 10^-5                                95

4.0 x 10^-5                                112

1.0 x 10^-4                                128

2.0 x 10^-3                                139

1.0 x 10^-2                                140

A. 1.0 x 10^-4

B. 1.0 x 10^-3

C. 1.0 x 10^-6

D. 1.0 x 10^-5

E. 1.0 x 10^-7

30. pH = pK_a when:

[A^-]/[HA] = 0

log ([A^-]/[HA]) = 1

[A^-] >> [HA]

[A^-] = [HA]

log ([HA]/[A^-]) = 1

31. In the conversion of A into D in the following biochemical pathway, enzymes

       E_A, E_B, and E_C have the K_m values indicated under each enzyme. If all of the

      substrates and products are present at a concentration of 10^-4 M and the

      enzymes have approximately the same V_max which step will be rate limiting?

                                         R_x1      R_x2     R_x3

                                                  A   ↔   B   ↔   C ↔ D

                                                          E_A        E_B       E_C

                                                           K_m =     10^-2M    10^-4M   10^-4M

A. Rx 3         

B. Rx 1         

C. Rx 2         

D. Rx’s 1 and 3

E. Rx’s 2 and 3       

32. Consider the following K_eq values with the appropriate ΔG^0’?

                                       K_eq                                        ΔG^0’

                        A           1.0                           1      28.53

                      B           10^-5                             2     -11.42

                        C           10⁴                          3      5.69

                        D           10²                          4       0

                        E           10^-1                         5      -22.84    

                                   

A K_eq of 10² would correspond to which of the following ΔG^0’ values?

28.53

-11.42

5.69

0

-22.84

33. For an enzyme that follows simple Michaelis-Menten kinetics, what is the

      value of V_max if V₀ is equal to 1 µmol minute^-1 when [S] = 1/10 K_m?

11.0 µmol minute^-1

1.10 µmol minute^-1

0.10 µmol minute^-1

10.1 µmol minute^-1

1.0   µmol minute^-1

Answer 1

Since deltaG is negative, the reaction is spontaneous and K > 1.

deltaG^o = - RT*ln(K)

=> - 20000 J = - 8.314*298*ln(K)

=> K = 3205

Since K>>1, forward reaction is favored and the equilibrium concentration of C will be much higher than both A and B.

Hence first four options are wrong

Also when A and B are mixed, the reaction will move rapidly towards the formation of C. Option (E) is correct.

#29:

Since the reaction velocity becomes constant at 140 mM/min, Vmax = 140 mM/min

Km is the substrate concentration at which the reaction velocity becomes half of the maximum velocity.

Half of maximum velocity = Vmax/2 = 140 / 2 = 70 mM/min

At 70 mM/min the concentration is 1.0 x 10^-5 M. Hence Km = 1.0 x 10^-5 (Answer)

Hence option (D) is correct.

#30: According to Hendersen equation for buffer solution:

pH = pKa + log[A-] / [HA]

=> pKa = pKa + log[A-] / [HA] [ pH = pKa]

Subtracting pKa from both sides:

=> 0 = log[A-] / [HA]

Taking power 10 to both sides

=> 10⁰ = [A-] / [HA]

[A-] / [HA] = 1

=> [A-] = [HA]

4th option is correct.