The enzyme requires the coenzymes FAD and TPP.
Use the letters from 1-4 as well as these: (S) substrate; (P) product;
(C) co-substrate coenzyme; (R) prosthetic coenzyme
Quiz 4 Key |
|
1-4. Use these types of enzymes: (O) oxidoreductase; (H) hydrolase; (I) isomerase;
(T) transferase; (Y) lyase; (G) ligase.
O converts Fe2+ to Fe3+ |
G forms a bond using ATP to provide energy |
H breaks a bond by adding H2O to it |
Y adds a functional group to a C=C bond |
H synthesizes a bond by removing H2O |
Y removes a functional group, |
I rearranges the structure of a |
T transfers a functional group from one |
5-8. For exam 4 we will study the enzyme that catalyzes this reaction.
The enzyme requires the coenzymes FAD and TPP.
Use the letters from 1-4 as well as these: (S) substrate; (P) product;
(C) co-substrate coenzyme; (R) prosthetic coenzyme
O,Y,T one type of enzyme this could be |
R FAD |
|||
I,G,H one type of enzyme this could not be |
R TPP |
|||
S pyruvate |
C CoA-SH |
|||
P CO2 |
P AcCoA |
|||
C NAD+ |
||||
9-10. Underline the value for the least favored (least spontaneous) reaction and circle the most favored. |
||||
ΔG°′ = – 2,000 J/mol |
Keq = 137 |
Keq = 0.62 |
ΔG°′ = 13,000 J/mol |
|
Keq = 137 |
ΔG°′ = 3,000 J/mol |
ΔG°′ = 3,000 J/mol |
Keq = 137 |
|
9-10. Underline the value for the least favored (least spontaneous) reaction and circle the most favored. |
||||
Keq = 0.062 |
ΔG°′ = 3,000 J/mol |
ΔG°′ = - 20,000 J/mol |
Keq = 1.37 |
|
ΔG°′ = 6,900 J/mol |
Keq = 0.298 = 0.30 |
Keq = 3200 |
ΔG°′ = –780 J/mol |
|
Comment: I added the complementary values to make the choices clearer, but I didn't expect you to do any calculations. Instead I hoped that the answers would be fairly obvious. What was clear was that most students are more comfortable with ΔG than with Keq. Oh yes, also note the units.