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7 Cards in this Set

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QUESTION #1:

Choose the best description of an enzyme:

(a) It makes a reaction thermodynamically favorable.

(b) It allows a chemical reaction to proceed extremely fast.

(c) It increases the rate at which a chemical reaction approaches equilibrium relative to its uncatalyzed rate.
ANSWER:

c
QUESTION #4:

Approximately how much does staphylococcal nuclease (Table 11-1, pg. 323) decrease the activation free energy (∆G ‡) of its reaction (the hydrolysis of a phosphodiester bond) at 25⁰C?
ANSWER:

At 25⁰C, every 10-fold increase in rate corresponds to a decrease of about 5.7 kJ/mol in ∆G ‡.

For the nuclease, with a rate enhancement on the order of 10¹⁴, ∆G ‡ is lowered about 14*5.7 kJ/mol, or about 80 kJ/mol.

Alternatively, since the rate enhancement, k, is given by k = e^(∆∆G‡_cat/RT) ,
ln (k) = ln (10¹⁴) = ∆∆G‡_cat / (8.3145 * (273 + 25))

Hence, ∆∆G‡_cat = 80 kJ/mol
QUESTION #5:

On the free energy diagram shown, label the intermediate(s) and transition state(s). Is the reaction thermodynamically favorable?
ANSWER: (PIC in back of book)

There are three transition states (X‡) and two intermediates (I).

* The three transition states are at the three "peaks" on the graph.
* The two intermediates are at the two "troughs" on the graph.

The reaction is *not* thermodynamically favorable because the free energy of the products is greater than that of the reactants.
QUESTION #8:

The covalent catalytic mechanism of an enzyme depends on a single active site Cys whose pK is 8. A mutation in a nearby residue alters the microenvironment so that this pK increases to 10. Would the mutation cause the reaction rate to increase or decrease? Explain.
ANSWER:

The active form of the enzyme contains the thiolate ion. The increase pK would increase the nuclophilicity of the thiolate and thereby increase the rate of the reaction catalyzed by the active form of the enzyme.

However, at physiological pH, there would be less of the active form of the enzyme and therefore the overall rate would be decreased.
QUESTION #11:

Explain why RNase A cannot catalyze the hydrolysis of DNA.
ANSWER:

DNA lacks the 2'-OH group required for the formation of the 2',3'-cyclic reaction intermediate.
QUESTION #13:

Explain why lysozyme cleaves the artificial substrate (NAG)₄ ~4000 times more slowly than it cleaves (NAG)₆.
ANSWER:

The lysozyme active site is arranged to cleave oligosaccharides between the fourth and fifth residues.

Moreover, since the lysozyme active site can bind at lease six monosaccharide units, (NAG)₆ would be more tightly bound to the enzyme than (NAG)₄, and this additional binding free energy would be applied to distorting the D ring to its half-chair conformation, thereby facilitating the reaction.
Duh!
QUESTION #14:

Lysozyme residues Asp 101 and Arg 114 are required for efficieny catalysis, although they are located at some distance from the active site Glu 35 and Asp 52. Substituting Ala for either Asp 101 or Arg 114 does not significantly alter the enzyme's tertiary structure, but it significantly reduces its catalytic activity.

Explain.
ANSWER:

Asp 101 and Arg 114 form hydrogen bonds with the substrate molecule (Fig. 11-19, pg. 343). Ala cannot form these hydrogen bonds, so the substituted enzyme is less active.