Enzymes II

Question 1

Reactions catalyzed by enzymes
occur at

Answer 1

the active site

Question 2

The active site is

Answer 2

a pocket or cleft in an enzyme that is the
reaction site
and binding site for the substrate

Question 3

do enzymes affect the equilibria

Answer 3

Enzymes increase the rate of the reaction;
they do not affect the equilibria

Question 4

negative Gibbs

Answer 4

reaction favors S to P
BUT, it does not mean that
the conversion will occur
at a detectable rate!

Question 5

Uncatalyzed raections

Answer 5

Uncatalyzed
reactions are often reactions are often
thermodynamically favorable…
but too slow

Question 6

how do enzymes work

Answer 6

Enzymes enhance reaction rates by lowering activation energies.

Question 7

specific ES complex

Answer 7

ES complex is stabilized by weak, non-covalent interactions: hydrogen bonds, ionic, hydrophobic, and Van der Waals interactions

When the weak interactions are formed, the free energy (binding energy) released stabilizes the ES complex AND
provides the energy used by enzymes to…..
lower Ea
give specificity

Question 8

Lock-and-Key Model

Answer 8

enzyme complementary to substrate
forms Es complex thats is so stable the reaction wont progress

Question 9

induced Fit Theory

Answer 9

changes in protein structure caused by a
substrate will bring the catalytic groups into proper orientation for reaction, whereas a non-substrate
will not.”

enzyme complementary to TS where optimal interaction occurs

Question 10

Enzyme kinetics

Answer 10

how the RATE of a reaction changes in response
to changes in the experimental parameters

substrate concentration affects the rate

Quantitatively this is expressed by the Michaelis-Menten equation:
V0 = Vmax [S] / Km + [S]

Question 11

Km calculated

Answer 11

the velocity is proportional to substrate concentration- first order- low [S]

velocity is INDEPENDENT of [S]- zero order- high [S]

Km = 1/2 Vmax

Question 12

Michaelis-Menten kinetics

Answer 12

Assumptions of Michaelis-Menton kinetics:

the concentration of substrate [S] is much greater than the concentration of enzyme [E]

• [ES] does not change with time…this is called steady-state

initial reaction velocities are used in the analysis, BUT the initial rate reflects the steady-state
rate of reverse reaction is ignored

Question 13

Michaelis-Menten kinetics: determining Vmax and Km

Answer 13

a double-reciprocal plot is used to generate
a straight line. ie Lineweaver-Burk plot

-1/Km is determined from
the x-intercept
1/Vmax is determined from
the y-intercept

Question 14

what does Km tell us?

Answer 14

Km is the Michaelis constant.
It is a characteristic of an enzyme substrate.
Km does NOT vary with enzyme concentration.

Question 15

km and affinity

Answer 15

ONLY if k2 is much smaller than k-1 (k2 is RATE-LIMITING), then Km reflects the AFFINITY (Kd) of the enzyme for that particular substrate.

Small Km = high affinity of enzyme for substrate, because a LOW concentration of substrate is needed to half-saturate the enzyme.

Big Km = low affinity of enzyme for substrate, because a HIGH concentration of substrate is needed to half-saturate the enzyme

Question 16

meaning of kcat

Answer 16

the rate-limiting step.
In a simple M-M reaction (one enzyme/one substrate and dissociation of EP fast), kcat = k2

represents the maximum number of substrate molecules converted to product per active site per unit time.

Question 17

meaning of kcat/Km

Answer 17

Specificity constant

compare catalytic efficiencies of different enzymes

determines the specificity for competing substrates

reflect the cellular environment, concentration of substrate encountered and the chemistry of the reaction.

Question 18

diffusion-controlled limit

Answer 18

Enzymes with kcat/Km values near the diffusion-controlled limit are said to have achieved catalytic perfection

When the [S] is «< Km,
V0 = kcat/Km [E] [S]
Since V0 depends on both the concentrations of [E] and [S], kcat/Km is a second-order rate constant.
 Units are M-1 sec-1
 The upper limit is set by the rate at which E and S diffuse together in a aqueous solution.
 This is called the diffusion-controlled limit (108 to 109 M-1 sec-1)

Question 19

The formation of a specific ES complex is
what distinguishes enzymes

Answer 19

The optimal interaction between enzyme and
substrate occurs when the ES complex reaches
transition state

When the weak interactions are formed, the free energy (binding
energy) released stabilizes the ES complex AND
provides the energy used by enzymes to
(1) lower the activation energies of reactions (CATALYSIS)
(2) give SPECIFICITY

Question 20

Initial velocity depends on substrate concentration- single substrate

Answer 20

SINGLE SUBSTRATE; however, many of the same
principles can be extended to multi-substrate system

Question 21

Steady-state kinetic analysis of bi-substrate reactions

Answer 21

obey M-M kinetics
when the concentration of one substrate is held
constant and the other is varied.

SEQUENTIAL
PING-PONG

Question 22

SEQUENTIAL

Answer 22

all substrates must bind before
product is formed

a. ordered
b. random

ternary complex
EAB formed

Question 23

DOUBLE-DISPLACEMENT (PING-PONG)

Answer 23

one or more products are released before
all substrates are added

no ternary complex

Question 24

steady state kinetics-limitations

Answer 24

Steady state kinetic measurements of an
enzyme usually only give the
Km, which may or may
not reflect the dissociation constant of ES complex, and

the kcat, which may be a combination of rate constants for several steps

Question 25

Pre- steady state kinetics

Answer 25

measure rate constants of individual
steps and detect transient intermediates

techniques/methods to make and
record measurements on a 1 to 10-7 SECOND timescale