CH 5 (L12)

Question 1

Steady staten for enzymatic reactions

Answer 1

When ES formation and dissociation is in equilibrium

Question 2

Vmax

Answer 2

What happens when you keep adding substrate?
Eventually E becomes saturated and Vo has reached approximation of its max (Vmax), this can be determined by the formula:

Vmax = k2[E]total

Substituting… we get Michaelis-Menten equation:

Vo = (Vmax[S]) / (Km + [S])

Question 3

Enter the catalytic constant

Answer 3

at high [S], the overall velocity of the reaction is approximating Vmax which is determined by [E]total

–> under these conditions, the rate constant observed is called Kcat (catalytic constant)

Question 4

kcat

Answer 4

represents the moles of substrates converted to product per second per mole of enzyme:
Smoles –> P/sec/Emol

Vmax = kcat[E]total

kcat = Vmax / [E]total

Question 5

kcat

Answer 5

• the catalytic constant tells us then, the maximum number of substrate molecules converted to product each second (aka turnover number)
• can also indicate efficiency
• to calculate Vmax from kcat, we need to know [E]

Question 6

is k2 = kcat?

Answer 6

K2 is a good approximation of Kcat
If it is a multi-step reaction with a rate-limiting step, the limiting step rate constant would equal kcat.

• kcat could be a combination of different constants
  Vmax = k2[E]total

Question 7

Km

Answer 7

km is the substrate concentration when the rate of the reaction is 1/2 vmax.
Km = ½ vmax

Km = k-1/k1 if the k2 is MUCH smaller than the others, (k2 can be ignored, not comparable to other values) → km = equilibrium constant for ES complex dissociation and association → km becomes a measure of affinity for E and S

higher Km = lower affinity (less substrate to reach ½ vmax)

Question 8

So far…

Answer 8

Km is a measure of the stability of the ES complex/
Kcat is similar to K2 (ES –> E + P) when the substrate is not limiting

In general, Vo = k[E][S], though what form that takes differs

Question 9

at low [S]

Answer 9

At low substrate concentration [S] → substrate as the limiting factor
We can ignore [s] in the denominator since it is much lower than km

• at low [S], Vo is equal to the catalytic constant (# of moles converted from S –> P/sec/mol) over the measure of affinity for ES complex times [E] and [S]

Will not be k2 because this is a rate-limiting step
→ v0=(kcat/km) ([E][S])

Question 10

kcat/km

Answer 10

• this ratio is important as it tells us (is a rate constant for) the formation of E+P from E+S when the reaction is limited by S encountering E
• Region a: high substrate concentration → v0 =kcat[E][S]
  (substrate no longer a limiting factor, enzyme is saturated)
• Region b: low substrate concentration → v0=(kcat/km) ([E][S])

Question 11

Catalytic proficiency comparison

Answer 11

• it is possible to asses how proficient an enzyme is by dividing rate constants for reaction in the presence of enzyme (kcat/km) by the rate constant of the reaction without enzyme (Kn)
• -> very few proficiencies are known because Kn is hard to measure

Table 5.2
Higher catalytic proficiency -> the one with the higher enzymatic rate constant (kcat/km)

Question 12

Km and Vmax

Answer 12

To obtain Km and Vmax from straight lines in graphs taken from calculated V0, we can use the double reciprocal lineweaver-burk plot

lower Km = higher ES affinity (k1/k2)

graph looks like:
Y axis: 1/Vo
X axis: 1/[S]
draw a longer x axis
draw a straight linear line
1/vmax: point hitting Y axis
-1/km: point at the x axis

What would we need to calculate Kcat from Vmax? [E]

kcat = Vmax / [E] total

Question 13

Multi-substrate reactions

Answer 13

• Complex that is formed from substrate A and B is EAB complex
• Products are P and Q
• sometimes it is sufficient to determine Km for each substrate in saturating amounts
  –> enzymes are not (permanently) altered by reactions
  E + A + B (EAB) –> E + P + Q

Question 14

Sequential reactions

Answer 14

• Sequential reactions require ALL substrates to be present before product is released, they can be ordered or random

ordered: obligatory order for the addition of substrates
random: random order for the addition of substrates

Question 15

Ping-pong model

Answer 15

Ping-pong model: product is released before all substrates are bound.
A alters the enzyme [E], B binds to the altered enzyme (F), the enzyme returns to normal

—> enzyme is modified in the middle of the reaction, return to its native state at the end of the reaction