Lecture 5 - Reaction Classes & Enzyme Kinetics

Question 1

What is homolytic vs heterolytic cleavage?

Answer 1

Homolytic - one electron remains with each atom

Heterolytic - the electron pair remains with one of the atoms

Question 2

What is a group transfer synonymous with?

Answer 2

Nucleophilic substitution - as in an acyl transfer, phosphoryl transfer, or glycosyl transfer

Question 3

What type of reaction is a dehydration reaction?

Answer 3

Elimination - eliminate a hydrogen and a hydroxyl on adjacent carbons to create a double bond

Question 4

What is an isomerization?

Answer 4

Most commonly forming an isomer that changes the location of a double bond - i.e. aldehyde / ketone conversation

Question 5

What is a rearrangement?

Answer 5

Intramolecular isomerization which changes the position of a functional group, or can even change the carbon skeleton

Question 6

What is a carbene?

Answer 6

A neutrally charged carbon with 2 bonds to hydrogen and a free electron pair, very reactive

Question 7

What functional groups work as electron sinks in enzymes to avoid unstable carbanion intermediates?

Answer 7

Vitamins B1 and B6

Question 8

What is the kinetic barrier vs the thermodynamic barrier?

Answer 8

Kinetic barrier - activation energy determines reaction rate

Thermodynamic barrier - the spontaneity of the reaction is determined by the gibbs free energy change, independent of path

Question 9

How does delta G relate to Keq?

Answer 9

deltaG = -RTln(Keq)

Question 10

In a rate law, how much the units work out?

Answer 10

The k term (rate constant) adopts whatever units needed to make the final rate M/s (molarity of product generated per sec)

Question 11

What is a zero-order reaction?

Answer 11

r = k, the rate is independent of reactant concentration and is constant

Question 12

What is a first-order reaction?

Answer 12

The rate is directly proportion to the concentration of one of the reactants

r = k[A], k has units 1/s

Question 13

What is a second order reaction?

Answer 13

Rate is directly proportion to >1 molar quantity.

r = k[A]^2, or r = k[A][B], k has units 1/M*s

Question 14

What is meant by saturation kinetics?

Answer 14

At a fixed enzyme concentration, there is a non-linear relationship between reaction rate (velocity) and reactant concentration in a first-order reaction. This is because an enzyme reaction is second-order (depends on concentration of both enzyme and substrate)

Question 15

What is the Michaelis-Menten Equation?

Answer 15

v = (Vmax*[S])/(Km + [S])

Question 16

What are the three assumptions in the derivation of the Michaelis-Menten Equation?

Answer 16

1. [S] far exceeds [E], so that [S] can be viewed as a constant
2. Initial velocity of reaction is measured, so that k4 (conversion of enzyme + product into enzyme-substrate complex) can be ignored. Reaction is reversible but not that far along.
3. Reaction is in steady state

Question 17

What does it mean for a reaction to be in steady state?

Answer 17

Rate of formation of enzyme substrate complex is equal to the rate of its dissociation (into E+S or E+P). Thus, [ES] is constant but difficult to measure.

Question 18

What is reaction velocity proportional to? How does this define an important term?

Answer 18

[ES]. The concentration of the enzyme substrate complex.

At 100% saturation, [ES] = [E]subT, the total enzyme. At this saturation, the reaction rate is equal to Vmax.

If [ES] < [E]subT, velocity is simply v.

Question 19

How is the Km derived?

Answer 19

With the assumption that the ES breakdown is by one of two first order reactions, with coefficients k2 for breakdown into E+S, and k3 for breakdown into E+P.

The sum of these two breakdown events is equal to the formation of ES in steady state.

Thus, k1[E][S] = (k2 + k3)[ES]
(k2+k3)/k1 = ([E][S])/[ES]

Question 20

Why is Km different than Kd (dissociation constant)?

Answer 20

Kd would only assume that the [ES] will return back to E + S if it dissociates. The only time Km approximates Kd is when the rate of product formation is incredibly low, even enzymatically. That is, k3 «&laquo_space;k2.

Km = for enzymes
Kd = for receptors (receptor saturation vs ligand)

Question 21

What is Km equal to?

Answer 21

The substrate concentration that gives 1/2 Vmax.

Question 22

How does the velocity vs [S] change in order with respect to S?

Answer 22

When [S] &laquo_space;Km, the velocity increases linearly with [S], and is thus first order ([E] is held constant)

When [S]&raquo_space; Km, velocity is not really affected by [S], and is thus zero order with respect to S (enzyme is rate-limiting).

Question 23

What is meant by “Unit” for velocity?

Answer 23

micromole / minute, its a term for velocity that is often used to describe Vmax

Question 24

What is meant by Specific Activity and how is it expressed?

Answer 24

When an enzyme is in a protein solution, and the molar concentration of the protein is unkonwn, its activity is expressed as a specific activity, which is a function of the total protein solution (in grams).

SA = Units (micromoles / minute, a velocity) per mg protein

micromol / (minutes * mg)

Question 25

How is Kcat defined?

Answer 25

It is the turnover number, and is calculated using Vmax for a particular enzyme. Vmax = Kcat * [E]T, Kcat has units 1/s The molar quantity of product formed per second, per molar quantity of enzyme (divide Vmax by [E]subT). More practically, it is the maximum number of molecules of substrate that an enzyme can convert to product per catalytic site per unit of time

Question 26

What is Kcat a function of?

Answer 26

Is it eh overall first-order rate constant for the conversion of ES to E+P, like k3 in the Michaelis-Menten derivation. If you add in the additional step of ES converting to EP before product release, kcat is equal to the addition of the forward reaction rate constants from ES to EP, and EP to E+P. Obviously, Kcat will be most influenced by the slowest step of the reaction.

Question 27

If substrate is in molar excess of enzyme, how does V relate to [E]?

Answer 27

v is directly proportional to [E] double [E], double the rate.