Lecture 9 - Enzymes

Question 1

What are the strengths and weaknesses of using enzymes to catalyse substances?

Answer 1

Strengths - almost always a 100% yield (no wasteful byproducts)

Weakness - Stereospecific: each enzyme typically reacts with one substrate and usually only one stereoisomer of that substrate (can also be a strength (above))

Question 2

The role of enzyme active sites

Answer 2

Binding energy from non-covalent interactions to bring the activation energy down

Covalent interactions (electrostatic, hydrogen bonds etc) to bring down the activation energy

Question 3

Cofactors: what are they, what do they do, and what are some examples?

Answer 3

Extra non-protein functional groups located near the active site that aid with catalysis

Essential ions, coenzymes, cosubstrates, and prosthetic groups

Question 4

Enzyme kinetics: what is the equation to work out the rate?

Answer 4

Δ[P]/ΔT = v = k[S]

This means that the change in the product concentration over the change in time is equal to the rate which is equal to the rate constant (k) of the concentration of substrate consumed ([S])

Question 5

Michaelis-Menton equation: what is it, what does it mean, and what does it assume?

Answer 5

An equation used to describe the hyperbolic nature of enzyme-catalysed reactions

E + S ⇌ ES -> E + P

The first reaction is a fast, reversible reaction where the Enzyme and Substrate bind and the last reaction is a slow, non-reversible reaction where the substrate is broken down and the product is formed

It assumes that product formation is the slowest step

Question 6

Michaelis-Menton equation: what is the actual full equation?

Answer 6

Vₒ = Vₘₐₓ[S]/kₘ+[S]

Vₒ is the velocity of the reaction
kₘ is the Michaelis-Menton equation
vₘₐₓ is the maximum velocity
[S] is the concentration of substrate

Question 7

kₘ: What is the equation of it, what does it mean, what does the lower or higher mean and what is it equal to?

Answer 7

kₘ = (k₋₁ + k₂)/k₁

k₂ is practically negligible so typically kₘ = k₋₁/k₁

A measure of the affinity of the enzyme for the substrate decomposition constant over the enzyme-substrate formation constant

The lower the kₘ value, the higher the enzyme’s affinity, and the tighter the substrate binding

Equals the concentration of substrate needed for 1/2 maximum velocity

Question 8

What does Vₘₐₓ mean and what is its relationship with enzyme concentration?

Answer 8

The maximum velocity when an enzyme is saturated with substrate

Proportional

Question 9

k꜀ₐₜ: what is it and how do you calculate it?

Answer 9

The catalytic constant/turnover number which acts as a measure of the number of molecules of substrate that are converted to product per second per active site

At saturating [S], vₘₐₓ = k꜀ₐₜ[Eₜ]

Question 10

Six classes of enzymes

Answer 10

Oxidoreductases
Transferases
Hydrolases
Lyases
Isomerases
Ligases

Question 11

Oxidoreductases

Answer 11

Catalyse oxidation-reduction reactions

Question 12

Transferases

Answer 12

Catalyse group transfer reactions

Question 13

Hydrolases

Answer 13

Catalyse hydrolysis

Question 14

Lyases

Answer 14

Lyse things and cleaves things and causes double bonds to form

Question 15

Isomerases

Answer 15

Catalyse isomerism reactions

Question 16

Ligases

Answer 16

Catalyse ligation using ATP

Question 17

What is the difference between a normal chemical reaction and a catalysed reaction

Answer 17

Chemical reaction - first-order linear change in concentration over time, the reaction is directly related to the rate constant k

Enzyme-catalysed reaction - Begins at first order but eventually becomes zero order (reaction becomes independent of [S]) at high [S] values (because there are a limited number of active sites)