SFP6+7 - Enzymes

Question 1

What is the equation to determine the reaction rate of a reaction?

Answer 1

∆R/P / ∆t (change in reactants or products over time)

Question 2

What tends to be the unit to measure the rate of the reaction?

Answer 2

concentration/time or mass/time (e.g - mM/min)

Question 3

What is k?

Answer 3

The rate constant which represents how fast a reaction is going

Question 4

A net decrease in ∆G is exothermic, what does this reaction mean?

Answer 4

It is spontaneous

Question 5

Which bond is broken in ATP?

Answer 5

Beta and gamma (-32kJ/mol)

Question 6

What is meant by half-life?

Answer 6

The amount of time for half the initial reactant to decrease by half

Question 7

What is the transition state theory?

Answer 7

Molecules in a chemical reaction need to overcome an energy barrier, known as the activation energy, to be converted from reactants to products. The transition involves passing through a high-energy state known as the transition state.

Question 8

What is the rate enhancement equation?

Answer 8

kEcat / kuncat
(Ecat = enzyme catalysed reaction)
(uncat= uncatalysed reaction)

Question 9

What is the relation / proportionality to k and ∆G?

Answer 9

Inversely proportional. If the activation energy decreases, the rate constant increases.

Question 10

This relationship has a log effect, what does this mean?

Answer 10

A small decrease in the activation energy will lead to a large increase in the rate constant

Question 11

The ∆G between the reactants and products is unchanged during the use of a catalyst. Enzymes stabilise the transition state which reduces the activation energy.

Question 12

What is the equation for ∆G?

Answer 12

∆G = ∆H - T∆S
(H = enthalpy = bond making/breaking)
(S = entropy = disorder/order)
(T = temperature)

Question 13

Changing ∆H, how is general acid-base catalysis carried out?

Answer 13

Acid = transfers H+ proton to the developing -ve charge to stabilise it

Base = accepts H+ proton from the devloping +ve charge to stabilise it

Question 14

Why is this done?

Answer 14

To prevent the devlopment of ‘naked’ charges

Question 15

How is electrocatalysis carried out by enzymes to stabilise charge?

Answer 15

The enzyme can have charged sidechains or metal cofactors which stabilise charge

Question 16

How does chymotrypsin act in electrostatic catalysis?

Answer 16

Its peptide backbone stabilises -ve charge on tetrahedral oxyanion

Question 17

What about metal ions - Zn2+?

Answer 17

Zn2+ acts as a metal cofactor in carbonic anyhydrase which stabilises the negative charge from oxygen

Question 18

How can enzymes change the ∆S?

Answer 18

By bringing the substrates into a closer proximity and correct orientation

Question 19

This changes the reaction from a bimolecular to a unimolecular reaction

Question 20

What is covalent catalysis?

Answer 20

Where the enzyme contains a reactive group that becomes temporarily covalently attached to part of the substrate (e.g - chymotrypsin)

Question 21

What is competitive inhibition and when can the enzyme still. function efficiently?

Answer 21

When an inhibitor binds to the active site of the enzyme. If the substrate concentration is high enough, the enzyme can still function as the substrate can outcompete the inhibitor.

Question 22

What happens to the Vmax during competitive inhibition?

Answer 22

It stays the same because the inhibitor does not directly affect the catalytic activity

Question 23

What happens to the Km during competitive inhibition?

Answer 23

It is increased. A high Km means a lower affinity for binding between the substrate and active site.

Question 24

What is Vmax and Km?

Answer 24

Vmax (maximum reaction rate) represents the maximum rate at which an enzyme can catalyze a reaction when it is saturated with substrate. Km is a measure of the affinity between the enzyme and the substrate. It represents the substrate concentration at which the reaction rate is half of the maximum reaction rate (Vmax/2).

Question 25

What is non-competitive inhibition?

Answer 25

The inhibitor binds to a site elsewhere to the active site, causing the enzyme to have a conformational change and reduce its catalytic activity

Question 26

What happens to the Vmax and Km during non-competitive inhibition?

Answer 26

Vmax is reduced because the catalytic activity is reduced. Km is unchanged because the affinity of the enzyme is unaffected.

Question 27

What is Kcat?

Answer 27

The turnover number

Question 28

What does the turnover number represent?

Answer 28

The turnover number is the number of substrate molecules converted to product per enzyme molecule per unit time. It is not a direct count of the number of molecules but rather a rate constant.

Question 29

E.g - Let’s say an enzyme has a kcat value of 100 s^-1. This means that, on average, each enzyme molecule can convert 100 substrate molecules into product per second when it is fully saturated with substrate.

Question 30

What is the equation for Kcat?

Answer 30

Kcat = Vmax / [E]

Question 31

What is the specificity constant?

Answer 31

The specificity constant, kcat/Km, provides a measure of the catalytic efficiency of an enzyme for a specific substrate. A higher specificity constant indicates a more efficient enzyme-substrate interaction and a faster rate of conversion of substrate to product. It is often used to compare and evaluate the efficiency of different enzymes or different enzyme-substrate pairs.