Topic 1 - Module 5

Question 1

Explain how enzymes speed up chemical reactions

Answer 1

Lowering the activation energy barrier, and accelerating a typically unfavorable reaction.

• binding substrates in the correct orientation
• providing catalytically active groups (side chains, metal ions etc)
• polarising the bonds and stabilising charged species (usually unstable)
• stabilising the transition state.

Question 2

Define transition states in terms of enzymes.

Answer 2

The transition state itself is a high energy, meta-stable species, hence, lowering free energy of this state is equivalent to lowering the activation energy (Ea)
Enzyme decreases the free energy of the substrate at the transitional state.

Question 3

Explain, using examples, the importance of enzymes in biology and medical fields of study.

Answer 3

Enzymatic catalysis is important to achieve adequate levels of fuel conversion needed over an appropriate timeframe necessary to sustain life.
Excessive/deficiency enzymatic activity results in disease.

Question 4

Describe different types of enzymes using examples

Answer 4

Kinase (transfer of phosphate, hexokinase); Phosphorylase (addition of free phosphate, glycogen phosphorylase);
Phosphatase (cleaves phosphate from a molecule, glucose-6-phosphate);
dehydrogenase (catalyse redox reactions, glyceraldehyde-3-phosphate dehydrogenase);
Mutase (catalyse shift of phosphate group within the same molecule, phosphoglycerate mutase);
Isomerase (structural isomerization, triase phosphate isomerase);
Hydratase (addition/removal of water, enolase);
Synthase (synthesis of a product, citrate synthase)

Question 5

Explain the difference between a reaction intermediate and a transition state

Answer 5

Intermediates are stable species, whereas the transition state is the unstable complex between reactant and product, showing higher energy.

Question 6

Describe how the enzyme concentration affects reaction rate

Answer 6

Enzyme concentration is likely to speed up the rate of reaction, by binding the substrate in the correct orientation relative to active groups.

Question 7

Define the following terms: binding energy, intial rate, Km and Vmax

Answer 7

Binding energy: the minimum energy required to disassemble a substrate-reactant complex into separate parts.
Initial rate: known as V0, (Vmax x S)/Km + S
Km: a constant that describes the reaction under steady-state conditions (not an equilibrium constant) - the amount of substrate at 1/2Vmax - and the affinity of the enzyme for its substrate

Question 8

What are the assumptions associated with the Michaelis-Menten steady-state descriptions of enzyme kinetics

Answer 8

• the conversion of ES to E + P is irreversible
• SS conditions ( [ES] remains constant, and its rate of formation = rate of dissociation)
• lots of substrate present, with little enzymes
• more substrate than the product assumed as initial conditions

Question 9

Describe the difference between irreversible and reversible inhibitors

Answer 9

Irreversible inhibitors will bind covalently to the active site, destroying the functional group needed for enzymatic activity or will form stable non-covalent complex with the enzyme.
Reversible inhibitors will bind reversibly to enzymes, temporarily blocking their activity and then dissociate afterwards.

Question 10

Describe the changes to Vmax and Km caused by competitive inhibitors

Answer 10

Apparent Vmax is unchanged, and apparent Km is larger, reflecting weaker binding affinities caused by the excess substrate concentration.

Question 11

Describe the changes to Vmax and Km caused by uncompetitive inhibitors

Answer 11

Apparent Vmax and Km are both reduced, as [ES] is reduced due to the presence of uncompetitive inhibitors

Question 12

Describe the changes to Vmax and Km caused by mixed inhibitors

Answer 12

Apparent Vmax is reduced, and the apparent Km is increased, both dependent on value of alpha and alpha’

Question 13

Describe the mechanisms of competitive inhibitors

Answer 13

EI and ES complexes are mutually exclusive, where the presence of the inhibitor will remove free enzyme [E] from the equation, reflecting apparently weaker binding affinities.

Question 14

Describe the mechanisms of an uncompetitive inhibitor

Answer 14

ESI complex has no activity, which reduces the [ES] - described by alpha’, hence apparent Vmax and Km are both smaller.

Question 15

Describe the mechanisms of mixed inhibitors

Answer 15

The enzyme is inactivated by the inhibitor, yet it is still able to bind substrates, this means that [ESI] (and [ES]) will reduce, while the overall [E] is still reduced.
Vmax is smaller, but Km is larger.

Question 16

list the properties of factors: alpha and alpha prime, then describe how they change apparent Vmax and Km

Answer 16

Alpha: the inhibition constant of inhibitors binding - affects Km;
alpha’ describes [ES] being reduced, and is dependent on how tightly inhibitor binds, as well as [I] - will reduce both Vmax and Km

Question 17

Describe the features of an allosteric enzyme

Answer 17

Allosteric enzymes regulate metabolic pathways by responding to changes in [molecules] in surroundings, this is done by reversible and covalent-binding to the substrate.

Question 18

explain how positive and negative allosteric modulators control enzyme activity

Answer 18

Modulators will cause a conformational change in the enzyme.
Positive ones will decrease Km, showing that the enzyme exhibits stronger binding affinity for the substrate - increasing rate of enzymatic activity; negative is the opposite.

Question 19

describe how different modulators stabilize different conformational states

Answer 19

Positive modulators increase affinity for substrate binding, this will stabilise the R-state, making the pure R-state more populated.
Negative modulators will decrease affinity for substrate-binding, stabilising the T-state.

Question 20

Describe how ATCase from E.Coli is an example of allosteric regulation

Answer 20

ATCase is an allosteric enzyme, positively modulated by ATP and negatively modulated by CTP. High levels of ATP in the bacteria indicate metabolic growth in the cell, leading to the need for more CTP present, which reflects lower K0.5, shifting the curve to the left showing the R-state being more populated.