MCBG Session 3 - Enzymes

Question 1

What is the function of an enzyme and how does it achieve this?

Answer 1

• To alter the rate of a reaction
• By providing an alternative pathway for the reaction with a lower activation energy (more molecules have sufficient energy), therefore accelerates the attainment of equilibrium.

Question 2

What are the key features of the ‘active site’ of the enzyme?

Answer 2

• Only occupies a small part of the enzyme
• Formed by different AA’s of the primary sequence (don’t have to be adjacent)
• Tend to be in clefts/crevices (to exclude water)
• Substrates bind weakly so that products can be readily released.

Question 3

What is the rate of a reaction of enzyme activity dependent on?

Answer 3

The substrate concentration.

Question 4

What is the Michaelis-Menton equation?
What is this equation telling us?
What is 1 unit for the velocity (Vo) of a reaction

Answer 4

Vo = Vmax x (S)/Km + (S)

• Vmax = maximum velocity (mol/min)
• S = substrate concentration (mol)
• Km = substrate concentration at half Vmax

Vo = velocity of reaction in standardised rate, e.g.: Per L, Per g of tissue etc.

• Telling us that rate of reaction is proportional to substrate concentration.
• 1 unit = the amount of substrate required to produce 1umol of product per minute under standard conditions

Question 5

How do you turn a Michaelis-Menton plot into a straight-line form?

Answer 5

Take double reciprocals and create Lineweaver-Burk plot.

Question 6

What is the difference between competitive and non-competitive inhibitors?

Answer 6

Competitive = Binds to active site, competes with substrate to reduce mount of ES complexes

Non-Competitive = Bind elsewhere, reduces concentration of functional enzyme (stops substrate binding)

Question 7

How can we distinguish between competitive and competitive inhibitors? (describe in terms of Km + Vmax)

Answer 7

• Effect of competitive inhibitors can be overcome by adding more substrate to outcompete inhibitor, therefore Km increases, Vmax stays the same.
• Adding more substrate in a non-competitive inhibitor will not overcome effect, therefore Km is unaffected, Vmax decreases.

Question 8

Note - need to be able to describe how rates of reaction vary as a function of enzyme and substrate concentration on simple velocity vs (S) graphs - see group-work for examples.

Answer 8

Ez.

Question 9

What does the X-axis intercept on a lineweaver burk plot indicate?
What does a shift to the right of the X-axis intercept indicate?

Answer 9

X axis intercept = 1/-Km

shift to the right = Km increases

Question 10

What does the Y-axis intercept on a lineweaver burk plot indicate?
What does a shift upwards on the Y-axis intercept indicate?

Answer 10

Y-axis intercept = 1/Vmax

Shift upwards = A decrease in Vmax

Question 11

What changes on a lineweaver burk plot would you expect if a competitive and non-competitive inhibitor was added to the reaction?

Answer 11

Competitive = Shift to the right for X-axis intercept, Same Vmax (as Km increases + Vmax is unchanged)

Non-competitive = Shift upwards in Y-axis intercept, Saem Km (as Vmax decreases + Km is unchanged).

(Competitive will cross the original plot line, non-competitive will not).