Practical D2

Question 1

If an enzyme is to be used to determine the concentration of substrate in a sample ….

Answer 1

(e.g. glucose oxidase is used to measure plasma glucose), then the substrate must be the limiting factor, and the concentration of substrate must be below Km, so that the rate of formation of product increases steeply with increasing concentration of substrate, so providing a sensitive assay for the substrate.”

Question 2

Km and Vmax are determined by ….

Answer 2

incubating the enzyme with varying concentrations of substrate; the results can be plotted as a graph of rate of reaction (v) against concentration of substrate ([S], and will normally yield a hyperbolic curve.

Question 3

The relationship between Km and Vmax is defined by ….

Answer 3

the Michaelis-Menten equation:

v = Vmax / (1 + (Km/[S]))

Question 4

The Lineweaver-Burk double reciprocal plot rearranges the Michaelis-Menten equation as:

1 / v = 1 / Vmax + Km / Vmax x 1 / [S] (eq from the picture)

plotting 1/v against 1/[S] give a straight line:

y intercept = 1 / Vmax

gradient = Km / Vmax

x intercept = -1/ Km

Answer 4

This is the most widely used method of linearising the data, and generally gives the best precision for estimates of Km and Vmax.

However, it has the disadvantage of placing undue weight on the points obtained at low concentrations of substrate (the highest values of 1/[S] and 1/v). These are the points at which the precision of determining the rate of reaction is lowest, because the smallest amount of product has been formed.

Question 5

The Eadie-Hofstee plot rearranges the Michaelis-Menten equation as:

v = Vmax - Km x v / [S] (eq in the picture)

plotting v against v / [S] gives a straight line:

y intercept = Vmax

gradient = -Km

x intercept = Vmax / Km

Answer 5

asdas This plot overcomes the problem of uneven spacing of points, and undue weight given to points at low concentrations of substrate. However, it has the disadvantage that v, which is a dependent variable, is used on both axes, and hence errors in measuring the rate of reaction are multiplied, resulting in lower precision of the estimates of Km and Vmax

Question 6

The Hanes plot rearranges the Michaelis-Menten equation as:

[S] / v = Km / Vmax + [S] / Vmax (eq in the picture)

plotting [S] / v against [S] gives a straight line:

y intercept = Km / Vmax

gradient = 1 / Vmax

x intercept = -Km

Answer 6

This plot overcomes the problem of uneven spacing of points, and undue weight given to points at low concentrations of substrate.

However, it has the disadvantage that [S] is used on both axes, and hence pipetting errors, which lead to errors in the true concentration of substrate available, are multiplied, resulting in lower precision of the estimates of Km and Vmax.

Question 7

An irreversible inhibitor causes ….

Answer 7

covalent modification of the enzyme, so that its activity is permanently reduced. Compounds that act as irreversible inhibitors are often useful as drugs that need be taken only every few days,

Question 8

The normal sequence of an enzyme reaction can be represented as:

Answer 8

E + S <=> E-S <=> E-P <=> E + P

where:

E = enzyme

S = substrate

E-S = enzyme-substrate complex

E-P = enzyme-product complex

P = product

Question 9

There are three main types of reversible inhibitor:

Answer 9

• competitive inhibitor
• non-competitive inhibitor
• uncompetitive inhibitor

Question 10

The normal sequence of an enzyme reaction in competitive inhibition can be represented as:

Answer 10

where:

E = enzyme

S = substrate

E-S = enzyme-substrate complex

E-P = enzyme-product complex

P = product

Question 11

1. Competitive inhibition:

Answer 11

This means that increasing the concentration of substrate will decrease the chance of inhibitor binding to the enzyme. Hence, if the substrate concentration is high enough the enzyme will reach the same Vmax as without the inhibitor. However, it will require a higher concentration of substrate to achieve this and so the Km of the enzyme will also be higher. Reacting the enzyme with a range of concentrations of substrate at different concentrations of a competitive inhibitor will give a family of curves as shown below:

Question 12

Competitive inhibition:

The Lineweaver-Burk double reciprocal plot for this set of data shows :

Answer 12

a series of lines crossing the y (1/v) axis at the same point

• i.e. Vmax is unchanged, but with a decreasing value of 1/Km (and hence a higher Km) in the presence of the inhibitor:

Question 13

2. Non-competitive inhibition

Answer 13

This means that increasing the concentration of substrate will not relieve the inhibition, since the inhibitor reacts with the enzyme-substrate complex. Reacting the enzyme with a range of concentrations of substrate at different concentrations of a non-competitive inhibitor will give a family of curves as shown below:

Question 14

Non-competitive inhibition:

The Lineweaver-Burk double reciprocal plot for this set of data shows…

Answer 14

a series of lines converging on the same point on the X (1/S) axis- i,.e. Km is unchanged, but Vmax is reduced:

Question 15

3. Uncompetitive inhibition

Answer 15

This is a very rare class of inhibition. An uncompetitive inhibitor binds to the enzyme and enhances the binding of substrate (so reducing Km), but the resultant enzyme-inhibitor-substrate complex only undergoes reaction to form the product slowly, so that Vmax is also reduced. Reacting the enzyme with a range of concentrations of substrate at different concentrations of an uncompetitive inhibitor will give a family of curves as shown below:

Question 16

Uncompetitive inhibition:

The Lineweaver-Burk double reciprocal plot for this set of data shows….

Answer 16

a series of parallel lines - both Km and Vmax are reduced:

Question 17

Which inhibitor should be used?

If the requirement is to increase the intracellular concentration of the substrate, then …

Answer 17

either a competitive or non-competitive inhibitor will serve, since both will inhibit the utilisation of substrate, so that it accumulates.

Question 18

if the requirement is to decrease the intracellular concentration of the product, then the inhibitor must be …

Answer 18

non-competitive. As unused substrate accumulates, so it will compete with a competitive inhibitor, and the final result will be a more or less normal rate of formation of product, but with a larger pool of substrate. Increasing the concentration of substrate does not affect a non-competitive inhibitor.

Question 19

The inhibitor constant, Ki, -

Answer 19

an indication of how potent an inhibitor is; it is the concentration required to produce half maximum inhibition.Plotting 1/v against concentration of inhibitor at each concentration of substrate (the Dixon plot) gives a family of intersecting lines.

Question 20

Ki: For a competitive inhibitor, the lines converge ….

Answer 20

above the x axis, and the value of [I] where they intersect is -Ki

Question 21

Ki: For a non-competitive inhibitor, the lines converge ….

Answer 21

on x axis, and the value of [I] where they intersect is -Ki