Bke2 Biochemistry Exercises

Suggested answers to Group exercise: Enzyme kinetics

What is meant by the term "enzyme kinetics"? What kind of information do kinetic studies give?

Answer:
Enzyme kinetics is the study of enzyme catalyzed reactions and gives information concerning the specificities and catalytic mechanisms of enzymes. We can use this kind of information for medically important purposes, like for identifying the specific problem in a metabolic disease, or finding a drug that is an inhibitor of a particular enzyme (and so kills a pathogen).
Write down the rate equation for a simple non-enzymatic reaction in solution (with a substrate S being converted to a product, P). What does the rate constant in this reaction indicate? Now, write down the rate equation for a simple enzyme-catalyzed conversion of S to P. What is the rate equation for this reaction? Describe in a graph, and with your own words, what happens to the reaction velocity when you add substrate to an enzyme at increasing concentrations. How is this different from what happens when enzyme is not there?

Answer:
The rate equation for a non-enzyme-catalyzed reaction (S --> P) is:

(change in [P])/(change in t) = - (change in [S]/(change in t) = v = k[S],

where v tells you how fast product appears, and k is the rate constant. The rate constant indicates how fast S is converted to P, for each molecule of S. So, if you add more S, the reaction gets faster.

The rate equation for an enzyme-catalyzed reaction (E + S <=> ES --> E + P) is:

v = k_cat * [ES]

As you add substrate, at first the velocity v increases linearly with increasing substrate concentration, just like in the non-enzymatic case. But, at a certain substrate concentration the increase in v starts to level off, and at a somewhat higher [S] the velocity v is constant. The enzyme is saturated. Because you depend on the enzyme, you have to take it into account.
The reaction velocity of a simple enzyme-catalyzed reaction, such as that in the previous question, is described by the Michaelis-Menten equation. Write down this equation. Under what conditions is this equation valid (i.e. what assumptions are made in deriving the equation)? By inspecting the equation, write down what the reaction velocity tends to as the substrate concentration becomes much larger than the K_m for that substrate. What is the velocity of the reaction when the substrate concentration is equal to the K_m?
What is the difference between a "big" K, like in K_m, and a "small" k, like in k_cat?

Answer:
The equation is derived under the assumptions that:
- [E]_tot << [S] but not necessarily saturated, so [S] will be more or less constant.
- The amount of product is negligible (these conditions give what we call an "initial velocity").
- The steady state condition is fulfilled: ES is formed as rapidly as it decomposes, i.e.:
vformation = k₁ [E] [S]
vbreakdown = k_-1 [ES] + k_cat [ES] = (k_-1 + k_cat) [ES]
vformation = vbreakdown ==> k₁ [E] [S] = (k_-1 + k_cat) [ES]

By collecting rate constants on one side, and concentrations on the other, and remembering that [E] = [E]_tot - [ES], the MM-equation can be derived by inserting the expression for [ES] into the rate equation v₀ = k_cat [ES]. For details, please refer to Horton.

Inspection of the MM-equation shows that
when [S] >> K_m, v₀ = V_max
when [S] = K_m, v₀ = V_max/2

"Small" k's are reserved for kinetic rate constants. The "big" K in K_m is similar to a dissociation constant (K_S), and, as a dissociation constant, is a ratio of kinetic rate constants (K_m = (k_-1 + k_cat)/k₁; K_S = k_-1/k₁).
K_m is frequently equated with K_S, the [ES] dissociation constant. However, there is usually a difference between those values. Why? Under what conditions are K_m and K_S equivalent?

Answer:
E + S <=> ES --> E + P

K_m = (k_-1 + k_cat)/k₁
K_S = k_-1/k₁ and thus
K_m = K_S when k_cat << k_-1
Assume that an enzyme-catalysed reaction follows Michaelis-Menten kinetics with a K_m of 1 uM. The initial velocity is 0.1 uM/min at 10 mM substrate. Calculate the initial velocity at 1 mM, 10 uM and 1 uM substrate. If the substrate concentration were increased to 20 mM would the initial velocity double? Why or why not?

Answer:
When
[S] is 10^-3 M, v₀ = V_max = 10^-7 M /min

[S] is 10^-5 M, v₀ = 9.1 x 10^-8 M /min (91% V_max)

[S] is 10^-6 M, v₀ = 5 x 10^-8 M /min (50% V_max)

[S] is 2 x 10^-2 M, v₀ = V_max, thus no increase in velocity.
Why are k_cat/K_m values useful to describe the specificity, or preference for different substrates, of an enzyme?

Answer:
The k_cat/K_m for any given substrate tells us about the behavior of the reaction when the concentration of that substrate is low compared to K_m (the usual situation in the cell). Suppose that two different substrates, A and B, are competing for binding and catalysis by the same enzyme. When the concentration of A and B are equal, the ratio of their conversion to product by the enzyme is equal to the ratio of their k_cat/K_m values. We can figure out why this is so: If the substrate concentration is low, the overall reaction is limited by the encounter of substrate S with the enzyme E, with the rate equation:

v₀ = (k_cat/K_m) [E]_tot [S]

In the case above [S] = [A] = [B], and so

v_0A/v_0B = (k_cat/K_m)_A / (k_cat/K_m)_B

k_cat/K_m is therefore often called the the specificity constant.

To see that the rate equation is v₀ = (k_cat/K_m) [E]_tot [S] for low substrate concentrations, start from the MM equation:

v₀ = V_max [S] / (K_m + [S] )

For [S] << K_m this reduces to v₀ = (V_max / K_m) [S]

Remember that V_max = k_cat [E]_tot, so

v₀ = (k_cat / K_m) [E]_tot [S] for low [S]

Note that the form of the rate equation at low [S] given in the book (eq. 5.28, page 128 in 2nd Ed., page 140 in 3rd Ed.) assumes that [E]_tot = [E], i.e. that a negligible amount of [ES] has been formed.

(The form of the equation as given in the book, with [E] rather than [E]_tot is actually valid at all substrate concentrations:

Start from the steady state assumption:

k₁ [E] [S] = (k_-1 + k_cat) [ES] (1)

Remember that the overall velocity of the reaction is given by

v₀ = k_cat [ES] (2)

From (1) we get the concentration of ES: [ES] = k₁ [E] [S] / (k_-1 + k_cat) = [E][S] / K_m

Insertion of [ES] into (2) then gives v₀ = (k_cat/K_m) [E] [S] (3)

This form of the equation is of less practical value, however, since [E] will vary with [S] in a way that also depends on the K_m for S:

[E] = [E]_tot - [ES]

[ES] = [E]_tot [S] / (K_m + [S])

[E] = [E]_tot - [E]_tot [S] / (K_m + [S]) =

[E]_tot ( 1 - [S] / (K_m + [S]) ) =

[E]_tot ( K_m + [S] - [S] ) / (K_m + [S]) ) =

[E]_tot K_m / (K_m + [S])

Inserting this expression for [E] into equation (3) above then gives back the normal form of the MM equation.)
You think you may have found a new inhibitor (I) of the HIV protease, so you perform a few experiments to characterize its behavior. You find that if you add an excess of synthetic substrate to the protease in the presence of the inhibitor, enzyme actitivity increases but only reaches 50% of the V_max observed in the absence of the inhibitor. The affinity of the substrate for the enzyme is unaffected by the inhibitor. Furthermore, if you treat the enzyme with the inhibitor, then dialyze the protein, you find that full activity can be restored. What class of inhibitor is I?

Answer:
Noncompetitive. A decreased V_max indicates that inhibition could be noncompetitive or uncompetitive, but the lack of effect on K_m indicates that the inhibition is noncompetitive. Recovery of activity after dialysis is typical of reversible inhibitors.
One of the effects of aspirin is to slow blood clotting by inhibiting the enzyme cyclo-oxygenase in blood platelets. The effects of aspirin wear off on the same time-scale as the platelets themselves are broken down (and new ones made). What kind of inhibitor is aspirin?

Answer:
Aspirin is an irreversible inhibitor of the cyclo-oxygenase. If it were a reversible inhibitor it would be released from binding to the enzyme when the levels of free aspirin in the blood decreases as it is cleared from the body. This rate is much faster than the life-time of blood platelets. Blood platelets do not have nuclei and therefore do not synthesize new protein. The activity is restored only when new platelets are formed with fresh and active cyclo-oxygenase.
Hexokinase catalyses the following reaction:

Which compound below would you choose as the best competitive inhibitor of the hexokinase reaction, and why?

Ribose Galactose

Fructose 6-deoxyglucose

Answer:
The correct answer is 6-deoxyglucose. Lacking the 6-OH group, yet retaining the remaining structural features of glucose, this molecule binds well to hexokinase but cannot be phosphorylated.
It is often convenient in experimental work to consider the Michaelis-Menten equation in terms of 1/v. Rewrite the equation in this form. Does it have the form of a straight line? If so, where does the line cross the x and y axes and what is its slope?

Answer:
1/v₀ = 1/V_max + K_m/V_max (1/[S])
This is a straight line with slope K_m/V_max and with the intercepts -1/K_m on the x-axis and 1/V_max on the y-axis.

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]


Ribose	Galactose

Fructose	6-deoxyglucose