Quantitative Enzymology calculation

Question 1

Review CHEM 201: An enzyme of molar mass 50.0 kg/mol catalyses reaction of a substrate to product according to Michaelis-Menton kinetics. Data below present initial reaction rates at the stated initial substrate concentrations. The enzyme concentration in each of the cases was 2.00 g/L.

Determine K_M, v_max, k_cat and η for this set of data below. (All 10^-4)

Plot the raw data directly as v s. [S]_o

Answer 1

If v_max = 0.0004 M s^-­1

then K_M is the value of [S]_o where v = v_max/2 = 0.0002, so K_M ~ 4 mM, and, because v_max= k_cat[E]_o , where [E]_o = (2g/L)/(50,000 g/mol) = 4x10^-­5 M,

k_cat = (4x10^-­‐4 M s^-­‐1)/( 4x10^-­‐5 M) = 10 s^-­‐1. So, η = k_cat/K_M = 10/0.004 = 2500 M^-­‐1 s^-­‐1.

Note: K_M. k_cat and η depend on the choice of asymptote, which has considerable uncertainty

Question 2

Review CHEM 201: An enzyme of molar mass 50.0 kg/mol catalyses reaction of a

substrate to product according to Michaelis-Menton kinetics. Data below present initial reaction rates at the stated initial substrate concentrations. The enzyme concentration in each of the cases was 2.00 g/L.

Determine K_M, v_max, k_cat and η for this set of data below. (All 10^-4)

b) Draw a Lineweaver-­‐Burk plot.

Answer 2

The vertical intercept = 2100 M^-­‐1 s = 1/v_max → v_max = 4.8 x 10-^­‐4 M s^-­‐1

slope = K_M/v_max = 11.24 s → K_M = (11.24 s)(4.8 x 10^-­‐4 M s^-­‐1)= 5.4 x 10^-­‐3 M

v_max= k_cat[E]_o , where [E]_o = (2g/L)/(50,000 g/mol) = 4x10^-­‐5 M

So, k_cat = (4.8 x 10- 4 M s^-­‐1)/( 4 x 10^-­‐5 M)

= 12 s^-1. And, η = k_cat/K_M = 12 s^{- 1}/(5.4 x 10^-­‐3 M)

= 2222 M^-­‐1 s^-­‐1

Question 3

Review CHEM 201: An enzyme of molar mass 50.0 kg/mol catalyses reaction of a

substrate to product according to Michaelis-Menton kinetics. Data below present initial reaction rates at the stated initial substrate concentrations. The enzyme concentration in each of the cases was 2.00 g/L.

Determine KM, vmax, kcat and η for this set of data below. (All 10-4)

c) Draw an Eadie-­‐Hofstee plot.

Answer 3

The values obtained for v_max, K_M. k_cat and η are the same as obtained from the

Lineweaver-­‐Burk treatment

Question 4

(a) At what substrate concentration would an enzyme with a k_cat of 30.0 s^-1 and a K_m of 0.0050 M operate at one-quarter of its maximum rate?
(b) Determine the fraction of V_max that would be obtained at the following substrate concentrations
[S]: 1/2K_m, 2K_m, and 10K_m

(c )An enzyme that catalyzes the reaction X⇌Y is isolated from two bacterial species. The enzymes
have the same V_max, but different K_m values for the substrate X. Enzyme A has a K_m of 2.0 μM,
while enzyme B has a K_m of 0.5 μM. The plot below shows the kinetics of reactions carried out
with the same concentration of each enzyme and with [X] 1 μM. Which curve corresponds to
which enzyme?

Answer 4

(a) Here we want to find the value of [S] when V₀ 0.25 V_max.

The Michaelis-Menten equation is V₀ = V_max[S]/(K_m + [S])
so V₀ = V_max when [S]/(Km + [S]) 0.25; or
[S] = 0.33K_m 0.33(0.0050 M) 1.7 10^-3 M
(b) The Michaelis-Menten equation can be rearranged to
V₀/V_max [S]/(Km + [S])
Substituting [S] = 1/2 K_m into the equation gives
V₀ /V_max = 0.5 K_m/1.5K_m = 0.33
Similarly, substituting [S] = 2K_m gives
V₀ /V_max 0.67
And substituting [S] = 10K_m gives
V₀ /V_max 0.91
(c) The upper curve corresponds to enzyme B ([X] is greater than the K_m for this enzyme),
and the lower curve corresponds to enzyme A. When the initial concentration of substrate
is greater than K_m, the rate of the reaction is less sensitive to the depletion of
substrate at early stages of the reaction and the rate remains approximately linear for a
longer time.

Question 5

A research group discovers a new version of happyase,
which they call happyase*, that catalyzes the chemical reaction
HAPPY⇌SAD
The researchers begin to characterize the enzyme.
(a) In the first experiment, with [E_t] at 4 nM, they find that the V_max is 1.6 μM s^-1. Based on this experiment,
what is the k_cat for happyase*? (Include appropriate units.)

(b) In another experiment, with [E_t] at 1 nM and [HAPPY] at 30 μM, the researchers find that V₀
300 nM s^-1. What is the measured K_m of happyase* for its substrate HAPPY? (Include appropriate
units.)
(c) Further research shows that the purified happyase* used in the first two experiments was actually
contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation, and the two experiments repeated, the measured V_max in (a) is increased
to 4.8 μM s^-1, and the measured Km in (b) is now 15 μM. For the inhibitor ANGER, calculate
the values of α and α’.
(d) Based on the information given above, what type of inhibitor is ANGER?

Answer 5

(a) Use the equation k_cat = V_max/[E_t] · k_cat = 1600 nM s^-1/4 nM 400 s^-1.
(b) Use the equation V_max = k_cat[E_t]. When [E_t] = 1 nM, V_max = 400 nM s^-1.
V₀/V_max = 300 nM s^-1/400 nM s^-1 = 3/4
Rearrange the Michaelis-Menten equation, substitute for V₀ /V_max, and solve for K_m.
V₀ /V_max [S]/(Km + [S])
3/4 = [S]/(Km + [S])
K_m [S]/3
In this experiment, the concentration of the substrate, HAPPY, was 30 μM,

so K_m = 10 μM.
(c) As shown in Table 69, V_max varies as a function of V_max/α’. Because V_max increased
by a factor of 3, 3. Similarly, K_m varies as a function of αK_m/α’. Given that K_m
increased by a factor of 1.5 when ANGER was removed (that is, the inhibitor decreased the observed K_m by 2/3) and α’ = 3, then α’ = 2.
(d) Because both α and α’ are affected, ANGER is a mixed inhibitor.

Question 6

Another enzyme is found that catalyzes the reaction
A ⇌ B
Researchers find that the K_m for the substrate A is 4 μM, and the k_cat is 20 min^-1.
(a) In an experiment, [A] = 6 mM, and the initial velocity, V₀ was 480 nM min^-1. What was the [E_t] used
in the experiment?
(b) In another experiment, [E_t] 0.5 M, and the measured V₀ = 5 M min^-1. What was the [A] used
in the experiment?
(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In an experiment with the same [E_t] as in part (a), but a different [A], an amount of Z is
added that reduces the rate V₀ to 240 nM min^-1. What is the [A] in this experiment?
(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic
perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic
perfection.

Answer 6

(a) Because [S] is much greater than (more than 1000-fold) Km, assume that the measured
rate of the reaction reflects V_max. Use the equation V_max = k_cat[E_t], and solve for [E_t].
[E_t] = V_max/k_cat = 480 nM min^-1/20 min^-1 = 24 nM.
(b) At this [E_t], the calculated V_max = k_cat[E_t] = 20 min^-1 *0.5 μ M = 10 μ M min^-1. Recall
that K_m equals the substrate concentration at which

V₀ = 1/2V_max. The measured V₀ is
exactly half V_max, so [A] K_m = 4 μ M.

(c) Given the same [Et] as in (a), V_max = 480 nM min^-1. The V₀ is again exactly half V_max
(V₀ = 240 nM min^-1), so [A] = the apparent or measured K_m. In the presence of an
inhibitor with α = 10, the measured K_m = 40 μM [S].
(d) No. k_cat/K_m = 0.33/(4 * 10^-6 M^-1 s^-1) = 8.25 * 10⁴ M^-1 s^-1, well below the diffusion controlled limit.

Question 7

Review: what is first-order reaction kinetics? Give two types of equation.

Answer 7

A simple unimoelcular reaction (only one reactant):

A→B

One reactant = first order reaction

velocity (v):

v= -d[A]/dt or v=d[B]/dt

first order rate equation: v = k[A], where k is the first-order rate constant

Recall that since there is only one reactant, this is a first order reaction whose rate, or velocity, can be described as the disappearance of the reactant over time or the appearance of the product over time. The reaction rate is represented by the slopes of these lines and is not constant.

Question 8

Explain the graph.

Answer 8

For this first-order reaction, the rate is directly proportional to the concentration of reactant. Here k is the first-order rate constant for the reaction. Although the rate constant does not change, the velocity of the reaction begins to decrease as the concentration of the reactant diminishes and the product accumulates. The rate of the reaction eventually slows to zero as the product of the initial forward reaction becomes a reactant for the reverse reaction. Equilibrium is the point where the forward and reverse reaction rates are equal, and the overall rate of the reaction is zero.

Question 9

How to describe this reaction in terms of kinetics? Can it apply to biological systems? Why or why not?

Answer 9

first-order reaction, because there is only one reactant.

The rate equation for an enzyme-catalyzed reaction with a single substrate is the same as the rate equation for the simple nonenzymatic reaction, but only when the concentration of substrate is so low that the enzyme has very little chance to convert it to product. Under these unique conditions, it appears that the reaction is first order with respect to substrate. However, this special case is much too limiting and rarely applies in biological and experimental systems. Consequently, biochemists must use a different model to describe the kinetics of biological reactions catalyzed by enzymes.

Question 10

What happens if there is a much greateer excess of subtrate than enzyme?

Answer 10

The kinetics of an enzyme-catalyzed reaction can be studied when the concentration of the enzyme is small compared to the concentration of the substrate. As long as the substrate is in large excess over enzyme, altering its concentration does not change the rate. This is quite different from the first-order reaction in the previous example, where the rate depended on the substrate concentration. Here, the rate does not depend on the substrate concentration, and the reaction is said to be zero order with respect to substrate.

Question 11

What is V_max?

Answer 11

Because the enzyme concentration is rate limiting, the rate is independent of substrate concentration. The zero-order rate constant is referred to as the maximal velocity, or V_max.

Question 12

Explain about k₁, k_-1, and k₂ in the image below

Answer 12

To account for this kinetic behavior caused by the substrate’s interaction with enzyme, the model for an enzyme-catalyzed reaction must include an enzyme-substrate complex, or ES complex. The reaction scheme is shown here. The rate constants k1 and k–1 are the forward and reverse rate constants for the formation of the ES complex, and k2 is the rate at which the ES complex converts the bound substrate into product. Notice that in order to simplify this model, we assume that the product formation step is irreversible (that is, there is no k–2).

Question 13

What is the Michaelis-Menten equation?

Answer 13

The Michaelis–Menten equation is limited to the initial rate v₀ so that we can ignore the possibility of the product accumulating and undergoing the reverse reaction.

Question 14

Explain how does the graph works.

Answer 14

This graph shows the reaction progress of an enzyme-catalyzed reaction. The Michaelis–Menten model makes a steady-state assumption, meaning that the concentration of the ES complex remains constant. This assumption is valid only when the concentration of substrate is much greater than the total enzyme concentration.

Question 15

What is the line-weaver Burke plot? What is the equation of the curve?

Answer 15

a linear equation with the form y = mx + b. The resulting plot is called a Lineweaver-Burk or double-reciprocal plot.

K_M and V_max can be readily obtained from the x and y intercepts of a Lineweaver-Burk plot. The y-intercept is 1/Vmax and the x-intercept is –1/K_M. Try varying the V_max and K_M on the following graphs.

Question 16

What is V_max and K_m?

Answer 16

V_max=maximum velocity seen at enzyme-saturating [S]

K_m = [S] at 1/2 V_max

K_M is a “lumped” rate constant incorporating all the rate constants for ES and P formation: k¹, k^–1, and k².

Km is also (rate of ES breakdown to S or P)/(rate of ES formation)

Km is also inveresely proportional to the fraction of E_t that is ES

Question 17

what is the inhibitor mechnisium used for?

Answer 17

decrease the enzyme’s abiliyt to bind substrate.

lower the enzyme’s catalytic actvity

Question 18

What is reversible inhibitor?

Answer 18

Reversible enzyme inhibitors bind and dissociate with their enzyme in a equilibrium process.

Question 19

At 20 ^oC, the following data were determined for enzyme-catalyzed hydrolysis of adenosine triphosphate (ATP). The initial concentration of enzyme was

[E]_o = 0.020 μM = 0.020×10^-6 M.

Draw on the grid found below an appropriate straight-line graph, consistent with the reaction
obeying Michaelis-Menton kinetics: v=v_max[S]/([S]+K_m)

ATP v (μM)

0. 60 0.81
1. 80 0.97
2. 4 1.30
3. 0 1.47
4. 0 1.69

Answer 19

Question 20

An enzyme has K_M = 3.90×10^-6M. In a solution with an initial substrate concentration of
0.0350 M, after one minute of reaction, the product concentration rose from zero to 6.20×10^-6M.
Assume Michaelis-Menton behavior, and that it takes > 1 hour for the system to reach equilibrium:

a) Determine v_max for the reaction.
b) Calculate the molar concentration of product after 3.00 min of reaction

c) At an initial concentration of substrate of 1.00×10^-5 M, estimate the initial rate of
reaction.

Answer 20

Question 21

The enzyme catalyzed conversion of a substrate at 25 ⁰C has a Michaelis constant of

0. 035 M. The rate of the reaction is 1.15x10^-3 M s^-1 when the substrate concentration is
1. 110 M. What is the maximum velocity of this enzymolysis?

Answer 21

Question 22

One transformation of the Michaelis-Menten equation is the
Lineweaver-Burk, or double-reciprocal, equation. Multiplying both sides of the Lineweaver-Burk equation
by V_max and rearranging gives the Eadie-Hofstee equation:

A plot of V₀ vs. V₀/[S] for an enzyme-catalyzed reaction is shown below. The curve labeled “Slope –K_m”
was obtained in the absence of inhibitor. Which of the other curves (A, B, or C) shows the enzyme activity
when a competitive inhibitor is added to the reaction mixture?

Answer 22

Curve A shows competitive inhibition. V_max for A is the same as for the normal
curve, as seen by the identical intercepts on the V₀ axis. And, for every value of [S] (until
maximal velocity is reached at saturating substrate levels), V₀ is lower for curve A than for the
normal curve, indicating competitive inhibition. Note that as [S] increases, V₀/[S] decreases, so
that V_max—that is, the V₀ at the highest (saturating) [S]—is found at the intersection of the
curve at the y axis. Curve C, while also having an identical V_max, shows higher V₀ values for
every [S] (and for every V₀/[S]) than the normal curve, which is not indicative of inhibition.
The lower Vmax for curve B rules out competitive inhibition.