Chapter 7 Reading Quiz and Homework

Which statement is true of studying the rates of enzyme‑catalyzed reactions?

It is called enzyme kinetics. It can involve determining how fast the substrate disappears as it is converted to product. It can involve following the appearance of product formed over time.

The protein catalase catalyzes the reaction 2⁢H2⁢O2⁡(aq)⟶2⁢H2O(l)+O2⁡(g) and has a Michaelis-Menten constant of KM=25 mM and a turnover number of 4.0×107⁢ s−1 . The total enzyme concentration is 0.013 μM and the initial substrate concentration is 6.32 μM . Catalase has a single active site. Calculate the value of Rmax (often written as Vmax ) for this enzyme. Calculate the initial rate, R (often written as V0 ), of this reaction.

Rmax = 𝑘2 × [E]t = (4.0×10^7 s−1) × (0.013μM) = 5.2×10^5 μM/s = 520mM/s R = 𝑅max[S]/𝐾M+[S] = (5.2×10^5 μM/s × 1 mM / 1000 μM) (6.32 μM × 1 mM / 1000 μM)/((25 mM)+(6.32 μM)(1 mM /1000μM)) = 0.13mM/s

What effect will a heterotropic activator have on a sigmoidal kinetic plot of V versus [S]?

The curve will shift to the left.

Feedback inhibition is important for the regulation of many metabolic pathways. Which of the statements are examples of feedback inhibition?

The production of tryptophan is halted by the presence of excess tryptophan. The beginning of glycolysis is inhibited by high levels of ATP in the cell.

What are characteristics of allosteric enzymes?

They may have binding sites for regulatory molecules that are separate from active sites. They undergo conformational changes as a result of modulator binding. They generally have more than one subunit.

Select the graph that correctly illustrates the effect of a negative modifier (effector) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed green curve represents the enzyme in the presence of the effector.

Reaction velocity is on the Y-axis. Substrate concentration is on the X-axis. Blue line is an s curve (sigmoid). Green dotted line is below the blue line.

An allosteric enzyme that follows the concerted mechanism (MWC model) has a T/RT/R ratio of 300 in the absence of substrate. Suppose that a mutation reversed the ratio. Note that R is the highly active form of the enzyme and T is the less active form of the enzyme. Select all of the effects this mutation will have on the relationship between the rate of the reaction (V) and substrate concentration ([S]) .

The enzyme would likely follow Michaelis-Menten kinetics. The plot of V versus [S][S] would be shaped like a hyperbola. The enzyme would be more active.

An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 2.35 mM⋅s−1. Calculate the reaction velocity, v0, for each substrate concentration.

V0 = ((Vmax * [S])/(Km + [S])) [S] = 4.00 mM V0 = 0.723 mM*s-1 [S] = 9.00 mM V0 = 1.18 mM*s-1 [S] = 13.5 mM V0 = 1.41 mM*s-1

On what basis are enzymes and proteins with allosteric properties different from those without allosteric properties? Allosteric enzymes have

different dependence on the substrate concentration.

Determine how feedback inhibition of enzyme 1 by a high concentration of product Z affects the concentrations of the different substrates for the reaction pathway shown. Assume that product Z diffuses out of the cell. W−→−−−−enzyme 1 X−→−−−−enzyme 2 Y−→−−−−enzyme 3 Z What happens if the concentration of Z decreases?

The concentration of X decreases . The concentration of Y decreases . The concentration of Z decreases . Enzyme 1 activity increases.

To obtain the turnover number of an enzyme (k2), one must

divide Vmax by the total enzyme concentration.

The effects of molecules other than substrate on allosteric enzymes are called

heterotropic effects.

The maximum velocity (Vmax) of an enzyme‑catalyzed reaction is

the rate observed when all enzyme active sites are saturated with substrate.

When the rate constant for dissociation of the enzyme-substrate complex (𝑘−1) is greater than the rate constant for conversion to product (k2), the KM is most analogous to

the 𝐾d.

Determine the value of the turnover number of the enzyme catalase, given that Rmax (Vmax) for catalase is 41 mmol⋅L−1⋅s−1 and [E]t equals 3.2 nmol⋅L−1. Catalase has a single active site.

turnover number: K2 = Rmax/[E]t = (41 *(10^6/1 mmol)/3.2 nmol/L) = 1.3 x 10^7 s-1

How do you find the KM from a double‑reciprocal, or Lineweaver-Burk, plot?

𝑥-intercept=-1/𝐾M

The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number=𝑅max[E]t=𝑘2 ⁡(Rmax is often written as Vmax) If the enzyme molecule has more than one active site, then [E]t is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that Rmax for carbonic anhydrase equals 249 μmol⋅L−1⋅s−1 and [E]t=2.48 nmol⋅L−1. Carbonic anhydrase has a single active site.

K2 = Rmax/[E]t = (249(1000 nmol/1 μmol)/2.48 nmol/L) = 1 x 10^5 s-1

One way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. Ribonuclease Substrate - RNA Turnover number (per s) - 100 Fumarase Substrate - fumarate Turnover number (per s) - 800 Lactate dehydrogenase Substrate - lactate Turnover number (per s) - 1,000 Urease Substrate - urea Turnover number (per s) - 10,000 How many molecules of RNA can one molecule of ribonuclease act on in 28.6 min ?

RNA molecules: 28.6 min x 60s/1 min = 1.72 x 10^3 s Turnover number: 1.72 x 10^3(1 x 10^2) = 1.72 x 10^5

Determine which of the graphs represents the relationship of reaction rate (velocity) and substrate concentration when the enzyme concentration of the non‑allosteric enzyme is constant. Determine how reaction rate (velocity) varies with substrate concentration.

Rate Increases additional substrate is added when substrate concentration is low. additional substrate is added when substrate concentration is high but is not yet saturating. Rate is unchanged substrate is added when enzyme is saturated with substrate.

Consider the reaction. 2X ⟶ Y + Z When the concentration of X is doubled, the reaction rate increases by a factor of 4. What is the order of the reaction? What is the rate equation? Calculate K when [X] = 0.23 M and the rate of the reaction is 0.0030 M/s.

Second-order rate = k[X]^2 v = k[X]^2 k = v/[X]^2 k = ((0.0030 M/s)/(0.23 M)^2) k = 0.057 M-1*s-1

The data in the table was collected for a certain enzyme‑catalyzed reaction. Use these data to determine the maximum rate of the enzyme (Rmax), often called the maximum velocity (Vmax ), and the Michaelis-Menten constant of the enzyme (KM). If you were to plot this data to graphically determine Rmax and KM using a Lineweaver-Burk plot, what would you plot? If you would use the original data in the table, enter the original value for your answer. Which graph would you use to graphically determine Rmax and KM? Determine the maximum rate of the enzyme (Rmax), often called the maximum velocity (Vmax ), and the Michaelis-Menten constant of the enzyme (KM).

Slope = Km/Rmax Y-intercept = 1/Rmax [S] = 5.0 μmol*L-1: Graph as 0.2 R = 22 μmol*L-1*min-1: Graph as 0.045 [S] = 20.0 μmol*L-1: Graph as 0.05 R = 65 μmol*L-1*min-1: Graph as 0.015 The graph is the one with a positive slope. Slope = 0.196. Y-intercept = 0.00609. Rmax = 1/y-intercept = 164 KM = slope*Rmax = 32

Suppose that your Ph.D. mentor asks you to compare the kinetic behavior of the two recently discovered peptidase isozymes PepA and PepB. Your mentor's initial studies identified a pentapeptide substrate called Tide5 that appeared to be able to distinguish the kinetic behaviors of PepA and PepB. All they had time to tell you was this: At low Tide5 concentration, the 𝑉o (initial velocity) of PepA was greater than PepB, but at high Tide5 concentration, the reverse was true. Before you start your kinetic assays, you decide it would be wise to ponder this small bit of information. You have no numbers to work with, but does it tell you something about the relative values of Km and Vmax for PepA and PepB? Select the statements that are consistent with the information that you were given.

The Km of PepB is higher. The Vmax of PepB is greater.

The amino acid asparagine can promote cancer cell proliferation. Treating patients with the enzyme asparaginase is sometimes used as a chemotherapy treatment. Asparaginase hydrolyzes asparagine to aspartate and ammonia. Considering the provided Michaelis-Menten curves for two different asparaginase enzymes, complete the passage. The arrow indicates the concentration of asparagine in the human body. Graph: V0 on the Y axis, [S] on the X axis. Asparaginase 1 increases in a straight line, steadly and then platuos. Asparaginase 2 increases in a curve before it platuos. Asparaginase 2 plautos earlier than Asparaginase 1. Asparaginase 1 gets to a higher V0 value faster than Asparaginase 2.

The Vmax of asparaginase 1 is faster than the Vmax of asparaginase 2. At the substrate concentration indicated by the arrow, asparaginase 1 reaction velocity is slower than asparaginase 2 reaction velocity. The KM of asparaginase 1 is greater than the KM of asparaginase 2. Considering the performance of the enzymes, asparaginase 2 would make a more effective chemotherapeutic agent.

You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. Wild-type Vmax = 100 μmol min-1 Km = 10 mM Mutant Vmax = 1 μmol min-1 Km = 0.1 mM Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which k−1 is much larger than k2, which of the following statements are correct? Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM.

The mutant version has a higher affinity for the substrate. The wild‑type version requires a greater concentration of substrate to achieve Vmax. V0 = 50 μmol min-1 The reaction equilibrium is reached once there is no net change in the concentration of the substrate or the product. Based on the data table and your initial velocity calculation, the reaction equilibrium is altered more in the direction of the product by neither version.

The hydrolysis of pyrophosphate to orthophosphate drives biosynthetic reactions such as DNA synthesis. In Escherichia coli, a pyrophosphatase catalyzes this hydrolytic reaction. The pyrophosphatase has a mass of 120 kDa and consists of six identical subunits. A unit of activity for this enzyme, U, is the amount of enzyme that hydrolyzes 10 μmol of pyrophosphate in 15 minutes. The purified enzyme has a Vmax of 2800 U per milligram of enzyme. When [S]>>𝐾M, how many micromoles of substrate can 1 mg of enzyme hydrolyze per second? If each enzyme subunit has one active site, how many micromoles of active sites, or [E]T, are there in 1 mg of enzyme? What is the turnover number, or kcat, of the enzyme?

U = 10 μmol/15 min U = 10 μmol/900s Vmax = 2800U/1 mg Vmax = 2800(10 μmol/900 s)/1 mg Vmax = (2800*10 μmol/1 mg*900 s) Vmax = 31.1 μmol⋅s−1⋅mg−1 [E]T = 6 μmol/ 120 mg [E]T = 5 x 10^-2 μmol⋅mg−1 Kcat = Vmax/[E]T Kcat = 31.1 μmol⋅s−1⋅mg−1/ 0.05 μmol⋅mg−1 Kcat = 31.1 s-1/0.05 Kcat = 622 s−1

A plot of V0 on the y-axis versus V0/[S] on the x-x-axis is known as an Eadie-Hofstee plot. Like the Lineweaver-Burk plot, the Eadie-Hofstee plot is derived from a transformation of the Michaelis-Menten equation. The Michaelis-Menten equation expresses the relationship between the following terms: V0, the rate of reaction; Vmax, the maximal rate; [S], the substrate concentration; and KM, the Michaelis constant. Rearrange the Michaelis-Menten equation to obtain the Eadie-Hofstee equation, giving V0 as a function of V0/[S]. In an Eadie-Hofstee plot, what is the significance of the 𝑥-intercept? The Eadie-Hofstee plot illustrated shows enzyme activity in the absence of an inhibitor. Adjust one or both points so that the line shows enzyme activity in the presence of a competitive inhibitor.

V0 = -Km(V0/[S]) + Vmax Vmax/Km Change the graph to V0 = 18; V0/[S] = 11

The Michaelis-Menten equation is an expression of the relationship between the initial velocity 𝑉0 of an enzymatic reaction and substrate concentration [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation to an expression from which the effect of [S][S] on the rate can be more readily determined. Match the condition (e.g., [S]=𝐾m) with the statement or statements that describe it.

[S] = 0.1 Km - Doubling [S] will almost double the rate. [S] = Km - Half of the active sites are occupied by substrate. [S] = 10 Km - About 90% of the active sites are occupied by substrate. [S] = 10 Km - Doubling [S] will have little effect on the rate. [S] = 0.1 Km - Less than 10% of the active sites are occupied by substrate. [S] = 10 Km - This condition will result in the highest rate.

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V0 for an enzyme‑catalyzed, single‑substrate reaction E+S ⇌ ES ⟶ E+P. The model can be more readily understood when comparing three conditions: [S]<<Km, [S]=Km, and [S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V0 where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme.

[S]<<Km [ES] is much lower than [Efree] [Efree] is about equal to [Etotal] [S]=𝐾m The rate is half of the maximum rate. [S]>>𝐾m Almost all active sites will be filled. Adding more S will not increase the rate. Not true for any of these conditions Increasing [Etotal] will increase Km