What Is the Ratio of S to Km When the Velocity of an Enzyme-catalyzed Reaction Is 80 of Vmax

10.E: Enzyme Kinetics (Exercises)

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10.1: Full general Principles of Catalysis

Q10.1a

One 24-hour interval in form about enzyme kinetics, Jack comes over to yous and asks, "I know enzymes are biological catalysts, just I do not understand how it works. Can you explain how enzymes make reactions get faster? And is it merely faster in 1 direction?"

S10.1a

Because the activation energy is the free energy colina between reactants and products, enzymes decreasing the size of the hill also decreases the amount of energy needed for reactions to go in either management. A smaller energy loma allows reactants and products to overcome the barrier quicker, resulting a faster reaction rate.

Q10.1b

If your educatee colleagues argues that a catalysts affects only the charge per unit of only 1 direction of a reaction. Explicate why he is correct or non.

S10.1b

False. Catalysts affect rate by providing an culling mechanism which has a lower transition state energy. It'southward incommunicable to lower transition state energy for only i direction of a reaction. (It'd be like making a loma shorter from the n, but keeping it the same summit from the s.)

Q10.1c

How does the enzyme catalysis affect both forward and opposite reaction?

S10.1c

The enzyme catalyst lowers the Gibb free energy of transition state, which reduces the activation free energy of both reactions. Therefore, it makes reactions occur faster.

Q10.2a

Given enzyme-catalyzed reaction k₁= 4x10⁶M^-1 due south^-ane , k_-ane=6x10⁴s^-one and k₂= 2.0x10³s^-one. Make up one's mind if the enzyme –substrate bounden follow the equilibrium or not ?

S10.2a

The dissociation constant yard_s= k_-1/ k₁=6x10⁴s^-i/ 4x10^{half dozen}1000^-ane s^-i= 0.15 M

The Michaelis constant thousand_K = thousand_-1+k_ii / k₁₌6x10⁴s^-1+ two.0x10^threesouthward^-one/ 4x10⁶M^-1 s^-1=0.155M

The ii constant are not equal. Therefore, the binding does non follow the equilibrium scheme.

Q10.2b

Is it appropriate to utilize the rapid equilibrium scheme to model the kinetics of a catalyzed reaction with the following rate constants?

\[ k_1 = 7 \times 10^7\ Chiliad^{-1}\ s^{-1} \]

\[ k_{-1} = 8 \times x^5\ s^{-1} \]

\[ k_2 = five \times 10^4\ s^{-ane} \]

S10.2b

Bold rapid equilibrium is appropriate when $k_{-1} \gg k_2$, so that K_G = K_S. In this instance, the values of thousand_-1 and k₂ are non dramatically different, then nosotros can calculate K_Southward and Grand_M and compare.

1. Calculate $K_M$:

\[ K_M = \dfrac{k_{-one} + k_2}{k_1} \]

\[ K_M = \dfrac{8 \times 10^five\ s^{-1} + 5 \times 10^4\ due south^{-1}}{7 \times ten^7\ M^{-one}\ south^{-1}} \]

\[ K_M = 0.01\ M \]

two. Calculate $K_S$:

\[ K_S = \dfrac{k_{-one}}{k_1} \]

\[ K_M = \dfrac{8 \times 10^5\ due south^{-1}}{7 \times 10^vii\ 1000^{-1}\ s^{-1}} \]

\[ K_M = 0.01\ M \]

Since $K_M = K_S$, it is advisable to presume rapid equilibrium.

Q10.2c

Given k₁= 7 x 10^half dozen M-1s-1, yard_-ane=6 ten ten^4 s^-1, g₃= 2 10 10^three s^-1, determine if the enzyme substance binding follow the equilibrium or steady state scheme?

\[ \brainstorm{align} K_s &= \dfrac{k_{-1}}{k_1} \\ &= \dfrac{6 \times 10^4\, s^{-1}}{7 \times 10^half dozen 1000^{-one}s^{-i}} \\ &=viii.eight \times 10^{-iii} Chiliad \stop{marshal}\]

\[\begin{marshal} K_M &= k_{-ane}+k_{2} \\ &= \dfrac{6 \times 10^iv due south^{-1} +2 \times 10^3s^{-1}}{ 7 \times 10^half dozen Grand ^{-i} s^{-1}} \\ &= eight.9 \times 10^{-3} G \end{marshal}\]

Q10.3a

The substrate Northward-acetylglycine ethyl ester tin can exist catalyzed by the enzyme carbonic anhydrase. This enzyme has a turnover rate of 30,000 s^-1. Determine how long it will take carbonic anhydrase to cleave the substrate.

S10.3a

Nosotros already know the turnover number ($k_{true cat}$). The amount of time necessary to cleave the substrate is the reciprocal of the turnover rate.

\[t= \dfrac{1}{grand} = \dfrac{1}{30,000 \;s^{-one}} = 3.33 \times 10^{-five}\]

\[(5 \;minutes)(vii.i \times 10^{-6}) = 3.v \times 10^{-5}\]

Q10.3b

RuBisCO is an enzyme in the Calvin wheel that fixes atmospheric carbon and has a turnover rate of 3.3 due south^-1. How long does it have RuBisCO to fix i molecule of carbon dioxide?

Spacefilling structure of RuBisCO created using Rasmol and the 8RUC file from the Protein Data Banking concern.

S10.3b

The turnover number is the number of molecules of substrate per unit time (when the enzyme is fully saturated). So simply accept the reciprocal to detect the time per molecule of substrate.

\[ \dfrac{1}{iii.3\ s^{-1}} = 0.xxx\ southward \]

Note: RuBisCO is a notoriously slow enzyme.

Q10.3c

The catalyze of acetylcholine has a rate 50000 s^-1 . Calculate the time for the enzyme to cleave i Ach molecule.

S10.3c

t= 1/k₂ =1/ 50000 = ii.0 x10^-5s

Q10.3d

Carbonic anhydrase, an enzyme that catalyzes the dehydration of carbonic acid to form carbonic acid, has the turnover rate of k_cat 4.0x 10⁵s^-1. Calculate how long does it take does it take for the enzyme to cleave one molecule carbonic acid?

S10.3d

The fourth dimension required for the enzyme to cleave one molecule carbonic acid:

t=ane/k_cat= 1/ 4x10⁵s^-ane=2.5x10^-vis = ii.5µs

Q10.3e

p-nitrophenyl acetate(PNPA) is catalyzed past chymotrypsin to yield p-nitrophenolate ion and acetate ion. The turnover rate of that enzyme is 40,000 s^-1. How long will information technology take for the enzyme to produce ane mole of Nitrophenyl acetate?

S10.3e

It takes i 2nd to convert all forty,000 molecule substrate into the product, so: $t=\dfrac{xl,000}{nine.5x10^5}$ = 3.8 sec.

ten.2: The Equations of Enzyme Kinetics

Q10.4

What is plotted on the x and y axes on a Lineweaver-Burk plot? Bear witness how to derive the equation for the plot from the equation

\[ v_0 = \dfrac{V_{max}[South]}{K_M + [Due south]} \]

and explain how V_max and K_M tin can be found from the graph'south intercepts. Hint: A Lineweaver-Burk plot is likewise sometimes called a double reciprocal plot.

S10.4

The x-centrality is 1/ν₀, and the x-axis is 1/V_max.

ane. Take the reciprocal of both sides of the equation.

\[ \dfrac{1}{v_0} = \dfrac{K_M + [S]}{V_{max} [S]} \]

\[ \dfrac{1}{v_0} = \dfrac{K_M}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}} \]

2. Set 1/[Southward] = 0 to find the y-intercept, and show that information technology relates to 5_max.

\[ Yint = \dfrac{1}{V_{max}} \]

\[ V_{max} = \dfrac{1}{Yint} \]

3. Set i/ν₀ = 0 to discover the x-intercept, and show that it relates to K_M.

\[ 0 = \dfrac{K_M}{V_{max}} Xint + \dfrac{1}{V_{max}} \]

\[ \dfrac{K_M}{V_{max}} Xint = - \dfrac{ane}{V_{max}} \]

\[ Xint = -\dfrac{1}{K_M} \]

\[ K_M = -\dfrac{1}{Xint} \]

Q10.4a

Derive the Michaelis-Menten equation by assuming rapid equilibrium.

\[ E + S \overset{K_{1}}{\rightleftharpoons} ES \overset{K_{2}}{\rightarrow} E + P\]

\[v_{0} = \dfrac{V_{max}[S]}{K_{M}+[Due south]}\]

S10.4a

$\dfrac{[E][Southward]}{dt} = K_{-1}[ES]$ $\dfrac{[ES]}{dt} = K_{i}[Due east][S]$ $\dfrac{[E][S]}{dt}=\dfrac{[ES]}{dt}$ $K_{-1}[ES]=K_{ane}[E][S]$ $K_{M}\dfrac{k_{-1}}{K_{1}}=\dfrac{[E][S]}{[ES]}$

given that $[East]_{0}=[E]+[ES]$

$K_{Thou}=\dfrac{[Eastward]_{0}-[ES][S]}{[ES]}$ $K_{Thousand}=\dfrac{[Due east]_{0}[S]-[ES][S]}{[ES]}$ $K_{M}\times[ES]=[Eastward]_{0}[S]-[ES][S]$ $K_{K}\times[ES]+[ES][S]=[Eastward]_{0}[S]$ $(K_{M}+[S])[ES]=[E_{0}][South]$ $[ES]=\dfrac{[E_{0}][S]}{K_{Chiliad}+[S]}$

since $\dfrac{d[P]}{dt}=k_{2}[ES]$

$v_{0}=\dfrac{K_{two}[E_{0}][S]}{K_{M}+[S]}$

since $v_{max}=k_{ii}[Due east]_{0}$

$v_{0} = \dfrac{V_{max}[S]}{K_{Thousand}+[South]}$

Q10.4b

We know that the Michaelis Menten derivation for the following reaction:

\[\ce{E + S \rightleftharpoons ES -> E + P}\]

However, what if the reaction took place in a different scenario whereby:

\[\ce{E + S \rightleftharpoons ES1 -> ES2 -> E + P}\]

What would be the respective Michaelis-Menten Ebe quation now?

S10.4b

This is the an outline for determining an expression for the rate of substrate conversion in the given case:

Prepare the reaction with charge per unit constants, bold $k_{-2}\approx k_{-3}\approx0$: \[\ce{E + S <=>[k_1][k_{-one}] ES_1->[k_2] ES_2 ->[k_3] Due east + P}\]
Set up the differential equations describing the reaction, i.e. the rate of change for each component with fourth dimension. The rate of substrate modify, for example, volition be \[\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{Due east}][\ce{S}] +k_{-i}[\ce{ES_1}]\]
Choose initial conditions and fix two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES₁ and ES_ii.
Brand the steady-state assumpt: presume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e.\[\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0\]
Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.

Q10.4c

Prove that $K_s$ equals the concentration S when the initial rate is one-half its maximum value.

S10.4c

We take: $v_{o}= \frac{V_{max}\left [ S \right ]}{K_{M}+\left [ S \right ]}$

Divided both said by i $\frac{1}{v_{o}}= \frac{K_{M}+\left [ S \right ]}{V_{max}\left [ S \right ]}$

$\frac{1}{v_{o}}= \frac{K_{M}}{V_{max}\left [ S \right ]}+\frac{\left [ S \right ]}{V_{max}\left [ S \right ]}$

$\frac{1}{v_{o}}= \frac{K_{M}}{V_{max}\left [ S \right ]}+\frac{1}{V_{max}}$

When the initial rate is half its maximum value: $v_{o}= \frac{V_{max}}{2}$

$\frac{2}{V_{max}}= \frac{K_{M}}{V_{max}\left [ S \right ]}+\frac{1}{V_{max}}$

$\frac{2}{V_{max}}-\frac{1}{V_{max}}= \frac{K_{M}}{V_{max}\left [ S \right ]}$

$\frac{1}{V_{max}}= \frac{K_{M}}{V_{max}\left [ S \right ]}$

$\frac{V_{max}}{V_{max}}= \frac{K_{M}}{\left [ S \right ]}$

$\left [ S \right ]= K_{M}$

Q10.5a

An enzyme that has a $K_m$ value of $4.vi \times 10^{-5}\; M$ is studied at an initial substrate concentration of 0.041 Grand. After a infinitesimal, information technology is establish that 7.3 uM of production has been produced. Calculate the value of Vmax and the amount of product formed after 4.5 minutes.

Q10.5b

An solution initially contains a catalytic amount of an enzyme with K_M = 1.5 mM, 0.25 One thousand of substrate, and no product. Afterward 45 seconds, the solution contains 25 µM of product. Find V_max and the concentration of product after ii.0 minutes.

Hint: [Due south] >> G_M

S10.5b

1. Notice the initial velocity.

\[ v_0 = \dfrac{25\ \mu M}{0.75\ min} \]

\[ v_0 = 33.three\ \mu G/min \]

ii. Use v₀, [S], and K_Thousand to solve for V_max.

\[ v_0 = \dfrac{V_{max} [Southward]}{K_M + [S]} \]

\[ V_{max} = v_0 \left ( \dfrac{K_M}{[S]} + ane \correct ) \]

\[ V_{max} = 33.iii\ \mu M/min \left ( \dfrac{1.5\ mM}{0.25\ One thousand} \times \dfrac{Chiliad}{1000\ mM} + one \correct ) \]

\[ V_{max} = 33.3\ \mu 1000/min \]

Notice that ν₀ = Five_max. Since [S] >>Chiliad_M, the reaction volition go along with a velocity of V_max for the residue of the two minutes.

3. Predict [P] after 2.0 minutes at the rate Five_max.

\[ [P] = V_{max} \times t \]

\[ [P] = 33.iii\ \mu M/min \times 2.0\ min \]

\[ [P] = 66.6\ \mu Chiliad \]

Q10.5c

A particular enzyme at a research facility is being studied by a group of graduate students. This enzyme has a M_m value of 5.0 X 10^{-half dozen} 1000. The students report this enzyme with an initial substrate concentration of 0.055 M. At one minute, 7 µM of product was made. What is the amount of product produced after 5 minutes. What is the V_max ?

S10.5c

This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program. 7.0 10 10^-half-dozen M = V_max (0.055 M) / (5.0 X 10^-6Grand + .055 M) V_max = 7.1 X 10^{-half dozen}M/min At five minutes the amount of production formed is:

Q10.5d

Calculate the value $v_{o}$ if an enzyme has $K_{M}= 5.4*10^{-4}$ , value $V_{max}= 48\mu Mmin^{-1}$ and $\left [ S \right ]= 0.001M$

S10.5d

Nosotros take: $v_{o}= \frac{V_{max}\left [ S \right ]}{K_{M}+\left [ S \right ]}$

$v_{o}= \frac{48*0.001}{5.4*10^{-4}+0.001}= 31 \mu Mmin^{-1}$

Q10.6a

Given the values, [Southward]/ten^-iv M three.0 4.vi 10.5 16.5

vo/10^-6 1000 * min^-1. 2.64 three.5 6.2 seven.8

construct a Lineweaver-Burk plot, and assuming Michaelis-Menten kinetics, calculate the values of Vmax, Km, and k2 using the constructed plot.

Q10.6b

The data below represents the data recorded after the hydrolysis of a substrate by an enzyme.

[Southward]/10-4 One thousand	2.i	4.2	nine.3	fourteen.2
vo / x-6 M · min-1	1.ii	3.1	vi.3	9.1

Calculate V_max, Chiliad_m and K_two using a Lineweaver-Burk plot. Assume Michaelis. Given [E}₀ is v.0 X 10^-6 M

S10.6b

(1/[Southward])/ten-3 Chiliad-1	(ane/vo )/ 10-5 Thousand-1 · min
iv.8	8.3
2.4	3.two
ane.1	ane.6
0.seven	ane.one

Figure: Lineweaver-Burk Plot The linear equation for the graph above is: y=176x+4.23X10^four To solve for V_max use: Intercept= i/Five_max 5_max= 1/(.4.23X10⁴) = 2.36 X 10^-v To solve for K_grand use Slope= 1000_m/V_max (176)( 2.36 Ten 10^-five) = 4.2 X 10^-5 To solve for Yard₂ utilize 1000₂ = V_max/[E}₀= (ii.36 Ten 10^-5)(v.0 X x^-6 M)= iv.72 min^-i

Q10.6c

Using the table below, summate the K_G, Five_max, and slope.

\[\dfrac{1}{V_{0}} (ten^{-3}\dfrac{sec}{M^{-1}})\]	\[\dfrac{ane}{[S]}(ten^{2}Thousand^{-1})\]
ii.ix	0.iii
3.2	one.2
4.4	1.eight
5.i	three.three

S10.6c

gradient=$\dfrac{5.ane\times10^{-3}-2.9\times10^{-three}}{iii.3\times10^{two}-0.3\times10^{ii}}=half dozen.73\times10^{-6}$ $6.73\times10^{-6}\times0.3\times10^{two}+\dfrac{i}{V_{max}}=2.9\times10^{-3}$ $\dfrac{ane}{V_{max}}=0.00270$ $V_{max}=370.62Ms^{-i}$ $6.73\times10^{-6}\times\dfrac{-one}{K_{M}}+0.00270=0$ $\dfrac{-1}{K_{M}}=-0.00943$ $K_{M}=106.044$

Q10.6d

Given the value $\left [ S \right ]$ = 0.00032 and $K_{M}= 3.5*10^{-5}$ . Find the ratio betwixt $v_{o}$ and $V_{max}$

S10.6d

$v_{o}= \frac{V_{max}\left [ S \right ]}{K_{M}+\left [ S \right ]}$

Divided both side past 1: $\frac{1}{v_{o}}= \frac{K_{M}}{V_{max}\left [ S \right ]}+\frac{1}{V_{max}}$

Divided both side by ane again: $v_{o}= \frac{V_{max}\left [ S \right ]}{K_{max}}+V_{max}$

$v_{o}= V_{max}\left ( \frac{\left [ S \right ]}{K_{max}} +1\right )$

$\frac{v_{o}}{V_{max}}= \frac{\left [ S \right ]}{K_{M}}+1$

$\frac{v_{o}}{V_{max}}= \frac{0.00032}{3.5*10^{-5}}+1= \frac{71}{7}= 10.14$

Q10.7a

From this graph determine the K_K and V_max?

S10.7b

From his graph we can see that the value Grand_M is 2. Then nosotros look to see where K_M is half. At that point, we see that K_Thousand/two is 1 and the x-value for that coordinate is 1. This means V_thousand _ax is 1.

Q10.7b

Neutral sphingomyelinase 2 converts sphingomyelin into ceramide and phosphocholine. Assume its Five_max is 35 µM min^-one. When yous provide 3 x 10^-5 M of sphingomyelin, you lot observe an initial velocity of vi.0 µM min^-i. Calculate the K_M for this reaction, rounding to 3 pregnant figures.

S10.7b

\[\dfrac{i}{V_{o}} = \dfrac{K_{M}}{V_{max}[S]}+\dfrac{1}{V_{max}}\]

\[\dfrac{one}{vi.0} = \dfrac{K_{M}}{(35)(30)}+\dfrac{1}{35}\]

\[\dfrac{1}{six}-\dfrac{i}{35} = \dfrac{K_{M}}{1050}\]

K_Grand = 145 µM

Q10.8a

Nosotros are given a 2d society equation: r=k[A][B]. The concentration of A is 0.05g and the concentration of B is two.5g. We decide this departure is bully enough to treat this equally a pseudo outset reaction? What concentration is held constant and why? Write the new equation.

S10.8a

We hold B constant because the concentration is so much larger, so information technology should be close to abiding for the reaction. The new equation would be r=k'[a]

Q10.9a

Name and briefly describe ii types of reactions that do not follow Michaelis-Menten kinetics.

S10.9a

Irreversible inhibition - the inhibitor binds covalently and irreversibly to the enzyme Allosteric interactions - the binding of effectors at allosteric sites (away from the active site) influence substrate binding.

Q10.9b

For the reaction mechanism below, how does the concentration of $C$ affect the concentration of $B$?

\[A\leftrightharpoons B\rightarrow C\]

S10.9b

Because the second state is irreversible (i.e., since arrow), it does not thing if you have a large or little concentration of $C$, information technology would non touch on $B$ and hence the kinetics of the reaction. However, the concentration of $A\0 would clearly affect the concentration of \(B$.

Q1

Speculate on how the catalytic rate constant tin can be determined from the spectrophotogram.
How can product be consistently produced if the charge per unit of change of the ES complex is 0?
How would the rate of product formation change if:
1. the substrate concentration were doubled?
2. the enzyme concentration were doubled?
3. The reaction was carried out in mono-deuterated water instead of H_iiO (comment qualitatively)?
Explicate the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin.
What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium.

S2

The catalytic rate constant can exist deduced from the graph by simply determining the gradient of the line where the reaction demonstrates 0-guild kinetics (the linear part).
This is pre-equilibrium kinetics in action. The ES complex is formed from E and Due south at a faster charge per unit than any other pace in the reaction. As soon as ES is converted to *ES, some other mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and East + Due south is constantly at equilibrium, and thus the change of either with respect to fourth dimension is 0.
1. No change.
2. 2-fold increase.
3. Because water is involved in the final, slowest step of the mechanism, deuterating the water would decrease the rate of the overall reaction from v- to xxx-fold.
Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine'due south O-H. This allows a facilitated nucleophilic attack of the hydroxyl oxygen on the substrates carbonyl grouping. Conversely, in the final footstep of the reaction, the jump serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate.
The graph would bear witness similar 0-society kinetics, merely the line would intercept the Y-axis at an absorbance of 0 instead of the 1:ane mole ratio of nitrophenolate to enzyme.

10.4: Multisubstrate Systems

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What Is the Ratio of S to Km When the Velocity of an Enzyme-catalyzed Reaction Is 80 of Vmax

10.1: Full general Principles of Catalysis

Q10.1a

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ten.2: The Equations of Enzyme Kinetics

Q10.4

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10.4: Multisubstrate Systems

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