lecture 12 - enzyme kinetics i

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  • 8/18/2019 Lecture 12 - Enzyme Kinetics I

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    Enzyme Kinetics I

    Maud Menten Leonor Michaelis

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    Enzyme Kinetics

    What is it?

    Enzymes are molecular machines – as with any machine or mechanistic

    process in nature, it is worthwhile to understand:

    the individual steps taken by the machine in its process

    the rate at which the machine can operate

    the environment in which the machine best operates

    + +

    Enzyme(E)

    Enzyme(E)

    Enzyme-Substrate

    Complex(ES)

    Substrate(S)

    Product(P)

    k1

    k-1

    k2

    k-2

    + +

    Enzyme(E)

    Enzyme(E)

    Enzyme-Substrate

    Complex(ES)

    Substrate(S)

    Product(P)

    k1

    k-1

    k2

    k-2

    Enzyme kinetics is a quantitative approach to understanding the functions of and

    mechanisms employed by enzymes.

    It is based upon determining the rates at which enzymes catalyze reactions

    and the environmental factors that influences these rates.

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    Why analyze enzyme kinetics?

    1. The kinetic analysis of enzyme behavior lends insight into the molecular

    mechanisms used by enzymes to convert substrate to product.

    2. Kinetic analyses provide both the experimental means and strategy

    towards understanding readily measurable quantities that can be used to

    describe enzyme behaviour. Quantities derived through enzyme kinetics are

    the universal language by which enzymes are compared by biochemists.

    3. Enzymes are used in many industrial processes or constitute a componentof useful products. Useful enzymes are identified by their kinetic properties

    e.g. their affinity for substrate and the efficiency with which they carry out

    reactions.

    4. In medicine: many chemicals living organisms may ingest either astherapeutic drugs or as toxins mediate their effects by binding to enzymes

    and altering the rate at which they catalyze reactions.

    5. The affinity of an enzyme for substrate and the rate at which it converts

    substrate to product is a consequence of its evolved role in the metabolicpathways that define life.

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    The kinetic analysis of enzymes preceded knowledge of their molecular

    nature

    pre 1900

    Pasteur posited that the special activities of living cells that could mediatechemical changes were a special property of life, inseparable from life.

    The concept of Vitalism.

    around 1900

    Eduard Buchner discovered that these chemical activities were the

    properties of molecules that could function separate from living

    organisms.

    The term “Enzymes” was coined and over the next few decades it was

    established that these molecules were proteins.

    post –  1900

    the emergence of theories that described the action of enzymes and their

    behavior

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    The rate of an enzyme catalyzed reaction (v) is dependent on substrate

    concentration [S]

      r  e  a  c

       t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]

      r  e  a  c

       t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]

    Led Victor Henri in 1903 to propose that reaction proceed via an

    interaction between enzyme (E) and substrate (S).

    ES complex

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      r  e  a  c

       t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]

      r  e  a  c

       t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]

    in 1913, Leonor Michaelis and Maud Menten proposed a general theory of

    enzyme activity:

    ax

    b + xy = 

    E + S ES E + Pk 

    -1 

    k 1 

    k -2 

    k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

    k -1 

    k 1 

    k -2 

    k 2 

    k -2 

    k 2 

    fast slow

    v =Vmax[S]

    Km + [S]v =

    Vmax

    [S]

    Km + [S]

    Vmax is the maximum velocity

    of the rate of reaction that occurs

    when enzyme is saturated.

    KM is a measure of the apparent

    affinity of the enzyme for substrate

    Vmax

    KM

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    Derivation of the M-M equation

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -2 

    k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -1 

    k 1 

    k -2 

    k 2 

    k -2 

    k 2 

    fast slow

    reaction rates are always measured

    as initial velocities, before there is

    appreciable depletion of substrate

    or formation of product

    approximation

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -2 

    k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -1 

    k 1 

    k -2 

    k 2 

    k -2 

    k 2 

    fast slow

    reaction rates are always measured

    as initial velocities, before there is

    appreciable depletion of substrate

    or formation of product

    approximation

    -the overall rate of rx (v) is limited by the conversion of ES → E + P  

    -2 factors influence this conversion: rate constant (k2) and concentration of ES, [ES]

    therefore, v = k2[ES] Equation 1

     

     

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    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    Two important assumptions are made during M-M analysis:1) substrate is in vast excess i.e. [S] >>[E]

    2) steady-state assumption, that ES is forming and breaking down

    at same rate.

    k1[E][S] = (k-1 + k2)[ES]

    collect terms on right side equation

    solve for [ES]

    k1[E][S] = k-1[ES] + k2[ES]

    formation breakdown

    k1[E][S] = k-1[ES] + k2[ES]

    formation breakdown

    [ES] =k1[E][S]

    (k-1 + k2)

    next page

     

     error in text pg 231 line 15 shouldread breakdown not formation

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    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    [ES] =k1[E][S]

    (k-1 + k2)

    since all 3 rate constants are on same side of equation,

    we can combine these into one term

    k1

    (k-1 + k2)=

    1

    KM

    (k-1 + k2)

    k1or KM =

    k1

    (k-1 + k2)=

    1

    KM

    (k-1 + k2)

    k1or KM = KM is called the Michaelis constant

    [ES] =

    [E][S]

    KM

    [E][S]

    KMEquation 2[ES] =

    [E][S]

    KM

    [E][S]

    KMEquation 2

    substitute KM into above equation

    next page

     

     

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    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    [ES] =[E][S]

    KM

    [E][S]

    KMEquation 2[ES] =

    [E][S]

    KM

    [E][S]

    KMEquation 2

    the total amount of enzyme, ET, is constant throughout experiment

    (i.e. the catalyst is not consumed).

    But it may be either free (not bound to substrate), E or bound

    to substrate, ES.

    therefore,

    [ET] = [E] + [ES]

    [E] = [ET

    ] - [ES]

    we can solve for [E]

    and substitute this definition of [E] into Equation 2, so

    [ES] =([ET]  – [ES])[S]

    KM

    next page

     

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    [ES] = ([ET]

     –[ES])[S]

    KM

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    [ES] =[ET][S]  – [ES][S]

    KM[ES] =

    [ET][S]  – [ES][S]

    KM

    [ET][S] = [ES]KM + [ES][S]

    [ET][S] = (KM + [S])[ES]

    [ES] =[ET][S]

    KM + [S][ES] =

    [ET][S]

    KM + [S]

    multiply through by [S]

    solve for [ET][S]

    collect terms

    solve for [ES]

    next page

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    [ES] =[ET][S]

    KM + [S][ES] =

    [ET][S]

    KM + [S]

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    remember Equation 1: v = k2[ES]then

    [ES] =v

    k2[ES] =

    v

    k2

    Now, the maximum rate (Vmax) occurs when all enzyme bound to substrate.That is, when [ET] = [ES]

    Under this condition, Equation 1 becomes: Vmax = k2[ET]

    v

    k2

    [ET][S]

    KM + [S]=

    vk2[ET][S]

    KM + [S]=v

    k2[ET][S]

    KM + [S]=

    substitute term for [ES]

    solve for v

    v =

    Vmax[S]

    KM + [S]

    substitute in Vmax

    the Michaelis-Menten equation

     

     

     

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      r  e  a  c   t   i  o  n   r

      a   t  e

     ,  v

    substrate concentration, [S]

      r  e  a  c   t   i  o  n   r

      a   t  e

     ,  v

    substrate concentration, [S]

      ax

    b + xy = 

    v =Vmax[S]

    Km + [S]v =

    Vmax[S]

    Km + [S]

    Vmax

    KM

    The Significance of Vmax, Km and other kinetic quantities

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    Vmax

    Vmax is the maximal velocity of the enzyme

    In general the rate of a reaction is dependent on the rate constant k2 and the

    substrate concentration. Therefore v = k2[ES]

    but, when substrate concentration is so high that all of the enzyme is saturated

    then the enzyme is operating as fast as it can and Vmax = k2[ET] where ET is

    the total Enzyme in the reaction ([ES]=[ET] when the enzyme is saturated).

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -2 

    k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -1 

    k 1 

    k -2 

    k 2 

    k -2 

    k 2 

    fast slow

    reaction rates are always measured

    as initial velocities, before there is

    appreciable depletion of substrate

    or formation of product

    approximation

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -2 

    k 2 

    fast slow

    E + S ES E + Pk -1 

    k 1 

    k -1 

    k 1 

    k -2 

    k 2 

    k -2 

    k 2 

    fast slow

    reaction rates are always measured

    as initial velocities, before there is

    appreciable depletion of substrate

    or formation of product

    approximation

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      r  e  a  c   t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]KM

    Vmax

    ½ Vmax

      r  e  a  c   t   i  o  n   r

      a   t  e ,  v

    substrate concentration, [S]KM

    Vmax

    ½ Vmax

    v =Vmax[S]Km + [S]

    v =Vmax[S]Km + [S]

    KMKM is the substrate concentration at which the enzyme is rate is half its maximal

    velocity (Vmax) or put another way:

    KM is the substrate concentration where the enzyme is half-saturated.

    On a Michaelis-Menten plot, KM equals the [S] at which the reaction rate

    is half-maximal (1/2 Vmax)

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    KM

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

     

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

     

    Km is actually a combined rate constant that describes the formation and breakdown

    of the ES complex

    in most reactions, k2 is usually much lower than k1

    therefore Km approximately reduces to:

    v =Vmax[S]

    KM + [S]

     

    (k-1 + k2)

    k1  KM = 

    (k-1 + k2)

    k1  KM =

    k1

    (k-1 + k2)=

    1

    KM

    k1

    (k-1 + k2)=

    1

    KM

      k-1 

    k1KM = 

    Note that when k1 > k-1, the formation of ES is favored and KM is small

    when k-1 > k1, the dissociation if ES is favored and KM is large

    that is, small KM indicates high affinity of enzyme for substrate

    high KM indicates low affinity of enzyme for substrate

    KM is a measure of the affinity of an enzyme for its substrate

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    KcatE + S ES E + P

    k -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    kcat is equivalent to the rate limiting step, e.g. k2 

    therefore, k2 = kcat

    and when all enzyme is bound to substrate,

    Vmax = k2[ET]

    therefore, k2 = = kcat Vmax

    [ET]

    kcat is also referred to as the turnover number (units of s-1

    )

    that is, the maximum number of substrate molecules per active site

    converted to product per second

    kcat is the turnover number

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    also known as the Specificity constant

    the parameters of KM or kcat alone are insufficient for comparing

    the overall efficiency of an enzyme

    a better measure is to calculate the ratio of kcat to KM 

    it is a measure of the overall rate of reaction where ultimately every

    reaction relies on an encounter of Enzyme with Substrate

    the upper limit of kcat/KM is defined by the rate at which E and S

    can diffuse together in an aqueous solution

    (about 108 to 109 M-1s-1)

    some enzymes have specificity constants approaching this value

    they are said to be “catalytically perfect” enzymes 

    Enzyme Substrate Kcat

    (s-1)

    KM

    (M)

    kcat /KM

    (M-1s-1)

    Catalase H2O2 4 x 107 1 x 100 4 x 107

    -lactamase penicillin 2 x 103 2 x 10-5 1 x108

    kcat

    KM

    kcat

    KM

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    Practical Determination of Kinetic Parameters in Laboratory

    Hypothetical case:

    You work for an industrial company that uses the enzyme Cellulase in the

    formulation of laundry detergents.

    It is your job to determine the kinetic attributes of 5 different Cellulase

    enzymes to assess their suitability for use as an additive in laundry detergent.

    Background

    Cellulase is an enzyme that breaks down Cellulose into glucose molecules

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    CO2 + H2O

    glucosecellulose

    cotton

    O2

    Energy

    Cellulases are enzymes that bind

    cellulose and chemically cleaves

    it back to free glucose molecules

    http://www.google.ca/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=LWsRCD4h98T0hM&tbnid=OF3weCUF0rk-mM:&ved=0CAUQjRw&url=http://science.howstuffworks.com/environmental/green-tech/energy-production/artificial-photosynthesis.htm&ei=txZKUsCoMofP2QW41oGIDQ&bvm=bv.53371865,d.b2I&psig=AFQjCNFIC1tONuAv9fhps7d45NE01NHU9A&ust=1380673566474132

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    light

    new shirt

    surface

    wear and tear after cellulase treatment

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    Your job

    Carry out kinetic analysis of 5 cellulase enzymes

    Cellulase 1Cellulase 2

    Cellulase 3

    Cellulase 4

    Cellulase 5

    -determine the most active enzyme of the five.

    M ll l ti it b i i f l t ti

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    time

       [  g   l  u  c  o  s  e   ]

    Measure cellulase enzyme activity by increase in free glucose concentration

    The Progress curve of reaction at defined concentration of cellulose, [S].

    y = m x + b

    m=d[glucose]/dt

    Enzyme + cellulose → Enzyme + glucose 

    The rate (v) is the increase

    in glucose liberatedfrom cellulose over time

    slope of the line = initial rate

    or velocity (v) of the reaction

    don’t confuse this with the 

    M-M plot, which has axes

    of rate vs [S]

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    Determination of reaction rates at varying substrate concentrations

    etc.

    [S]=4[S]=3

    [S]=2

    [S]=1

    time

            [      g        l      u      c      o      s      e        ]

    multiple progress curves

    etc.

    [S]=4[S]=3

    [S]=2

    [S]=1

    time

            [      g        l      u      c      o      s      e        ]

    multiple progress curves [S] v

    1 0.001

    2 0.002

    3 0.004

    4 0.007

    5 0.010

    6 0.013

    Next, plot Rates vs [S]

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    Plotting substrate concentration vs rate – The Michaelis-Menten curve

    [substrate], [S]

    rate, v

    o

    o

    o

    o

    o

    oo

    o

    o

    o

    o

    oo o o

    oo Vmax

    ½ Vmax

    KM

    v =Vmax[S]

    KM + [S]

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    So,

    Having generated a Michaelis-Menten Plot specific for the Cellulase #1:

    You now know the Vmax for the enzyme.

    You can now determine KM  for the enzyme since KM is the substrate concentration

    at 1/2Vmax

    Also, you can calculate Kcat since you know how much enzyme you used in the assay

    and

    Kcat =

    Therefore you can also calculate the specificity constant since

    Specificity Constant = Kcat/KM

    Vmax

    [ET]

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    Enzyme kcat

    (s-1)

    KM

    (M)

    kcat /KM

    (M-1s-1)

    Cellulase 1 4 x 104 1 x 10-2 4 x 106

    Cellulase 2 1 x 102 4 x 10-4 3 x 105

    Cellulase 3 2 x 105 4 x 10-2 5 x 106

    Cellulase 4 4 x 101 4 x 10-3 1 x 104

    Cellulase 5 2 x 103

    2 x 10-5

    1 x108

    Therefore, you would report that Cellulase #5 is the most active enzyme

    Note that Cellulase #1 has a higher turnover number (kcat) than Cellulase #5

    But Cellulase #5 has a much higher affinity for substrate (KM).

    Overall, based on the specificity constant (kcat/KM) Cellulase #5 is the most active

    enzyme.

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    Understand the meanings of

    Vmax

    the relationship between Vmax and KM

    the meaning of Km in terms of substrate affinity

    the turnover number, Kcat

    Specificity constant Kcat/KM

    the general process for determining Vmax and KM in the lab

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

    E + S ES E + Pk 

    -1 

    k 1 

      k 2 

    fast slow

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    Recommended Reading

    Stryer 8th Edition

    Ch 8 Enzymes: Basic Concepts and Kinetics Pgs 216-245.

    Stryer 7th Edition

    Ch 8 Enzymes: Basic Concepts and Kinetics Pgs 219-248.

    If you would like to look at and discuss your exam:

    Email me ([email protected]) with BIOL 2023 in subject line.

    List 3 days/times you can meet and I will pick one that fits my schedule.

    I cannot respond to emails that do not conform to above specs.