Chemistry 301 Introduction to Biochemistry

Study Guide :: UNIT 4

Catalysis

Overview

The thousands of biochemical reactions that go on in vivo are not random events, but rather are controlled by biological catalysts called enzymes. Besides their effect on biological order and control, enzymes have evolved so that the reactions they catalyze can occur quickly under the mild conditions that exist in vivo. Enzyme‑catalyzed biochemical reactions can be written as follows:

Substrate(s)   → Enzyme   Product(s)

Enzymes, as you learned in Unit 3, are one of the subclasses of proteins. Enzymes are made up of stereospecific amino acids (remember that only L‑amino acids are found in proteins); therefore, enzymes themselves are stereospecific. This characteristic means that the enzyme will only work on a particular anomer. For example, the enzyme glucose oxidase, will only oxidize D‑glucose, but not L‑glucose. In addition, most enzymes have a small, non‑protein group attached either covalently or noncovalently. These attached groups, which are the actual sites of catalysis, are called “coenzymes.”

In Unit 4, we discuss the structure and some functions of enzymes and co‑enzymes.

The unit is divided into five lessons:

Lesson 1 Activation Energy

Lesson 2 Enzyme Catalysis

Lesson 3 Enzyme Kinetics

Lesson 4 Enzyme Inhibition

Lesson 5 Control of Enzymes

Learning Objectives

After completing this unit, you should be able to:

  1. Define “enzyme” and “coenzyme,” and explain the relationship between the two.
  2. Describe the six classes of enzymes.
  3. Explain the specificity of enzymes.
  4. List the types of chemical reactions involved in enzyme catalysis.
  5. Define “enzyme kinetics,” “enzyme inhibition,” and “Michaelis‑Menten enzyme.”
  6. Use graphical techniques to analyze and describe simple, unimolecular enzyme activities.
  7. Explain the difference between Michaelis‑Menten and allosteric enzymes.

Glossary

allosteric enzyme

a regulatory enzyme whose affinity for its substrate is affected by the presence or absence of other molecules

apoenzyme

protein portion of an enzyme (i.e., lacking a coenzyme)

enzyme

protein which catalyzes a biochemical reaction in vivo at 37°C, 0.1 M salt, and room pressure; frequently an enzyme name ends in ‑ase (e.g., amylase and carbonic anhydrase)

holoenzyme

complete active enzyme (i.e., protein + coenzyme)

enzyme classification

assignment of an enzyme to one of six groups, depending on the type of chemical reaction which the enzyme catalyzes

first order kinetics

rate of reaction is directly proportional to the concentration of starting materials (i.e., substrate)

inhibition

alteration in an enzyme’s activity, usually caused by modification of the enzyme active site, so that substrate cannot bind to the enzyme, or substrate can bind but cannot be converted to product, or product cannot be released

Michaelis‑Menten kinetics

simple mathematical description of a first‑order enzyme reaction (Leonor Michaelis was British and Maud Menten was Canadian)

Lesson 1: Activation Energy

Overview/Objectives

After completing this lesson, you should be able to:

  1. Demonstrate an understanding of the role of catalysts in chemical reactions.
  2. Define activation energy and discuss the role of catalysts in activation energy.
  3. Define transition state.
  4. Demonstrate an understanding of the reversibility of reactions and how the energy difference of the products and reactants determine how a reaction will proceed.

Readings and Activities

  1. Read “Introduction” and “Activation Energy” on page 82 of the textbook.
  2. You can also watch the three video lectures on Catalytic Mechanism:

Commentary

The general definition of a catalyst is a substance that alters the rate of a reaction, but is not changed in the process. As we learned in Unit 2, the Gibbs Free Energy (G) of a reaction is the thermodynamic potential of a reaction, meaning the amount of maximum or reversible work that may occur in a reaction at a constant temperature and pressure.

Looking at the graph on page 82 of the textbook, it can be seen that as a reaction proceeds, the total Gibbs energy or change in energy (ΔGTOTAL) for the substrates to products is the same whether the reaction involves a catalyst or not. However, if you look at the two slopes of the graph above the line denoting the free energy of the substrates, you will see that the slope of a catalyzed reaction is smaller compared to the slope of the reaction if it was uncatalyzed. This means that the catalyst does not change the overall energy of the reaction, but allows that reaction to proceed faster (ΔG°). What the catalyst does, however, is it lowers the activation energy of the reaction, meaning that it enhances the ability of the substrate molecules to get the energy needed to enable the reaction to occur.

Study Questions

  1. Define the terms “activation energy” and “transition state.”
  2. How does a catalyst influence the overall energy of a reaction? How does a catalyst affect the activation energy of a reaction?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 2: Enzyme Catalysis

Overview/Objectives

After completing this lesson, you should be able to:

  1. Explain what an enzyme is, and list its catalytic properties.
  2. Explain the role of coenzymes, and distinguish between cosubstrates and prosthetic groups.
  3. Explain what the “enzyme active site” is.
  4. Describe the chemical basis of substrate specificity, and list the chemical forces involved in substrate binding.
  5. Explain the difference between a substrate and an inhibitor.
  6. Demonstrate an understanding of the general principles behind enzyme classification.
  7. Describe the six chemical mechanisms by which enzymes catalyze biochemical reactions.

Readings and Activities

  1. Read “General Mechanisms of Action,” “Substrate Binding,” “Enzyme Flexibility,” “Active Site,” and “Chymotrypsin” (pages 83–87 of the textbook).
  2. You can also watch four video lectures on Enzyme Catalysis:

Commentary

Enzymes

Enzymes were one of the first classes of biomolecules to be studied. In the nineteenth century, it was known that biological systems contained something that could speed up some organic reactions. It was also known that these speed‑up factors were pretty specific, so the theory of enzyme catalysis as a lock‑and‑key mechanism was put forward. This model postulates that the enzyme has a relatively rigid three‑dimensional active site that will accept and chemically change only the one molecule (or key) which fits the site. More recently, the induced‑fit model of enzyme action has been suggested. In this model, a substrate molecule binds to an enzyme and a structural change occurs in the enzyme, thereby beginning the catalysis process.

We know from Unit 2 that the forces that hold a protein’s three‑dimensional structure in place are pretty strong—as long as solution conditions, such as pH, temperature, and salt concentration, do not change. So, it is unlikely that a monomeric enzyme will alter significantly when a substrate binds. The issue (lock‑and‑key vs. induced‑fit) becomes complicated because some enzymes are multimeric (usually dimeric or tetrameric), and so substrate binding may shift the relative geometry of the subunit proteins. We can conclude that the lock‑and‑key model is probably the predominant one in substrate‑enzyme recognition, and that the stereospecificity of enzymes for their substrates is determined mostly by the pre‑existing geometry of the active site. But the mechanism of action (i.e., how an enzyme effects the change from substrate to product) may result from alterations in the enzyme’s tertiary or quaternary structures.

The active site of an enzyme is the place, usually a cleft on the enzyme surface, where a substrate binds and is converted to product. The binding forces holding an enzyme and substrate together can be ionic, electrostatic, hydrophobic, or some combination. Less commonly, these forces are covalent. Covalent bonds are less common because covalent forces require more energy, and also because release of product is more difficult. If a product cannot be released, the enzyme is unavailable for further catalytic action.

As mentioned in Lesson 1 the general definition of a catalyst is a substance that alters the rate of a reaction, but is not changed in the process, and this is the case for enzymes. Frequently, an enzyme will undergo some change (e.g., a redox change) during the conversion of substrate to product, but always the final step will be a return to the original enzyme form. Suicidal enzymes are rare!

Coenzymes are found in the active site of most, but not all, enzymes. The coenzyme may be a metal ion (e.g., Zn2+ in carbonic anhydrase), a metal ion complex (e.g., the heme moiety in cytochrome c), or a small organic group (e.g., nicotinamide adenine dinucleotide [NADH] in alcohol dehydrogenase). Coenzymes exist to make an “active site” more chemically reactive. The amino acids that make up proteins are polar or non‑polar, or are organic ions. Metal ions or redox species such as NADH are more active—both in binding properties and in reactivity—than are any of the amino acids. The forces holding a coenzyme in place can be ionic, electrostatic, hydrophobic, covalent, or any combination of these types. A covalently bound coenzyme is called a “prosthetic group.” A noncovalently bound coenzyme is called a “cosubstrate.” This distinction is particularly important during the regeneration of the coenzyme for another round of catalysis.

A substrate is an organic biochemical starting material which is converted to product biomolecules by enzyme action. As noted above, the substrate fits closely in ionic charge, three‑dimensional structure, and hydrophobicity with the active site of its enzyme. However, an interesting feature is that the fit between enzyme and substrate, while close, is rarely perfect.

If we think closely about what an enzyme does, this “non‑perfect” fit makes sense. The product must be released speedily for the enzyme to qualify as a catalyst. If the enzyme and substrate are a perfect match, it is energetically more difficult to get release of product. In fact, some biochemical inhibitors fit better into the active site than the substrate. These “inhibitors” are not released, and so prevent binding and conversion of the actual substrate. (Malonate as an inhibitor of succinate dehydrogenase is a good example.)

An inhibitor may also bind to an area of the enzyme distant from the active site, and affect the active site indirectly. An inhibitor may bind to a metal ion coenzyme to prevent its redox cycling. Or, an inhibitor may alter the charge or geometry of an active site. There are almost as many ways an inhibitor can inhibit as there are chemical binding mechanisms.

Look at the example of chymotrypsin in the textbook (pages 85–87). This digestive enzyme is an important component in pancreatic juice for the breakdown of proteins and polypeptides. This enzyme is a protease that cleaves specific peptide bonds, in particular those adjacent to certain amino acids, one of these being phenylalanine. The substrate binding site is in a region of the protease known as the S1 pocket. Preferred substrates include hydrophobic amino acids like phenylalanine. Binding of phenylalanine or another hydrophobic amino acid induces a conformational change in the enzyme, in particular of three amino acids in the S1 pocket (aspartic acid, histidine, and serine) in the active site, referred to as the catalytic triad.

The shift in position of the negatively charge aspartic acid towards histadine favours the movement of a proton from the serine hydroxyl group. This results in a reactive alkoxide ion, which performs a nucleophilic attack on the peptide bond on the carboxyl side of phenylalanine, causing the peptide bond to break. One side is freed, but the phenylalanine is covalently‑linked to the serine, so a second reaction involving water is needed to release the phenylalanine‑serine bond.

More detail about the reaction is available in the textbook, but it is important in reading this commentary that you note the importance of specificity in enzyme function and can demonstrate an understanding of the detailed steps involved in catalysing the reaction. Imagine that a protease inhibitor would likely bind the S1 pocket of chymotrypsin more readily than a hydrophobic amino acid, which would result in the inability of this enzyme to catalyze peptide bond breakage and therefore act in protein digestion.

Enzyme Classification

This section of the commentary is to provide some information on how enzymes are classified. To bring some order into the confusion that surrounded enzyme naming, the International Union of Biochemistry established an Enzyme Commission to recommend systematic nomenclature. All enzymes are now given an Enzyme Commission (EC) number, a systematic descriptive name, and a trivial name. When writing or talking about an enzyme, a biochemist lists all three descriptors. We will use EC 2.7.3.2, (ATP:creatine N‑phosphotransferase), creatine kinase as an example:

Enzyme Commission (EC) number is EC 2.7.3.2
Systematic descriptive name is ATP:creatine N‑phosphotransferase
Trivial name is creatine kinase

The reaction catalyzed by creatine kinase is:

ATP+creatine  → creatine kinase  ADP+phosphocreatine

In an EC designation, the first number indicates the kind of reaction catalyzed by the enzyme. There are six classes:

  1. oxidation‑reduction
  2. transfer of a functional group from one molecule to another
  3. hydrolysis
  4. elimination of functional groups (usually and ) to form a C$\ce{=}$C double bond
  5. isomerization (very rare)
  6. formation of a bond with ATP hydrolysis providing the energy (i.e., X + ATP → Y + ADP + PO43-)

Each of the six classes is divided into subclasses. As we have seen, the first number indicates the kind of reaction; the second number indicates the enzyme’s first subclass (see below); the third number indicates its second subclass; and the final number is an arbitrary but unique serial number within the second subclass. This is summarized below:

EC 1st number: the kind of reaction catalyzed by the enzyme
EC 2nd number: (1st subclass) the atoms or bonds acted on by the enzyme
EC 3rd number: (2nd subclass) the enzyme type that catalyzes the specific reaction on the specific type of atom or bond
EC 4th number: an arbitrary but unique serial number for each enzyme within subclass 2

The systematic, descriptive name should adequately describe the reaction that is catalyzed. The trivial name is short and appropriate for everyday use. The trivial name is arbitrary—you can name your newly discovered enzyme after your dog if you like, although this is discouraged—and protocol asks that you end the trivial name with ‑ase. Having done all this, you have unambiguously identified the new enzyme. Now you are free to use only the trivial name (creatine kinase in the example).

Enzyme Catalysis

We know that the bonding forces holding substrate and enzyme together are ionic (electrostatic), H$\ce{-}$bonding, hydrophobic, or some combination of these (see Lesson 1). Covalent forces may also be involved. Since an enzyme, by definition, increases the rate of reaction, the binding forces are an important part of the transformation process. The methods by which enzymes achieve the transformation of substrate to product are divided into six categories that reflect these bonding forces:

  1. Acid‑base catalysis, which involves the transfer of a hydronium ion to or from the substrate, so that bond breaking or rearrangement within the substrate is facilitated. Note that the enzyme must be returned to its original form by the end of the reaction, so the hydronium ion transfer is temporary. A good example of this type of process is hydrolysis of RNA by RNase A. Two his residues in the active site of RNase A act as both proton donor and proton acceptor, in cleaving the RNA.
  2. Electrostatic or ionic catalysis. Electrostatic interactions participate in enzymatic catalysis in two ways:
    1. A charged substrate can be guided to the active site by the charge distribution of the enzyme’s surface amino acids. A good example of this type of process is the anion superoxide interacting with positively charged amino acids on the surface of the enzyme superoxide dismutase.
    2. An electrostatic interaction occurs between substrate and the amino acids within the active site. Electrostatic interactions are considerably stronger in the hydrophobic pocket formed in the active site than they are in aqueous solution. Thus, the chemical reactivity of the substrate is enhanced.

  3. Metal ion catalysis (very common). Transition metal ions within the active sites of enzymes are strongly bound to the enzyme. Alkali and alkaline earth metal ions are more loosely bound to the enzyme. Both types of metal ions serve to orient a substrate within the active site, to neutralize a negative charge on the substrate, or both. Metal ions also electrostatically shield or stabilize negative charges on reaction intermediates. Metals that can undergo redox changes (e.g., Mn2+ and Fe3+) usually act as redox catalysts. A good example of this last class is the selenium‑containing enzyme glutathione peroxidase, which catalyzes the redox reaction:

    glutathione (reduced) + organic peroxide → glutathione (oxidized) + alcohol + water

  4. Proximity and substrate orientation catalysis. This type of catalysis is important for bimolecular reactions. The three‑dimensional nature of the active site of an enzyme allows two substrates to be oriented properly, and brought into closer proximity than is possible in bulk solution. Thus, the reaction will occur more efficiently than in bulk solution, where reactant‑reactant contact is random.

  5. Covalent catalysis. The formation of a temporary covalent bond between substrate and enzyme enhances the reactivity of a portion of the substrate which is otherwise of low reactivity. Since substrates and enzymes are organic molecules, it is not surprising that such reactions are nucleophilic (or electrophilic):

    S + E → SE

    promoting a subsequent electrophilic (or nucleophilic) conversion to product:

    SE → P + E

    A good example of this type of process is the conversion of acetoacetate to acetone via a Schiff base intermediate. See the Wikipedia page Acetoacetate decarboxylase for an illustration of the mechanism.

    This reaction occurs when diabetic individuals, lacking insulin, break down lipids rather than sugars for energy. As you will see in Unit 10, the lipid product, acetoacetate, is produced at a faster rate than it can be metabolized, and so a side reaction produces acetone.

  6. Transition state binding. The charge distribution on a substrate roughly complements the charge distribution on the enzyme’s active site. If the substrate is relatively large, this not‑quite‑perfect match distorts the substrate for a better charge fit. The distorted substrate, called the transition state, is more readily cleaved to product than the substrate itself. A good example of this type of process is lysozyme cleaving polysaccharides. Unless the polysaccharide is a hexamer or larger, the interaction between enzyme and substrate is insufficient to effect polysaccharide cleavage.

Study Questions

  1. What is the active site of an enzyme? What forces are involved in the interaction between an enzyme and its substrate? Why are covalent interactions not usually involved?
  2. Distinguish between a coenzyme, a prosthetic group, and a cosubstrate.
  3. List the six mechanisms of enzyme catalysis.
  4. Leucine aminopeptidase is an enzyme that catalyzes the hydrolysis of peptide bonds. The data below indicates the relative rate of hydrolysis of three different dipeptides. Explain the differences in rates of hydrolysis.

    Substrate

    Relative Rate of Hydrolysis

    L‑leu‑L‑leu

    100

    D‑leu‑gly

    0

    gly‑L‑leu

    10
  5. Malonate is an inhibitor of succinate dehydrogenase. What reaction does succinate dehydrogenase catalyse? Why would malonate be an inhibitor of this reaction?
  6. Superoxide dismutase has Cu2+ as a coenzyme. Look up the formula for superoxide (the substrate for this enzyme), and explain why copper is an appropriate choice for coenzyme. What does “dismutase” mean?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 3: Enzyme Kinetics

Overview/Objectives

After completing this lesson, you should be able to:

  1. Describe the effect of an enzyme on the energetics of a biochemical reaction.
  2. Sketch a plot, for a Michaelis‑Menten enzyme, of the rate of product production vs. concentration of substrate, and explain its shape.
  3. Define the enzyme constants KM and Vmax.
  4. Explain the relationship between the KM of an enzyme and its affinity for its substrate.
  5. Explain why the kinetic analysis is important in the study of enzymes.
  6. Use the Michaelis‑Menten equation to determine KM and Vmax from experimental data.
  7. Explain the significance of KM and Vmax in terms of an enzyme’s ability to catalyze a biochemical reaction.
  8. Identify a non‑Michaelis‑Menten enzyme from the plot of (1/Vmax ) vs. (1/[S]) (double reciprocal or Lineweaver‑Burk plot).

Readings and Activities

  1. Read “Enzyme Parameters,” “Vmax and Kcat,” “Km,” “Perfect Enzymes,” and “Lineweaver-Burk Plots” (pages 87–91 of the textbook).
  2. You can also watch the four video lectures on Enzyme Catalysis:

Commentary

Enzymes + Substrate

Enzyme kinetics is the study of the rates of biochemical reactions. An understanding of enzyme kinetics sheds some light on how substrates bind, how inhibitors work, and how the conversion of substrate to product occurs. We should note right at the beginning of this lesson that enzymes cannot change the direction of a reaction, only the speed with which it occurs.

The general form of a biochemical reaction is as follows:

E + S ↔ ES → P + E

Writing the reaction in this way indicates that the substrate (S) forms some kind of activated complex (called a transition state or ES) with an enzyme. When product (P) is released, the enzyme reverts to its original form and is available to catalyze the reaction of a second molecule of substrate. The diagram below indicates that the substrate is at a higher free energy than the products, but that there is an energy barrier preventing the conversion of substrate to product. An enzyme, by forming an “activated complex” with the substrate (ES), reduces the barrier between substrate and product.

As an example, consider the last donut you ate. It was converted by enzymes to CO2 and H2O (well, maybe a little fat, but that is an issue for a later unit). Donuts sitting in a box or even dropped into coffee do not turn before your eyes to carbon dioxide and water. The carbohydrates that make up the donut are at a higher free energy than CO2 and H2O, but are quite stable in the absence of an enzyme because of the energy barrier between donut and carbon dioxide/water. You have the enzymes to reduce the energy barrier (as in the above graph) and convert the donut to CO2 and H2O.

Rates of Reaction

At low concentrations of substrate, the rate of a biochemical reaction is proportional to the amount of substrate available. However, because enzymes are at much lower concentration than substrate (at least 10–30 times lower) in vivo, it is theoretically possible to have all the enzyme molecules in the ES form. When this occurs, no matter how much substrate is added to a solution, the reaction cannot proceed any faster. The rate determining step will become how fast ES can dissociate into product. This situation is shown in the diagram below.

The shape of this graph (i.e., the rate of product formation vs. the concentration of substrate) will be similar for all enzymes. However, the numbers on the axes will be different. Some enzymes (such as carbonic anhydrase) turn substrate into product so quickly that it is virtually impossible to saturate them (i.e., reach the flat part of the graph). Some enzymes (such as superoxide dismutase) are very slow. The object of studying enzyme kinetics is to find a way to characterize the enzymes more exactly than “very fast” or “moderately slow.” Michaelis and Menten worked out a way to do this. We will now introduce two useful expressions derived from the Michaelis‑Menten equation below.

Vmax and KM

Michaelis‑Menten kinetics produce two numbers to characterize an enzyme: Vmaxand KM. Vmax is the maximal rate of product formation catalyzed by the enzyme in question. KM is the substrate concentration when the rate of reaction is 1/2 maximal. KM can also be thought of as the concentration of substrate that will half saturate the enzyme. When a low concentration of substrate saturates the enzyme, the enzyme is said to have a “high affinity” for that substrate. If a high concentration of substrate is needed to saturate the enzyme, then the enzyme is said to have a “low affinity” for its substrate. “So what,” you say? A few study questions will illustrate the biochemical consequences of enzymes with different values for Vmax and KM.

Enzyme Kinetics

Some students find the mathematics of enzyme kinetics a bit difficult. It is not really important to work through the mathematics, unless you are planning to be an enzymologist. What is important is that you understand why the mathematics is important. The mathematics provides two constants, KM and Vmax, which characterize an enzyme’s kinetic abilities. To understand the significance of these constants, and to find out how to calculate them, it is necessary to work through some of the mathematics. Keep in mind that the mathematics is not an end in itself, but a means of calculating enzyme constants.

The textbook does a really good job of simplifying the mathematics to identify exactly how these constants are derived. It also offers an excellent explanation of the Lineweaver-Burk plot. It is important to understand the meaning of KM and Vmax as well as Kcat. It is also important to understand the difference between the Michaelis‑Menten and Lineweaver‑Burk graphs.

The rate of reaction for the simple S → P reactions graphed earlier is similar to that of any kinetic reaction:

This equation means that, for any reaction, there is a tendency for the substrate to form an ES complex. The rate at which this formation will occur is described by the rate constant, k1. Once ES is formed, it can dissociate back into substrate and free enzyme (described by the rate constant, k-1), or it can go on to product and free enzyme (described by the rate constant, k2). Enzyme kinetics differs from general kinetics in ignoring the back reaction of P + E to give ES, because such a reaction is so unlikely.

It is not easy to measure rate constants directly. The only things that are easy to measure experimentally are substrate concentration [S] and the initial rate of reaction (V0) at that concentration. (A rate of reaction would be, for example, how many mMoles of product are produced per second.) So, Michaelis and Menten worked out a rate equation which involved only constants (Vmax and KM) and easy‑to‑measure variables (V0 and [S]):

v 0 = V max  [S] K M + [S]

It is difficult to solve this form of the Michaelis‑Menten equation directly to determine the values of KM and Vmax, because if one plots initial rate of reaction (V0) vs. substance concentration [S], the curve is hyperbolic. It is thus difficult to extrapolate constant values and behaviour of enzyme‑substrate reactions. However, if we take the reciprocal of this equation, it is relatively easy to understand the type of reaction that occurs: reciprocal values of a hyperbolic curve yield a straight line. The reciprocal values of the equation above are:

1 v 0 = K M V max   1 [S] + 1 V max

that is, y = mx + b

But y = mx + b is the general equation for a straight line.

If the enzyme is a true first‑order enzyme (i.e., a true Michaelis‑Menten enzyme) the plot of (1/V0) vs. (1/[S]) gives a straight line, just as a plot of (y vs. x) gives a straight line. This reciprocal graph is called a “Lineweaver‑Burk plot.” The (1/V0) intercept is (1/Vmax) and the (1/[S]) intercept is −(1/KM). Graphing sets of experimental data (1/ V0 and 1/[S]) allows the direct calculation of KM and Vmax.

Not all enzymes obey Michaelis‑Menten kinetics. The mathematics of non‑Michaelis‑Menten enzymes will not be discussed in this course. However, you should be able to recognize from a Lineweaver‑Burk plot what is and what is not a Michaelis‑Menten enzyme. A true first‑order enzyme will always produce a straight‑line plot; a non‑Michaelis‑Menten enzyme will produce convex or concave curves when converted to double‑reciprocal Lineweaver‑Burk plots. A Michaelis‑Menten plot is shown on page 87 of the textbook. A Lineweaver‑Burk plot is shown on page 90.

Study Questions

  1. Consider the three graphs below. Assume that the units of the axes are the same for each graph, and note that Vmax and KM are marked for each enzyme.

    #

    1. Suppose the system is suddenly flooded with S. If the biochemical object is to get rid of S as quickly as possible, which enzyme would be best?
    2. If E2 and E3 were both competing for the same S (i.e., S is the same for both, but the identity of P will be different), which enzyme will capture virtually all of the S under normal physiological conditions?
    3. If the biochemical object is to remove S, but not to overload the system with P, which enzyme is best?
  2. Consider the experimental data obtained for the hydrolysis of penicillin by β‑lactamase. (β‑lactamase is the enzyme that penicillin‑resistant bacteria use to inactivate this drug.)

    Penicillin (M)

    Amount hydrolyzed (moles/min)

    0.1 × 10–5

    0.11 × 10–9

    0.3 × 10–5

    0.25 × 10–9

    0.5 × 10–5

    0.34 × 10–9

    1.0 × 10–5

    0.45 × 10–9

    3.0 × 10–5

    0.58 × 10–9

    5.0 × 10–5

    0.61 × 10–9

    Is β‑lactamase a Michaelis‑Menten enzyme? What are the values of KM and Vmax for β‑lactamase? Which portion of this substrate concentration range is most useful for calculating these values? Why?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 4: Enzyme Inhibition

Overview/Objectives

After completing this lesson, you should be able to:

  1. Describe how competitive, noncompetitive, and uncompetitive enzyme inhibitors affect enzyme activity.
  2. Determine, from Lineweaver‑Burk (or double‑reciprocal) plots, whether an enzyme inhibitor is competitive, noncompetitive, or uncompetitive.
  3. Determine, from Lineweaver‑Burk (or double‑reciprocal) plots, the inhibitor constant, KI.

Readings and Activities

  1. Read “Enzyme Inhibition,” “Competitive Inhibition,” “No Effect on Vmax,” “Increased KM,” “Non‑Competitive Inhibition,” “Uncompetitive Inhibition,” and “Suicide Inhibition” (pages 91–96 of the textbook).

Commentary

Enzyme activity can be altered by a number of factors: pH, temperature, salt concentration, chemicals competing with the substrate for the active site, or substances which alter enzyme conformation. In vivo, the latter two effects are the most important, and so they are considered in this lesson.

Inhibitors are of three types: competitive, non‑competitive, and uncompetitive. These three can be distinguished by their effect (or lack of effect) on the KM and Vmax of an enzyme.

Competitive inhibitors

Competitive inhibitors usually resemble the normal substrate chemically. The inhibitor and normal substrate compete for the active site; hence, a competitive inhibitor acts by reducing the concentration of free enzyme available to bind the substrate. In the presence of large amounts of substrate, the effect of the inhibitor is negligible, because it is swamped by the excess substrate. (The enzyme is statistically more likely to interact with the chemical in highest supply.) Competitive inhibition is, therefore, indicated when KM is increased and Vmax is unaffected. Note the figure showing competitive inhibition on page 91 of the textbook.

Non-competitive inhibitors (mixed inhibitors)

Non‑competitive inhibition results when the inhibitor binds irreversibly to the active site of an enzyme. No amount of added substrate will dislodge the bound inhibitor. Non‑competitive inhibition may be recognized when KM is unaffected and Vmax is decreased.

Pure non‑competitive inhibition is rare. A figure of non‑competitive inhibition is shown on page 94 of the text.

Uncompetitive inhibitors

When an inhibitor binds to ES but not to E, the inhibition is uncompetitive. The catalytic activity of the enzyme is reduced, and the enzyme is locked in its enzyme‑substrate complex, thereby increasing the apparent substrate binding. This type of inhibition may be recognized when KM is reduced and Vmax is decreased.

As the readings show, KM varies according to the enzyme used. KM values also depend on the substrate used, and on factors such as temperature and pH.

The capacity of an inhibitor to dissociate from an enzyme can be determined from the inhibition constant, KI. To determine KI easily, we calculate the slope of a Lineweaver‑Burk plot of inhibitor action:

slope= K M V max  1+  [I] K I

Once we have determined the slope, we can plot slope vs inhibitor concentration. The point at which the line intercepts the x‑axis is −KI.

Study Questions

  1. How are Vmax and KM affected for each of the types of enzyme inhibition: competitive, non-competitive, and uncompetitive?

  2. Indicate the type of inhibition occurring in each of the three graphs below and justify your answer.

    #

    Adapted from: Enzyme Inhibition lineweaver-burk plots.gif, Bizz1111, 2016, CC0.

  3. Draw Michaelis-Menten plot of an enzyme in the presence of a competitive and a non-competitive inhibitor.
  4. What is suicide inhibition? Give an example.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 5: Control of Enzymes

Overview/Objectives

After completing this lesson, you should be able to:

  1. Explain the difference between Michaelis‑Menten enzymes and allosteric enzymes.
  2. Describe the value of allosteric enzymes as control points in metabolism.
  3. Sketch graphs illustrating the effect of allosteric regulators on an enzyme’s allosteric activity.

Readings and Activities

  1. Read “Control of Enzymes,” “Allosterism,” “Covalent Control of Enzymes,” “Other Enzyme Control Mechanisms,” and “Ribozymes” (pages 96–100 of the textbook).
  2. You can also watch three video lectures on Enzyme Regulation:

Commentary

Normally, substrate concentrations are maintained at relatively low levels in vivo. Look at the graph of rate of product formation vs. concentration of substrate in Lesson 4. (This graph describes Michaelis‑Menten enzymes.) Note that at low [S], the rate of product formed is directly proportional to [S]. This relationship makes sense if the biochemical object is to prevent the accumulation of substrate. But in a complex system, in which each product is the substrate for the next reaction, how can overall control be exerted? In other words: if the final product of a multistep process is not to be used immediately, how can the body prevent the earlier reactions from continuing to overload the system with the final product?

There is a class of enzymes whose catalytic activity, unlike that of Michaelis‑Menten enzymes, can be regulated by factors other than substrate concentration. These are the “allosteric enzymes.” Allosteric enzymes are always multimeric. Each linked set of reactions in vivo (metabolic pathway) will usually contain two allosteric enzymes: one near the beginning of the sequence of reactions and one at the end. The allosteric enzyme’s ability to bind substrate is affected by the presence or absence of a third molecule, often its own product or a product further down the pathway. A set of graphs of rate of reaction vs. [S] for an allosteric enzyme is shown below.

Graph 1 looks just like a Michaelis‑Menten enzyme, while Graphs 2 and 3 indicate that something is causing slower product formation for the same [S]. Look at the example given in the textbook (page 97); that is, the conversion of carbamoyl phosphate to N‑carbamoyl aspartate. This process is the first committed step (i.e., a step in which the product cannot be used for anything else) in the six‑step synthesis of the pyrimidine, cytosine triphosphate. The first committed step of a multistep reaction is the usual place to find an allosteric enzyme. The allosteric enzyme controlling the reaction of carbamoyl phosphate and aspartate is ATCase:

carbamoyl phosphate + asparate  → ATCase  N−carbamoyl asparate

Look again at the graph on page 97 of the textbook.

Line 2 (green) Only the substrates and the enzyme are present.

Line 3 (blue) There is decreased production of product, for a given [S], compared with Graph 2. Excess pyrimidine end product (CTP) is present. CTP will bind to ATCase, lowering the affinity of the enzyme for aspartate and carbamoyl phosphate. Therefore, less N‑carbamoyl aspartate (product) will be produced.

Line 1 (purple) There is increased production of product, for a given [S], compared with Graph 2. Large amounts of ATP (a purine derivative) are present. ATP can also bind to the enzyme. This factor increases ATCase’s affinity for substrate, and so more product will be formed.

Although there are many allosteric enzymes, not all of them are well understood. However, it seems likely that all are influenced by external molecules that affect the active site. These effects are probably indirect, occurring through changes in the quaternary structure of the enzyme.

Study Questions

  1. List and define the four general mechanisms of enzymatic control. Are some enzymes controlled by more than one mechanism?
  2. Outline the allosteric control of the enzyme HMG‑CoA reductase, which is involved in the synthesis of cholesterol.
  3. What is feedback inhibition and what type of enzymatic control is it? Give an example of feedback inhibition of an enzyme.
  4. What is a zymogen? What kind of enzymatic control operates here? Why is this beneficial?
  5. What is a ribozyme?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Answers to Unit 4 Study Questions

Lesson 1: Activation Energy

  1. Activation energy is the reduced energy barrier of a reaction (also called free energy of activation) to reach its transition state. Enzymes do not reduce the overall energy of a reaction, but they lower the activation energy of a reaction. This means that the molecules can more easily reach the energy necessary for the reaction to occur. The transition state is the substrate-enzyme activated complex. This state corresponds to the highest potential energy of a reaction or the state at which the reaction will be driven to reactants or products depending on the energy levels of each.
  2. A catalyst does not change the overall energy of a reaction and any reaction will proceed to equilibrium given enough time. A catalyst lowers the activation energy of a reaction and helps the reaction proceed so it lowers the activation energy of a reaction.

Lesson 2: Enzyme Catalysis

  1. The active site of an enzyme usually a cleft on the enzyme surface where the substrate binds and is converted to product. The binding forces holding an enzyme and substrate together are ionic, electrostatic, hydrophobic, or a combination of these. Covalent bonds may occur, but are not common because these bonds require more energy and also because it would be harder to release the product and the enzyme would not be able to catalyse another reaction.
  2. A coenzyme is molecule that makes the active site of an enzyme more chemically reactive. A coenzyme may be a metal ion or a metal ion complex or a small organic group. A prosthetic groups is a covalently bound coenzyme. A cosubstrate is a noncovalently bound coenzyme.
  3. The six mechanisms of enzyme catalysis are: acid-base catalysis, electrostatic or ionic catalysis, metal ion catalysis, proximity and substrate orientation catalysis, covalent catalysis, and transition state binding.
  4. Leucine aminopeptidase hydrolyzes the peptide bond on the amino end of a leu. It will act most efficiently at cleaving the peptide bond between two leu residues. Note that the difference in rates can also be due to the small size of gly in gly‑L‑leu. It is possible that leu aminopeptidase requires a larger amino acid as the carboxylic side of the peptide bond. What additional dipeptide would you use to check this possibility? D‑amino acid is not recognized as a substrate by the enzyme; hence the zero rate.

  5. Succinate dehydrogenase catalyzes the conversion of succinate to fumarate. Succinate contains a C$\ce{-}$C single bond and fumarate contains a C$\ce{=}$C double bond. The enzyme removed two hydrogen atoms, hence “dehydrogenase.” Malonate has a similar structure to succinate, and so will fit in the active site of succinate dehydrogenase. However a C$\ce{-}$C bond from which hydrogen atoms can be abstracted does not occur in malonate.

  6. Superoxide is an anion‑radical: O2−. Binding between positively charged copper and negatively charged superoxide seems reasonable. In addition, copper is a metal capable of undergoing oxidation‑reduction reactions. “Dismutase” means that two superoxide anion‑radicals are involved in the reaction. One is reduced to hydrogen peroxide and one is oxidized to molecular oxygen. Copper (or iron) would be reasonable coenzymes for superoxide dismutase.

Lesson 3: Enzyme Kinetics

  1. For the graphs of enzyme kinetics shown:
    1. Either E1 or E2 would get rid of S. “Flooded” indicates high concentrations of S, that is, the right‑hand side of the graph. Since both E1 and E2 have the same (high) Vmax both would be better than E3 which has a lower Vmax.
    2. E3, which has a lower than E2, will capture virtually all the substrate. Note that physiological concentrations of substrate are normally low.
    3. E2 would be best for removing S without overloading the system with P. This enzyme has the lowest affinity for S of the three enzymes (i.e., the highest KM). Therefore, the slowest rate of production of P will result.
  2. For these data, the plot of 1/v vs. 1/[S] is linear; therefore, β‑lactamase displays Michaelis‑Menten behaviour. KM = 5.2 × 10–6 M; Vmax = 6.84 × 10–10 moles/min. The lowest concentrations of S are the most useful for these calculations, because the reciprocal of the Michaelis‑Menten equation (1/v vs. 1/[S]) is the plot actually used to calculate KM and Vmax. The lower the value of [S], the greater its prominence in the reciprocal plot.

Lesson 4: Enzyme Inhibition

  1. Competitive – KM increased, Vmax unaffected. Non‑competitive – KM unaffected, Vmax reduced. Uncompetitive – KM reduced, Vmax reduced.
  2. Graph A – competitive because KM is increased and Vmax is unaffected. Graph B – uncompetitive because KM and Vmax are both reduced. Graph C – non‑competitive because KM is unaffected and Vmax is reduced.

  3. A Michaelis‑Menten plot of an enzyme in the presence of a competitive and a non-competitive inhibitor would look like this:

    #

    Adapted from: Enzyme inhibition. In Biology, OpenStax CNX, Rice University. Creative Commons Attribution 4.0 License. https://cnx.org/contents/GFy_h8cu@9.85:MnC6GuJi@7/Enzymes

  4. Unlike the other types of inhibition, suicide inhibition involves irreversible binding of the inhibitor to the enzyme by covalent bonding. Like competitive inhibition, the suicide inhibitor usually resembles the substrate; but, unlike competitive inhibition, the suicide inhibitor is chemically reactive in the active site and remains there due to the formation of a covalent bond. An example of a suicide inhibitor is the antibiotic penicillin. It binds the bacterial enzyme D‑D transpeptidase, which is needed for cell wall synthesis (peptidoglycan specifically), and this binding stops the transpeptidase from working. This is how the antibiotic is effective. It stops the activity of the enzyme and the bacterial cell cannot divide and reproduce because it cannot synthesize the cell wall.

Lesson 5: Control of Enzymes

  1. The four mechanisms of enzymatic control are:
    • Allosteric control (allosterism): The small molecules that bind the enzyme and affect its activity are not substrates, and they bind in a location separate from the active site. These molecules may activate or inhibit an enzyme.
    • Covalent modification: when enzymes are synthesized in an inactive form and activation of the enzyme requires that covalent bonds in them be cleaved
    • Access to the substrate: also called substrate-level control. This involves inhibition of the enzyme by its product.
    • Control of enzyme synthesis/breakdown: This type of control occurs at the level of gene expression (transcription). This can be through signal transduction resulting in a transcriptional activator binding to the promoter region of the gene encoding the enzyme, or through the binding of other activators that induce transcription of the gene encoding a particular enzyme.

    Yes, some enzymes may be controlled by more than one mechanism.

  2. HMG‑reductase is allosterically controlled by cholesterol. Binding of cholesterol inhibits the enzyme. Cholesterol is an end product of the pathway, but is not a direct substrate for this enzyme. Therefore an abundance of cholesterol inhibits further synthesis of cholesterol by binding to and inhibiting an enzyme that is active earlier in the synthesis pathway.
  3. Feedback inhibition is where a product of a synthetic pathway binds to and inhibits an enzyme early in the pathway to prevent further synthesis of this product if it is not needed. Feedback inhibition is a form of allosteric control. An example of feedback inhibition is the allosteric control of HMG‑reductase by cholesterol and another example is the binding of the nucleotide CTP to the enzyme aspartate transcarbamoylase (ATCase).
  4. A zymogen is an inactive form of an enzyme. Sometimes enzymes are synthesized in an inactive form and they must be activated by the cleavage of some covalent bonds. This is called covalent modification of enzymes. This is beneficial as a way to control enzymes and keep them off until they are needed.
  5. Ribozymes are a special form of enzymes. Unlike most enzymes, ribozymes are not proteins. They are RNAs. A ribozyme is a catalytic RNA capable of cutting RNA (self-splicing). Ribosomes can also be considered to be ribozymes because they catalyse the formation of peptide bonds.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.