BIOCHEMISTRY AND MOLECULAR BIOLOGY Problem Unit Two ...

SIU School of Medicine

BIOCHEMISTRY

Enzymes/Membrane Transport

BIOCHEMISTRY AND MOLECULAR BIOLOGY Problem Unit Two 1999/2000 Enzymes/Membrane Transport Copyright 1999, E.C. Niederhoffer. All Rights Reserved. All trademarks and copyrights are the property of their respective owners.

Faculty: P.M.D. Hardwicke

Module 1:

Enzyme Kinetics

Module 2:

Clinical Enzymology

Module 3:

Membrane Transport Problem Unit 2 - Page 1


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Faculty:

Dr. Peter M.D. Hardwicke Biochemistry & Molecular Biology Office: 210 Neckers Bldg. email: [email protected] Telephone: 4 5 3 - 6 4 6 9

Learning Resources:

ESTIMATED WORK TIME: 40 hours. A. This study guide is provided in two forms: printed and electronic (cowritten and produced by Dr. E.C. Niederhoffer, Biochemistry and Molecular Biology). It is best viewed in electronic form as a pdf file which can be read on your computer using Adobe Acrobat Reader. See Appendix I for an introduction on how to view a pdf file. The pdf file can be downloaded from the biochemistry server (http://www.siu.edu/departments/biochem) and Acrobat Reader can be downloaded free from Adobe’s web page (http:// www.adobe.com/acrobat). They should also be installed on the student computers. There are a number of advantages to using the electronic version including color, a hypertext index, and hypertext links within the text. Hypertext links in the text body are in blue underlined characters (such as this). Clicking on these will lead to a jump to the linked material for further details. The destination material is indicated by red underlined characters (such as this). (Clicking on the black double arrows in the menu bar will allow you to “hyperjump” back and forth.) This and other study guides are provided to help you focus on the topics that are important in the biochemistry curriculum. These are designed to guide your studying and provide information that may not be readily available in other resources. They are not designed to replace textbooks, and are not intended to be complete. They are guides for starting your reading and reviewing the material at a later date. B. Textbooks: 1. Devlin, Textbook of Biochemistry with Clinical Correlations, 4th ed. ('97), Wiley-Liss. Core text for Biochemistry & Molecular Biology. 2. Champ & Harvey, Lippincotts Illustrated Reviews: Biochemistry, 2nd ed. (‘94), Lippincott. Efficient presentation of basic principles. 3. Murray et al., Harper's Biochemistry, (24th ed.) ('96), Appleton & Lange. An excellent review text for examina-


Problem Unit 2 - Page 2


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tions. 4. Marks, Marks, and Smith, Basic Medical Biochemistry: A Clinical Approach, (‘96), Williams & Wilkins. Good basic presentation with clinical relevance. 5. Cohn and Roth, Biochemistry and Disease, (‘96), Williams & Wilkins. A good bridge between the basic sciences and clinical medicine. 6. Garrett and Grishham, Biochemistry, 1st ed., (‘95), Saunders College Publishing. 7. Garrett and Grishham, Molecular Aspects of Cell Biology, 1st ed., (‘95), Saunders College Publishing. Most texts of biochemistry have sections on enzymes and membrane transport. The content of the subject is much the same from text to text; the differences are basically in style and rigor. The Study Guide, Pretest, and Post Test in the Problem Unit will set the level of rigor expected of you. Read the sections on enzymes and membrane transport in several texts. What differences there will be between these texts and the Study Guide will be helpful to you in gaining perspective on the subject. Additional material can be found on the web at the National Institutes of Health (http://www.nih.gov), the National Library of Medicine (http://www.nlm.nih.gov), and the free MEDLINE PubMED Search system at the National Library of Medicine (http://www3.ncbi.nlm.nih.gov/PubMed/). C. Journals/Reviews. You may find worthwhile reading in some of the more popular journals and review series (see also the searchable SIU-SOM database). These resources typically contain specific articles involving enzymes and membrane transport. Suggestions for journals include American Family Physician, Journal of Biological Chemistry, Nature, Science, and Scientific American (and SA’s Science and Medicine). Excellent reviews may be found in the Annual Review of Biochemistry, Cell and Developmental Biology, Genetics, Medicine, and Microbiology. D. Lecture/Discussions Especially recommended for those who have not had biochemistry and for those who have questions.

Evaluation Criteria:


A written examination will be scheduled. Answers to questions and the solving of problems will be judged against the learning resources. Examples of exam questions are given in the Problem Sets. The pass level is 70%. Problem Unit 2 - Page 3


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Module 1: Enzyme Kinetics Introduction:

Biochemical reactions that occur in living cells are in most respects ordinary chemical reactions. What makes these reactions unique is that they proceed very rapidly at relatively low temperatures (physiological temperature, 37°C or 98.6°F, is low when compared to the chemical reactions used in industrial processes). These low temperature reactions have accelerated rates because of the action of very efficient catalysts which we refer to as enzymes. Different cells have different numbers and types of enzymes. The exact number of enzymes in any cell remains unknown, but there are certainly thousands. There may be as many as 100,000 in some unicellular organisms. Enzymes, then, are necessary for the normal functioning of cells. Disease states may be caused by the absence or alteration of an enzyme (i.e., a genetic disease), the introduction of an enzyme inhibitor (i.e., a bacterial toxin, a chemical, etc.), the overproduction of an enzyme, or the introduction of a foreign enzyme (i.e., viral infection). Enzymes are popular therapeutic targets. Pharmaceuticals are frequently enzyme inhibitors. Enzyme assays are important in diagnosis. A diminution or increase in enzyme activities in tissues and fluids is indicative of the various causes of disease listed above. In addition, injury may release tissue enzymes to the blood and thus, their concentrations in the blood can be used to locate the site of injury. Temperature (fever) affects the rate of enzyme-catalyzed reactions, as do other physical factors. For these reasons, it is necessary to have an understanding of enzymes, how they act, how they are made, and how their activity is controlled.

Objectives:

1. After reading a passage from a medical journal, monograph, or textbook that describes a clinical investigation of changes in enzyme activities, or a clinical assay procedure, or which gives a molecular description of an enzyme, answer questions about the passage (which may involve drawing inferences or conclusions) or use the information given for the solving of a problem. Examples of the kinds of passages about which understanding is required are found within this Study Guide. Examples of the types of questions and problems to be solved are included in the Problem




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Sets. In order to accomplish objective 1, you will need to be able to do the following, which are also objectives: 2. Define the terms in the Nomenclature and Vocabulary list and use them properly in answering questions concerning this module.

Nomenclature and Vocabulary:

Activation energy Active site Binding site Catalytic site Cofactor Dissociation constant Enzyme Enzyme-substrate complex Free energy of reaction Isoenzyme kcat

Activator Allosteric enzyme Catalyst Coenzyme Competitive inhibition Enzyme assay Enzyme inhibitor First-order reaction Induced fit Isozyme Km

Lineweaver-Burke plot

Maximal velocity (Vmax)

Mechanism-based inhibitor

Michaelis constant (Km)

Michaelis-Menten equation Monomeric enzyme Noncompetitive inhibition Ordered Product inhibition Protomer Saturation velocity Specific activity Substrate Suicide substrates Unit of enzyme activity

Microsomal enzyme Negative allosteric effector Oligomeric enzyme Positive allosteric effector Prosthetic group Random Sequential mechanism Steady state Subunit Turnover number Vmax

Vitamin Zymogen

Zero-order reaction

3. Enzymes have systematic and trivial names that must be learned. Describe the action of given enzymes from their systematic names and the action of those enzymes that have the more obvious trivial Faculty: P.M.D. Hardwicke



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names (i.e., alkaline phosphatase, urease, ribonuclease). [The action of those enzymes with less obvious trivial names (i.e., aldolase, enolase, trypsin) will have to be picked up as your studies proceed.] 4.

Explain the concept of the active site of an enzyme.

5. Discuss the effects of the following on enzyme activity and understand the origin of the relationships in terms of enzyme catalyzed reactions: Activators Coenzymes Competitive inhibitors Enzyme concentration Irreversible inhibitors Noncompetitive inhibitors pH Products Substrate concentration Temperature 6. Discuss the criteria that must be met for a valid enzyme assay and for using an enzyme as a reagent. 7. Describe the construction and interpretation of a linearized form of the Michaelis-Menton equation (Lineweaver-Burk plot) and be able to evaluate the pertinent constants from it. 8. By plotting v against [S], show the relationship of brain hexokinase and liver glucokinase to the normal blood glucose concentration. For this plot assume the Vmax values are identical for the two enzymes. 9. Discuss the rationale for design of mechanism-base inhibitors and their therapeutic applications. 10. Use the material covered on the above objectives and previous objectives to solve new, related problems such as those given in the Prestest and Post Test.

Key Words:


Binding sites Coenzymes Enzyme inhibitors Kinetics

Biochemistry Enzyme activation Enzymes



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Pretest: A Pretest is available for those interested in assessing their current knowledge base and conceptual understanding. You may hyper-jump to Study Guide-1. 1.

Serum alkaline phosphatase is a phosphomonoesterase. The pH optimum of its activity is at pH 9.0. Free -SH groups inhibit the activity of this enzyme, while magnesium ions enhance it. There are two possible ways of measuring the activity of phosphatase: first, determination of phosphate liberated by the enzyme per unit of time, and second, determination of the organic part of a phosphoric acid ester used as substrate, which is liberated under the same conditions.

Sample questions: a. What can be inferred about this enzyme from the terms alkaline phosphatase and phosphomonoesterase? answer b. The anticoagulants EDTA and citrate may not be used in this assay. Why? answer c. Must the reaction be done at exactly pH 9.0? Why? answer 2.

Skeletal muscle aldolase, which catalyzes the reverse aldol condensation of D-fructose-1,6-diphosphate into dihydroxyacetone phosphate and D-glyceraldehyde-3-phosphate and vice versa, has a molecular weight of 150,000 and contains four major subunits. Acidification of the enzyme causes it to dissociate into the subunits, which are inactive; upon neutralization, the subunits reassemble quickly and spontaneously to reform the active enzyme. Aldolase contains free -SH groups, some of which are essential for catalytic activity. Sample questions: a. Enzymes with multiple subunits are _____________ enzymes. answer b. With this enzyme, an intact ___________ structure is necessary for activity. answer c. From the information given, what can be inferred about the active site of the enzyme? answer d. The assay procedure for this enzyme uses the enzymes triose phosphate isomerase and glyceraldehyde-3-phosphate dehydrogenase which catalyze the reactions respectively: dihydroxyacetone phosphate → glyceraldehyde-3-phosphate Faculty: P.M.D. Hardwicke



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glyceraldehyde-3-phosphate + NAD+ → 3-phosphoglycerate + H+ + NADH These enzymes are used because the appearance of NADH can be easily measured spectrophotometrically. How much of the two enzymes must be added? answer 3.

Enolase, which catalyzes the dehydration of 2-phosphoglycerate to convert it into phosphoenolypruvate, has an absolute requirement for a divalent cation (Mg2+ or Mn2+), which complexes with the enzyme before the substrate is bound. The enzyme is strongly inhibited by fluoride. Sample questions: a. It is likely that F- is what kind of inhibitor? answer b. Enolase catalyzes a critical reaction in the energy producing scheme. Fluoride is essential in small amounts for bones and teeth but has been used as a rat poison. Explain. answer c. What is the inferred role of divalent cations in this reaction? answer 4.

"The conversion of glycine to glyoxylate and NH3 is catalyzed by an enzyme purified by Ratner et al. This enzyme, which is present in liver and kidney, is the same enzyme as D-amino acid oxidase. It has a high Km and is therefore thought not to play a major role in glycine degradation." (From "Non-ketotic Hyperglycinemia" by W. L. Nyhan in The Metabolic and Molecular Bases of Inherited Disease.) Sample questions: a. What is an enzyme? answer b. What does it mean that glycine oxidase is "the same enzyme as D-amino acid oxidase?" answer c. Why does high Km indicate that the enzyme does not "play a major role in glycine degradation?" answer 5.

"Trypsin is an example of a large class of enzymes with a reactive serine at the catalytic site. Trypsin is most active in the pH range 7 to 9. Calcium ion increases the stability and the activity of trypsin solutions and is a necessary component for the complete conversion of trypsinogen to trypsin by autocatalytic activation."

The conversion of bovine trypsinogen into trypsin is effected by release of a single peptide, Val-Asp-Asp-Asp-Asp-Lys, from the amino terminal of the zymogen, accompanied by a conformational change. Trypsinogen is stable at pH 3, but when it is disFaculty: P.M.D. Hardwicke



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solved in neutral or slightly alkaline solution, conversion to trypsin occurs. The rate of this activation increases rapidly as the trypsin formed begins to catalyze the activation reaction." [From "Zymogens of Proteolytic Enzymes" by B. Kassell and J. Kay, Science, 180, 1022-1027 (1973)]. Sample questions: a. Explain the meaning of the phrase "reactive serine at the catalytic site". answer b. What is calcium ion called with respect to trypsin? answer c. What is meant by the phrase "from the amino terminal of the zymogen"? answer d. What is the significance of the fact that when trypsinogen "is dissolved in neutral or slightly alkaline solution conversion to trypsin occurs"? answer e. What is the significance of the fact that "the rate of this activation increases rapidly"? answer f. The small intestine, into which trypsinogen is released and in which trypsin acts, is slightly alkaline; but what is the pH optimum of trypsin? answer g. What is a zymogen? answer 6.

The following statements are taken from the article "The Isolation and Properties of Phenylalanine Hydroxylase from Human Liver" by S.L.C. Woo, S.S. Gillman, and L.I. Woolf, Biochem. J., 139, 741-749 (1974), but not in the order in which they originally appeared. "Phenylketonuria is a genetically determined disease in which the enzyme phenylalanine 4-hydroxylase (EC 1.14, 16.1) is absent or has very low activity . . . The phenylalanine hydroxylase activity of the liver was confined to a single protein of molecular weight of approximately 108,000 . . . It seems to consist of two polypeptide chains, each with a molecular weight of approximately 54,000 . . . By using the doublereciprocal plots, the apparent Km values for phenylalanine were 3.5 x 10-4 M for the full-term infant and 3.8 x 10-4 M for the adult preparations, respectively. For the synthetic cofactor, apparent Km values were 6.8 x 10-5 M and 6.6 x 10-5 M, respectively . . . The activity of the enzyme was assayed, by using various concentrations of phenylalanine, in the presence and absence of 0.5 mM p-chlorophenylalanine . . . Iron-chelating and copper-chelating agents inhibited human phenylalanine hydroxylase. Thiol-binding agents inhibited the enzyme but, as with the




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rat enzyme, phenylalanine both stabilized the human enzyme and offered some protection against these inhibitors." Sample questions: a. The Km value of the enzyme from human fetal liver for phenylalanine was 3.18 x 10-4 M. How does the affinity of the enzyme for the substrate change with development? answer b. What is the reaction catalyzed by phenylalanine 4-hydroxylase? answer c. Is this a systematic (I.U.B.) or trivial name? How can you tell? answer d. How can a synthetic cofactor have a Km value? answer e. The apparent Km of the enzyme from human fetal liver for the synthetic cofactor was 7.15 x 10-5 M. What changes with development can be inferred from this information? answer f. Is this a monomeric enzyme? Explain. answer g. What kind of inhibitor would you guess they found p-chlorophenylalanine to be? Why? answer h. What does the next to the last sentence tell you? answer i. What does the last sentence tell you? answer

Answers to Pretest Questions


1.

a. The enzyme catalyses the hydrolysis of monoesters of phosphoric acid and has a pH optimum above 7. b. Magnesium ions are required as cofactors, and chelators such as EDTA and citrate would tie them up, making them unavailable to the enzyme. c. No. As a matter of fact, the Bessey-Lowry method for alkaline phosphatase uses a pH 10.25 buffer. Since the enzyme activities are small, the assay should be done close to the optimum pH to keep the reaction time reasonably short, and the assays must always be done at the same and a constant pH; but the pH chosen does not have to be exactly at the optimum.

2.

a. Oligomeric b. Quaternary c. It can be inferred that the active site contains a sulfhydryl Problem Unit 2 - Page 10


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group, i.e., the amino acid cysteine. d. They must be in large excess so that the two reactions they catalyze are essentially instantaneous, making the reaction catalyzed by aldolase the only rate-limiting step. Neither can NAD+ be limiting.


3.

a. Since it is so unlike the substrate, it must be either a noncompetitive inhibitor or perhaps an irreversible inhibitor. b. Simply, fluoride strongly inhibits a key enzyme for energy production. c. The negatively charged phosphate group of the substrate probably coordinates with the Mg2+ or Mn2+ ions, i.e., electrostatic interaction of substrate to the active site.

4.

a. A catalyst for a biological reaction which is usually a protein. b. It means that the enzyme has both activities. [Actually, since glycine does not have an asymmetric carbon atom, it is neither D nor L (or both, depending on how you look at it); so it could be considered a D amino acid where R = H.] c. A high Km means that it takes a high substrate concentration to achieve one-half maximum velocity (a low affinity of the enzyme for the substrate). Although we are not given the Km or the physiological concentration of glycine, we can assume that the latter is far below the Km and that there is really a Damino acid oxidase that also happens to work on glycine, but very poorly.

5.

a. The R groups of serine is one of the groups at the site where reaction takes place and is somehow more reactive than the R groups of other serines in the protein. b. Cofactor c. This means "from the end of the protein chain that is terminated with a free alpha-amino group". d. Trypsinogen by itself must have some proteolytic activity. Conversion must come from splitting off of a hexapeptide (proteolysis), not by ionization alone. e. Trypsin is a better enzyme for this conversion, so as trypsin is made from trypsinogen, the conversion goes faster and faster. Trypsin acts at the carboxyl end of lysine. f. pH 7-9 g. A zymogen is an inactive precursor of an enzyme (a proenzyme). Problem Unit 2 - Page 11


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6.


a. Decreases b. CO2H

CO2H H2 N

H2 N

CH

CH CH2

CH2

NADPH + H+

O2

NADP+

H2O OH

phenylalanine

tyrosine

c. Systematic. It has an Enzyme Commission number after it. d. One can determine the concentration that gives 1/2 of the Vmax for any substrate, coenzyme, or synthetic coenzyme that binds with the enzyme and is required for reaction to take place. e. The affinity of the enzyme for the cofactor increases with development. f. No. It is a dimeric enzyme, i.e., it contains 2 subunits. g. Competitive. It is a derivative of both its substrate and the product. Hence, it would likely bind at the active site but not reaction would take place since it cannot be hydroxylated at the 4 position. h. The enzyme requires iron and copper ions for its activity. When they are tied up and unavailable to the enzyme, no reaction occurs. i. There is a -SH group at the active site because reagents that react with -SH (sulfhydryl or thiol) groups inactivate the enzyme; but if the active site is occupied by a substrate molecule, the thiol-binding agents cannot get to the -SH group.




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STUDY GUIDE-1 I. Where are enzymes located?

Enzymes are present in the cytosol, in all organelles (e.g., the nucleus, mitochondria, ribosomes, etc.), and in membranes. They are found singly and in multienzyme complexes.

II. What are catalysts and how do they work?

Enzymes are proteins (or ribonucleic acids, ribozymes) that are catalysts. As catalysts: 1.

They are unchanged in the overall reaction, but may be temporarily modified during intermediate steps.

2.

They are effective in small amounts.

A unit of enzyme activity is the amount of enzyme that will catalyze the transformation of 1 µmole of substrate/minute at a stated temperature and pH (usually optimal for both). One katal (or kat) = conversion of 1 mole substrate per second. Specific activity is the number of enzyme units/mg protein = µmoles of substrate reacted/minute/mg protein.


3.

They do not affect the equilibrium of a reversible chemical reaction. The function of a catalyst is to speed up the process in either direction, i.e., enzymes affect only the kinetic and not the thermodynamic properties of a reaction. Therefore, the same chemical equilibrium will be reached with or without the enzyme, although it may not be reached in an reasonable time without the enzyme.

4.

They exhibit specificity in their ability to accelerate chemical reactions, although the degree of specificity varies greatly. Enzymes involved in digestion are generally rather nonspecific. Many others are very specific in that they act only with a single substrate or with a very limited number of chemically similar compounds. (The substrate is the reactant in an enzyme-catalyzed reaction.) For example, pancreatic lipase is a digestive enzyme, an esterase, which will catalyze the hydrolysis of glycerides without much specificity in terms of the fatty acid, while acetylcholine esterase is an enzyme that rather specifically catalyzes the hydrolysis of acetylcholine. Some proteases are specific for peptide bonds involving certain amino acids; others are nondiscriminating about the nature of the amino acids forming the peptide bond. Problem Unit 2 - Page 13


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Enzymes, like other catalysts, lower the activation energy of a reaction. Consider the general reaction: Substrate (S) → Product (P) In any given population of molecules, there is a range of energies in a Boltzmann distribution (Fig. 1, curve A). If temperature is increased, a new distribution is set up (Fig 1, curve B). (The areas under the two curves are identical.)

Molecules must have a certain energy level before they can react, i.e., before S can be converted into P, they must be activated. The amount of energy required is called the activation energy. (∆E, Figure 2; energies exceeding that given by the solid vertical line in Figure 1.) It is seen that the distribution that pertains at the higher temperature has a larger number of molecules with the necessary activation energy than does the low temperature distribution. Consequently, there will be a proportionately larger number of molecules undergoing reaction per unit time in the case of the high temperature distribution.




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∆E

= Eact = energy of activation for S

∆G

= the free energy of reaction, i.e., the difference in free energy of S and P = activation of S (or P) determines the rate of the reaction. ∆G (sometimes termed ∆F) determines the position of equilibrium

S* Eact

P

There are two ways to accelerate a chemical reaction. The reaction can be heated, which increases the energy of the molecules and hence, increases the percentage of molecules with the required energy of activation (compare dark stippled area of Figure 1 with lightly stippled area). However, living cells of homeothermic animals such as man operate in a very limited temperature range so that application of heat energy cannot be used as a means of speeding up chemical reactions. Catalysts provide the second way to accelerate a chemical reactions; they lower the activation energy, i.e., the energy barrier that substrates must overcome before they can be converted into products (dotted vertical line, Figure 1 versus the dashed vertical line). Enzymes do this by providing the reaction a different route. The enzyme (E) reacts with the substrate to form an enzyme-substrate complex or compound (ES); in a second step, ES (which now can also be considered to be an enzyme product complex or compound) dissociates to regenerate the enzyme and release the product (P). binding E+S

catalysis ES

release

release EP

E+P binding

Each of these reactions has its own activation energy, which is much Faculty: P.M.D. Hardwicke



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lower than that of the uncatalyzed reaction (Figure 3).

Because catalysis is very fast compared to the other steps, it is assumed that the rate of reaction is determined by the rate of binding and the rate of release and the reaction is usually written: E+S

ES

E+P

The ∆G and the equilibrium constant of the reaction is unchanged. However, the Eact is lowered, which also has the effect of increasing the proportion of molecules that are in an activated state. For example, Eact for the decomposition of H2O2 (2H2O2 → 2H2O + O2 ) is about 18,000 cal.mol-1 for an uncatalyzed reaction and = 2,000 cal.mol-1 in the presence of the enzyme catalase. Urease catalyzes the following reaction: O H2N-C-NH2 + H2O → CO2 + 2NH3 Urea Urease lowers the activation energy for the reaction to less than 10,000 cal.mol-1. Eact for the acid-catalyzed reaction is = 25,000 cal.mol-1. What this means for the reaction can be seen from the fol-




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lowing table: Table 1: Eact, cal.mol-1

k(first-order rate constant)(sec-1)

half-time

10,000

7.7 x 105

9 X 10-6 s

15,000

1.7 x 102

0.004 s

25,000

8.0 x 10-4

145 min

As will be discussed in more detail later, kinetic analysis is undoubtedly the most useful tool for the determination (assay) of enzyme activity and for investigation of the specific means of reaction. Therefore, we will review kinetics briefly.

III. What are zeroorder and first-order reactions?

Consider the reaction S → P. The reaction rate is the amount of S reacted in a unit of time or the amount of P formed in a unit of time. n d[P] d[S] rate ( v ) = – ----------- = ----------- = k [ S ] dt dt

n = order of reaction Enzymes have either first-order, fractional order or zero-order kinetics depending on the substrate concentration. x a a-x c

= = = =

the amount of S reacting in time t the initial amount of S the amount of S remaining unchanged at time t. concentration of S undergoing change at time t time

S

P

zero

a

zero

t

a-x

x

A first-order reaction depends only on the substrate concentration (for example, radioactive decay). dc – ------ = k 1 [ c ] dt (c is to the first power and, hence, the reaction is first order) Faculty: P.M.D. Hardwicke



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If integrated within limits of a - x and a, then 1 a k 1 = --- ln ----------t a–x All units except 1/t cancel. Hence, the unit for k1 is 1/t. For example, if for a certain reaction with first-order behavior k1 = 0.001 sec-1, this means that each succeeding second finds 1/1000th of the remaining reactant [S] converted into product (Figure 4).

A zero-order reaction is independent of substrate concentration. If the reactant is present in sufficient excess that, for all practical purposes, its concentration remains constant during the measured course of a reaction, than that reaction rate is constant and the amount of product formed depends only on the time elapsed (Figure 5). dc – ------ = k 0 dt (c is to the zero power, i.e., [c]i = 1 and the reaction is, therefore, zero-order) If this equation is integrated within limits of a-x and a, then ko = x/t and the unit for ko = moles x sec-1.




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IV. What factors affect enzyme-catalyzed reactions?

The following factors affect enzyme-catalyzed reactions:

V. What is the effect of substrate concentration?

When doing enzyme assays, i.e., determining the amount of enzyme activity, enzyme solutions are added to substrate solutions and the initial rate is measured. Under these conditions, the concentration of product (P) is zero and the equation becomes.

1. 2. 3. 4. 5. 6. 7.

Concentration of substrate Concentration of enzyme Concentration of cofactors (activators and coenzymes) Concentration of inhibitors Temperature pH Concentration of allosteric effectors

E+S

ES

EP

E+P

Initial Rate v = dc/dt

In general, one enzyme, molecule (polypeptide chain) can combine with one molecule of substrate, although there are exceptions with oligomeric enzymes. Thus, as the ratio of substrate molecules to enzyme molecules increases, the initial rate (velocity) increases in proportion to the substrate concentration ([S]). This proportionality decreases with increasing substrate concentration, i.e., the reaction order becomes less than unity and continues to decrease to zero order in the limit of very high substrate concentration. Under conditions of a large excess of substrate, essentially all enzyme molecules are tied up in an enzyme-substrate complex or compound. At this point, there are, for all practical purposes, no free enzyme molecules that can combine with more substrate molecules, so the velocity will not increase no matter how many more substrate molecules are added. The curve depicting the effect of substrate concentration on reaction rate is given in Figure 6. Maximum rate (Vmax) (Saturation velocity)

Zero order v = k0

Vmax/2

First order v = k1[S] Km

[S] Figure 6




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This effect can be likened to a crew rowing a boat. Two crew members can row the boat faster than one, and three faster than two. However, a point is reached when all crew positions are filled, there are no more available oars, and no more crew can be put in the boat. At this point the boat is going at its maximum speed, and no matter how many additional crew are standing on the shore, watching, the boat cannot go faster because those not in the boat cannot increase its speed.

VI. What is the importance of Km? What is the basis of enzyme assays?

The enzyme concentration ([E]) also enters into the rate equation but, when it is held constant as above, the numerical value of [E] merges into the rate constant (k). When [S] is held constant and [E] is varied, the substrate concentration becomes part of the proportionality constant and the initial rate is proportional to the enzyme concentration, i.e. dc v = – ------ = k [ E ] dt Therefore, if the substrate concentration is in saturating amounts (large excess) and [E] is doubled, the rate is doubled, etc. (Figure 7). This fact is the basis for enzyme assays (determination of the amount of enzyme activity).

Michaelis and Menten defined a constant that is a dissociation constant for the enzyme-substrate complex (ES). It is a rapid equilibrium constant and can be defined mathematically as follows: k –1 [E][S] K m = ------= ----------------[ ES ] k1 Faculty: P.M.D. Hardwicke



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Where k1 E+S

k2 ES (EP)

E+P

k-1 Km, termed the Michaelis constant, is characteristic of the enzyme just as melting point, boiling point, etc. are characteristic of organic compounds. It was defined by Michaelis and Menten as a measure of the affinity of the enzyme for the substrate (KS). However, this is true only when k-1 >> k2 which is not always the case. It can be shown that (see Appendix) V max V max [ S ] - = ----------------v = --------------------Km Km + [ S ] 1 + ------[S] which is the equation for any surface catalysis. If one solves for Km, then V max  –1 K m = [ S ]  ----------- v  and Km = [S] when Vmax/v = 2, i.e., v = 0.5Vmax (Figure 6) Thus, Km is the substrate concentration that gives an initial rate that is 1/2 of the maximum rate of a given enzyme concentration. The Michaelis-Menten equation relates the experimentally determined initial velocity (v) of an enzyme-catalyzed reaction to the saturation, limiting or maximal velocity at infinite substrate concentration (Vmax ) and to the Michaelis constant (Km). To find Km, which has the units of concentration, several plots can be used. The most common is the Lineweaver-Burke plot (Figure 8) which is based on a reciprocal form of the Michaelis-Menten equation. Km Km + [ S ] 1 1 1 --- = --------------------- = ------------ × -------- + -----------V max [ S ] V max V max [ S ] v




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It should be noted here that these derivations and equations are for single substrate reactions and that biochemical reactions often involve two and three substrates reacting together. Multisubstrate, multiproduct enzyme-catalyzed reactions are more complicated to treat mathematically. For our purposes we should simply understand that these more complex reactions are characterized by sequential mechanisms. Substrates and products may add to or be released from the enzyme in a random (phosphorylation of glucose by ATP as catalyzed by hexokinase) or ordered (NAD+ and NADP+ requiring dehydrogenases) manner, respectively. One special case, where the first substrate adds and the first product releases before the second substrate adds and the second product releases, is denoted as a Ping-Pong reaction and is characteristic of serine proteases and transaminases.

1/v

slope = Km/Vmax

-1/Km 1/Vmax 1/[S] Figure 8 As stated earlier, Km is often an indicator of the dissociation constant for the ES complex. The importance of Km to metabolic control which is important in the etiology of several diseases and in treatment of some conditions can be seen in the following examples. Consider the situation with brain hexokinase and liver glucokinase (also denoted hexokinase D or hexokinase IV), two enzymes that catalyze the phosphorylation of D-glucose by ATP to yield D-glucose-6phosphate, a process involved in the transport of glucose into cells. Brain hexokinase has a Km for glucose of [E] by experimental design therefore [S]t >>> [ES] and since [S]t = [S] + [ES] then [S]t = [S] = concentration of free substrate [E] = [E]t - [ES] = concentration of free enzyme With these equations and definitions, it is possible to derive the rate equation for any single-substrate system where only initial rates (v) are measured. k1 [E] [S] = k-1 [ES] + k2 [ES] k1 ([E]t - [ES]) [S] = (k-1 + k2) [ES] k –1 + k 2 ( [ E ] t – [ ES ] ) [ S ] [E][S] ------------------= ---------------------------------------- = ----------------- = K m k1 [ ES ] [ ES ] It is convenient to rearrange the equation as follows (solving for [ES]): [ E ]t  ----------- – 1 [ S ] = K m  [ ES ]  Faculty: P.M.D. Hardwicke



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[ E ]t [ S ] – [ S ] ------------------------------- = Km or [ ES ] [ E ]t [ S ] [ ES ] = --------------------Km + [ S ] The measured rate (v) is the rate of formation of product, i.e., v2 = k2 [ES] = v k2 [ E ]t [ S ] Therefore, v = ----------------------Km + [ S ] This is the same as the Michaelis-Menten equation except that Km is a steady state constant (that is, it has a different definition here). You may wish to think of Km as a commitment constant. As a dissociation constant (k-1/k1), it would be a measure of enzyme-substrate affinity. However, it is a true affinity constant only when k-1 >> k2. Km is a characteristic constant for the enzyme and independent of the concentration of enzyme used in its determination. Why this is so can be seen from the following graph.

Post Test 1.

Definitions a. What is meant by the order of a reaction? answer b. Define Km, Ks, kcat. answer c. Km is all too frequently equated with Ks. In fact, in most reactions there is an appreciable disparity between the values




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for Km and Ks. For the reaction A → B define conditions under which Km = Ks. Describe conditions under which this is not true. answer d. What is the steady-state approximation, and under what conditions is it valid? answer 2.

True or False. If False, explain why they are false. a. At saturating levels of substrate, the rate of an enzyme catalyzed reaction is proportional to the enzyme concentration. answer b. The Michaelis constant Km equals the substrate concentration at which v = 1/2 Vmax. answer c. The Km for a regulatory enzyme varies with enzyme concentration. answer d. If enough substrate is added, the normal Vmax of an enzymecatalyzed reaction can be attained even in the presence of a noncompetitive inhibitor. answer e. The Km of some enzymes may be altered by the presence of metabolites structurally unrelated to the substrate. answer f. The rate of an enzyme-catalyzed reaction in the presence of a rate-limiting concentration of substrate decreases with time. answer g. The sigmoidal shape of the v versus (S) curve for some regulatory enzymes indicates that the affinity of the enzyme for substrate decreases as the substrate concentration is increased. answer


3.

a. The _____________ of a reaction is the numerical relationship between substrates and products. answer b. The rate constant ____________ of an enzyme-catalyzed reaction is a measure of the catalytic efficiency at saturating levels of substrate. answer c. _______________ inhibitors do not alter the Vmax of an enzyme-catalyzed reaction. answer d. The sigmoidal shape of the v versus [S] curve for some regulatory enzymes results from a _______________ effect of substrate on the substrate binding sites. answer e. For an enzyme whose Km can be regulated, the presence of a _____________ effector increases the level of substrate required to attain a given reaction rate. answer

4.

Assume that an enzyme catalyzed reaction follows MichaelisProblem Unit 2 - Page 42


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Menten kinetics with a Km of 1 x 10-6 M. If the initial reaction rate is 0.1 µmol/min at 0.1 M, what would it be at 0.01 M, 10-3 M, and 10-6 M? answer 5.

Consider the reaction HCO

CH2OH

H

C

OH

C

O

OH

C

H

OH

C

H

H

C

OH

H

C

OH

H

C

OH

H

C

OH

CH2OPO3 2-

CH2OPO3 2-

glucose-6-phosphate → fructose-6-phosphate which is catalyzed by phosphoglucose isomerase. a. What is its stoichiometry? answer b. What is the simplest representation of this reaction in terms of S,E, and P? answer c. What are S,E, and P in this reaction? answer 6.

A more general form of an equation for an enzyme catalyzed reaction is: k1 E+S

k2 ES

k-1

k3 EP

k-2

E+P k-3

Consider the essentially irreversible reaction represented by the freeenergy diagram below.




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a. Using the letters indicated in the diagram, relate each of the rate constants in the Equation above to the energy-level difference that determines it. answer b. Which rate constant limits the rate of formation of product? answer c. Does Km approximately equal Ks for this enzyme? answer 7.

To study the dependence of the rate of an enzyme-catalyzed reaction on the substrate concentration, a constant amount of enzyme is added to a series of reaction mixtures containing different concentrations of substrate (usually expressed in mol/L). The initial reaction rates are determined by measuring the number of moles (or µmoles) of substrate consumed (or product produced) per minute. Consider such an experiment in which the initial rates in Table 2 were obtained at the indicated substrate concentrations. Table 2: Initial rates at various substrate concentrations for a hypothetical enzyme-catalyzed reaction [S] (mol/L)

v (µmol/min)

2.0 X 10-1

60

2.0 X 10-2

60

2.0 X 10-3

60

2.0 X 10-4

48

1.5 X 10-4

45

1.3 X 10-5

12

a. What is Vmax for this reaction? answer Faculty: P.M.D. Hardwicke



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b. Why is v constant above substrate concentrations of 2.0 x 10-3 M? answer c. What is the concentration of free enzyme at 2.0 x 10-2 M substrate concentration? answer 8.

Using the experimental procedure described in Problem 7, the data in Table 3 were obtained for an enzyme in 10-ml reaction mixtures. Use numerical (not graphical) calculations in answering the following questions. Table 3: Initial rates at various substrate concentrations for a hypothetical enzyme-catalyzed reaction. [S] (mol/L)

v (µmol/min)

5.0 X 10-2

0.25

5.0 X 10-3

0.25

5.0 X 10-4

0.25

5.0 X 10-5

0.20

5.0 X 10-6

0.071

5.0 X 10-7

0.0096

a. What is Vmax for this concentration of enzyme? answer b. What is the Km of this enzyme? answer c. Show that this reaction does or does not follow simple Michaelis-Menten kinetics. answer d. What are the initial rates at [S] = 1.0 x 10-6 M and at [S] = 1.0 x 10-1 M? answer e. Calculate the total amount of product made during the first five minutes when [S] = 2.0 x 10-3 M. Could you make the same calculation at [S] = 2.0 X 10-6 M? answer f. Suppose that the enzyme concentration in each reaction mixture were increased by a factor of 4. What would be the value of Km? of Vmax? What would be the value of v at [S] = 5.0 x 10-6 M? answer 9.


The Km of a certain enzyme is 1.0 X 10-5 M in a reaction that is described by Michaelis-Menten kinetics. At a substrate concenProblem Unit 2 - Page 45


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tration of 0.10 M, the initial rate of the reaction is 37 µmol/min for a certain concentration of enzyme. However, you observe that at a lower substrate concentration of 0.010 M the initial reaction rate remains 37 µmoles/min. a. Using numerical calculations, show why this tenfold reduction in substrate concentration does not alter the initial reaction rate. answer b. Calculate v as a fraction of Vmax for [S] = 0.20 Km, 0.50 Km, 1.0 Km, 2.0 Km, 4.0 Km, and 10 Km. answer c. From the results in (b), sketch the curve relating v/Vmax to [S]/Km. What is the best range of [S] to use in determining Km or investigating the dependence of v on [S]? answer 10. The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions such as the synthesis of DNA. This hydrolytic reaction is catalyzed in E. coli by a pyrophosphatase that has a mass of 120 kDa and consists of six identical subunits. Purified enzyme has a Vmax of 2800 units per milligram of enzyme. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10 µmol of pyrophosphate in 15 minutes at 37°C under standard assay conditions. a. How many moles of substrate are hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km? answer b. How many moles of active site are there in 1 mg of enzyme? Assume that each subunit has one active site. answer c. What is the turnover number of the enzyme? answer

Answers to Post Test 1.


a. Reactions can be independent of the concentration of substrate (0-order), directly dependent on substrate (1st-order) or dependent on substrate concentration raised to some higher power (2, etc.). b. Km = Michaelis constant. It is the substrate concentration that gives 1/2 Vmax. That is, it is the concentration of substrate at which half the active sites are filled. Km is also related to rate constants of the individual steps in the reaction.



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k –1 + k 2 K m = ------------------k1 Ks = dissociation constant of the ES complex for k1 E+S

ES k-1

At equilibrium the rate of formation of ES must equal its rate of dissociation to E + S. Therefore you should be able to show that k –1 K S = ------k1 In the case where k2 is much smaller than k-1, Km approaches k-1/k1 = Ks. Under these conditions, the Km determined kinetically is equal to the dissociation constant, Ks. kcat = the turnover number and is equal to k2 in the Michaelis Menten mechanism. The turnover number of an enzyme is equal to the number of moles of substrate converted to product per minute per mole of enzyme present when the enzyme is fully complexed with substrate. c.

A+E

EA

E+B

If k2 >k-1 d. The steady state approximation assumes that the concentrations of the intermediates in a reaction do not change while the rate of product formation is being measured. This holds for the early stages of a reaction, after the ES complex has formed and before appreciable changes have occurred in either the substrate or product concentrations. 2.

a.

True. Vmax = k2[E]t

b. True c. False. The value of Km is independent of enzyme concentraFaculty: P.M.D. Hardwicke



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d. e. f. g.

3.


tion for almost all enzymes. False. A non-competitive inhibitor cannot be overcome by substrate concentration. True. This occurs in regulatory enzymes. True. False. The initial increasing slope of the curve shows that binding of the first substrate molecule increases the affinity of the enzyme for subsequent substrate molecules.

a. Stoichiometry b. kcat c. Competitive d. homotropic e. negative

4. [S], M

5.

v(µmol/min

0.1

0.1

0.01

0.09999

0.001

0.09990

0.000001

0.050

a. glc-6-P

Km = 1 x 10-6 M

fru-6-P

or S

P k1

b.

Vmax

E+S

k2 E+P

ES k-1

k-2

Notice that k-2 is included for a reversible Rx. c. S = glucose-6-P E = phosphoglucose isomerase P = fructose-6-phosphate 6. Faculty: P.M.D. Hardwicke

a.

The rate constant for each step is inversely related to the difProblem Unit 2 - Page 48


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ference between the energy level of the reactants and the highest energy barrier between the reactants and the products of that step. In terms of the letters in the Figure, k1 is determined by b-a k-1 is determined by b-c k2 is determined by d-c k-2 is determined by d-e k3 is determined by f-e, and k-3 is determined by f-g. b. k2 corresponds to a much higher energy barrier than the other forward rate constants and therefore must limit the rate of product formation. c. Since k2 is small relative to k-1 , Km approximates Ks for this enzyme. Therefore, Km is a measure of affinity for substrate. 7.

a.

Vmax = 60µmol/min

b. v is constant because it has reached Vmax; the enzyme is saturated with substrate. c. The concentration of free enzyme is negligible because all of the enzyme is in the ES form. 8.

a.

Vmax = 0.25 µmol/min

b. For a reaction obeying Michaelis-Menten kinetics, Vmax and Km are simply constants relating v to [S]. Km can be calculated by substituting Vmax and any pair of v and [S] values at v < Vmax. For example, at [S] = 5.0 x 10-5 M and v = 0.20 µmol/min the equation becomes 0.25 0.20 = ------------------------------Km 1 + --------------------–5 5.0x10 Solve:

Km= 1.25 X 10-5 M

c. If the reaction follows simple Michaelis-Menten kinetics, then the Michaelis-Menton equation should relate v to [S] over a wide range of [S]. This can be tested by determining whether the equation yields the same value of Km at several different values of [S] and v < Vmax. Under the conditions of this problem, the same value, Km = 1.3 x 10-5 M, is obtained at [S] = 5.0 x 10-6 M, v = 0.071 µmol/min and at [S] = 5.0 x




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10-7 M, v= 0.0096 µmol/min. Therefore, Michaelis-Menten kinetics are obeyed. d. At [S] = 1.0 x 10-6 M 0.25 0.25 v = ------------------------------ = --------------- = 0.018 µmol/min –5 1 + 13 1.3x10 1 + --------------------–6 1.0x10 e. At [S] = 2.0 x 10-3 M, v = Vmax = 0.25 µmol/min. Since 0.25 µmole is much less than the amount of substrate present (2.0 x 10-3 mole/liter x 10-2 L x 106 µmol/mol = 20 µmol) the reaction can proceed for five minutes without significantly changing the substrate concentration. Thus, 0.25 µmol/min x 5 min = 1.25 µmol At [S] = 2.0 x 10-6 M, 0.25 0.25 0.25 ---------- = 0.033 µmol/min v = -----------------------------= ---------------= –5 7.5 1 + 6.5 1.3x10 1 + --------------------–6 2.0x10 During 5 minutes at this rate, 0.033 µmol/min x 5 min = 0.17 µmol of product would be produced. However, this value exceeds the total amount of substrate present (2.0 x 106 mol/L x 10-2 L x 106 µmol/mol = 0.020 µmol). Clearly, during the-5 min reaction, [S] and therefore v would decrease significantly. Calculation of the exact amount of product made would require integration of a differential equation; this amount obviously cannot exceed 0.020 µmole. f. Km is independent of enzyme concentration, since a change in [E] does not affect the three rate constants, k1, k2, and k3. Hence Km would remain equal to 1.25 x 10-5 M. Since Vmax = k3[E]o, increasing the enzyme concentration by a factor of 4 increases Vmax by a factor of 4. Therefore, Vmax = 1.0 µmole/min. At [S] = 5.0 x 10-6 M, 1 1 1.0 - = ---------------- = ------- = 0.28 µmol/min v = -----------------------------–5 3.6 1 + 2.6 1.3x10 1 + --------------------–6 5.0x10 9.


a.

Since both substrate concentrations are well above Km, you can assume that Vmax = 37 µmol/min. Then Problem Unit 2 - Page 50


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37 37 - ≈ 37 µmol/min. v = ------------------------------ = -----------------------------–3 –5 1 + 1.0x10 1.0x10 1 + --------------------–2 1.0x10 Therefore, at [S] 1.0 x 10-2 M, v still is equal to Vmax. b. From the Michaelis-Menten equation, the following relationships can be calculated: [S], Km

v, Vmax

0.20

0.17

0.50

0.33

1.0

0.50

2.0

0.67

4.0

0.80

10.0

0.91

c. When you plot these values you will be able to see that the best range of [S] for studying the dependence of v on [S] is in the neighborhood of Km or below it, since changes in [S] below Km cause greater changes in v than do changes in [S] above Km. Therefore, when using graphic methods to determine Km and Vmax, several measurements should be made at [S] well below Km. 10. a.1 'unit' here = 10 µmole/15 min at 37ºC = 10/15 µmole min-1 Faculty: P.M.D. Hardwicke



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= 10/15.60 µmole s-1 When [S] >> Km, Vo = Vmax Here Vmax = 2800 units mg-1 = 2800 x 10/15.60 = 31.3 µmol s-1 mg-1. NOTE. THIS IS BY DEFINITION THE SP. ACT. b. 1mg of enzyme = 10-3/MW moles = 103/MW µmoles = 103/120000 = 1/120 µmoles. But each mole of enzyme has 6 active sites. Therefore 1 mg of enzyme ≡ 1/120 x 6 = 1/20 = 0.05 µmoles active site. c. The units of sp. act. are, µmoles (unit time)-1 (mg enzyme)-1. Here µmoles s-1 mg-1 is used, and the value is 31.3 µmole s1 mg-1. The turnover number, kcat, is effectively the activity in terms of (µmole active site)-1, instead of (mg enzyme)-1 as in sp. act., and its units are, µmoles s-1 (µmole active site)-1 = s-1. We know from '(b)' that there are 0.05 µmoles active site per mg of enzyme, i.e., µmole of active site = 1/0.05 = 20 mg of enzyme, and that the sp. act. (activity per mg of enzyme) = 31.3 µmoles s-1 mg-1. Therefore the kcat (activity per mmole of active site, i.e. its activity per 20 mg of enzyme) is = 31.3 x 20 = 626 s-1.

Problem Set 1.

In the cases of severe liver damage, an enzyme EL is released into the blood. After severe exercise, an isozyme from muscle, EM, is found in the blood. EL and EM can be differentiated since they have different kinetic constants. The Km of the liver enzyme is 3 x 10-4 M; the Km of the muscle enzyme is 7 x 10-5 M. Data from assays on an unconscious patient's blood are given below. Ten microliters of blood was used in each assay.




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S (M)


v (µmol/min/10 µl

3 x 10-2

990

3 x 10-3

909

1 x 10-3

769

7 x 10-4

700

3 x10-4

500

1 x 10-4

250

7 x 10-5

190

3 x 10-5

91

1 x 10-5

32

a. Is the patient likely to be suffering from a liver disease or had she been exercising too strenuously? answer b. Explain your reasoning for your answer to part a. answer 2.

Define the following: a. a unit of enzyme activity answer b. Km answer c. steady-state conditions answer d. oligomeric enzyme answer


3.

Under what experimental conditions does an enzyme-catalyzed reaction follow zero-order kinetics? answer

4.

Indicate the effects of substrate concentration, enzyme concentration, temperature, inhibitors or activators on enzyme activity by labeling correctly both axes of the graphs given shown below.



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Your choice for axes are: Energy, [E], Temperature, [S], 1/[S], 1/v, v. (You may use the same label on more than one graph.) answer 5.


An enzyme-catalyzed reaction was assayed at several substrate concentrations. Two data points which fell on the LineweaverBurk plot are v = 41.7 µmol S/min when [S] = 5 x 10-4 M and v = 16.7 µmol S/min when [S] = 5 x 10-6 M. Place the two points on a line on the accompanying graph. a. Determine the value of Km and Vmax in the correct units. answer When an inhibitor was added, the velocities fell to 1/2 their uninhibited values. b. Plot the inhibited line on the graph. answer c. Is the inhibitor competitive or non-competitive? On what evidence did you base your decision? answer



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6.


Fumarase (L-malate hydrolase) catalyzes the reversible hydration of fumarate to L-malate. The fumarase from pig heart has been crystallized (M.W. 197,000 Da) and consists of four subunits that can be dissociated into inactive monomers under relatively mild conditions. Substrate can induce reformation of tetramers with complete recovery of the activity. The subunits have a molecular weight of 48,500 Da and each contains three free -SH groups. Fumarase requires no cofactors. Kinetic studies implicate the participation of a pair of groups on the enzyme (one acidic, one basic) with pKa's of 6.2 and 6.8. These groups have been postulated to be two imidazole groups of histidine residues, one in the imidazole form and one in the imidazolium form. The reverse reaction is stereospecific for L-malate and only Lmalate is produced from fumarate. CO2CO2-

H C

HO

C

H

+ H2 O C -O

2C

H

H

H CO2-

a. For fumarase, an intact ___________ structure is necessary. answer b. Predict graphically the effect of pH on the activity of fumaFaculty: P.M.D. Hardwicke



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rase. At what pH would you expect maximal enzyme activity? answer c. Malonate, -OOC-CH2-COO- is an inhibitor of fumarase. What type of inhibitor would you expect malonate to be? answer d. Using a Lineweaver-Burk plot, graphically illustrate the effect of malonate on the kinetics of fumarase. Be sure to label all parts of the graph as well as the inhibitor data. answer 7.

a. What is meant by an allosteric enzyme? answer b. What is the difference between the active site and the regulatory site of an allosteric enzyme? answer c. How does an allosteric inhibitor produce its effect on an enzyme? answer

8.

a. The activation energy for a non enzyme-catalyzed reaction is _____________ than the activation energy for the same reaction catalyzed by an enzyme. answer b. Why would increasing the temperature of an enzyme-catalyzed reaction from 25°C to 37°C increase the rate of the reaction? answer c. Why would increasing the temperature to 60°C probably cause a decrease in the rate of the enzyme-catalyzed reaction? answer

9.

a. If the concentration of substrate is 10-3 M and the concentration of enzyme is 10-8 M, what would be the effect of the observed rate of production of product if the enzyme concentration were doubled? (Assume Km is 10-6 M.) answer b. What would be the effect on the observed rate if the substrate concentration were doubled to 2 x 10-3 M but the enzyme concentration remained at 10-8 M? answer

10. Answer the following with true or false; justify your answer in each case. a. The initial rate of an enzyme-catalyzed reaction is independent of substrate concentration. answer b. If enough substrate is added, the normal Vmax of an enzymecatalyzed reaction can be attained even in the presence of a noncompetitive inhibitor. answer c. The rate of an enzyme-catalyzed reaction in the presence of a rate-limiting concentration of substrate decreases with time. Faculty: P.M.D. Hardwicke



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answer d. The sigmoid shape of the v-versus-[S] curve for some regulatory enzymes indicates that the affinity of the enzyme for substrate decreases as [S] is increased. answer 11. Two forms of isocitric dehydrogenase exist in mammals. One which is NAD+ specific and found only in mitochondria and a NADP+ specific enzyme found in both cytosol and mitochondria CH2-COOH

HO

HC

COOH

C

COOH

H isocitric acid

CH2-COOH + NAD+ or NADP+

H

C

H

+ H+ + NADH or NADPH

C-COOH O α-ketoglutaric acid

(isocitric acid; pK1 = 3.2, pK2 = 4.8, pK3 = 6.4) NADP+ specific isocitric dehydrogenase catalyzes the decarboxylation by formation of an unstable enzyme bound chelate of Mn2+ and a αketo acid intermediate. The free energy change for formation of aketo glutaric acid under physiological conditions is: ∆G˚ = -5 kcal/ mol. AMP regulates the enzymatic activity by reducing Km for isocitrate by 10 fold. Only {isocitrate2-} was found to be the substrate form that binds to the enzyme. a. Mn2+ is an example of a(an) answer b. NAD+ is an example of a(an) answer The reaction velocity was found to be 4th order with respect to isocitric acid indicating high cooperativity. c. The fact that "high cooperativity"is found indicates that this enzyme is a(an) ______________ enzyme. answer The NAD+ specific enzyme has a molecular weight of 330,000 Daltons and is made up of 8 identical subunits. d. The NAD+ specific enzyme is an example of a(an) ______________ protein and the level of structural organization for the 8 identical subunits is termed the structure of the protein. answer e. AMP is said to act as a(an) answer Reaction velocity is decreased in presence of ATP which acts by binding at the same site that NAD+ binds. Faculty: P.M.D. Hardwicke



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f ATP is said to act as a(an) ____________ answer g.Production of NADH during the course of the reaction would be expected to ____________ the reaction velocity and NADH would be an example of a(an) ____________. answer h. What is the significance of ∆Go = -5 kcal/mole in terms of: 1. The desire of the reaction to go as written? (One sentence answer). 2. The activation energy of the reaction? answers i. If isocitrate ionization was the only important controlling factor for the pH-activity profile of the reaction. (See characteristics at the beginning of this exam). Using the information given, construct an accurate pH-activity profile for the enzyme catalyzed reaction. Use graph paper. answer 12. Zero order kinetics in an enzyme catalyzed reaction only occurs when we have: a. a high specific activity b. an isozyme present c. high substrate concentration d. a high Km e. high enzyme concentration f. a high turnover number answer 13. The redox pairs NAD+

NADH and NADP+

NADPH are well suited for use in coupled clinical enzyme assay systems: a. because of their acid-base properties. b. because of their distinctly different absorbance properties of their oxidized and reduced species. c. because of their occurrence in cells. d. because of their ability to take the place of enzyme reactions. e. because they are coenzymes. answer 14. The success of a measurement of an enzyme activity using a coupled enzyme assay depends on having a. The NAD+ Faculty: P.M.D. Hardwicke

NADH dehydrogenase reaction. Problem Unit 2 - Page 58


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b. the second reaction as non-limiting. c. A non-buffered reaction medium. d. a colorimeter adequate in the visible region of the spectrum. e. a trained M.D. to supervise. answer 15. In measuring the rate of a coupled reaction one must know: a. how to control the humidity in the sample chamber. b. where the lag phase ends. c. the maximum absorbance of the unknown enzyme. d. the rate of absorbance change during the preincubation period. e. the total absorbance change throughout the assay. answer 16. The levels of LDH isozymes in the blood are indicative of certain disease states: Some of the isozymes present will react with a particular substrate while others will not. Thus, one can experimentally measure the activity of these particular isozymes while other isozymes are also present in the sample. In particular, this assay makes particular use of: a. the colligative properties of the solution. b. the substrate specificity of the isozyme of interest. c. the preincubation phase of the isozyme of interest. d. the tertiary structure of the isozymes of interest. e. the total activity of all enzymes species. answer 17. Let's say the following assay was established to measure E1 levels in serum. E1 A

B + NAD

B

+

E2

(1)

NADH + H+ + C

(2)

a. List the solution conditions you would have to control to make this a valid assay. answer b. List the species to equation 1 and 2 whose concentrations




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you would have to manipulate in order to make a valid assay. answer c. One of your technicians ran the assay under conditions which, with normal serum levels of E1, the Vmax for equation (1) turned out to be about equal to Vmax for equation (2). State what is wrong with the assay. answer d. In one sentence state what you would do to rectify the problem. answer 18. Clinical data for enzymes are often reported in terms of international units. What is an enzyme unit of activity and what relationship does it bear to specific (enzyme) activity? answer 19. What is meant by the term "zero order" reaction as it pertains to enzymes? answer 20. Why is the preincubation phase of a coupled enzyme assay necessary? answer 21. Which one of the following relationships best describes how reaction velocity can be used to indicate the level of enzyme in blood serum as developed for clinical labs? a.

k –1 + k 2 K m = ------------------k1

[S] b. ln ---------- = – kt [ S0 ] c.

Km 1 1 1 --- = ------------ -------- + -----------[ S ] V v V max max

d. Vmax = k [E0] V max e. when Km >> [S], v = ------------ [ S ] Km f.

[S0] - [Sf ] + [ES]

answer

ANSWERS TO PROBLEM SET 1.


a. Liver b. The [S]Vmax/2 is close to the Km of the liver enzyme



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2.


a. The amount of enzyme which converts 1 µmole of substrate to product per min (at 25/37oC) [ E ] SS [ S ] SS k cat + k –1 b. K m = [ S ] Vmax ⁄ 2 = ---------------------- = --------------------------k1 [ ES ] SS c. Under steady state conditions, S is converted into P at a constant rate, the [S] and [P] vs time plots are linear, and [ES] is constant d. An enzyme with a quaternary structure, i.e., made up of more than one subunit (protomer, monomer). The subunits may be the same (e.g., a homodimer), or different (e.g., a heterodimer).

3.

Saturating substrate ([S] >> Km)

4.

Upper L: vo vs [S] Upper R: Vmax vs [E]tot Lower L: vo vs Temperature Lower R: vo vs [S] for a Km-type allosteric enzyme showing +ve homotropy. 5.


a. Km = 8.3 x 10-6 M, Vmax = 45.5 µmoles min-1



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b. 1/v

slope =

KmI/VmaxI Km/Vmax

1/VmaxI -1/KmI = -1/Km

1/Vmax 1/[S]

c. Non-competitive (Km same, Vmax smaller) 6.

a. Quaternary (oligomeric) b. pH 6.5 c. Competes with fumarate because of similar structure, therefore a competitive inhibitor d. 1/v slope =

-1/KmI -1/Km

+malonate KmI/VmaxI -malonate

Km/Vmax 1/Vmax = 1/VmaxI 1/[S]


7.

a. They are enzymes whose kinetic properties cannot be accounted for by the Michaelis-Menton model. b The active site is where the substrate binds to the regulatory enzyme; the regulatory site is where the effector molecules bind. These two sites are different. c. An allosteric inhibitor typically binds and stabilizes the enzyme in an inactive or less active (conformation) state.

8.

a. Greater b. Yes c. Yes

9.

a. [S] >> Km Therefore, vo = Vmax Therefore, the velocity is doubled when [E]tot is doubled Problem Unit 2 - Page 62


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b. None, because the enzyme is already saturated with substrate 10. a. b. c. d.

F F T F (positive Km-type)

11. a. Cofactor b. Coenzyme c. Allosteric (positive Km-type) d. e. f. g. h. i.

Oligomeric Quaternary +ve heterotropic effector -ve heterotropic effector Equal, coenzyme

1. If NAD+/NADP+ and NADH /NADPH and isocitrate and a-ketoglutarate all at 1 M concentration are mixed at 1 atm pressure, since the free energy change under standard conditions for the forward (L

R) reaction is negative,

NAD+/NADP+ and isocitrate will be converted into NADH/NADPH and α-glutaric acid. (Also, at equilibrium, the ratio of products to reactants (Keq), will be greater than 1, i.e., the equilibrium lies to the right.) 2. The standard free energy change in the L

R direc-

tion is the difference between the standard free energies of activation in the forward and reverse directions (∆G0' = ∆G*for - ∆G*rev) j. 12. c 13. b 14. b 15. b 16. b




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17. a. The standard conditions of assay for E1 must be met (pH, ionic strength, cofactors, temperature, etc), so the results may be compared to the range of normal values. b. [A] must be at a sufficiently high concentration to saturate E1. [NAD+] must be at a sufficiently high concentration to saturate E2. E2 must be present at sufficiently high levels of activity for it not to be a limiting factor in the assay c. The rate of the second reaction must be much greater than that of the first reaction (E2 at high activity), so that only the amount of E1 limits the rate of the reaction, i.e., Vmax2 must be >> Vmax1. d. Increase the activity of E2 present in the assay medium, so that it is much greater than that expected for E1. 18. The IU is the amount of enzyme converting 1 µmole of S into P per min (at a given temperature, pH etc). The specific activity = IU per mg total protein (it increases as the enzyme is purified) 19. When the enzyme is saturated with substrate, v0 = Vmax = kcat Et = a constant for that assay, i.e., the rate = a constant = k, therefore it is not dependent on [S], and zero order kinetics in [S} are observed. 20. To take into account: Breakdown of NADH/NADPH due to a. Its intrinsic chemical instability b. The effect of other factors in our serum sample on NADH levels besides the enzyme we want to measure. 21. d. Vmax = k [Eo] (i.e., Vmax = kcat [E]tot)




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Module 2: Clinical Enzymology Objectives:

1.

Be able to define or identify the correct from the incorrect use of each of the following terms as they relate to enzymes and catalysis. Vmax coenzyme Km zero order reaction saturation velocity specific activity active site substrate specificity isoenzyme enzyme assay

2.

Be able to describe the two general ways in which enzymes are used clinically and be able to supply a specific example of each.

3.

Be able to explain why the conditions of pH, temperature, substrate concentration and inhibitors must be controlled in enzyme assays.

4.

One of the most useful enzyme assay procedures makes use of the redox pair: NAD+/NADH. Describe those features of the NAD+/NADH interconversion that make it well suited for clinical measurements.

5.

Explain the concept of coupled reactions. In your answer be sure to comment on each of the following: Why is it useful to couple reactions? What is meant by preincubation and lag phase? What aspects of the coupled reactions one must be concerned with in order to assure an accurate measurement?

a. b. c.


6.

Give an example of an organ specific enzyme which can appear in blood plasma and describe a clinical procedure that uses the NAD+/NADH (or NADP+/NADPH) couple in its assay.

7.

Several enzymes found in blood plasma originate from specific tissues or organs. Give examples of 4 different enzymes that belong to this category. State the conditions leading to their eleProblem Unit 2 - Page 65


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vated concentration in blood plasma. 8.

Elevation of concentrations. a. Describe the relationships between enzyme and substrate concentrations which provide the best evaluation of enzyme activity (see goal 3). b. Differentiate between enzyme concentration and enzyme activity. c. Depending on the spectrophotometric equipment available, a number of approaches are possible for evaluating reaction rate from a given assay system. Describe three ways that reaction rate may clinically be evaluated and give the advantages and disadvantages of each.

9.

Given a passage from a journal of text concerning the principles and uses of enzymes in clinical measurements (which may be either a clinical investigation or enzymology description), answer questions about the passage. (Answers may involve drawing inferences or conclusions based on principles mastered in BIOCHEM PU02-1 and/or in this domain.)

10. Be able to answer questions such as those given in the Clinical Problem and the Self Test. In addition, be able to define and correctly use the words in the NOMENCLATURE and VOCABULARY and KEY WORD lists (if abbreviation, know what the letters stand for). Be able to answer questions such as those in the Problem Sets and the Practice Exam. CLINICAL PROBLEM: Consider the following clinical problem with a focus on the biochemistry of clinically important enzymes. (Glucose-6-phosphate Dehydrogenase Deficiency. p. 298, Biochemistry: A Case Oriented Approach, Montgomery, Dryer, Conway, Spector, 1974]) A 29-year-old previously healthy Iranian physician, S.I., was admitted to the hospital because of generalized myalgia of 4 days duration. His major complaint was preceded, in sequence, by a temperature of 102 degrees Fahrenheit (38.9 ˚C), dark urine, anorexia, nausea, and ultimately scleral icterus. He had been in contact with jaundiced patients in the course of his hospital work but he denied exposure to hepatotoxins in any form, including injections or blood transfusions. The significant findings of the physical examination were deep jaundice, a moderately tender liver palpable 2 cm below the right costal margin, and no splenomegaly. Faculty: P.M.D. Hardwicke



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The immediately relevant hematologic and liver function studies are summarized in Table 1 Please note that a nonstandard unit (U/dl) has been provided by this particular clinical laboratory. A red blood cell (RBC) survival study with sodium chromate-51 Cr showed an initial precipitous drop in counts over a 3-day period leading to an extrapolated Cr half-life of 6 days. Findings from paper electrophoresis of the serum protein were normal except for a moderate decrease in albumin. Biopsy of the liver demonstrated a marked inflammatory reaction involving the portal areas, with occasional areas of focal necrosis in the lobule. Myeloid metaplasia was also noted in the biopsy specimen, most probably the result of intermittent, lifelong hemolysis. After careful questioning, the patient described repeated episodes of malaise, myalgia, and severe fatigability that had accompanied relatively minor respiratory infections and that probably represented undiagnosed episodes. The color generated during the methemoglobin reductase test for glucose-6-phosphate dehydrogenase estimated visually was compatible with complete absence of the enzyme. This suspicion was supported later by quantitative assay. Further discussion of the case is contained on pp. 299-302 in Montgomery et al., Biochemistry, 1st edition. Table 4: Laboratory Data for Patient S.I. Test Values


S.I.*

Normal

Hemoglobin

15.1 (9.2) (g/100 ml)

14-18 (g/100 ml)

Hematocrit

47.0 (29) (%)

40-54 (%)

Reticulocytes

12.4 (%)

0.5-1.5 (%)

Alkaline phosphatase

8.0 (U/dl)

5-13 King Armstrong units/dl

AST (or SGOT)

620.0 (U/dl)

15-40 Karmen units/dl

ALT (or SGPT)

1220.0 (U/dl)

5-40 Karmen units/dl

Direct bilirubin

35.5 (mg/100 ml)

0.0-0.1 (mg/100 ml)

Total bilirubin

57.0 (mg/100 ml)

0.2-1.0 (mg/100 ml)



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Table 4: Laboratory Data for Patient S.I. Test Values

S.I.*

Normal

Serum iron

247.0 (µg/100 ml)

75-175 (µg/100 ml)

LDH

985.0 (U/dl)

90-200 IU/dl

Glucose-6-phosphatedehydrogenase activity

Absent

* Figures in parentheses indicate lowest value measured during hospitalization.


Acid Phosphatase Albumin Alkaline Phosphatase ALT (or SGPT) Assay α-Amylase β-Amylase Antibiotics Asparaginase AST (or SGOT) Assay Blood Urea Nitrogen (BUN) Ceruloplasmin Cholinesterase Coupled Assays CPK Assay Creatinine levels Functional Plasma Enzymes Galactosemia Glucose-6-phosphatase deficiency γ-Glutamyl Transferase (GGT) α-Hydroxybutyrate Dehydrogenase (α α-HBD) Hyperlipoproteinemias Inborn errors of metabolism Lag Phase LDH Lipoprotein Lipase (LPL) Lovostatin Lower Limit Normal (LLN) Non-functional plasma enzymes Pancreatic Lipase Phenylketonuria Plasminogen Preincubation Scoline Sorbitol Dehydrogenase Sphingolipidoses Streptokinase Upper Limit Normal (ULN)

Key Words:

Binding Sites Biochemistry Coenzymes Enzyme Inhibitors Enzyme Tests Enzymes




Self-Test

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1.

Describe the function of a catalyst.

2.

How do enzymes differ from other chemical catalysts?

3.

Why are enzymes necessary in the biological system?

4.

What are isoenzymes and how are they useful clinically?

5.

How may enzymes be used to provide useful clinical information?

6.

What is the function of the coenzyme NAD+ and NADP+ in enzyme reactions?

7.

Distinguish between lag phase and preincubation.

8.

What reaction parameters affect the enzyme-catalyzed reaction?

9.

What is a coupled enzyme reaction?

10. How can hemolysis affect the enzyme assay? 11. What is the significance of a-hydroxybutyrate dehydrogenase? 12. In what clinical conditions would you expect AST to be greater than ALT and vice versa? 13. How would the clinical laboratory's use of a nonstandard unit (U/dl) impact on your interpretation? 14. What is (are) the true physiological substrate(s) for alkaline phosphatase?




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STUDY GUIDE-2 I Diagnostic

Enzyme levels are often altered in disease. There are two basic situations: 1.

Where the abnormal level of enzyme is in fact the fundamental cause of the disease. Many inherited diseases are due directly to an insufficiency or complete absence of an enzyme activity. Such inborn errors of metabolism derive from a fault at the genetic level, i.e., the condition is passed from one generation to the next through transmission of the parental DNA. For example, a mutation at a codon coding for an amino acid critical for the function of the enzyme may affect its Km or kcat, reducing its efficiency, or completely inactivating it.

There are many examples of such conditions. E.g.,

Glucose-6-phosphatase deficiency

Ordinarily, this enzyme is only expressed in liver and kidney calls. It allows glucose-6-phosphate, produced by breakdown of liver or kidney glycogen under conditions where blood glucose is low, to be converted into glucose that can then be released into the bloodstream for by other tissues, particularly the brain and muscles. If its activity is too low, a severe hypoglycemia results.

Phenylketonuria

Here, the enzyme activity (phenylalanine hydroxylase) which ordinarily converts the amino acid phenylalanine to tyrosine is missing, and phenylalanine accumulates, with damaging consequences.

The Sphingolipidoses

Sphingolipids are catabolized in lysosomes. The enzymes carrying this out have to work in a specific sequence, so that if only one of them is absent, its substrate accumulates with pathological consequences, even if all the prior and subsequent enzymes in the pathway are functional. (See Module 4, PU6).

Galactosemia

In this condition, the enzyme activity galactosyl-1-phosphate uridyl transferase is absent. This enzyme is important for the conversion of galactose into glucose, which is then metabolized by the glycolytic pathway, pentose phosphate pathway, and glycogen synthase. Thus, galactose cannot be utilized, and accumulates as galactose-1-phosphate, producing many symptoms. Enzyme assays allow diagnosis of these genetically determined diseases. Such diagnosis often allows intervention to ameliorate the con-




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dition. E.g., in galactosemia, galactose intake can be restricted. Lactose contains galactose, so that removal of dairy foods from the diet is useful. Again, in phenylketonuria, restriction of phenylalanine intake is useful, for example, by reducing dairy food consumption, particularly cheese. 2) Where the abnormal enzyme activity is a consequence of disease or trauma. For example, a permeability barrier may be damaged secondary to some disease process, allowing high levels of some enzyme activity to leak into the blood; or, a tissue which normally produces an enzyme may be damaged or destroyed, thereby reducing the enzyme activity.

Determination of Whether an Enzyme Activity Level is Abnormal

This is done by assaying under a fixed, well-defined standard set of conditions, that are used in all hospitals, so that we can COMPARE our measured activity directly with the normal activity as assayed under the same conditions. Thus, our assay must be carried out at exactly the same pH, temperature, and ionic strength as the standard assay, and all the necessary cofactors and ingredients must be present at their standard concentrations. We determine whether our assay value lies within the range defined by the Upper Limit Normal (ULN) and the Lower Limit Normal (LLN) which depend on the particular enzyme.

Sampling

1.

Sometimes a specific tissue biopsy has to be taken; e.g., skeletal muscle in the case of a suspected muscular dystrophy.

2.

Blood is the most convenient tissue to sample, if possible. If blood is used, certain precautions have to be taken. If the sample is being used for diagnosis of a suspected disease associated with blood clotting, the blood plasma obtained by centrifuging down the blood cells has to be used. This requires the use of anti-coagulants agents such as EDTA or citrate (which chelate Ca2+ needed for clotting) or heparin. These compounds interfere with many enzyme assays, and are avoided if at all possible. Thus, serum is the most convenient sample form. This lacks the clotting factors, and is obtained by allowing clot formation and retraction to take place in the sample, when the straw-colored supernatant is collected. Additionally, because the RBC's contain enzymes which are sometimes the same as those we are trying to assay, it is most important to avoid hemolysis when handling blood samples. It is necessary to distinguish between 'Functional Plasma Enzymes' such as lipoprotein lipase, which are present under normal circumstances in the blood, which have a real physiologic role, and 'Non-functional plasma enzymes',




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which only appear in the blood in pathological conditions.

Measurements Units

The International Enzyme Unit (I.U.): typically, the amount of enzyme converting 1 mmole of substrate into produce min-1, usually at 37 or 25ºC. (However, if the enzyme is of intrinsically low activity, the units might be, say, nmoles hr-1.) The Specific Activity: effectively, enzyme units mg-1. Typically, µmoles min-1 mg-1. Note that if we give the activity in terms of µmole-1 instead of per mg-1, we obtain a quantity identical with kcat, i.e., µmole min-1 µmole-1 = min-1, the units of a first order rate constant. Thus, if we know the molecular weight of our enzyme, we can interconvert kcat and the specific activity, MW SpAct ,pureEnzyme × MW k cat = --------------------------------------------------------------------- (  1µmole = ------------ mg )  1000  1000 In practice, we always work under conditions where the assayed enzyme should be completely saturated with its substrate ([S]>>Km), so that Vmax is determined, and the measured activity is directly proportional to the amount of enzyme added. Students are sometimes confused by the fact that when they are introduced to Vmax, they see the equation, Vmax = kcat [E]T, where Vmax is typically in units of mM min-1, kcat is in min-1, and [E]T is in mM (i.e., the units are concentration per unit time), while in practice Vmax is often given in, say, µmoles min-1, i.e., in absolute amounts of substrate converted or product formed. This is because if we multiply the above equation for Vmax by the volume of the system, we obtain Vmax = kcat ET, where Vmax is typically in µmoles min-1, kcat is in min-1, and ET is in µmoles.




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Vmax µmoles min-1 or mM min-1

ET or [E]T (usually µl of serum) Thus, our measured activity is directly proportional to the amount of enzyme added, i.e., the plot of activity against [E]T or ET is linear. In this way, the specific activity and kcat are given simply by the slope of the plot. It is important to remember that serum is a very complex mixture of enzymes, proenzymes, transport proteins, immunoglobulins, cofactors etc, and many trace molecules. Thus, in general, activities based on serum samples are given in units per unit volume of blood for the purposes of comparison with normal enzyme levels, say per ml or per 100 ml, and the amount of sample is quantitated in terms of the volume added to the assay mixture. If 10 µl, of serum give a Vmax of, say, 2 µmoles min-1, then 1 ml of blood will give 200 µmoles min-1, which we can compare with the range of normal values based on activity per ml blood. Since ET or [E]T is directly proportional to Vmax, provided we know there is nothing intrinsically wrong with the enzyme (no genetic defect), we ought to be able to calculate the absolute amount of enzyme present in our sample, V max E T = ---------------- mg , SpAct

V max or E T = ------------µmoles k cat

However, in general, unless we have other data, we cannot distinguish between low total amounts of good (functional) enzyme, and a low activity due to the poor intrinsic activity of a defective enzyme. Thus, we make our measurements under steady-state conditions (S → P at a constant rate), with sufficient substrate to keep the enzyme saturated, so that our measured velocity is the maximal velocity (Vmax), and our progress curve is linear, with a slope proportional to the specific activity/kcat of the enzyme. In practice, we can either folFaculty: P.M.D. Hardwicke



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low the decrease in substrate concentration or the increase in product concentration. The assay has to be carried out under exactly the correct conditions -pH, ionic strength, temperature, correct levels of coenzymes etc, so as to allow comparison with tabulated data. A very common and convenient method used to determine the concentration of a chemical species is near ultraviolet or visible spectrophotometry. If radiation of a wavelength corresponding to the energy needed to excite the molecules is absorbed, the intensity of the radiation is reduced by its passage through the sample. The amount of radiation absorbed at a particular wavelength λ (units nm) is measured by a quantity called the Absorbance, which is on a logarithmic scale. Absorption bands I Aλ

I0 I0 = initial intensity I = intensity after passage through sample

λ nm

Aλ = log(I/I0)λ = -log(I0/I)λ Often there is a very simple relationship between the absorbance, and the concentration of the species, which can be summarized in the Beer-Lambert Law, Aλ = ελcl, where ελ is the extinction coefficient for the molecule at wavelength λ, c is its concentration, and l is the pathlength. Usually, l = 1 cm. Thus, the rate of change of the absorbace is directly proportional to the rate of change in concentration. ε λ ∆c ∆A λ ----------- = ----------∆t ∆t i.e., the rate of the reaction is effectively given by the rate of change of the absorbance. Sometimes a substrate or product itself may absorb light, and we can Faculty: P.M.D. Hardwicke



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follow the reaction directly. This can be done in particular if NAD+/ NADH or NADP+/NADPH are involved. (These are cosubstrates in many oxidation-reduction reactions.) This is because the reduced forms absorb light at 340 nm, while the oxidized forms do not.

NAD+/NADP+ NADH/NADPH Aλ

200

250

300 350 λ nm

400

Thus, in an enzyme catalyzed reaction, AH2 + NAD+ → B + NADH + H+ we will see an increase in A340 as the absorbing species NADH is formed from the non-absorbing species NAD+, while in A + NADH → BH2 + NAD+ we will observe a decrease in A340. Our sample is usually serum, so that many enzymes and other extraneous molecules including enzymes that use NADH/NADPH will be present during the assay, which may cause oxidation/reduction of the NADH or NAD+ that we add to the assay medium. Also, our monitored molecule may be unstable, and also contribute to any change in A340 - this is particularly true of NADH or NADPH. Suppose we have oxidation of NADH to NAD+,




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serum added excess ‘A’ added Rates ~ slopes A340

preincubation phase

Time We can circumvent these potential problems by allowing the assay to proceed at first without having any substrate (A) added to the assay mixture. This is called ‘preincubation’. After we have determined our background rate of change of A340, we then add our specific substrate. Only the enzyme that uses that substrate will bind it and convert it into product, so that any increase in the reaction rate after we add the specific substrate can only be due to the enzyme we are interested in. I.e., we ‘isolate’ our particular enzyme by making use of the fact that only that enzyme can use the specific substrate we add to form the specific product. We then substrate the ‘background’ rate from the total rate seen after adding the specific substrate. True rate = Rate after adding specific substrate - Preincubation background rate

Synthetic Chromogenic Substrates

For a lot of enzyme activities we may wish to measure, NAD+/ NADP+ may not be a cosubstrate, but many synthetic (chromogenic) substrates that either absorb light or give products that absorb light are now available, e.g., p-nitrophenolphosphate (pNPP). This is used in the determination of phosphatase activities. OO2N

P O-

phosphatase + H2 O O2N

O

p-nitrophenolphosphate p-nitrophenolate anion O2N absorbs maximally at 400 nm

OH

+ PO4 3

p-nitrophenol O-

+ H+

A yellow color develops as the pNPP is hydrolyzed, the increase in the intensity of which is followed at 400 nm. In this case, we simply Faculty: P.M.D. Hardwicke



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measure the rate of change in A400 before and after adding our sample (serum), and substract the background rate. I.e. although we correct for instability of the artificial substrate by following background breakdown of the pNPP, we do not strictly actually do a ‘preincubation’ of the serum in this situation. True rate = Rate after adding serum - Rate before adding serum

Coupled Assays

Sometimes there is no appropriate synthetic substrate available for the enzyme we wish to assay, and the enzyme does not itself use NADH/NADPH. However, sometimes the product of the enzyme (E1) we wish to assay is a substrate of an enzyme (E2) using NADH/ NAD+. E1 A

B

B + NADH + H+

E2

BH2 + NAD+

or E1 A

BH2

BH2 + NAD+

E2

B + NADH + H+

Provided the two sets of assay media are compatible, we can combine the two assays into a single system, so that as B/BH2 is formed from A by E1, it is converted into BH2/B by E2 with the concomitant oxidation/reduction of NADH/NAD+, i.e., we can indirectly follow the E1 activity using the rate of change of A340. We have coupled the two reactions. We must ensure: 1.

As usual, the pH, temperature, etc, correspond to standard assay conditions.

but in addition, 2.

The only limiting factor in the overall reaction is the amount of E1.

Thus,




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(i) the concentration of the initial substrate A must be saturating for E1. (ii) the concentration of NADH/NAD+ must be saturating for E2. (iii) the second enzyme, E2, must be present at sufficiently high levels that the overall rate will not depend on the second reaction, i.e., the second reaction is not limiting. The assay medium initially contains all of the components EXCEPT (1) the specific substrate (A), and (2) the serum sample containing the enzyme whose activity we want to measure (E1). Suppose we have a reaction where in the second step NADH is oxidized to NAD+. As with all NADH/NADPH assays using serum as the sample, we first add the serum and allow the system to preincubate to determine any background breakdown of NADH due to instability of the NADH and extraneous enzymes in the serum. After we have determined this rate, we add the specific substrate for E1 (A), and measure the increase in rate of change of A340 over the background preincubation rate. Because only E1 can use A with the coupled oxidation of NADH, this increase is specific for E1. We then subtract the background preincubation rate from the rate after A has been added. Note that after we have added A, the reaction at first accelerates over a period of time called the ‘lag phase’. The part of the progress curve whose slope we use for the determination of activity is the linear phase after the lag phase has ended. The lag phase arises because after we have first added A, although E1 is now saturated with its substrate, and that reaction is going at its maximum rate, the steady state level of B, the product of E1 and substrate for E2 has not yet been attained. This takes a finite period of time to be generated. The second reaction (catalyzed by E2), which is the reduction of B with the concomitant oxidation of NADH, will increase in rate until the concentration of its substrate B has achieved its steady state level. In such a coupled reaction we might see the following progress curve-




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I

II preincubation III lag Add serum

A340

Add excess ‘A’

IV Steady-State linear Depletion V of NADH

Time We can divide the time course into the following sections: I.

Substrate A and serum both not yet added. Low background breakdown of NADH due to its chemical instability.

II. Preincubation. Serum sample added. Preincubation begins. Background breakdown of NADH due to non-specific factors in the serum as well as intrinsic instability of NADH. III. Lag Phase. A saturating amount of A, the specific substrate for E1, has been added. Only E1 can use this, so any increase in rate must be due to the activity of E1. The reaction accelerates during the lag phase until the steady state concentration of B has been reached. IV. Steady-State, Linear phase. For every molecule of A converted into B by E1, a molecule of NADH is oxidized by E2. Under these conditions, the rate of the reaction catalyzed by E1 is the same as the rate of the reaction catalyzed by E2, so we can measure the activity of E1 by the rate of fall of A340 during this phase after we have subtracted the rate observed during the preincubation phase. V. Depletion of NADH occurs.

Some of the Commonly Determined Enzyme Activities and Other Biochemical Quantities: the Meaning of Some of the Acronyms on a Lab Report. Albumin


This transport protein is made by the liver, and serum levels of albumin are one of the indications of liver dysfunction. Since gamma globulin is synthesized by the immune system, while albumin is made Problem Unit 2 - Page 79


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by the liver, the albumin/globulin ratio is sometimes measured, being decreased in liver disease.

BUN, Blood Urea Nitrogen

Used to monitor renal function. It may be elevated in renal disease, and in states of dehydration.

Creatinine levels

Used along with BUN to monitor renal function. Creatinine is formed by spontaneous (non-enzymatic) breakdown of creatine phosphate in the muscles, NH2+ -O

C

CH2 C

N

O

CH3

H N

O

spontaneous

P N H

HN

C

O-

N

O-

creatine phosphate

C

O

+ PO4 3-

CH2

H3C

creatinine

and is normally excreted only by the kidney into the urine. Serum levels of creatinine are raised in renal insufficiency. The rate of formation is proportional to muscle mass, and is normally at a constant rate, which allows the creatinine level in a urine sample to be used as a measure of the time over which the urine was produced.

Ceruloplasmin

This is a serum protein (an α2-globulin) that has two functions. Firstly, it carries 90% of the Cu in the blood. Secondly, it acts as a ferroxidase, oxidizing Fe2+ to Fe3+. This is important in the blood, because the major Fe carrier, transferrin, can only bind Fe in its ferric (Fe3+) state. Ceruloplasmin thus acts to link Cu and Fe metabolism. Its main diagnostic use is as an indicated of Wilson's disease (a genetically determined hepatoreticular degeneration), when there is a failure of the liver to excrete Cu into the bile, and Cu accumulates in the organs of the body. Although an abnormality in ceruloplasmin is not the primary cause of Wilson's disease, its levels are decreased in that condition.

ACP, Acid Phosphatase

A non-specific phosphatase with an acid (~pH 4) pH optimum. High levels are present in the prostate gland. Serum levels are raised in ~80% of cases where prostate cancer has metastasized. Prostate acid phosphatase is inhibited by tartrate, unlike most other acid phosphatases, which are inhibited by formaldehyde (e.g., those of RBC's, bone, liver and spleen). Often assayed using the chromogenic synthetic substrate p-nitrophenyl phosphate, which gives a yellow color absorbing at 400 nm as it is hydrolyzed. The physiological




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function of this enzyme is unclear.

ALP, Alkaline Phosphatase

Another non-specific phosphatase, but with an alkaline pH optimum (~10). Also assayed with p-nitrophenylphosphate. It has less specific locations than acid phosphatase. It is present in high levels in liver, bone (osteoblasts), and the intestine. The hepatic content of ALP is increased in cholestasis (obstruction of the bile duct), and can give increased serum concentrations. It is released into the blood by active osteoblasts, and can therefore indicate active bone formation. Thus, it is at high levels in infants of 1-2 years and in the early teens. It is an important indicator of bone pathology, and is a valuable guide to the activity of Paget's Disease. It indicates bone remodelling due to bone resorption. The serum activity is often elevated in bone cancer. The physiological function of this enzyme is unclear.

The Amylases

These are digestive enzymes produced in two tissues, namely the salivary glands and the exocrine pancreas. There are tissue-specific isozymes, which can be distinguished by the effect of inhibitors, and by mobility on electrophoresis. β-amylase is an exoglycosidase, cleaving the α (1 → 4) glycosidic links in starch by hydrolysis, working from the non-reducing end of the glucose chain. α-amylase is an endoglycosidase, cleaving the internal α (1 → 4) glycosidic linkages randomly. The most important clinical use is in detection of acute pancreatitis (where serum levels of >10 x ULN may be seen). Increased serum levels are seen in mumps, and some other conditions, such as peptic ulcer, gallstones and intestinal obstruction.

Pancreatic Lipase

This is a digestive enzyme of the exocrine pancreas released into the duodenum that hydrolyses the ester links at the 1 and 3 positions of triglycerides in the presence of colipase, bile and Ca2+ to give 2monoglyceride and the fatty acids from the 1 and 3 positions (see Module 3, PU6). O O

H2 C O CR1 2H2O O CH2OH + R1COOH + R3COOH R2 C O CH O R2 C O CH H2 C O CR3

triglyceride


CH2OH

2-monoglyceride



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This may be a better indicator for diagnosing pancreatitis than amylase.

LPL, Lipoprotein Lipase

This enzyme is bound to the lumen of the capillaries, and catalyses the breakdown of triglyceride to 2-monoglyceride and R1COOH and R3COOH, (like pancreatic lipase). (It is discussed in detail in PU 6, Modules 3 and 4.) O O

H2 C O CR1 2H2O O CH2OH + R1COOH + R3COOH R2 C O CH O R2 C O CH H2 C O CR3

CH2OH

triglyceride

2-monoglyceride

Its levels are low in Type I diabetes (juvenile onset), and lipoprotein lipase activity may be greatly reduced in Type I hyperlipoproteinemia, an autosomally recessive inherited disease.

CK, Creatine Kinase/ CPK, Creatine Phosphokinase

This enzyme of the muscle sarcoplasm catalyzes the transfer of a phosphoryl group from creatine phosphate to ADP to regenerate ATP used up in muscle contraction (creatine phosphate is the immediate energy reserve in muscle.)

-O

C O

H2 C

NH2+ O N

C

N H

P O

O- + ADP -

creatine phosphate

-O

C

H2 C

NH2+ N

C

NH2 + ATP

O

creatine

The active enzyme consists of a dimer. The two subunits (protomers, monomers) may be of two types: M and B. Therefore, there are 3 types of active isozyme., MM in skeletal muscle MB 30% of the CK in the heart BB Brain and Thyroid Usually, most of the CK in serum is from skeletal muscle (MM), and raised levels of MM are seen after strenuous exercise. High levels of the MB isozyme are associated with myocardial infarction (MI). If seen in young boys, high levels of MM are consistent with Duchenne Faculty: P.M.D. Hardwicke



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muscular dystrophy (DMD).

LDH, Lactate Dehydrogenase

This catalyzes the interconversion of ketopyruvate and L-lactate, O

O

OC

HO

OC

+ NAD+

CH CH3

C

O

+ NADH + H+

CH3

lactate

ketopyruvate

The active enzyme is a tetramer, which is made up of two types of subunit, M and H. There are thus 5 types of isozyme, which are distributed differently among the tissues,

}

Heart, RBC, kidney and brain (inhibited by pyruvate, use α-hydroxybutyrate as a substrate)

LDH3 H2M2

Pancreas, lung, spleen, lymph, adrenal and thyroid glands

}

Skeletal muscle, liver (not inhibited by pyruvate, cannot use α-hydroxybutyrate as a substrate)

LDH1 H4 LDH2 H3M

LDH4 HM3 LDH5 M4

The different isozymes represented by the different combinations of H and M have evolved to suit the requirements of a particular tissue. The inhibition of the H4 and H3M isozymes by the substrate pyruvate is an excellent example of a negative homotropic effect. In skeletal muscle, where isozymes insensitive to pyruvate are present, lactate may accumulate without disastrous results, apart from a leg cramp. However, in heart muscle, any form of cramp would be fatal, and so the isozymes in that tissue cannot in practice exceed a certain activity to prevent the lactate levels rising too high.

α-HBD, α-Hydroxybutyrate Dehydrogenase


An important difference between the LDH isozymes is that H4 and H3M show higher activity with α-hydroxybutyrate as the substrate than with the normal substrate, lactate, while the other isozymes do not.



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O


O

OC

HO

OC

+ NAD+

CH

C

O

CH2

CH2

CH3

CH3

+ NADH + H+

α-ketobutyrate

L-α-hydroxybutyrate α-HBD

Thus, if we use α-hydroxybutyrate as the substrate, we are effectively only assaying heart or RBC LDH - we can distinguish between the isozymes both by the degree of inhibition by pyruvate, and their activity towards α-hydroxybutyrate dehydrogenase. This is very important in practice, as elevated serum levels of LDH occur under many situations, so that total serum levels of LDH are not very useful. Raised levels of the heart and RBC isozymes (H4 and H3M) occur in myocardial infarction and sickle cell hemolytic crises. Since the H4 and H3M isozymes are present in RBC's, care has to be taken when sampling for diagnosis of myocardial infarction that hemolysis does not occur.

The Transaminases

These enzymes catalyze the transfer of an α-amino group from an αamino acid to an α-keto acid. NH3+ H

C R1

CO2-

O

+

NH3+

O C R2

CO2-

C R1

CO2-

+ H

C

CO2-

R2

Frequently, α-ketogluarate is the acceptor, when glutamate is formed,




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NH3+

O C NH3+ H

C

CO2-

CO2-

CH2

H

CO2-

C CH2

R1

+ CH2

R1

CO2-

α-amino acid

AST, Aspartate Transaminase/GOT, Glutamate Oxaloacetate Transaminase (SGOT = Serum GOT)

α-KG

CO2-

CH2

O

+

C

CO2-

α-keto acid

glutamate

This catalyzes the transfer of the amino group of apartate to α-ketoglutarate, NH3+

O NH3+ H

C

C

CO2-

CH2 CO2-

L-Asp

CO2-

CH2

+ CH2 CO2-

α-KG

H

O CO2-

C CH2

CO2-

oxaloacetate

C

CO2-

CH2

+ CH2 CO2-

L-glutamate

AST/GOT is found in many tissues, both in the cytosol and mitochondria of the cells. In myocardial infarction and liver damage/disease, it leaks out of the damaged cells so that the SGOT rises.

ALT, Alanine Transaminase/GPT, Glutamate Pyruvate Transaminase (SGPT = Serum GPT)


This catalyzes the transfer of the amino group of alanine to α-ketoglutarate,



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NH3+

O C NH3+ H

C

CH2

CO2-

CH3

CO2-

H

CO2-

C CH2

+ CH2

CH3

CO2-

L-Ala

CO2-

CH2

O

+

C

CO2-

α-KG

ketopyruvate

L-glutamate

ALT/GPT is a cytosolic enzyme particularly concentrated in the liver cells, but is less abundant in the liver than AST/GOT. In situations where the entire liver is affected, such as cirrhosis, cancer, or hypoxia, SGOT levels are raised more than SGPT levels; but if the liver cell plasma membrane is mainly affected, as in viral hepatitis, SGPT is higher than SGOT (because it is mainly cytosolic enzymes which leak out of the damaged liver cells). The levels of CK, AST (GOP), and α-HBD have characteristic time courses in the serum after a myocardial infarction (MI). CK Relative

Enzyme Activity

AST

α-HBD

5 Time, Days

GGT, γ-Glutamyl Transferase

Sorbitol Dehydrogenase


10

This is located in the ER of liver cells (where its function is not clear), and in the plasma membrane of renal tubular cells, where it may be involved in the transport of certain amino acids in the kidney. It is used as an indicator of some hepatobiliary diseases, when it is released into the blood. Increased serum levels are seen after liver damage.



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Cholinesterase

A non-specific pseudocholinesterase (sometimes called butyrylcholinesterase or serum cholinesterase) of unknown physiological function is secreted by the liver into the blood. (This must be distinguished from the specific cholinesterase, acetylcholinesterase, which mediates the termination of the acetylcholine signal at both nicotinic and muscarinic cholinergic synapses by hydrolyzing acetylcholine into acetic acid and choline.) The liver enzyme is important clinically because we rely on it to hydrolyze succinylcholine ('Scoline'), a depolarizing cholinergic blocker used to obtain relaxation during surgery, into succinic acid and choline. In some patients (~0.05%), there is a genetic defect, and the liver enzyme is inactive, so that recovery from scoline is greatly delayed, giving rise to a condition called 'scoline apnea'. Thus, it is important to screen patients before surgery, to make sure this condition does not exist.

Other Enzyme Activities

Clearly, if a specific condition is suspected, then other assays may be carried out, e.g., for glucose-6-phosphatase if glucose-6-phosphatase deficiency is likely, etc, etc. Click on the hypertext link for a listing of some popular enzyme-based diseases and syndromes.

II. Therapeutic

The major use of our knowledge of enzymes is in the treatment of disease by enzyme inhibitors. There are many examples of this, e.g., the use of lovostatin, an inhibitor of cholesterol synthesis in the treatment of certain hyperlipoproteinemias. Occasionally, an enzyme may be used directly, e.g., the use of asparaginase to destroy asparagine, O

NH2 C

O

H2O

C

CH2

+ NH4 +

CH2

CH H3 N+

O-

CH CO2-

L-Asn

H3 N+

CO2-

L-Asp

in the treatment of adult leukemias. Note that streptokinase, used to remove bloodclots in the legs, is not only not a kinase, but is in fact not an enzyme at all. It is a small protein which binds to plasminogen in the blood, and is able to activate the zymogen to its plasmin activity without cleaving it by proteolysis.




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Many antibiotics are effectively enzyme inhibitors, e.g., penicillin is a suicide substrate for the enzyme involved in last step in bacterial cell wall biosynthesis.

Post Test 1.

All of the following enzymes are markers for liver disease EXCEPT a. alanine transaminase (ALT) b. aspartate transaminase (AST) c. creatine phosphokinase (CPK) d. sorbitol dehydrogenase e. lactate dehydrogenase (LDH) answer

2.

The isozymes of lactate dehydrogenase a. demonstrate the evolutionary development of this enzyme b. range from monomers to tetramers c. differ only in a single amino acid d. exist in five forms, depending upon the content of M and H monomers e. are forms of the enzyme that differ in activity but not in electrophoretic mobility answer

3.

An individual suffering chest pain went to the local hospital emergency room for treatment. Analysis of the patient’s blood showed elevated levels of the enzyme lactate dehydrogenase (LDH) and predominance of the H4 isozyme. Give one possible explanation for the increased levels of LDH. Explain the biochemistry behind your medical diagnosis. answer

Answers to Post Test 1.


c This enzyme catalyzes the transfer of a phosphoryl group from creatine phosphate to ADP to regenerate ATP used up in muscle contraction (creatine phosphate is the immediate energy reserve in muscle). The enzyme comprises two subunits, M and B,



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which forms three isozymes. MM is found in skeletal muscle, MB found in the heart, and BB found in brain and thyroid.


2.

d The LDH tetramer may be composed of subunits characteristic of heart or muscle, in any possible proportion, making five possible tetramers.

3.

The patient may have had a heart attack, causing damage to heart tissue. LDH molecules of the H4 variety leaked out from the damaged heart tissue into the blood.



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Module 3: Membrane Transport Objectives:



1.

Describe the permeability properties of lipid bilayer membranes.

2.

Be able to describe the difference between simple diffusion and facilitated passive diffusion through biological membranes.

3.

Be able to distinguish between facilitated passive diffusion and active transport

4.

Be able to distinguish between primary and secondary active transport..

5.

Describe symport and antiport transport. Be familiar with how coupled transport processes are used a) at the plasma membrane to move sugars and amino acids with Na+ and b) at the mitochondrial membrane to shuttle metabolites in and out of the mitochondrion.

6.

After reading a short passage from a medical journal or textbook be able to interpret the data and draw conclusions about the significance of the data.

7.

Understand the terms in the Nomenclature and VocabularyY list as well as the Key Words list. Be able to answer questions and problems similar to those on the Practice Exam.

Active transport

Antiport

Ca2+-ATPase Coupled cotransport Electrochemical gradient

Cotransport Digitalis Facilitated diffusion

Flux (J) Ion channel Ligand-gated channel

H+-pump Ionophore Membrane potential

Mobile carriers Net flux Passive mediated diffusion Permeable Primary active transport Secondary active transport Symport

Na+,K+-ATPase Passive diffusion Passive facilitated diffusion Permeability Saturation kinetics Simple diffusion Voltage-gated channel



Key Words:


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Adenosine Triphosphatase, Sodium, Potassium Biochemistry Biological Transport Cell Membrane Permeability Ion Channels Membrane Potentials Membranes Permeability



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STUDY GUIDE-3 I. Membrane Transport

Biological membranes are constructed around the phospholipid bilayer. This is a 2-dimensional micelle, with the non-polar acyl chains of the phospholipids forming an oily internal film sandwiched between two polar hydrophilic surfaces formed by the head groups of the phospholipids, together with the solvating medium of water and counterions. The phospholipid bilayer represents a barrier to the free diffusion of solutes, acting to separate the internal contents of the cell from the external medium, and to form internal compartments between different functional parts of the cell. Because of the high electrical resistance of the hydrocarbon chains in its interior, the bilayer also acts as a capacitor, separating electrical charges across the membrane. Operation of three phenomena, the Donnan effect, the differential permeability of the membrane to ions and the Na+, K+pump combine to give the interior of the cell a negative potential relative to the outside. There is an electrical potential gradient across the plasma membrane tending to pull positively charged solutes into the cell, and push negatively charged solutes out.

II. The Electrochemical gradient

When there is a difference in concentration of an uncharged solute between adjacent regions with no barrier between them, diffusion allows net transport (flux) of the solute to occur from the region where it is initially at high concentration to the region where it is initially at low concentration, until the difference in concentration has disappeared. This process is thermodynamically spontaneous, and its ∆G is negative. Free diffusion is described by Fick's Laws, the details of which do not concern us here, except to note that an uncharged solute will flow down its concentration gradient, and that the rate of transport depends on the size of the concentration gradient of the solute between the two regions, i.e., the difference in concentration. If there is a barrier between the regions, the permeabilty of that barrier to the solute will determine the rate of flow of molecules from the high concentration compartment to the low concentration compartment. (Strictly, we should use the term 'chemical potential' instead of 'concentration', but for us the difference is not important. The term 'net' is used because diffusion is a random (stochastic) process, with molecules moving in both directions. Some will be moving from the low to the high concentration region, but more will be moving from the high to the low. It is this excess which represents the net transport from high to low.) If the molecule possesses an electric charge, the presence of the mem-




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brane potential has to be taken into account, since the negative charge on the internal surface of the cell tends attract positive ions into the cell, and repel negative ones, whilst the relatively positive outer surface attracts negatively charged solutes and repels positively charged solutes. Thus, for a charged molecule there is a combination of two effects to be taken into account in determining its direction and rate of net movement across a biological membrane: its concentration gradient across the membrane; and its electrical potential gradient across the membrane. These may act in the same direction and reinforce one another, or oppose one another. The net driving force can be easily calculated by converting the concentration driving force into electrical units (using the Nernst equation), and adding or subtracting it from the electrical driving force depending on its sign. A charged molecule moves spontaneously down its electrochemical gradient, until its electrochemical potential is the same on both sides of the membrane (∆G < 0). (We will usually use the general term 'electrochemical gradient', remembering that for uncharged molecules this simply means concentration gradient.)

III. Simple Passive Diffusion Across Phospholipid Bilayers

very non-polar small uncharged polar

O2 N2 benzene CO2 H2O urea glycerol

small charged polar

acetate

large polar

sucrose

inorganic ions

K+ Ca2+ Na+ Cl-

Fig. 1 Permeability of a Phospholipid Bilayer to a Selection of Solutes In this case, the direction and rate of net transport is determined solely by the electrochemical gradient of the particular solute across the bilayer, and the permeability of the bilayer towards that solute. Net transport of solute occurs spontaneously down its electrochemiFaculty: P.M.D. Hardwicke



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cal gradient until its electrochemical potential is the same on both sides of the membrane, with a decrease in the free energy of the system (∆G < 0). The molecule diffuses across enclosed in a cavity formed by kinks and defects in the packing of the acyl chains. The permeability of the membrane to a solute depends on a number of factors, one of the most important of which is its solubility in the non-polar hydrocarbon interior of the bilayer. Thus, very non-polar uncharged molecules, which dissolve well in the oily interior of the membrane, move easily down their concentration gradient: i.e., the activation free energy for the process is low. The membrane is quite permeable to small uncharged polar molecules, such as urea, glycerol or water itself. The presence of an actual electrical charge on a polar molecule greatly lowers its permeability, but if the molecule is small, and the formal charge can be reversibly lost and regained on the two surfaces of the membrane, quite high permeabilites can be seen (e.g., acetate anion->acetic acid-> acetate anion). Phospholipid bilayers are very impermeable to large polar molecules (e.g., sucrose). The membrane is most impermeable towards inorganic ions; these cannot lose their charge, and the activation free energy for their movement is very high. Movement of a solute across membranes is quantitated in terms of its net flux, J, which is the rate of net transport per unit area of membrane. As with rates in enzyme kinetics, we always measure the initial rate, Jo, since obviously the rate falls off as the electrochemical gradient is dissipated. For an uncharged solute moving across a given membrane, the electrochemical gradient is just the concentration gradient, which for a membrane of a defined thickness we can estimate by the concentration difference of the solute between the two sides of the membrane. As Fig. 2 shows, the initial flux does not reach a saturating (constant) value as the concentration difference across the membrane is increased, but shows a linear increase with ∆G.

JM J0

facilitated hyperbolic

JM/2

linear KM

simple

∆c

Fig. 2 Simple and Facilitated Passive Diffusion across a membrane Faculty: P.M.D. Hardwicke



Facilitated (Mediated) Passive Diffusion and Active Transport

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Clearly, since the cell needs to admit and expel an enormous number of different solutes in a highly selective way for many purposes, special mechanisms must have evolved to control movement of molecules and ions across biological membranes. These all use proteins, which, since they all act to lower the activation free energy for movement, have the characteristics of enzymes. (Only O2 and CO2 move across cell membranes without special proteins being involved-there are even specific channels for water.) A fixed, limited, number of protein molecules, often located in special regions of the membrane are involved. These are usually very specific for the solute whose transport they are catalyzing, and must be regulatable to control the flux. We can divide the transport proteins into two broad classes: facilitated passive diffusion and active transport.

Facilitated (Mediated) Passive Diffusion

Here the driving force for the transport process is just the electrochemical gradient for the solute, which moves spontaneously down its electrochemical gradient. ∆G is negative, and the process is passive just as with the simple case, because we do not have to do anything to the system to make the net transport occur. Because there are only a limited number of transporter protein molecules, the flux will reach a limiting value as the concentration difference of the solute across the membrane is increased and all the protein molecules become occupied in transporting the solute; i.e., we have saturation kinetics, as shown in Fig. 2. The protein effectively catalyzes the transport of the solute and we get equations for J0 and ∆c very similar to the Michaelis -Menten equation for v0 and [S]. ∆c J 0 = J M --------------------∆c + K M and the J0 vs. ∆c curve looks just like the hyperbolic curve for an ordinary Michaelis-Menten enzyme. Thus, Lineweaver-Burke plots can be made, just as with soluble enzymes Since the binding sites on the protein molecules which catalyze transport can only accomodate the particular solute or structurally similar molecules, facilitated (mediated) passive diffusion thus differs from from simple (passive) diffusion by: 1) Speed and specificity 2) Saturation kinetics




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3) It can be chemically inactivated (by reagents that react with the transporter) 4) It can be competitively inhibited. +I

-I

1/J0

1/JM -1/KM -1/KIM

1/∆c

Fig. 4 Lineweaver-Burke plots can be used to identify competitive inhibitors of facilitated diffusion.

Sub-Types of Facilitated Diffusion

Uniport:-A single type of solute moves down its electrochemical gradient

X

Uniport Gap Junctions (Allows ions and small molecules to move from one cell to another)

The GLUT class of glucose transporters e.g. in red blood cells, muscle, adipocytes

Ion Channels

Ligand Gated

Symport and Antiport Coupled Cotransport

Voltage Gated

In symport, two different types of solute are transported in the same direction across the membrane In antiport, two different types of solute are transported in opposite directons




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x y


x y Symport

Transport of 'X' depends on the simultaneous or sequential transport of 'Y' in both symport and antiport mechanisms

Antiport

E.g., the redblood cell anion exchanger (antiport) of HCO3 for Cl-. (Symport and Antiport also occur in active transport, see below). -

Ionophores provide good examples of facilitated diffusion.

These are organic molecules, often from microorganisms, that mediate facilitated diffusion of ions across membranes. They sometimes have antibiotic properties. The driving force for net solute flow is again the electrochemical gradient of the solute. There are two classes of ionophore: 1. Mobile Carriers. These are characterized by having: (1) A nonpolar outer surface that dissolves well in the hydrocarbon interior of the membarne bilayer; (2) A hydrophilic interior with a polar surface that can solvate the transported ion in its unhydrated form. The cavity is often very specific for a given type of ion. Both the complexed and uncomplexed forms of this type of ionophore can diffuse back and forth across the membrane bilayer. Some, like valinomycin (specific for K+) and the nactins (also mainly K+), catalyze a uniport type of process which is therefore electrogenic (since net charge is transported across the membrane). Others, such as nigericin (mainly a K+ carrier) and monensin (mainly Na+), lose a proton on binding the ion on one side of the membrane, and bind a H+ when they release the ion on the other side: thus, they catalyze an electroneutral exchange (antiport) of K+ or Na+ for H+. Mobile carriers can only function when the interior of the membrane is in its liquid crystalline state, i. e., the membrane is fluid and above its transition temperature. 2. Channel Formers. These do not depend on the membrane being in a fluid state. They often show only poor selectivity, with very high transport rates compared to mobile carriers. Hydrophobic groups on the outside of the channel allow it to be inserted into the bilayer, whilst its interior is lined with polar groups which let ions pass through the channel in their hydrated forms, so that a pore is formed in the membrane. E. g., gramicidin A.

Active Transport


Here the electrochemical gradient for the process is unfavorable -in the absence of the transporter, the ∆G is positive (> 0), and transport will not occur spontaneously in the direction needed. The electrochemical gradient opposes the desired movement. To get movement Problem Unit 2 - Page 97


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against the electrochemical gradient, we have to couple the transport process to a second process with a sufficiently negative ∆G to make the ∆G for the total coupled process negative. ∆Gtot(< 0)) = ∆Gtrans (> 0) + ∆Gcoupled process (more negative than ∆Gtrans is positive) In order for two processes to be coupled, there must be an obligatory relationship between them, i.e., one cannot happen with out the other also happening. The free energy of hydrolysis of ATP is often coupled to active transport, either directly or indirectly. Thus, in active transport, the cell has to invest energy to accomplish the movement of the solute in the desired direction. There are two forms of active transport:

Primary (Direct) Active Transport

Secondary (Indirect) Active Transport


In this, an energy source is used directly to push the solute against its electrochemical gradient. There are three main forms: 1.

Where the free energy of hydrolysis of ATP is coupled to the transport process. Examples are the P-type ATPases, in which a phosphorylated intermediate is formed, such as the Na+, K+ATPase in the plasma membrane, and the Ca2+, H+-ATPases in the plasma membrane and SER. The V-type H+ transporting ATPases of endosomes and lysosomes represent another class, where a phosphorylated intermediate is not formed.

2.

Where the transport of H+ against their concentration gradient out of the mitochondria is coupled to passage of electrons through Complexes I, III, and IV of the electron transport chain in the inner mitochondrial membrane.

3.

Where light energy is captured and used to move H+ across membranes, as in bacteriorhodopsin.

In this, an electrochemical gradient is first established by some energy requiring primary active transport process, such as the Na+ gradient generated across the plasma membrane by the Na+, K+-pump, and this is then coupled to movement of a second solute against its electrochemical gradient through a coupled cotransporter working in either the Symport or Antiport mode, i.e. the movement of one solute down its electrochemical gradient is coupled to the movement of a second solute against its electrochemical gradient. An example of secondary active transport across the plasma membrane is the uptake of glucose or amino acids from the gut lumen against their electro-



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chemical gradients in symport with Na+ moving back down its electrochemical gradient into the intestinal epithelial cell. Another example is the movement of Ca2+ out of a heart muscle cell against its electrochemical gradient in antiport with Na+ moving into the cell down its electrochemical gradient. This is catalyzed by the Na+-Ca2+ exchanger, 3Na+out + Ca2+in

3Na+in + Ca2+out

(Since digitalis inhibits the Na+, K+ pump, which thus reduces the Na+ gradient across the sarcolemma of the muscle cell, less Ca2+ are pumped out, and the concentration of Ca2+ in the muscle cell rises. This increases the contractile force of the heart muscle cell, and is the basis for the positive inotropic effect of the cardiac glycosides.) The H+ gradient produced by the H+ pumping complexes of the electron transfer chain is coupled to the secondary active transport of many solutes across the inner mitochondrial membrane against their electrochemical gradient in the symport or antiport modes, as H+ move down their electrochemical into the mitochondrial matrix. (Movement of OH- out of the mitochondria in symport or antiport also is used, when we remember that movement of OH- out is equivalent to movement of H+ in.) Both primary active transport, and therefore secondary active transport as well, are inhibited by metabolic poisons interfering with ATP production or H+ pumping in the mitochondria. Secondary active transport is blocked by ionophores which dissipate the driving Na+ or H+ electrochemical gradient.

Post-Test 1.


Selective permeability of cell membranes is achieved, in part, by active transport systems. Active transport differs from passive transport in that it a. requires energy but no transport carrier b. depends primarily on diffusion and osmosis c. requires a carrier but no energy d. requires energy, usually in the form of phosphate anhydride bonds Problem Unit 2 - Page 99


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e. is necessarily associated with pinocytosis answer 2.

The best described cotransport systems involving glucose and sodium are a. uniports b. two oppositely oriented uniports c. two uniports oriented in the same direction d. antiports e. symports answer

3.

Ionophores include a. pore-forming antibiotics that act by passive transport b. carrier antibiotics that act by active transport mechanisms c. pore-forming and carrier antibiotics that act by active transport and passive transport respectively d. pore-forming and carrier antibiotics that act by passive transport and active transport, respectively answer


4.

All of the following substances freely diffure across biological lipid bilayer membranes EXCEPT a. carbon dioxide b. malate c. nitric oxide d. oxygen e. urea answer

5.

The sodium dependent transport of glucose in the kidney is an example of a. antiport b. group translocation c. passive transport d. primary active transport e. secondary active transport answer

6.

What type of plot represents substrate concentration versus rate Problem Unit 2 - Page 100


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of transport by simple diffusion? a. bell-shaped curve b. rectangular hyperbola c. signoidal curve d. straight line answer 7.

All of the following are true statements regarding the sodium/ potassium-ATPase EXCEPT a. It consists of two α-subunits and two β-subunits b. It forms an aspartyl-phosphate high-energy bond during translocation c. It forms an aspartyl-phosphate low-energy bond during translocation d. It translocates three sodium ions per ATP e. It translocates three potassium ions per ATP answer

8.

Which of the following is a true statement regarding cardiac glycosides such as digitalis, which are used in the treatment of heart failure? a. They inhibit calcium-ATPase b. They activate calcium-ATPase c. They inhibit sodium/potassium-ATPase d. They activate sodium/potassium-ATPase e. They inhibit translocation by the plasma membrane sodiumcalcium antiport protein answer

9.

The enzyme responsible for maintaining the pH inside of lysosomes is a class a. P ATPase b. V ATPase c. F ATPase d. M ATPase answer

Answers to Post Test Faculty: P.M.D. Hardwicke




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1.

d Active transport is the passage of material against a concentration gradient and requires the expenditure of energy, often the hydrolysis of ATP or PEP.

2.

e

3.

a

4.

b Malate is a polar compound that does not cross biological lipid bilayers by simple diffusion. The inner mitochondrial membrane contains several specific transport proteins that translocate this substance.

5.

e Secondary active transport uses a source of chemical energy such as a sodium gradient that is produced by primary active transport. The cotransport of sodium and glucose in the same direction is called symport; this is an example of secondary active transport.

6.

d The rate of simple diffusion is a linear function of the substrate concentration.

7.

e Three potassium ion are NOT translocated per ATP, only two potassium ions are translocated per ATP.

8.

c Cardiac glycosides bind to the exterior surface of the sodium/ potassium-ATPase and inhibit the enzyme. As a result, the intracellular concentration of sodium is increased somewhat. This intracellular sodium exchanges for extracellular calcium by a process that is mediated by an antiport protein. The higher intracellular calcium ion concentration resulting from this process is thought to augment cardiac muscle contraction.

9.

b This class of enzymes is responsible for maintaining the pH inside of lysosomes and secretory vessicles. These enzymes lack a phosphorylated enzyme intermediate. The class V enzymes are composed of from three to five different polypeptide subunits.



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Practice Exam The two hour exam for Biochemistry PU02 will be composed of multiple choice questions. The pretest, post test, and problem sets in this Problem Unit provide examples of short answer questions to check your knowledge base while the (25) questions below are actual questions of the "Board Type" given to a previous MSI class. You may wish to time your use of this Practice Exam to one hour or less in order to pace yourself for the exam. For each of the following multiple choice questions, choose the most appropriate answer. 1.

Which one of the following statements about the maximum velocity (Vmax) of an enzyme catalyzed reaction is true? a. Vmax is unaltered in the presence of a noncompetitive inhibitor. b. Vmax is directly proportional to enzyme concentration. c. Vmax is directly proportional to substrate concentration. d. Vmax is inversely proportional to the measured rate (v). e. Vmax is directly proportional to Km, being higher for enzymes with higher Km values for their substrates. answer

2.

Ingested glucose is phosphorylated in the liver by both hexokinase and glucokinase. Hexokinase has a Km for glucose of about 1 x 10-5 M and a glucokinase has a Km for glucose of about 1.5 x 10-2 M. The glucose concentration available to liver cells is approximately equal to the concentration in the blood, and this value is about 135 mg% - or 0.75 x 10-2 M. When the glucose level falls from 135 mg% to 80 mg%, the in vivo rate of phosphorylation of glucose in the liver may change because: a. the rate of the reaction catalyzed by hexokinase will fall but that of glucokinase will be little affected. b. the rate of the reaction catalyzed by glucokinase will fall but that of hexokinase will be essentially unaffected. c. the rates of both enzymes will decrease. d. the rates of both enzymes will not be affected much. e. the rate of glucose will increase but that of hexokinase will remain about the same.




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answer 3.

For any enzymic reaction, when the ratio [ES]/[E] increases: a. the velocity of the reaction will increase. b. the velocity of the reaction will decrease. c. the Km is increased. d. the turnover number decreases. e. an allosteric effector must be present. answer

4.

In the presence of effector A an enzyme displays sigmoidal behavior when initial velocity is plotted against substrate concentration. With increasing concentrations of A the curve is displaced to the left (i.e., towards the ordinate). Effector A is thus: a. a positive allosteric effector. b. a competitive inhibitor. c. a negative allosteric effector. d. a non-competitive inhibitor. e. an uncompetitive inhibitor. answer

5.

The activity of enzyme 1 (E1) is to be measured in a coupled assay system utilizing enzyme 2 (E2) according to the scheme: E1 A+B

E2 P1

P2 (measured product)

Where A and B are substrates for enzyme 1, P1 is the product of the reaction catalyzed by enzyme 1, and P2 is the product of the action of enzyme 2 on P1. Which of the following conditions must be satisfied to have a valid assay for enzyme 1? a. The concentration of substrate A must be much greater than substrate B in order to have the rate of the reaction dependent on substrate B. b. The activity of enzyme 2 must be much greater than enzyme 1 in the assay system. c. The rate of reactions of enzyme 1 and enzyme 2 must be adjusted so that a steady state concentration of P1 will be maintained during the reaction. d. The formation of P1 must be detectable. e. None of the above. answer Faculty: P.M.D. Hardwicke



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6.


The initial velocity of a reaction at various initial substrate concentrations was determined with an enzyme of connective tissue metabolism from a HEALTHY patient. The results obtained were as follows: [S] (mol/L)

v (µmol/min)

1.0 x 10-3

65

5 x 10-4

63

1 x 10-4

51

5 x 10-5

42

3 x 10-5

33

2 x 10-5

27

1 x 10-5

17

5 x 10-6

9.5

1 x 10-6

2.2

5 x 10-7

1.1

The best estimates from this data with respect to this enzyme are: a. Km = about 5 x 10-5 M, Vmax about 63 µmol/min. b. active sites are unoccupied on the great majority of enzyme molecules when the substrate concentration equals 3 x 10-5 M. c. Km = about 3 x 10-5 M, Vmax about 65 µmol/min. d. Km = about 2 x 10-5 M, Vmax about 130 µmol/min. e. the enzyme's active sites are approximately half-saturated at 1 x 10-4 M. answer 7.


The role of most vitamins in metabolism is: a. to serve as coenzymes or precursors of coenzyme. b. to serve as precursors of essential amino acids. c. to serve as key intermediates in the citric acid cycle. d. to build up resistance to bacterial and viral infections. Problem Unit 2 - Page 105


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e. to "spark" metabolism. answer 8.

All of the following statements regarding diagnostic enzymology are true EXCEPT: a. serum enzymes are usually measured by their activity rather than by the absolute concentration of enzyme molecules. b. enzyme activity should be measured with the substrate concentration at less than 10 times the Michaelis constant to ensure reaction rate linearity. c. multiple point determinations yield superior accuracy and precision over two-point determinations in kinetic analyses. d. the presence of cofactors in the reagent may increase the enzyme activity. e. enzymes generally show higher activity at 37°C than at 30°C. answer

9.

If in a Lineweaver-Burk plot the 1/v intercept is 2 x 10-3 min per mole and the 1/[S] intercept is -2 x 104 M-1, what is the value of Km? a. 5 x 10-3 M b. 5 x 10-4 M c. 5 x 10-5 M d. 0.5 x 104 M e. None of these. answer

10. What is the reaction rate in a simple enzymatic system, if the substrate concentration(s) is much less than Km? a. It is maximal. b. It would be too slow to measure. c. It is virtually proportional to substrate concentration. d. It would be reduced by addition of more substrate. e. It would be uninfluenced by the addition of more substrate. answer 11. What is one good reason why initial velocities are employed in enzyme assays? Faculty: P.M.D. Hardwicke



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a. Substrate inhibition is minimized. b. Substrate activation is minimized. c. Errors in technique are minimized. d. Product inhibition is minimized. e. The sensitivity of the assay is increased. answer 12. Which one of the following applies to competitive enzyme inhibition? a. Vmax is decreased from the uninhibited value. b. Km is unchanged from the uninhibited value. c. Vmax is the same.as the uninhibited value. d. Velocity is independent of substrate concentration. e. Km is decreased from the uninhibited value. answer 13. Enzymes affect the rate of a chemical reaction by: a. decreasing the free energy of the reaction. b. increasing the free energy of the reaction. c. lowering the energy of activation of the reaction. d. raising the energy of activation of the reaction. e. displacing the equilibrium constant. answer 14. Ingested glucose is phosphorylated in the liver by both hexokinase and glucokinase. Hexokinase has a Km for glucose of about 1 x 10-5 M and glucokinase has a Km for glucose of about 1.5 x 10-2 M. The glucose concentration available to liver cells is approximately equal to the concentration in the blood, and this value is about 135 mg% or 0.75 x 10-2 M. Assuming that the ATP concentration and other factors are optimal, the velocity of the glucokinase reaction will be about: a. 75% of Vmax. b. 66% of Vmax. c. 50% of Vmax. d. 33% of Vmax. e. 16% of Vmax.




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answer 15. The fact that an enzyme-catalyzed reaction is first order with respect to substrate at low substrate concentration but becomes zero order with respect to substrate at high substrate concentration is explained by: a. the catalytic constant. b. reversibility of formation of the enzyme-substrate complex. c. three point attachment. d. specificity. e. saturation. answer 16. If a competitive inhibitor of an enzyme is added to the assay mixture, the effect on a Lineweaver-Burk plot is to: a. decrease the slope. b. increase the slope. c. increase the slope and decrease the Y intercept. d. increase the Y intercept and decrease the slope. e. decrease both slope and the Y intercept. answer 17. The effect of pH on enzyme-catalyzed reactions can be related to all of the following EXCEPT: a. the influence of pH on the equilibrium. b. the need for certain forms of ionizable groups on the substrate. c. the need for certain forms of ionizable groups in the enzyme. d. the general effect of pH on protein structure. e. the influence of pH on the interaction of hydrophobic residues in the protein. answer 18. All of the following statements about enzymes are true EXCEPT: a. enzymes reduce the activation energy needed for a reaction to occur. b. pH affects enzyme activity by determining the degree of ionization of catalytic groups. c. turnover number is a measure of the maximal activity on a molecular level.




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d. the Michaelis constant Km is equal to the maximal velocity Vmax/2. e. induced fit may occur in nonallosteric enzymes. answer Answer the following questions using the key outlined below: A. If 1, 2, and 3 are correct B. If 1 and 3 are correct C. If 2 and 4 are correct D. If only 4 is correct E. If all four are correct 19. A sigmoidal substrate saturation curve implies: 1. that the enzyme must possess more than one subunit. 2. that the enzyme does not obey the Michaelis-Menten Equation. 3. a slower reaction velocity than hyperbolic kinetics. 4. that over some concentration range the change in enzyme activity will be greater than the change in substrate concentration. answer 20. For a certain enzyme, an irreversible, non-protein inhibitor is found that completely blocks enzymatic activity. The enzyme is reacted with the inhibitor until the enzymatic activity is exactly 1/2 its starting value, and the excess inhibitor is removed. Which of the following statements concerning the kinetic properties of this modified enzyme preparation are true? 1. Km remains unchanged. 2. Vmax remains unchanged. 3. The plot of v vs [S] is identical to the plot for the unmodified enzyme at 1/2 the enzyme concentration. 4. The kinetic behavior cannot be predicted from the information provided. answer 21. A homotropic allosteric effector: 1. is usually a substrate for the enzyme. 2. always gives positive cooperativity. 3. binds to an allosteric site which is also an active site. Faculty: P.M.D. Hardwicke



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4. binds to an allosteric site which is not an active site. answer 22. A non-competitive inhibitor causes: 1. an apparent decrease in Vmax. 2. no apparent change in Km. 3. the formation of an EIS complex. 4. a family of parallel lines in 1/V vs 1/[S] plots. answer 23. The Km for glucokinase is approximately 5 mM, while the Km for hexokinase is about 0.01 mM. At normal blood glucose (M.W. = 180) concentrations (80-100 mg/100 ml): 1. the activity of hepatic glucokinase is equal to one-half Vmax, while the hexokinase activity is near Vmax. 2. the activity of hepatic glucokinase is approximately 0.01 Vmax, while that of hexokinase activity is 100 Vmax. 3. glucokinase activity will be unaffected by glucose 6-phosphate levels, while hexokinase activity may be reduced. 4. the activity of extrahepatic glucokinase will be variable, depending upon the intracellular glucose concentration. answer 24. In enzyme assays, it is preferable to measure initial velocities in order to: 1. avoid substrate inhibition. 2. avoid product inhibition. 3. increase the sensitivity of the assay. 4. avoid the reverse reaction. answer 25. In clinical isozyme determinations on plasma, which quantity(ies) is/are significant? 1. Enzyme levels (concentration). 2. Relative amounts of the isozymes in the plasma. 3. Isozyme distribution pattern of various tissues. 4. Number of subunits of the enzyme. answer




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Answers to Practice Exam Questions 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25.

B A B A C D C E E C B B A

2.B 4.A 6.C 8.B 10.C 12.C 14.D 16.B 18.D 20.B 22.A 24.C

APPENDIX I: Using Acrobat Reader with pdf Files Portable Document Format (PDF) files can be read by Acrobat Reader, a free program which can be downloaded from the Adobe Web site (http://www.adobe.com/acrobat). If Acrobat Reader is installed on your system, it will automatically open simply by doubleclicking on the pdf file that you wish to read.

Acorbat Window

The document will be displayed in the center of your window and an index will appear at the left side of the screen. Each entry in the index is a hypertext link to the associated topic in the text. Using hypertext links in a pdf document is exactly like that in a web page or html document. When you place the cursor over a hypertext link, it changes to a hand with the index finger pointing to the underlying text. Clicking the mouse causes the text window to jump to that location. The index does not change. Magnification may need to be adjusted using the menu option in the lower part of the screen




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to optimize the view and readability. The best magnification is usually around 125%. Subheadings in the index can be viewed by clicking on the open diamonds to the left of appropriate entries to cause them to point downwards. Clicking again will close the subheadings lists.

Hypertext links

Hypertext links in the text (not in the index) are indicated by blue underlined text. The cursor should change to a hand with the index finger pointing to this text when it passes over it. Clicking will cause the text page to move to the associated or linked text which will be highlighted in red underlined text. Red underlined text is not a hyperlink, only a destination.

How to back up to a previous window:

If you wish to return to a previous text window after following a hypertext link, use the black double solid arrow key at the top of the Acrobat window (or use the key equivalent “command - “). Acrobat keeps a record of your last 20 or so windows so that multiple steps back can be made by repeating the command.

Links to web sites

A number of url links to web sites are located in the pdf file and appear in blue underlined type starting with http:// (e.g. http:// www.som.siu.edu). Clicking on these should open a web browser such as Netscape and take you to those web sites. You may need to resize the Acrobat Window to view the web browser window displayed underneath it.

COMMENTS I hope that you find this pdf file useful. Comments on how to make it better would be greatly appreciated. Please notify me in person or by email ([email protected]) of any errors so that they can be removed. The online version on the Biochem server can be easily updated.

