Download 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
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
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-
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 2
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BIOCHEMISTRY
Enzymes/Membrane Transport
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:
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 4
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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
Problem Unit 2 - Page 5
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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:
Faculty: P.M.D. Hardwicke
Binding sites
Coenzymes
Enzyme inhibitors
Kinetics
Biochemistry
Enzyme activation
Enzymes
Problem Unit 2 - Page 6
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 7
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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
Problem Unit 2 - Page 8
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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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 9
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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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Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
6.
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 12
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Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 14
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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
Problem Unit 2 - Page 15
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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-
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 16
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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
Problem Unit 2 - Page 17
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Enzymes/Membrane Transport
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
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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 <0.1 mM and liver glucokinase reaches 1/2 of its maximal velocity at [glucose] of ~5 mM.
[Please note that glucokinase (a monomeric enzyme) displays sigmoidal kinetics (it does not follow Michaelis-Menten kinetics) with a
Hill coefficient of 1.5 and therefore we denote a K0.5 instead.]
Because the Km is the substrate concentration that gives 1/2 Vmax
and the normal concentration of glucose in the blood is about 4.5
mM (80 mg/100 ml), the normal concentration in the blood is about
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45 times the Km for brain hexokinase. Thus, it is saturated (zeroorder) at all normal blood glucose concentrations. Because the brain
is almost totally dependent on blood glucose for its energy supply, it
is very efficient in glucose utilization, hexokinase making glucose
always available at an essentially constant rate, a very necessary condition.
On the other hand, 4.5 mM approximates the K0.5 for liver glucokinase. It is acting within the region of the v versus [S] curve in which
the rate would change with any change in glucose concentration; a
situation in keeping with the physiological role of the liver in controlling blood glucose levels.
However, this says nothing about the total amount of glucose converted to glucose-6-phosphate and thus taken up respectively by the
two tissues. What determines that is the relative Vmax's, the relative
rates of the conversions catalyzed by the two enzymes at 4.5 mM glucose, and the relative concentrations of the two enzymes in their
respective tissues.
To cite another example, drinking methanol can be fatal. Methanol
itself is not toxic, but it is converted by alcohol dehydrogenase into
formaldehyde, which is fatal. (Alcohol dehydrogenase of the liver, an
enzyme requiring NAD as a coenzyme, normally oxidizes ethanol to
acetaldehyde.) One treatment for methanol poisoning is to give large
amounts of ethanol. If one were to ask, "Why get the patient
drunk?", the answer would be "because the Km of alcohol dehydrogenase for ethanol is lower than the Km for methanol". Thus, flooding
the system with ethanol can competitively inhibit oxidation of methanol until it is excreted, an example of the use of an alternative substrate as a competitive inhibitor.
VII. What is the effect
of the concentration of
coenzymes and activators?
As will be discussed later, many enzymes require inorganic ions for
activity, coenzymes that are cosubstrates and act as acceptor or donor
molecules, and/or coenzymes that are cocatalysts. The concentration
of these cofactors, if limiting, will affect the rate of the enzyme-catalyzed reaction. Hence in any determination of enzyme activity, the
concentrations of required cofactors must be in amounts that are in
excess of the stoichiometric requirements.
There are four terms used to refer to the nonprotein components of
an enzyme that influence the rate of the catalyzed reaction. They are
as follows:
Cofactor--nonprotein component of an enzyme (which can be a
metal ion or a small organic molecule) required for catalytic activity
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by the enzymes. The component can be loosely or tightly linked to
the enzyme.
Prosthetic group--small organic compound more or less tightly
linked to the protein and required for efficient performance of catalytic function.
Coenzyme--low molecular weight, dialyzable, organic compound
required for the action of some enzymes. A hazy distinction is drawn
between coenzyme and prosthetic group with those "tightly bound"
to enzyme termed a prosthetic group. (Vitamins are coenzymes that
must be obtained through the diet.)
Coenzymes may be structurally altered in the course of reaction but
in subsequent reactions (with other enzymes) the original structure is
regenerated. This distinguishes coenzyme from substrates.
Activators are not required for enzyme activity but can increase
enzyme activity by putting the enzyme molecule in the proper state
to combine with the substrate or remove inhibitors.
VIII. What are competitive and noncompetitive inhibitors?
What is their effect on
enzyme-catalyzed reactions?
Studies of the effects of enzyme inhibitors on the kinetics of enzymecatalyzed reactions are a most important part of biochemistry and
pharmacology.
As has been discussed above, enzymes are catalysts. Therefore, as the
reaction proceeds, the concentration of products continuously and
stoichiometrically increases at the expense of the disappearance of the
corresponding substrates, while, ideally, the total concentration of
enzyme (in all its possible forms) remains fixed and invariant. Thus,
a consideration of the rate behavior or kinetics of such reactions furnishes a powerful tool, not only for their determination, but also for a
definition of their properties and possible modes of action. On the
molecular level, we are always dealing with interrelations between
structure and function. Kinetics provides us with the most important
(and with impure preparations, the only means of coming to grips
with enzyme function).
Selective inhibitions by natural or synthetic compounds (antimetabolites) forms the base of a broad approach to pharmacology and chemotherapy. Antimetabolites will be considered in the module on
metabolic integration and regulation.
Before we discuss inhibitors and their action, we must first discuss in
more detail the mechanism of enzyme action. Each enzyme molecule
has an active site. This active site is the part of the enzyme molecule
that interacts directly with the substrate, determining both the speci-
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ficity (high affinity for specific compounds) and the catalytic function
of the active protein molecule. The active site (or active center) is frequently spoken of in terms of binding sites that hold the substrate in
a specific conformation and the catalytic site where reaction takes
place. Active site = binding sites + catalytic site.
It is reasonable to consider the active site as composed of several
amino acid residues (at least 3 to explain stereospecificity) brought
together in the specific, active, geometrical pattern by the folding of
the polypeptide chain, and thus being determined by weak (secondary and tertiary) bonds as well as the primary amino acid sequence.
Enzymes are often much, much larger than their substrates; at least
they are much larger than the part of the substrate with which they
interact. Therefore, only a small portion of the enzyme can be in
contact with the substrate. Sometimes part of the protein molecule is
dispensable, sometimes not. The bulk of the enzyme may be required
to bring the active site in contact with the substrate. It certainly is
required in most cases to put the reactive groups into proper juxtaposition to each other.
Emil Fischer pictured a "lock and key fit" between the enzyme and
the substrate. Koshland and others have suggested that some
enzymes change their shape on combination with their substrate, i.e.,
the substrate induces the enzyme to conform to the binding geometry
("induced fit" hypothesis).
Reactive groups frequently found in the catalytic site are the CH2OH group of serine, the -CO2H group of aspartic and glutamic
acids, the imidazole group of histidine, and the -CH2SH group of
cysteine. The groups involved can be determined by kinetic analysis
(use of multiple substrates and inhibitors and investigation of pH
dependence), chemical modification (reaction with group specific
reagents and specific labeling of the enzyme with substrate or substrate analogues), and X-ray analysis of crystalline ES complexes or
compounds.
There are two general types of reversible enzyme inhibition; competitive and non competitive.
In competitive inhibition, the inhibitor competes with the substrate
for the active site of the enzyme. Thus, competitive inhibitors are
usually structurally similar to the substrate. Product inhibition is
quite common.
In competitive inhibition, inhibition depends on the relative concentrations of the inhibitor and substrate. One such example is the use
Faculty: P.M.D. Hardwicke
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of sulfa drugs to block the synthesis of folic acid.
CO2-
H2 N
p-aminobenzoate
O
H2 N
S-NH2
O
sulfanilamide
Sulfa drugs, such as sulfanilamide, which are structurally similar to
the B vitamin, p-aminobenzoic acid, are effective competitive inhibitors in the synthesis of 7,8-dihydropteroate, and because some microorganisms make their own folic acid, another B vitamin, provide an
effective means to prevent their growth.
In this case, the initial reactions are
E+S
ES
E+I
EP
E+P
EI (no further reaction)
The Michaelis-Menten equation becomes
V max [ S ]
v = -------------------------------------------[I]
K m  1 + ------ + [ S ]


Ki
[E][I]
where K i = --------------[ EI ]
from which it can be seen that, if [S] >> [I], the effect of the inhibitor
can be overcome.
To test for competitive inhibition, Km is determined in the absence of
the inhibitor, then increasing concentrations of inhibitor are added
and a Lineweaver-Burke plot is made (Figure 9).
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Increasing [I]
1/v
[I] = 0
-1/{Km(1 + [I]/KI)}
1/Vmax
1/[S]
Figure 9
An increasing slope without a change in the y-intercept (without a
change in Vmax which remains constant because high concentrations
of S can overcome the effect of the inhibitor) is evidence for competitive inhibition.
In noncompetitive inhibition, the initial reactions are
E+S
ES
EP
E+I
EI
E+P
Both Ki’s are the same.
ES + I
ESI
where in the formation of ES and EI, respectively, S and I are reacting
with independent sites.
EI + S
ESI is also possible. In the
Lineweaver-Burke plot, as [I] increases, both the slope and the yintercept increase (Figure 10).
V max [ S ]
v = ------------------------------------------------[I]
( K m + [ S ] )  1 + ------

Ki 
Faculty: P.M.D. Hardwicke
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slope = (Km/Vmax)(1 + [I]/KI)
Increasing [I]
1/v
1/Vmax(1 + [I]/KI)
[I] = 0
1/[S]
Figure 10
-1/Km
Some inhibitors are true competitive inhibitors, but many others are
neither true competitive nor noncompetitive inhibitors. Hence,
other kinds of inhibition (uncompetitive) have been hypothesized
and treated mathematically in attempts to explain the observed Lineweaver-Burke plots.
IX. What is the effect
of irreversible inhibition?
There are at least two classes of inhibitors that form covalent or very
tight linkages with the enzyme and effectively prevent catalysis.
In many enzymes, sulfhydryl (-SH) groups are required for activity
and the reaction of heavy metal ions (e.g. silver, mercury, lead, etc.)
with sulfhydryl groups on these enzymes render them irreversibly
inactive. The effect of this type of inhibitor on enzyme kinetics is to
decrease the apparent Vmax without affecting the Km. Consequently,
the kinetic plots that occur for this class of irreversible inhibitors are
identical to those observed for noncompetitive reversible inhibitors.
This has led to confusion as to the differences between reversible
noncompetitive and irreversible inhibition. Reversible noncompetitive inhibitors can be distinguished from irreversible inhibitors by
dialyzing the enzyme preparation containing the inhibitor. Full
enzyme activity will be regenerated on dialysis of the enzyme preparation containing a non-competitive inhibitor but not with an enzyme
preparation containing an irreversible inhibitor.
Reversible noncompetitive inhibition:
E+S
ES
I
EI + S
Faculty: P.M.D. Hardwicke
E+P
I
EIS
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Irreversible inhibition:
E+S
ES
E+P
I
EI
In a number of diseased states, the activity of one or more enzymes is
often greater than needed to maintain normal metabolic balance. In
the past few years, a strategy for drug design has developed in which
irreversible inhibitors are used to maintain control over the metabolic
imbalance. These irreversible inhibitors are referred to as "suicide"
inhibitors or "mechanism-based" inhibitors. The rationale for this
therapeutic strategy is to design a molecule that looks very much like
the substrate, but when the enzyme binds this molecule and begins
the catalytic process, it creates a species that is chemically more reactive than the normal reaction intermediate. This intermediate must
be reactive enough to covalently modify a residue in the active site or
a bound coenzyme rather than form a product metabolite molecule as
does the normal substrate molecule.
Let us consider the action of the suicide substrate gabaculine on the
pyridoxal phosphate requiring bacteria enzyme, GABA transaminase.
CO2-
NH3+
GABA
CO2-
NH3+
gabaculine
The carbon skeleton of GABA forms part of the structure of gabaculine and the enzyme will accept gabaculine into its active site. The
enzyme tries to transaminate gabaculine and in the process it irreversibly modifies its pyridoxal moiety. Gabaculine renders inactive every
enzyme molecule that takes part in this reaction.
Recent examples of target enzymes for which suicide substrates have
been designed include: (a) aromatase, the key enzyme in the interconversion of androgenic and estrogenic steroid sex hormones;
(b) GABA transaminase, an enzyme target for antiepileptic drugs;
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and (c) DOPA decarboxylase, an enzyme involved in regulation of
blood pressure.
Enzyme
pyridoxal phosphate
CO2-
CO2-
NH3
H
NH+
+
2-
H
OH
3 OPO
N+
H
CH3
CO2-
:B
NH2+
2-
O-
3 OPO
N+
H
CH3
Knowledge of the mechanistic details of the enzyme-catalyzed reaction is used to design the irreversible inhibitor (mechanism-based
substrate) and cause the enzyme to self destruct (thus the name suicide inhibitor). In principle, if the suicide substrate is designed so
that only the targeted enzyme can begin catalysis to produce the reactive species, then a drug with maximum specificity and minimized
side effects should result. Any class of enzymes including proteases
and phosphatases, in which a normal intermediate can be re-routed
to mechanism-based auto destruction, should be amenable to targeting. In point of fact, the action of several classes of therapeutic agents
in current use can be explained in molecular terms by the suicide substrate analysis.
X. What is the effect
of temperature?
Faculty: P.M.D. Hardwicke
As with other chemical reactions, the rates of enzyme-catalyzed reactions are increased by increases in temperature. However, because
enzymes are proteins, thermal denaturation of the protein with
increasing temperature will decrease the effective concentration of the
enzyme. The result on an enzyme-catalyzed reaction is depicted in
Figure 11. The upper limit for most human enzymes is 40 to 50˚C
(normal body temperature 37˚C).
Problem Unit 2 - Page 30
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XI. What is the effect
of pH on enzyme-catalyzed reactions?
BIOCHEMISTRY
Enzymes/Membrane Transport
Changes in pH will change the ionization of the enzyme that is, of
course, amphoteric; for example,
H+
+
E+S
H+
+
ES (EP)
EH+
EHS+
E+P
If the substrate possesses ionizable groups, as many do, its ionic state
will change with changes in pH; for example, focusing on the substrate ionization.
H+
+
HE + S-
E- + H + SH
H+
+
HES- (HEP-)
HESH+
H+
+
HE + P-
E- + H + PH+
Thus, as the ionic nature of either the substrate or the enzyme is
changed from that form that combines with the other, the reactions
will be slowed and a curve such as that in Figure 12 will result.
To illustrate one example of how a bell-shaped pH profile might
occur, let's consider an example in which the enzyme has a group at
the active site which must be positively charged in order for substrate
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to bind, and the pK of the group is, e.g., 8.0. Furthermore, let us say
that the substrate has to be negatively charged in order for it to be
attracted to the enzyme active site and the pK of the group on the
substrate is, e.g., 4.0.
At low pH the substrate is protonated (SH) and will not want to bind
to the enzyme that also is in its protonated form (EH+). As the pH is
increased we remove the proton from substrate to form S-, which is
the form that will bind to enzyme. Thus, we will see an increase in
enzymatic activity proportional to the fraction of substrate molecules
in the S- form. At pH 6 we should have essentially 100% of the substrate as S- and maximum catalytic activity will be observed. As we
go to pH values higher than 6, we begin to remove the proton from
the group at the enzyme active site. There will be a decrease in activity proportional to the amount of unprotonated enzyme present
because the unprotonated form will not bind substrate (S-). The
result is the bell-shaped pH profile seen in Figure 12.
The pH optimum may be either narrow or broad. As the pH
increases or decreases above or below the optimum pH, the activity of
the enzyme (the velocity of the enzyme-catalyzed reaction) will
decrease because:
a.
There will be decrease in enzyme substrate binding due to repulsion or to loss of charge on either the substrate or the enzyme.
b. the catalytic functional groups will be in the wrong ionization
state.
c. The enzyme may become denatured.
XII. What are allosteric enzymes and
allosteric effectors?
Faculty: P.M.D. Hardwicke
Allosteric enzymes are enzymes whose kinetic properties cannot
be accounted for by the Michaelis-Menten model. The activity of
allosteric enzymes is altered by regulatory molecules, and there is a
cooperative binding of substrates.
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With allosteric enzymes, an inhibitor or an activator (termed negative and positive effectors, respectively) interacts at a site other than
the active site on a multi-subunit protein. These effector molecules
are quite different in shape and structure in comparison with the substrate molecules. The sites at which the allosteric effectors bind are
called regulatory or allosteric sites.
Allosteric control is dependent on binding sites for the effectors and
on the existence of at least two conformational states of the enzyme,
one in which the binding sites have a high affinity for the substrate,
and another in which they bind weakly, if at all, with the substrate.
One model of the action of allosteric enzymes is that binding of the
effector molecules stabilizes the enzyme in one of its possible conformation. Binding of a negative effector stabilizes the enzyme in an
inactive state, and binding of a positive effector stabilizes the enzyme
in its active state. This can be seen diagrammatically in Figure 13.
Sometimes allosteric enzymes have regulatory subunits and catalytic
subunits. This kind of inhibition is of the greatest physiological significance because of its importance in regulation of biochemical reactions. The effector molecules can be hormones, the immediate
results of the action of a hormone (with polypeptide hormones the
result is usually cyclic AMP), or intermediates or the end-product of
a pathway. Several examples of the different kinds of control involving allosteric enzymes will be covered as you progress through basic
Faculty: P.M.D. Hardwicke
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metabolism.
XIII. Why do enzyme
assays?
The thing you as a physician will be most interested in is the enzyme
concentration. This parameter varies from patient to patient and is
important in diagnosis. In some cases, a low concentration is significant (e.g., erythrocyte glucose-6-phosphate dehydrogenase is
decreased in some cases of non-spherocytic hemolytic anemia), and
in some cases a high concentration is important (e.g., serum creatine
phosphokinase is increased following a myocardial infarction).
Enzyme concentration is not reported in milligrams per liter or milliequivalents per liter, for this type of measurement cannot be done
on blood samples, or other impure fluids. Therefore, enzymes are
measured by the work they do, i.e., the amount of substrate converted to product (the activity).
Enzyme results are reported in activity units and it is extremely
important to know the methodology used and the normal values for
that method. Because there are several methods for assaying most
enzymes, the activity of the enzyme will depend on the methodology
chosen by the particular hospital; and results from one hospital to
another cannot necessarily be compared unless identical methods are
used. This problem is evident at hospitals into which patients are
transferred from other hospitals.
It is hoped that, in the future, some uniformity can be achieved in
reporting enzyme activities so that the problem of comparing results
from one institution to another can be overcome. Two attempts to
achieve uniformity were by the use of international units and the
katal. An international unit (IU) is what was defined earlier as a unit,
i.e., the amount of enzyme activity that will catalyze the conversion
of one micromole of substrate to product in one minute, while one
katal equals the conversion of one mole of substrate per second. As
has been discussed, the pH, temperature, and other factors affecting
enzyme activity must be specified and carefully controlled in determining units; their control is the responsibility of the laboratory. It
may be many years before a uniform system of reporting enzyme
results can be implemented; until then remember that you cannot
necessarily compare results from one institution to another. You
must compare the results from any hospital with the normals established for that hospital.
Now, let us consider how enzymes get into the serum. The concentration of an enzyme in serum is a result of (a) its rate of release from
the tissue and (b) the rate at which it is removed from the circulation.
There is some leakage of enzymes into serum from normal cells.
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When a cell is damaged, it releases enzymes into the serum at a rate
faster than it can be removed, and therefore, an increase in the serum
concentration is noted. The nature and the amount of the enzyme
released will depend on the type of cell injury that has occurred.
Some mild degree of cell injury, such as is caused by progressive muscular dystrophy, will increase plasma membrane permeability. This
increased permeability will result in a partial loss of soluble enzymes
from the cytoplasm to the serum, in turn resulting in a moderate
serum elevation.
Cell necrosis, such as is caused by myocardial infarction, acute hepatitis and acute pancreatitis, will result in a complete loss of all cellular
enzymes, including those in organelles such as mitochondria. This
complete release of cellular enzymes results in high serum levels.
An increased number of cells with an accompanying increased production of enzymes, such as occurs in neoplastic disease, can give
moderate to high serum levels of enzymes.
To take a brief look at how enzyme assays are used in diagnosis, we
can consider the following. Some enzymes are fairly localized; that is,
only specialized tissues have a high concentration. Amylase is an
example of this type of enzyme. It is found in high concentration
only in the pancreas, and therefore, a high serum amylase concentration reflects specific pancreatic damage. Other enzymes such as lactic
dehydrogenase are present in almost all tissues, and thus, total LDH
analysis will provide less information about specific tissue damage.
However, as we will see in a later section, determination of the LDH
isoenzyme pattern, although more difficult to do than an ordinary
enzyme assay, can be an important diagnostic tool.
XIV. How does one
assay an enzyme?
Specific activity can be determined in one of two ways. In the "onepoint method," the reaction is allowed to proceed for a time and then
the enzyme is denatured by acid, base, or heat. The reaction is then
analyzed by a chemical or physical method to see how far it has gone.
In the other method, data are obtained continuously without stopping the reaction. This is usually done through physical methods
such as spectrophotometry and polarimetry, although titration may
also be used. The rate is then calculated and converted to specific
activity.
Before a procedure can be used as an assay, it must be shown that the
enzymic reaction is linear over the time period studied and that the
reaction rate is proportional to the amount of enzyme present.
Faculty: P.M.D. Hardwicke
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An example of an enzyme assay is the determination of alkaline phosphatase in human serum. The phosphatases are generally located in
the cytoplasm of the cell; alkaline phosphatase is a microsomal
enzyme. (Microsomes are fragments of endoplasmic reticulum isolated after tissue homogenization and differential centrifugation.)
Phosphatases catalyze the hydrolysis of phosphate esters with liberation of inorganic phosphate.
Clinical assay procedures make use of synthetic substrates. For example, the Bessey-Lowry method used p-nitrophenyl phosphate which,
in the presence of the enzyme, undergoes hydrolysis to yield p-nitrophenol and inorganic phosphate. The rate of reaction is determined
by measuring the change in absorbance at 405 nm, the absorbance
maximum for the product.
Product is measured because an excess of substrate is used. Small
changes in concentration can be determined with much, much less
error when the initial concentration is zero than when the initial concentration is some large value as it would be if disappearance of substrate were to be measured. also, measurement of product formation
is more desirable than measurement of substrate disappearance
because the fluids assayed contain many enzymes and, hence, substrate disappearance may occur via other reactions.
The normal values vary according to the method used. With the
Bodansky method the normal adult serum activity is 1.5 to 4.0
Bodansky units. With the King-Armstrong method, the normal
range is 1.35 to 4.0. These are both arbitrary units. In International
Units (Bessey-Lowry method), the normal range is 20 to 85 mU/ml.
It is essential that the serum sample be properly processed. Great
variation may occur in the assay results if certain precautions are not
observed. Things that may alter activity are storage (even under
refrigeration), losses or increases in dissolved carbon dioxide, and
addition of excessive oxalate or fluoride to the specimen.
The level of alkaline phosphatase is higher in growing children than
in adults. There are many other physiologic and pathologic conditions in which there is an increase in the activity of serum alkaline
phosphatase and still others that lead to lower than normal adult levels.
The use of glucose oxidase to determine glucose in the blood and
urine is an example of the use of an enzyme for assay and of the use of
coupled enzyme reactions. The reactions are as follows:
glucose oxidase
Faculty: P.M.D. Hardwicke
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D-glucose + O2 → D-gluconolactone + H2O2
peroxidase
H2O2 + reduced dye → oxidized dye + 2H2O
(colorless)
(green)
The intensity of the green color is proportional to the original glucose concentration.
XV. How are enzymes
named?
There are four ways of naming enzymes. Three give the enzyme a
trivial name. The first enzymes discovered were given names that
have no relationship to their activity; these are largely digestive
enzymes (e.g., pepsin, trypsin, chymotrypsin). Another way to name
enzymes is to name the substrate and add the suffix :"ase" (e.g., maltase, fructose-1,6-disphosphatase, fumarase, ribonuclease, urease) or
the reaction alone can be named and the suffix "ase" added (e.g.,
invertase). A third way names the substrate and the reaction catalyzed and adds the suffix "ase" (e.g., malic dehydrogenase, glutamicoxaloacetic transaminase, hexokinase). (Kinases are enzymes that
transfer a phosphate group from ATP to a substrate.)
There is now an international system of enzyme classification that has
specific and systematic rules for naming enzymes. This method
assigns to each enzyme a number that is an indication of its activity.
The six main groupings are as follows:
1.
2.
3.
4.
5.
6.
oxidoreductases (oxidation-reduction reactions)
transferases (transfer of functional groups)
hydrolases (hydrolysis reactions)
lyases (addition to double bonds)
isomerases (isomerization reactions)
ligases (formation of bonds with ATP cleavage)
The international systematic name describes the reaction, substrates,
and products in question. The trivial name is used in general reference where the longer systematic name would be cumbersome.
As an example of the difficulty, the commonly used names for the
two clinically significant transaminases described above are
glutamate-oxaloacetate transaminase (glutamic-oxaloacetic transaminase) (SGOT) and glutamate-pyruvate transaminase (glutamic-pyruvic transaminase) (SGPT). (S stands for serum). More meaningful
common names for these particular enzymes (to be discussed in a
later domain) would be aspartate aminotransferase (AST) and alanine
aminotransferase (ALT), respectively.
Faculty: P.M.D. Hardwicke
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Abbreviations are another matter. In many cases two or more abbreviations are used for the same enzyme. The names and abbreviations
used in this module are ones that are, more or less, in general usage.
However, we will have to live with the confusion in nomenclature
and abbreviations, with the fact that biochemists and pathologists
may use different names for the same enzyme, and with the fact that
names and abbreviations may change from one hospital laboratory to
another.
XVI. Do enzymes
have prosthetic groups?
Sometimes an organic molecule is covalently linked to the protein
molecule (the apoenzyme) forming a conjugated protein (the holoenzyme). In this case, the nonpeptide part of the protein is called a
prosthetic group. The addition of a prosthetic group occurs after initiation of protein synthesis and usually considered as one form of
post-translational modification.
The prosthetic group of an enzyme generally has the same function as
that of a coenzyme. Popular examples of prosthetic groups include
FAD of the flavoenzymes and TPQ (trihydroxyphenylalanine
quinone) of serum amine oxidase. However, proteins other than
enzymes have prosthetic groups, for example, the heme group found
in hemoglobin.
XVII. What are isoenzymes?
Some proteins are associations of several polypeptide chains. Each
individual chain is termed a subunit; thus, a subunit is any polypeptide chain in the completed functioning protein that is not covalently
bound via a peptide linkage to other peptide units and can thus be
readily separated from other subunits. A protomer is a subunit in a
protein containing a finite number of identical subunits. Oligomers
are combinations of similar or different subunits to form the totally
functioning protein. Allosteric enzymes have such a structure. Those
enzymes consisting of a single polypeptide chain are termed monomeric enzymes. Those enzymes consisting of two or more polypeptide chains are termed oligomeric enzymes.
Some enzymes occur in multiple molecular forms that catalyze the
same reaction in the same organism. These forms are known as
isoenzymes, or more simply as isozymes, i.e. isoenzymes are different proteins within an organism with similar enzymic activity. The
most thoroughly studied of these enzymes is lactate dehydrogenase
(LDH), which can occur in five hybrid forms (LDH1 to LDH5), each
of which catalyzes the reaction:
Faculty: P.M.D. Hardwicke
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Enzymes/Membrane Transport
CO2
C
CO2
O
+ NADH + H+
HO C
CH3
H
+ NAD+
CH3
pyruvate
lactate
When serum of a tissue extract is subjected to electrophoresis on any
usual electrophoretic medium (agar, starch acrylamide gel, cellulose
acetate), five zones of LDH activity are seen.
These LDH isoenzymes are tetramers (M.W. about 135,000 daltons)
that can be split into four monomeric subunits (H and M subunits)
(M.W. about 32,000).
There are other enzymes of clinical importance that exist as isoenzymes (for example, creatine phosphokinase, CPK1, CPK2, and
CPK3).
Appendix: Derivation of Km and
Relation of Km to [E]
The simplified equation for an enzyme-catalyzed reaction is:
k1
E+S
k2
E+P
ES
k-1
k-2
However, in an enzyme assay, only the initial rate is measured so that
in the limit of t = 0, the concentration of product is zero. Therefore,
there is no back reaction and the equation can be simply written.
k1
E+S
k2
ES
E+P
k-1
Michaelis and Menten and Henri defined a dissociation constant for
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 39
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Enzymes/Membrane Transport
ES
k –1 + k 2
and when k2 << k-1
K m = ------------------k1
k –1
[E][S]
= ----------------then K m = ------[ ES ]
k1
However, k2 is not always negligibly small in comparison with k-1.
An exact derivation (made by Briggs and Haldane), in which we do
not have to assume k2 << k-1, involves the assumption that the reaction is in steady state; i.e., we assume that, under the conditions
under which the rate of reaction is measured, the concentrations of S
and ES are constant. In other words, rate of formation of ES = rate of
disappearance of ES. Thus, for the mechanism given:
v1 = v-1 + v2
k1 [E] [S] = k-1 [ES] + k2 [ES]
[E]t = total enzyme concentration
[S]t = total substrate concentration
but [S]t >>> [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 ] 
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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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Problem Unit 2 - Page 41
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Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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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
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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.
Faculty: P.M.D. Hardwicke
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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Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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.
Problem Unit 2 - Page 46
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Enzymes/Membrane Transport
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, Km = Ks
Conditions where Km not equal Ks
If k2 is approximately equal to k-1
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.
Enzymes/Membrane Transport
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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Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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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.
Faculty: P.M.D. Hardwicke
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
Problem Unit 2 - Page 51
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BIOCHEMISTRY
Enzymes/Membrane Transport
= 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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 52
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BIOCHEMISTRY
S (M)
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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.
Problem Unit 2 - Page 53
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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
Problem Unit 2 - Page 54
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BIOCHEMISTRY
6.
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 55
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Enzymes/Membrane Transport
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
Problem Unit 2 - Page 56
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 57
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Enzymes/Membrane Transport
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 59
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
a. Liver
b. The [S]Vmax/2 is close to the Km of the liver enzyme
Problem Unit 2 - Page 60
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BIOCHEMISTRY
2.
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
a. Km = 8.3 x 10-6 M, Vmax = 45.5 µmoles min-1
Problem Unit 2 - Page 61
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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]
Faculty: P.M.D. Hardwicke
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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Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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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)
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 64
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 66
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Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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)
Problem Unit 2 - Page 67
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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.
Nomenclature and
Vocabulary:
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 68
SIU School of Medicine
Self-Test
BIOCHEMISTRY
Enzymes/Membrane Transport
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?
Faculty: P.M.D. Hardwicke
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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
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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+,
Faculty: P.M.D. Hardwicke
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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
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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-
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 78
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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
Faculty: P.M.D. Hardwicke
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
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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
Faculty: P.M.D. Hardwicke
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
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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
Faculty: P.M.D. Hardwicke
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
Enzymes/Membrane Transport
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)
Faculty: P.M.D. Hardwicke
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
Faculty: P.M.D. Hardwicke
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.
Problem Unit 2 - Page 86
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 87
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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,
Problem Unit 2 - Page 88
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
which forms three isozymes. MM is found in skeletal muscle,
MB found in the heart, and BB found in brain and thyroid.
Faculty: P.M.D. Hardwicke
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.
Problem Unit 2 - Page 89
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
Module 3: Membrane Transport
Objectives:
Nomenclature and
Vocabulary:
Faculty: P.M.D. Hardwicke
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
Problem Unit 2 - Page 90
SIU School of Medicine
Key Words:
Faculty: P.M.D. Hardwicke
BIOCHEMISTRY
Enzymes/Membrane Transport
Adenosine Triphosphatase, Sodium, Potassium
Biochemistry
Biological Transport
Cell Membrane Permeability
Ion Channels
Membrane Potentials
Membranes
Permeability
Problem Unit 2 - Page 91
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
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-
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 92
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 93
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 94
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Facilitated (Mediated)
Passive Diffusion and
Active Transport
BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 96
SIU School of Medicine
BIOCHEMISTRY
x
y
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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-
Problem Unit 2 - Page 98
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
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
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 101
SIU School of Medicine
Faculty: P.M.D. Hardwicke
BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Problem Unit 2 - Page 102
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 103
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 104
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BIOCHEMISTRY
6.
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
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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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 106
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Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 107
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Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 108
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Problem Unit 2 - Page 109
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BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 110
SIU School of Medicine
BIOCHEMISTRY
Enzymes/Membrane Transport
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
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 111
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BIOCHEMISTRY
Enzymes/Membrane Transport
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.
Faculty: P.M.D. Hardwicke
Problem Unit 2 - Page 112