Astrakhan State Medical University
Астраханский государственный медицинский
университет
T.B.Vorobyeva, D.M.Nikulina
Т.Б.Воробьева, Д.М.Никулина
THE MANUAL
ON BIOCHEMISTRY
РУКОВОДСТВО К ЗАНЯТИЯМ
ПО БИОХИМИИ
Part 1
Часть 1
ASTRAKHAN - 2016
АСТРАХАНЬ – 2016
UDC 577.1(076)
LBC 28/902
V75
T.B.Vorobyeva, D.M.Nikulina. The manual on biochemistry. Part 1.
– Astrakhan, Astrakhan State Medical University, 2016, 87с.
English editor S.S.Uzhakina, senior lecturer of Department of Foreign languages, ASU
The training guide contains information which is necessary for individual and group work of students in the classroom and at home.
Reviewers: V.R. Gorst, Doctor of Medicine, Professor of Physiology
Department, ASMU
P.V.Loginov, Candidate of Biology Sciences, Associated
Professor of Chemistry Department, ASMU
T.S. Kirillova, Doctor of Philology, Professor, Chief of
Foreign Languages Department, ASMU
ISBN 978-5-4424-0140-0 (основной)
ISBN 978-5-4424-0192-9 (часть 1)
 T.B.Vorobyeva, D.M.Nikulina.
 Astrakhan State Medical University, 2016
2
RULES ON SAFETY FOR STUDENTS
WHEN WORKING IN LABORATORIES
1. Only the students who have passed the training on handling
of chemicals and biopsy materials are allowed to work in the labs.
2. In the laboratories of the department it is allowed to work
only in a lab coat with long sleeves. Long hair should be under the
cap.
3. All procedures while performing work for the preparation
of solutions, the measuring reagent (transfusion) should only be
conducted in the workplace on the table (or in the exhaust case). It is
prohibited to carry out any manipulation of the reagents above the
floor.
4. You must be very careful when working with strong acids
and alkali solutions (10% and above). Special care and caution is
needed when working with concentrated acids and alkalis and other
aggressive substances (acetic anhydride, phenol, perhydrol, etc.).
The admeasurement of these reagents should be performed using an
automatic pipette (not mouth, avoid chemical burns of the oral cavity). Avoid getting these chemicals on the skin (burns) and clothes
(damage of fabrics). Work with these substances should be carried
out in rubber gloves. Upon dilution of strong acids to prevent splashing should add acid to water and not on turnover.
5. Spilled alkali you need to add sand or sawdust. Then sand
or sawdust to clean that place to pour the much diluted acid (hydrochloric, acetic).
If acid is spilled, sand (not sawdust) is to be added and then
sand should be removed and soda should be added, then soda should
be removed. After that the place should be washed with plenty of
water (wear rubber gloves).
6. When heated test tubes are boiling and the content of it may
release to a distance of 2-3 m, it is especially dangerous in case of
contact of boiling spray with face and eyes. You should take care
that the hole of the tube not to be directed at yourself or anyone else
in the lab. You cannot close a hot solution tube to cool down.
3
7. When working with the centrifuge the lid should be closed
tightly. The speed of rotation can be increased only gradually. Open
the lid only after a complete stop of the rotor.
8. All questions of safety regarding the work, you should immediately ask to the teacher or lab assistant.
4
THE CONTENT OF THE CLASS
I. Theoretical part.
I.1. The control of initial level of knowledge.
I.2. Discussion of the basic matters of topic.
I.3. Test control.
II. Practical part.
II.1. Laboratory works.
II.2. Independent work with the methodological material.
II.3. The presentation of research results.
III. Final part.
III.1. Analysis of the obtained results.
III.2. Individual report on the laboratory work.
III.3. Assignment for next class.
III.4. Individual work of the teacher with students
MATERIAL FOR SELF-STUDY
I. LITERATURE
1. U. Satyanarayana //Biocnemistry. Books and allied (p) Ltd.
8/1 Chintamoni Das Lane, Kolkata, 700009 (India), 2004.
2. Robert.K.Murray, D.K.Granner, P.A.Mayes, V.W.Rodwell
//Harper’s Illustrated Biochemistry.26-th edition. McGraw-Hill
Companies, Inc., USA, 2003.
3. David E. Metzler, Carol M. Metzler, David J. Sauke // Biochemistry: the chemical reactions of living cells. (Edition 2-nd),
2001.
4. A. Lehninger, D. Nelson, M. Boyle Cox //Principles of biochemistry (4th ed.), 2005.
II. LECTURE NOTES
5
ESTIMATION CRITERIA OF STUDENT
KNOWLEDGE
The student rating:
EXCELLENT - with in-depth knowledge of program material, demonstrated by: a) comprehensive answer; b) literate fluency in
theory when solving the logical problems and practical tasks; C) familiarity with additional literature.
WELL - with sure knowledge and competent presentation of
program material, the lack of significant inaccurate-scribed in the
answer, the correct application of theoretical positions in solving
practical tasks.
SATISFACTORY - with knowledge only of the main provisions of the program material and incomplete mastery of its parts,
characterized by the discrepancies of formulations is insufficient
competent and consistent response, difficulty in solving problems
UNSATISFACTORY - in the absence of knowledge for a
large portion of the program material, with considerable errors in the
answer, the inability to use the theoretical material in the solution of
logical problems.
6
THE GENERAL BIOCHEMICAL
PROCESSES IN LIVING ORGANISMS
SECTION I. STRUCTURE AND PROPERTIES OF PROTEINS
Class 1
THE BIOLOGICAL ROLE AND THE PHYSICAL-CHEMICAL
PROPERTIES OF PROTEINS. METHODS OF EXTRACTION
AND FRACTIONATION OF PROTEINS
The aim of the lesson:
1. To revise physical and chemical properties and to know biological role of proteins.
2. To know the principle of separation of proteins into fractions
by the salting out method.
Initial level of knowledge:
- acids and bases in organic chemistry;
- chemical properties of a carboxylic group;
- the mechanism of nucleophylic replacement at a trigonal atom
of carbon;
- chemical properties of the amino group: the basicity and nucleophylicity;
- structure, amphotericity and solubility of amino acids;
- colloidal solutions and their properties.
Main topics.
I.2. The elemental composition of proteins.
Classification and the nomenclature of amino acids that build proteins.
Functions of proteins.
Specific physical and chemical properties of proteins.
Methods for the isolation and purification of proteins.
7
II.1. Work. SALTING-OUT OF PROTEINS
Salting-out is the precipitation of proteins by high concentrations of neutral salts: NaCl, (NH4) 2 SO4, etc.
Principle of the method.
Reaction of salting-out is caused by dehydration of the protein
macromolecules with a simultaneous neutralization of electric
charge. For salting-out of various proteins the certain concentration
of the same salts is required. At salting-out a protein usually does
not lose its native properties. It can, for example, again show catalytic activity. Salting-out method allows to obtain proteins in crystalline form and to separate the protein mix into fractions, namely: proteins of blood serum - albumins (А) and globulins (G). Globulins,
having a large molecular weight, are precipitated in a semi-saturated
(50%), and albumins are precipitated in saturated (100%) solution of
ammonium sulfate.
Sodium chloride precipitates globulins at full saturation
(100%) in solution; besides it additional acidification is required for
deposition of albumin.
Salting-out of proteins is a reversible reaction, since the precipitate of the protein can be dissolved again after reducing the concentration of salts by dialysis or by dilution with water.
For the detection of protein in solution universal biuret reaction is used.
The biuret reaction chemistry. The biuret reaction reveals in
the protein a peptide (-CO-NH-) bond. In alkaline medium the solution of the protein upon interaction with copper ions becomes blueviolet colour, and the products of its partial hydrolysis (peptides) pink colouring. The biuret reaction is able to provide substances
which contain at least two peptide bonds. The biuret reaction due to
the formation of biuret complex, as a result of connection of copper
with peptide group of protein. In alkaline medium the lactim forms
are formed from the lactam forms. The biuret reaction proceeds
schematically as:
8
H R2 O
H R4 O
| | ||
| | ||
Cu(OH)2+2NaOH
H2N-CH-C-N-CH-C-N-CH-C-N-CH-C-N-CH- .....
| ||
| | ||
| |
R1 O
H R3 O
H R5
polypeptide (lactam form)
R2 OH
R4 OH
| |
|
|
H2N-CH-C=N-CH-C=N-CH-C=N-CH-C=N-CH- ....
| |
| |
|
R1 OH
R3 OH
R5
polypeptide (lactim form)
R2
|
R1-HC-C=N-CH-C=N-CH-R3
| |
|
|
H2N O
O C-ONa
||
Cu  — ―N
↑
¦
|
R4-CH
|
|
|
C -ОNа
|
||
└ ―――— N
|
CH-R5
Individual amino acids do not
give positive biuret reaction.
The exception of: histidine and
asparagine, with which the biuret reaction is carried out under
the condition of high concentrations in solution.
СOONa
|
CH-NH2
|
CH2
| N
COONa
|
CH-NH2
|
CH2
N
N
Cu
N
copper complex with histidine
in the case of biuret reaction
biuret copper complex
purple
9
NOTE: when setting the biuret reaction, you cannot add an
excess of copper sulphate, as a blue precipitate of cupric hydroxide
masks violet colour biuret protein complex.
Practical procedure.
Reagents
The protein solution
NaCl powder
Sated solution of
(NH4)2SO4
The precipitate
1% solution of
CH3COOH
Powder (NH4)2SO4
test tube 1
20 dr.
To complete
saturation
-
test tube 2
20 dr.
20 dr.
FILTRATION
Globulins
Globulins
To filtrate 1 add
10-20 dr.
-
To complete
saturation
FILTRATION
The precipitate
Albumins
Albumins
CONTROL - BIURET REACTION
5 dr.
5 dr.
Filtrate 2
10% solution NaOH
5 dr.
5 dr.
1% solution CuSO4
1 dr.
1 dr.
RESULTS
CONCLUSION:
III.2. Control questions.
What are the factors that keep proteins in solution?
Why do proteins form colloidal solutions?
What is the basis of the reaction of protein precipitation?
10
What are the physical and chemical properties of proteins that allow
them to isolate?
What is the salting out?
What reactions can reveal protein in the solution?
What color does protein (protein-free) solution have at carrying out
of biuret reaction?
Which individual amino acids can give positive biuret reaction?
What is the difference between albumins and globulins?
III.3. Home assignment.
1. Coagulation of the proteins is:
A. formation of separate amino acids
B. formation of precipitate
C. formation of the supermolecular complexes
2. Formation of precipitate of the protein is called:
A. coagulation
B. hydrolysis
C. electrophoresis
D. chromatography
3. Isoelectric point is:
A. the total negative charge of the protein molecule
B. pH, at which a protein is electrically neutral
C. pH value equal to 7
D. temperature, which is optimal for the functioning of the protein
4. Match the protein and its main function.
A. Myosin
1. protective
B. Immunoglobulin
2. transport
C. Hemoglobin
3. contractile
D. Casein
4. trophic
5. Salting-out is the precipitation of the proteins by:
A. only salts of mineral acids
11
B. salts of heavy metals
C. salts of light metals and ammonium
D. any salts
6. Match the protein and the conditions of its precipitation.
A. Albumin 1. sated solution of ammonium sulfate
B. Globulin 2. saturated solution of sodium chloride
3. semi-saturated solution of ammonium sulfate
4. saturated solution of sodium chloride in acidic medium
Material for self-study. I. 1.pp.45-46, 64-67
2.pp.21-29; II.
Class 2
STRUCTURE OF PROTEIN MOLECULES.
METHODS OF PURIFICATION OF PROTEINS
The aim of the lesson:
1. To know the structural organization of the protein molecule as
a basis for the functioning of all proteins.
2. To know features of structure of fibrous proteins.
3. To master methods of protein purification from impurities example dialysis.
Initial level of knowledge:
- amino acids: structure, classification;
- covalent and non-covalent bonds: characteristics, examples;
- physical and chemical properties of proteins.
Main topics.
I.2. The structure of globular proteins (primary, secondary, tertiary,
quaternary structures).
Features of the molecular structure of fibrous proteins.
The relationship of structure and function of proteins.
12
Specific difference of the primary structure and antigenic properties
of proteins.
Methods of protein purification (chromatography, electrophoresis,
dialysis, etc.).
Protein hydrolysis (types, conditions, products of incomplete and
complete hydrolysis).
II.1. Work. DIALYSIS OF PROTEINS
The principle of the method.
The dialysis is based on the ability of semi-permeable membranes to pass through the pores of low-molecular substances and
detain macromolecular colloidal particles. Dialysis is the convenient method of protein purification. During dialysis the colloidal solution is placed in the cellophane bag and immersed in distilled water. Molecules of salts, sugars and other low-molecular substances
can easily diffuse through the membranes, and the colloidal substances remain in the bag.
A)
B)
1
2
3
Fig.1. Options of system for dialysis of the proteins. A) a dialyzer from a glass cylinder with the bottom of the semi-permeable
membrane. B) a dialyzer from the cellophane bag. 1-holder, 2dialyzing solution, 3-dialyzate.
13
Practical procedure.
Pour 5 ml of the salt solution of the protein in cellophane bag
(dialyzer). Ship in a beaker with distilled water for 1 hour. Then to
perform reactions on chlorides and protein with small portions of the
dialyzate (outer fluid) and make sure that the mineral salts diffused
in the outer container and the protein remained in the contents of the
bag.
Test for protein (biuret reaction)
Reagents (dr.)
test tube 1
(dialyzing solution)
The tested liquid
5
10% solution NaOH
5
1% solution CuSO4
1
RESULTS:
The tested liquid
10% solution HNO3
1% solution AgNO3
RESULTS:
Test for chlorides
10
1
1
CONCLUSION:
III.2. Control questions.
What is dialysis?
What is the dialyzate and dialyzing liquid?
What reactions are used to control the dialysis?
How is dialysis used in medicine?
14
test tube 2
(dialyzate)
5
5
1
10
1
1
III.3. Home assignment.
1. Name amino acids
НО-
NH2-CH-COOH
-СН2-СН-СOОН
|
NH2
NH2-CH-C00H
|
CH2-OH
(CH2)3-CH2-NH2
NH=C-NH-(СH2)3-CH-COОН
NH2
NH2
2. Match the amino acids and the properties of their radicals.
A. Arginine
1. has negative charge at pH=7
B. Glutamate
2. contains sulfur in its radical
C. Serine
3. has positive charge at pH=7
D. Cysteine
4. contains hydroxyl group in its side chain
3. Choose the appropriate characteristics for the protein structure.
A. Secondary structure 1. the order of sequence of amino acids in
the polypeptide chain
B. Tertiary structure
2. the native structure of protein
C. Quaternary structure 3. complex of some polypeptide chains
D. None
4. the conformation of polypeptide chain as α–
helix
4. Write the structural formula of each tripeptide and answer the
questions:
A. Val-His-Leu
B. Pro-Glu-Arg
C. Asp-Gly-Thr
D. Ala-Tyr-Asn
15
What types of the bonds may be formed between the amino acid radicals of each peptide?
What peptide contains only hydrophobic amino acids?
What peptide contains the smallest amino acid?
What peptide contains the cyclic amino acid?
What peptide contains an imino acid?
Material for self-study. I. 1.pp.46-64
2.pp.14-20, 30-36, 40-47; II.
Class 3
CLASSIFICATION OF PROTEINS. CHEMISTRY OF
SIMPLE AND CONJUGATED PROTEINS. METHODS
OF PROTEIN PRECIPITATION
The aim of the lesson:
1. To know a classification of the proteins, the characteristics of
individual groups of proteins and peptides.
2. To master methods of protein precipitation.
Initial level of knowledge:
- structure and properties of proteins;
- biological role of proteins.
Main topics.
I.2. Characteristics of the main groups of simple proteins.
The principle of classification of conjugated proteins.
Characteristics of natural peptides.
The content of proteins and amino acids in the blood in norm and at
the pathology.
I.3. Reference work.
SUMMARY QUESTIONS FOR A TEST
"THE STRUCTURE AND PROPERTIES OF PROTEINS»
16
1. Character of proteins. The elemental composition, the smallest
structural unit, the molecular weight of proteins.
2. Classification of amino acids that build proteins, based on the polarity of radicals; examples of nonpolar, polar, positively and negatively charged amino acids.
3. Amphotericity and solubility of amino acids. Functional groups of
amino acids involved in the formation of covalent, ionic, electrostatic and hydrophobic bonds of the protein molecule.
4. The structure and name of amino acids derivatives of propanoic,
butyric, pentanoic, hexanoic, succinic and glutaric acids.
5. The structure and name of hydroxy- and sulfur-containing, homoand heterocyclic amino acids.
6. Write the tripeptide: а) glycil-valil-isoleucine;
б) leucil-tyrosil-histidine; в) alanil-methionil-tryptophan, etc.
7. Name the tripeptide, for example:
H3C-CH-CH2-CH-CO-NH-CH-CO-NH-CH-COOH
|
|
|
|
CH3
NH2
CH3
CH2
|
CH2-S-CH3
-CH2-CH-CO-NH-CH-CO-N
|
|
NH2
(CH2)3
|
CH2NH2
|
COOH
8. Factors that hold a protein in solution. Isoelectric point of proteins. Methods of protein precipitation, value for medicine. Reversible and irreversible precipitation. Examples.
9. General physical and chemical properties of proteins. Properties
of colloidal solutions, different from the properties of true solutions.
17
Methods for the isolation and purification of proteins. Dialysis, value for medicine.
10. Methods for detection of proteins in solution and study of the
qualitative composition of the protein. Hydrolysis of protein, types
of hydrolysis, intermediate and end-products. The use of protein hydrolyzates in medical practice.
11. Proteins - the basis of vital processes. Levels of structural organization of the protein molecule. The primary structure, bonds forming it. The dependence of biological properties of proteins and the
nature of the amino acid sequences in polypeptide chain (HbA,
HbS).
12. The conformation of peptide chains in proteins: secondary and
tertiary structures, links forming these structures. Reversible and irreversible denaturation of proteins; the reasons and value.
13. The supreme form of protein molecule structure - quaternary
structure. Cooperative changes of the protomer conformation (for
example, hemoglobin and myoglobin).
14. The dependence of the functional properties of proteins and their
conformation. The ability to specific interactions as a basis for the
functioning of all proteins. Antigenic properties of proteins.
15. Biological functions of proteins. The differences of the body’s
protein composition. The change in protein composition during ontogenesis and illnesses.
16. Classification of proteins. Chemistry of conjugated proteins. The
most important representatives of simple and conjugated proteins in
the human body.
17. Structural proteins. Classification, structure, distribution in tissues. The self-assembly of multimolecular protein structures on the
example of the collagenic fibers.
II.1. Work 1. PRECIPITATION OF PROTEINS BY BOILING
The principle of the method is based on the denaturation of
proteins by heating their solutions at temperature above 5О-6О0С
that leads to the loss of solubility, especially in isoelectric point.
18
Almost all of the proteins coagulate and precipitate when
heated in neutral and subacid environment. In strongly acidic and alkaline environment proteins denature when they are boiled, but do
not coagulate and can give a precipitate only when adding a sufficient amount of a neutral salt (NaCl, ammonium sulfate, etc.). The
stability of the protein in solution depends on the acquisition of (+)
charge in the case of a strongly acidic environment and increasing (-)
charge in an alkaline environment.
+H+
+
H3N -CH-COOH
|
R
the charged molecule
cule
"+"
рН = 2-3
-H+
+
-
H3N -CH-COO
|
R
zwitterion form
of amino acid
рН = 4-9
(amphoteric ion)
H2N-CH-COO
|
R
the charged mole-
"-"
рН = 9-10
With decreasing рН the dissociation of carboxylic group is
suppressed and the molecule becomes positively charged. With increasing pH there is separation of the bound proton of the amino acid and the molecule acquires a negative charge. A more complete
and fast sedimentation occurs when the isoelectric point. Isoelectric
point is the рН value at which the total electric charge of proteins is
zero and, hence, the protein in an electric field loses mobility, i.e.
remains at the start. For most proteins isoelectric point corresponds
to a subacid environment (in the range from 5.5 to 6.9).
So, important role in the denaturation of proteins by heating
plays the hydrogen ion concentration (i.e. the certain reaction of environment) and the presence of salts.
Practical procedure.
Test tubes
Reagents (dr.)
1
2
3
4
5
1% solution of
10
10
10
10
10
egg protein
1% solution
2
10
10
19
CH3COOH
Sated solution
of NaCl
10% solution
NaOH
Reaction of environment
-
-
-
4
-
-
-
-
-
4
BOILING
Neutral Slightly Strong- Strongly
acidic ly acid acidic, +
electrolyte
Alkaline
RESULTS:
CONCLUSION:
Work 2a. PRECIPITATION OF PROTEINS BY HEAVY METAL
SALTS
Principle of the method.
Precipitation of proteins by salts of heavy metals unlike salting-out occurs when concentration of salts is small. Proteins in the
interaction with salts of heavy metals (lead, copper, silver, mercury,
etc.) adsorb them, forming salt-like and complex compounds, soluble
in plenty of these salts (excluding salts AgNO3 , HgCl2), but insoluble in water.
Dissolution of the precipitate in excess of salts is called adsorptive peptization. This phenomenon occurs due to the occurrence
of the same (+) charge on the particles of the protein. The ability of
the proteins to tightly bind heavy metal ions in the form of insoluble
sediment in water is used as an antidote in cases of poisoning by
salts of mercury, copper, lead, etc. Usually proteins of milk and eggs
use immediately after poisoning while these salts are still in the
stomach and did not have time to absorption. Following the intro20
duction of the protein should be caused vomiting to purge the poison
from the body.
Practical procedure.
Reagents (dr.)
а) Solution of egg protein
1% solution CuSO4
5% solution Pb(CH3COO)2
5% solution AgNO3
RESULTS:
1
5
2
-
Test tubes
2
5
2
-
3
5
2
b) continue experience in the same tubes (add)
1% solution CuSO4
5-10
5% solution Pb(CH3COO)2
5-10
5% solution AgNO3
5-10
RESULTS:
CONCLUSION:
Work 2б. PRECIPITATION OF PROTEINS BY CONCENTRATED MINERAL ACIDS
Principle of the method.
Concentrated mineral acids cause denaturation of proteins
and form complex salts of protein with acids. When all mineral acids
are in excess, except for nitric, the precipitated protein solutes
again. Qualitative reaction with concentrated HNО3 underlies the
quantitative determination of proteins in urine by the method of
Roberts - Stolnikov-Brandberg.
Practical procedure.
21
stage I
stage II
Reagents (dr.)
Test tubes
1
2
1
2
Concentrated HNO3
10
10
Concentrated H2SO4
10
10
Solution of protein
10
10
Note: the protein solution is laid gently on acid on the side of the
test tube inclined at an angle of 450.
RESULTS:
CONCLUSION:
1
2
Fig.2. Laying of the sample in a concentrated acid: 1 - solution
of protein; 2 - acid.
III.2. Control questions.
What are the factors that keep proteins in solution?
What is the reaction of protein precipitation?
What is coagulation of proteins? Types of coagulation.
What is denaturation of protein? Types of denaturation.
22
Under what conditions do such processes as coagulation with denaturation, coagulation without denaturation, denaturation without coagulation occur?
What is adsorptive peptization?
How is adsorptive peptization used in medicine?
III.3. Home assignment.
1. Choose the wrong answer. Collagen contains:
A. hydroxylysine and hydroxyproline
B. many hydrophobic amino acids
C. 30% glycine
D. a lot of cysteine
E. hydrophobic amino acids
2. Desmosine is included into:
A. collagen
B. elastin
C. alpha-keratin
D. beta-keratin
3. Match the protein and the function of protein.
A. fibrinogen
1. contractive element
B. pepsin
2. structural element
C. prolactin
3. regulative molecule
D. albumin
4. protective agent
E. alfa-keratin
5. catalyst
F. none
6. transportation
4. Fetal hemoglobin has such chains as:
A. 4 alfa
B. 2 alfa and 2 beta
C. 2 alfa and 2 gamma
D. 2 alfa and 2 delta
E. 2 alfa and 2 epsilon
5. What type of talassemia is incompatible with life?
23
A. α
B. β
C. γ
D. δ
E. ε
6. Specify the correct order of stages of collagen synthesis and
formation of collagenic fibers.
A. Oxidative deamination of ε-amino groups of lysyl and hydroxylysyl residues to aldehydes
B. Assembly of collagen fibers in quarter-staggered alignment
C. Formation of triple helix
D. Formation of intrachain and interchain S-S bonds in extension
peptides
E. Cleavage of signal peptide
F. Cleavage of amino- and carboxy-terminal propeptides
G. Formation of intra- and interchain crosslinks via Schiff bases and
aldol condensation products
H. Hydroxylation of γ-prolyl residues and some γ-lysyl residues
7. Excretion of hydroxyproline in urine is decreased at old age because:
A. activity of collagenase decreases
B. the number of cross links in collagen increases
C. activity of hydroxyproline oxydase increases
D. all of the above.
Material for self-study.: I. 1.pp.53, 60, 67-72
2.pp.32-33, 37-39, 535-540; II.
SECTION II. ENZYMES
Class 4
24
STRUCTURE AND BASIC PROPERTIES OF ENZYMES
The aim of the lesson:
1. To revise the material on the physical and chemical properties
of proteins.
2. To know modern views of the structure of enzymes and to
study their specific properties.
Initial level of knowledge:
- structure and physical and chemical properties of proteins;
- the concept about catalysts.
Main topics.
I.2. The concept of catalysis.
Chemical nature of enzymes.
Properties of enzymes.
The structure of the enzymes (functional sites, their structure).
Factors determining the activity of enzymes.
Vitamins as coenzymes.
The concept of isoenzymes.
Multimolecular enzyme systems.
II.1. Work 1. ASSAY OF TERMOLABILITY OF SALIVARY
AMYLASE
Principle of method and chemism.
Salivary amylase (α-1,4-glycosidase) catalyzes the hydrolysis
of α-1,4-glycoside bond of starch and glycogen to the disaccharide
maltose with intermediate formation of dextrins of various size.
Unsplit starch with iodine gives a dark blue colouring, amylodextrins - purple, erythrodextrins - red - brown and maltose - yellow
(the color of a water solution of iodine, i.e. staining with iodine is
not given). End-products of hydrolysis of starch - maltose and glucose - have a free aldehyde group and can be found out by the
Trommer’s reaction which is based on the redox reaction. By heating
the aldehyde group is oxidized to gluconic acid, and copper of cupric
hydroxide (blue) is restored in a cuprous hydroxide (yellow). On further heating red cuprous oxide is formed:
25
H
|
C=O + 2CuSO4 + 5NaOН 
|
R
O
║
C–O–Na + 2CuOH + 2Na2SO4 + 2H2O
|
R
One of the specific properties of enzymes is thermolability,
i.e. the sensitivity of the enzyme to temperature. For many enzymes
the maximum enzymatic velocity is observed at 38-400C. This temperature is called the temperature optimum. When heated above
700С they lose their properties of biological catalysts. The degree of
inactivation depends on the duration of thermal exposure. The slowdown and cessation of enzyme reaction occurs due to thermal denaturation of the protein molecule of the enzyme. At low temperature
the enzymes do not denature, but the reaction rate is sharply reduced
by decreasing the kinetic energy of the molecules. Action of enzyme
is judged by the disappearance of substrate or the appearance of
products of reaction.
Practical procedure.
Collect about 2 ml of native (undiluted) saliva in a clean test
tube and boil for 5 minutes, cool.
Test tubes
Reagents (dr.)
1
2
3
1% starch solution
10
10
10
Native saliva diluted 1:10
10
Boiled undiluted saliva
10
Distilled water
10
Place in the thermostat at 380С for 10 min. From the contents of
each test tube cast on 5 drops in other test tubes.
а) Iodine test (for starch)
Test tubes
Reagents (dr.)
1-a
2-a
3-a
The test solution
5
5
5
26
Solution of iodine in potassium iodide
RESULTS:
1
1
b) Trommer’s reaction (glucose and maltose)
Reagents (dr.)
1-b
2-b
Test solution
5
5
10% solution NаОН
5
5
1% solution CuSO4
3
3
RESULTS:
1
3-b
5
5
3
CONCLUSION:
Work 2. ASSAY OF INFLUENCE OF pH ON ENZYME ACTIVITY AND DETERMINATION OF рН OPTIMUM
Principle of method and chemism.
There are individual optimum рН for different enzymes, where
they are most active. For example, optimum рН of pepsin (an enzyme of gastric juice) – 1.5-2.0; arginase (a liver enzyme) is 9.5. For
salivary amylase optimum рН – 6.8-7.2; in the sour and alkaline medium the activity decreases due to the interaction of protein molecule
with ions of the solution.
Chemism of the reactions see above (Work 1).
Practical procedure.
Dilute saliva in 10 times. Take 5 test tubes and prepare
ture in accordance with table.
Test tubes
Reagents (dr.)
1
2
3
4
5
5
5
5
Solution 1
8
6
5
3
Solution 2
27
a mix-
5
5
1
1% solution NaCl
7
9
10
12
14
рН
5.8
6.9
7.4
7.9
8.5
0.5% solution of starch
7
7
7
7
7
Diluted saliva
7
7
7
7
7
0
Mix, put in a thermostat at a temperature of 38 С
for 10 min.
Iodine solution
1
1
1
1
1
Color
The products of hydrolysis of starch
CONCLUSION:
Work 3. ASSAY OF SPECIFICITY OF SALIVARY AMYLASE
Principle of the method.
Enzymes have specificity as they can catalyze only certain
chemical reactions. The specificity of enzyme action is the absolute,
relative (wide) and stereo chemical. It is determined by the fact that
only some strictly certain functional groups included in enzymes,
can participate in the formation of the enzyme - substrate complexes.
Salivary amylase accelerates the hydrolysis only of polysaccharides, having no effect on disaccharides. Salivary maltase accelerates the hydrolysis of the formed disaccharide maltose and has no
effect on sucrose.
Chemism of the reaction see in Work 1.
Practical procedure.
Reagents (dr.)
test tube 1
1 test tube 2
Saliva diluted 1:5
5
5
1% starch solution
10
1% solution of sucrose
10
Placed in a thermostat at a temperature 380С for 10 min
28
Fehling’s reagent
9
Heat to boiling and boil 1 min
9
RESULTS:
CONCLUSION:
III.2. Control questions.
How and why does the activity of enzymes depend on the temperature?
What is the temperature optimum of the enzymes?
How and why does the activity of enzymes depend on pH (graph)?
What enzymes are active at sour; neutral; an alkaline environment?
What is the specificity of enzymes?
What types of enzyme specificity do you know? Examples.
Specify qualitative reaction for starch.
III.3. Home assignment.
1. Choose the wrong answer. Enzymes differ from inorganic catalysts because they:
A. have specificity
B. work at usual temperature and pressure
C. are active only at the presence of the metal ions
D. decrease the energy barrier of reaction
2. Match the enzyme and the kind of specificity.
A. Tripsin
1. absolute
B. Lipase
2. wide
C. Arginase
3. stereochemical
D. L-fumarase
3. Choose the correct answer. pH affects to enzyme activity as:
A. the direction of the reaction may be influenced by the H+
29
B. the ionization state of dissociating groups on the enzyme may be
modified
C. the ionization state of the substrate may be modified
D. all the above
4. Choose the correct answer. At low temperature enzyme activity
decreases because:
A. enzyme denatures
B. enzyme coagulates
C. kinetic energy of the reacting molecules decreases
D. all the above
5. What is typical for conjugated enzyme?
A. consists only of the amino acids
B. consists of some polypeptide chains
C. has coenzyme and apoenzyme
D. has the allosteric site
6. Choose the wrong answer. Isoenzymes are:
A. complexes of some enzymes, which control the reactions of one
biochemical process
B. enzymes that catalyze the same reaction
C. enzymes which may be separated from each other by electrophoresis
D. enzymes which have genetically determined differences in amino
acid sequence
Material for self-study: I. 1.pp.85-91, 94-96
2.pp.49-51, 54-55, 57, 63-64; II.
Class 5
MECHANISM OF ENZYME ACTION.
KINETICS OF ENZYME REACTIONS
30
The aim of the lesson:
to know the current views on the mechanism of action of enzymes and the kinetics of enzyme reactions.
Initial level of knowledge:
- protein nature of enzymes;
- structure of conjugated enzymes;
- structure of enzyme molecule;
- kinetics of chemical reactions.
Main topics.
I.2. Modern ideas about the mechanism of enzyme action (Fischer's
and Koshland’s hypothesises).
Multifunctional catalysis.
Kinetics of enzyme reactions. Michaelis constant.
Dependence of enzyme activity on various factors.
Determination of enzyme activity: the principle, value in medicine.
II.1. Work 1. ESTIMATION OF  - AMYLASE IN SALIVA
Principle of the method.
The method is based on determination the least amount of amylase (maximum dilution of saliva), a fully breaks down all of the
added starch. Activity of salivary amylase is expressed by the
amount (in ml) of 0.1% solution of starch which is cleaved with 1 ml
of undiluted saliva at a temperature of 380С for 30 minutes. Normal
amylase activity is 160 - 320. Amylase activity is denoted by A
380/30´. This method is widely used to determine the amylase activity of blood and urine.
Practical procedure.
Pour 1 ml of water in each of 10 test tubes. Add 1 ml of saliva
diluted in 10 times in the 1-st test tube. Shake the content of the 1-st
test tube and transfer 1.0 ml of liquid using a pipette to the 2-d test
tube. Also shake content and 1.0 ml of that transfer to the 3-d test
tube, etc. up to 10-th test tube. From 10-th test tube pour 1.0 ml of
the mix.
31
1 ml of saliva
1 ml of mix
1 ml of water
I
II
n
III
Fig.3. Dilution of saliva.
To all test tubes add 1.0 ml of water and 2.0 ml of 0.1% starch solution, shake, stir and put in the thermostat at 380С for 30 min. After
incubation cool the test tubes with tap water, add 1 drop of 0.1% of
iodine solution and stir. Note the color in each test tube and, using
the table to do the calculation.
1
2
3
4
Test tubes
5
6
7
8
9
10
Dilution of
saliva
Coloring solution with
iodine
CONCLUSION:
Calculation: mark the test tube where the hydrolysis of starch
was completed with the least amount of enzyme and by quantity of
32
undiluted saliva (А) in the test tube to calculate the amylase activity
(X) as follows:
A ml of saliva splits 2.0 ml of 0.1% starch solution
1 ml of saliva splits X ml of starch solution.
In the 3-d test tube, for example, A =1/80 ml.
Work 2. ESTUMATION OF AMYLASE (DIASTASE) IN
URINE
Principle of the method.
The method is based on determination the time required for
complete splitting of starch in the presence of 1 ml of urine. The
amount of the enzyme splitting 2 mg of starch within 15 minute is
taken as one unit of amylase activity in urine. Amylase activity is
expressed by the number of units in 1 ml of urine.
Reference values: 1-2 UNITS,
by Volgemut’s method 16 - 64 UNITS.
The urine of healthy people has a low amylase activity compared with the saliva amylase. Determination of -amylase activity
in urine and blood serum is widely used in clinic in the diagnosis of
diseases of pancreas. In the first days of the disease amylase activity
increases in urine and blood serum of ten and more times, and then
gradually returns to norm. In renal insufficiency amylase in urine is
absent.
In the childhood the increase of amylase activity is observed
in endemic parotitis which indicates that a simultaneous lesion of the
pancreas by the mumps virus. The flu virus also affects the pancreas,
but less frequently.
Practical procedure.
Drip on a dry Petri dish n different places 1 drop of 0.1% iodine solution in potassium iodide (8-10 drops). In a test tube add 2
ml of 0.1% starch solution containing 2 mg of starch, 1 ml of 0.85%
sodium chloride solution and place the test tube in a thermostat at
370С for 2 minutes. In 2 minutes add in a test tube 0.5 ml of urine,
mix and note the starting time of the reaction. Then every 2-3
minutes transfer a drop of a mix from the test tube on a Petri dish in
33
a drop of iodine solution until the appearance of yellow (or colorless) color and note the reaction time in minutes. Activity of urine
amylase is calculated by the formula:
15
Х=
Т ∙ 0.5
where X – amylase activity in 1 ml of urine;
15 - the time, required to complete splitting of 2 mg of starch
in min;
0.5 - the amount of urine, taken in a reaction mix in ml;
T - time of reaction in minutes.
RESULT:
CONCLUSION:
III.2. Control questions.
What reaction does amylase catalyze?
Specify the intermediate and end-products of hydrolysis of starch.
What is normal urine diastase equal to?
What diseases does amylase activity in urine increase (adults and
children) in?
When is amylase in urine not detected?
Write the formulas of vitamins В2 and niacin; specify the corresponding coenzymes.
III.3. Home assignment
1. Choose the correct answer. Michaelis constant of hexokinase is
less than one of glucokinase. That reflects its:
A. higher affinity for glucose
34
B. less high affinity for glucose
C. resistance to denaturing agents
D. kind of the specificity
2. Choose the correct answer. The international unit of enzyme activity is:
A. the amount of enzyme that catalyzes transformation of 1 mol of
substrate per second
B. the amount of enzyme that catalyzes transformation of 1 mmol of
substrate per minute
C. the number of units of enzyme activity per mg of enzyme protein
3. Choose the wrong answer. What substances may be as coenzymes?
A. Na+
B. Ca2+
C. Fe3+
D. Zn2+
4. Choose the correct answer. The velocity of enzyme catalyzed reaction:
A. increases as the enzyme concentration decreases
B. increases as the substrate concentration increases
C. doesn’t depend on the temperature
D. is the highest at saturation of enzyme with substrate
5. Choose the correct answer. Multimolecular enzyme system is:
A. set of enzymes, located in one compartment of cell
B. group of enzymes catalyzed same reaction in different tissues
C. complex of enzymes catalyzed the reactions of one biochemical
process
D. group of enzymes with the same structure of active site.
Material for self-study: I. 1.pp.87-89, 96-99, 103, 105-110
2.pp.51-54, 67-70; II.
35
Class 6
REGULATION OF ACTIVITY OF ENZYMES.
CLASSIFICATION, NOMENCLATURE AND
THE PRACTICAL APPLICATION OF ENZYMES
The aim of the lesson:
to know the methods of regulation of enzymes in vivo, the principles of quantitative characteristics, classification, the nomenclature
of enzymes, their application in medicine.
Initial level of knowledge:
- the structure of holoenzymes;
- specificity of enzyme action;
- the mechanism of formation of the enzyme-substrate complex;
- dependence of enzyme activity on various factors.
Main topics.
I.2. The factors regulating the activity of enzymes in vivo.
Activation of enzymes, types.
Types of enzyme inhibition.
Classes of enzymes, examples.
The nomenclature of enzymes.
Sections of medical enzymology, examples.
Immobilized enzymes.
Enzymes of blood plasma.
II.1. Work 1. ASSAY OF INFLUENCE OF ACTIVATORS AND
INHIBITORS ON AMYLASE ACTIVITY IN SALIVA
Principle of method and chemism.
Activators stimulate the action of enzymes, but do not participate in the reaction. Activators of certain enzymes are: for saliva
amylase - sodium chloride; for pepsin - ions Н+ hydrochloric acid;
for lipase - bile acids; for adenosine triphosphatase - Мg++, Mn++.
Inhibitors break the action of enzymes down to a complete
stop the reaction. Inhibitors are often intermediate or end products of
the reactions of any biochemical process. Examples: DFP (diiso36
propylfluorophosphate) for cholinesterase and trypsin; hydrocianic
acid - for cytochrome oxidase; malonic acid - for succinate dehydrogenase; copper sulfate - for saliva amylase.
Activators and inhibitors alter the activity of the enzyme, by
affecting its active or allosteric site.
Chemism of reactions see in work 1 (class 4).
Practical procedure.
Reagents (dr.)
Saliva diluted 1:40 (ml)
Н2О
1% solution NaCl
1 % solution CuSO4
1% solution of starch
Solution of iodine
Н2О (ml)
RESULTS:
Test tubes
1
2
3
1
1
1
2
2
2
5
5
5
Incubate at room temperature for 2 min
1
1
1
2
2
2
CONCLUSION:
Work 2. ESTIMATION OF AMYLASE ACTIVITY IN BLOOD
SERUM
Principle of the method.
The method for estimation of amylase is based on the determination of the unsplit remainder of the starch and intermediate
products of its hydrolysis (dextrins). In norm 25-40 mg of glucose is
formed from the digested starch. Amylase (diastase) of blood serum
is activated significantly in diseases such as diabetes mellitus, illnesses of the pancreas, infectious parotitis and some others.
37
Practical procedure.
Recently taken blood serum in a volume of 0.1 ml (gain micropipette) is heated in a closed test tube in a thermostat at 380С for
1-2 minutes. Then add 0.25 ml of 0.1% solution of starch and put
back in the thermostat at the same temperature for 15 mines, shaking
occasionally. After incubation add one drop of 0.025N solution of
iodine and note the color. The amount of glucose corresponding to
the amount of split starch can be found in the following table:
PAINTING
Glucose in mg
About 20
25-35
35-40
40-60
Above 60
Blue
Bluish-green
Color of bile
Yellow
Orange-red
RESULT:
CONCLUSION:
III.2. Control questions.
Specify the role of activators and inhibitors of enzyme reaction.
What substances can act as activators and inhibitors of enzymes?
Specify the normal content of amylase in human blood.
Specify the clinical value of determination of diastase in the blood.
What is an inhibitor of a saliva amylase?
What is an activator of a saliva amylase?
III.3. Home assignment
1. Choose the correct answer. In vivo activation of the enzymes is
carried out by:
A. allosteric regulation
B. chemical modification of the enzyme
38
C. partial proteolysis
D. compartmentation
E. all the above
2. Choose the correct answer. Allosteric enzymes always have the
following properties:
A. they are conjugated enzymes
B. they have one binding site
C. the catalyze the end reaction of the pathway
D. they are the regulatory enzymes in the metabolic pathway
3. Match the medicine and a manner of action.
A. Penicillin
1. competitive inhibitor
B. Acetylsalicylic acid 2. noncompetitive inhibitor
C. Sulfonamides
3. “suicide” inhibitor
4. Choose the wrong answer. The primary enzymopathy is:
A. phenylketonuria
B. galactosemia
C. albinism
D. gastritis
E. hemophilia
5. Choose the correct answer. Tripsinogen transfers to active form
by:
A. phosphorylation
B. changing of the primary structure
C. changing of the secondary structure
D. changing of the tertiary structure
E. adenylation
6. Choose the correct answer. Competitive inhibitor:
A. changes the primary structure of enzyme
B. changes the specificity of enzyme to its substrate
C. stabilizes the enzyme-substrate complex
39
D. uses phosphorylation for binding to active site of the enzyme
E. none of the above
Material for self-study: I. 1. pp.86, 90-94, 99-105, 110
2. pp.49-50, 55-59, 67-70, 72-79; II.
Class 7
COLLOQUIUM ON SECTION "ENZYMES"
Questions for colloquium
1. The history of discovery and study of enzymes.
2. The similarity (difference) of enzymes and inorganic catalysts.
3. General characteristic of enzymes. The biological role of enzymes
and place of enzymology in biology and medicine.
4. Chemical nature of enzymes. Simple and complex enzymes. Multimolecular enzyme systems.
5. The structure of enzymes. Active and allosteric sites.
6. Cofactors and prosthetic groups of enzymes: metal ions and coenzymes. Chemical structure and functions of coenzymes.
7. Isozymes as an example of biochemical polymorphism.
8. The dependence of enzyme activity on the concentration of enzyme and substrate. Vmax and KM of enzymes.
9. The dependence of enzyme activity on рН.
10. The dependence of enzyme activity on temperature.
11. The specificity of action of enzymes. Kinds of enzyme specificity.
12. The structure of allosteric enzymes. The allosteric transition.
13. Hypotheses of action of enzymes as biological catalysts.
14. The modern theory of enzyme catalysis. Mechanism of enzyme
action.
15. Regulation of enzyme activity in vivo.
16. Activation of enzymes. Types of activation. Proenzymes.
40
17. Inhibition of enzymes. Reversible and irreversible inhibition;
competitive, noncompetitive and uncompetitive inhibition.
18. The principles of qualitative and quantitative estimation of enzyme activity. The units of activity and amount of enzymes
19. Enzymopathology. Hereditary (primary), secondary enzymopathies.
20. Enzymodiagnosis. Enzyme distinctions of organs and tissues.
Diagnostic significance of plasma enzymes.
21. Classification and the nomenclature of enzymes.
22. Enzymotherapy. Enzymes and enzyme inhibitors as pharmaceutical drugs. Immobilized enzymes.
Practical tasks
1. Does solubility of proteins change …. or no?
- at addition of ammonium sulfate to a solution;
- at addition of nitric acid to a solution;
- after a solution was boiled.
If yes, when reversion of a primary condition of protein solution is
possible?
2. Why are uncooked eggs or milk used at poisoning with salts of
heavy metals and alkaloids? Does the positive medical effect depend
on the volume of the food substances used as antidote?
3. Determine, what substance is dissolved in water if this solution:
opalescent in luminous beam, gives blue-violet colouring at addition
of copper sulfate, and forms sediment at boiling.
4. Determine the direction of migration (to the cathode or the anode)
for dipeptide Asp-Leu at рН=1 and рН=10.
5. Histones are precipitated at simple change of рН up to 10 (without
special removal of hydrated cover). Explain this fact.
6. Pepsin of gastric juice (рН=1.5) has the isoelectric point about 1,
i.e. is much lower, than other proteins. What functional groups are
dominant in pepsin? What amino acids have more such groups in the
structure?
41
7. By development of the ways of separation and purification of enzymes of R.Vilshtetter has found that after using the filtration
through the semi permeable membrane some enzymes lose their activity. These data have allowed renewing to vitalistic suggestions at
the nature of the enzymes, connecting their activity with vital force
of an integral cell. What actually has taken place with a molecule of
enzyme, why has it lost the specific activity?
8.At increase and decrease of the temperature against the optimum
value, the enzyme activity slows down to full inactivation. Why is
the body temperature decreased when the operation on the heart is
done? Why is material for disinfection processed with high temperature?
9. The doctor prescribes sulfanilamide to a patient with infectious inflammatory disease (for example, with pneumonia caused by
pneumococcus) and specifies that it is to be taken strictly according
to the scheme- at first the shock dozes are taken and then the supporting dozes are taken. How can you explain it from the biochemical point of view?
10. The researcher has two enzymatic preparations with identical
KM, but different Vmax at different concentration of the same substrate. What conclusion is it possible to make about these preparations? Are they the samples of the same enzyme or enzyme
isoforms? Prove the answers.
11. Methanol and antifreeze that contains it are very toxic: the intake
of only 30 ml of methanol can lead to death. The high toxicity of
methanol is not caused mostly methanol itself, but it is caused by a
product of its metabolism - formaldehyde in which it is oxidized under the action of liver enzyme alcohol
dehydrogenase. One of the methods of treatment of methanol poisoning is the prescription of ethanol per os or intravenously in high
doses to a patient. A healthy person is intoxicated in this case. Why
is this treatment efficient? Give explaination
12. When enzyme solution is heated it loses its catalytic efficiency
gradually. It is caused by denaturation of the native enzyme molecule which accepts the conformation of a chaotic ball when its ther42
mal energy increases. At incubation of solution of hexokinase at
45оС enzyme loses 50% of activity, and at presence of very big concentration of one of its substrate - glucose - only 3% of activity. Explain, why.
13. The transformation of pyruvate to lactate is a convertible reaction. But М4-form of LDH has higher affinity to pyruvate and Н4form has the same one to lactate. What direction, mainly, does the
reaction go in skeletal muscles in and what does in a myocardium
in? Which tissues are capable to form and secrete lactate in blood
and which one, on the contrary, absorb it from blood and transform
in pyruvate (and further into other metabolites)?
14. What enzymatic tests distinguish:
- diseases of liver from diseases of myocardium;
- diseases of myocardium from diseases of skeletal muscles?
15. Below you can see the speed of enzymatic reactions at equal
concentration of a substrate without inhibitor (0), at presence of inhibitors 1 and 2 (inhibitor concentration is constant).
S in mm
1
2
3
5
10
20
30
V in
(0) 30 55
65
80 110 110 110
μg / min (1) 15 28
38
55 80 110
110
(2) 15 27,5 32,5 40 55
55
55
Which of these inhibitors is competitive and which is noncompetitive (allosteric)? What is Michaelis constant equal to?
III.3. Material for self-study: I. chapter “Enzymes”; II.
SECTION III. VITAMINS
Class 8
CLASSIFICATION, STRUCTURE AND BIOLOGIC ROLE
OF VITAMINS. METHODS FOR THE DETECTION OF VITAMINS
43
The aim of the lesson:
to know the role of vitamins in metabolism, their classification,
characteristics of each vitamin.
Initial level of knowledge:
- general concepts about vitamins, their biological value;
- symptoms of some hypovitaminoses;
- routes of entry into the body.
Main topics.
I.3. Reference work
SUMMARY QUESTIONS FOR A TEST
"VITAMINS"
1. History of the study of vitamins.
2. The definition of "Vitamins", the functions of vitamins, their participation in the metabolism.
3. Classification of vitamins, lettering, empirical name.
4. Fat-soluble vitamins: biological role, sources, formulas of vitamins A and D3.
5. Water-soluble vitamins: biological role, sources; formulas of vitamins B1, B2, B3, B5, B6, Н, C.
6. Vitamin-like substances: the name, biological role; formulas of lipoic acid, choline, inositol, CoQ, para-aminobenzoic acid.
7. The concept of alimentary and secondary avitaminoses, hypo-and
hypervitaminoses.
8. The reasons for the development of hypovitaminoses.
9. Diseases associated with an excess, a deficiency or lack of vitamins in the body. Specify their main symptoms.
10. Coenzyme functions of vitamins; formulas of NAD+ (NADP+),
FAD (FMN), PPh, TDP.
11. The concept of antivitamins and the mechanism of their action.
12. Structural analogues of vitamins - medicines, their application.
13. Provitamins. Provitamin A.
14. To name the substances according to the formula. For example:
44
СООН
НОНС
НОНС
N
СНОН
СНОН
СНОН
CНОН
II.1. Work 1. QUALITATIVE REACTION FOR VITAMIN В2
Riboflavin is the part of the prosthetic group of flavin enzymes - flavoproteins which are coenzymes flavin adenine dinucleotide (FAD) and flavin mononucleotide (FMN). Flavoproteins are involved in the reaction of dehydrogenation, i.e. the removal of electrons and protons from the substrate. They participate in НАДН oxidation in the chain of biological oxidation of -amino acids, ketoacids and other substrates.
CH2-(CHOH)3-CH2OH
|
N N
H3C═O
+Zn+2HCl
Н3СNH
N ║
O
Oxidized riboflavin (В2)
(yellow)
H3C-
CH2-(CHOH)3-CH2OH
|
N NH
═O
H3C-
NH
NH ║
O
Reduced riboflavin (В2)
(colorless)
Principle of the method and chemism.
The oxidized form of vitamin В2 is a yellow fluorescent in ultra-violet light substance. Reaction to vitamin В2 is based on its ability to restore easily. Thus the solution of vitamin В2 with yellow colour, becoming first pink colour due to the formation of intermediate
compounds, and then fades, because the reduced form of vitamin В2
is colourless.
Practical procedure.
45
To the 10 drops of solution of vitamin В2 add 5 dr. of concentrated hydrochloric acid and lower the grain of the metal zinc.
RESULT:
Work 2. QUALITATIVE REACTION ON NIACIN
Vitamin В3 is the part of coenzymes of pyridine-dependent enzymes: nicotinamide adenine dinucleotide (NAD+) and nicotinamide
adenine dinucleotide phosphate (NADP+). These coenzymes are
parts of dehydrogenases and participate in many oxidation-reduction
reactions. The lack of niacin in the diet causes pellagra.
Principle of the method.
Vitamin В3 at heating with a solution of copper acetate forms
a poor soluble dark blue precipitate of copper salt of nicotinic acid.
Practical procedure.
Before a test 3% solution of vitamin В3 is to be shaken. Then
take 20 dr. of it and heat to boiling; this turbid solution becomes
transparent. Shake 5% solution copper acetate and pour 20 dr. of it
to the heated solution of vitamin В3. Then the contents of the test
tube bring to a boil and immediately cool under cold running tap water.
RESULT:
Work 3. QUALITATIVE REACTION FOR VITAMIN C
Ascorbic acid is associated with the glutathione system and is
involved in tissue oxidation-reduction processes. It is necessary for
hydroxylation of steroid hormones, synthesis of hydrofolic acid, activation of dopamine hydroxylase.
Principle of the method.
Ascorbic acid is easily oxidized and restores various substances (potassium ferricyanide, methylene blue, molecular iodine, 2, 6 dichlorphenolindophenol, etc.).
46
1.
O
OH
|
O=C-CHO=COH-CH-CH-CH2OH + 2K3Fe(CN)6 + 2KOH →
O
OH
|
→ O=C-CO-CO-CH-CH-CH2OH + 2K4Fe(CN)6 + 2H2О
2. 3K4Fe(CN)6+4FeCl3
→
Fe4[Fe(CN)6]3+12KCl
Berlin blue
Practical procedure.
To 5 drops of 1% solution of vitamin C pour 1 dr. of 10% solution NaOH and 1 dr. of 5% solution potassium ferricyanide, stir;
add 3 dr. of 10% solution of hydrochloric acid and 1 dr. of 1% solution of ferric chloride.
RESULT:
III.2. Control questions.
What coenzymes contain vitamin В2?
How to identify the oxidized and reduced forms of riboflavin?
What reactions does ascorbic acid take part?
What is the principle of the determination of ascorbic acid based on?
What coenzymes contain vitamin В3?
What disease develops in the absence of niacin?
III.3. Home assignment.
1. Complete the table
Vitamin
Vitamin
(name)
(structure)
Coenzyme
(name)
47
Coenzyme
(structure)
2. Choose the wrong answer. Vitamins:
A. are synthesized by intestinal microflora
B. are precursors of coenzymes
C. perform specific cellular functions
D. mainly are the dietary components
E. are required in quantities not less 3 g daily
3. Match the vitamin and the coenzyme.
A. B1
1. NADP
B. B2
2. FAD
C. B3
3. H4folate
D. B5
4. CoA
E. Bс
5. TDP
4. Match the vitamins and their functions.
A. Riboflavin
1. coenzyme of transaminases
B. Pyridoxine
2. coenzyme of carboxylases
C. Thiamine
3. coenzyme of oxidases
D. Biotin
4. coenzyme of dehydrogenases of alpha-ketoacids
E. Ascorbic acid 5. coenzyme of oxidoreductases
5. Choose the wrong answer. Folic acid:
A. is synthesized by intestinal microflora
B. contains in the leaves of plants
C. is formed from p-aminobenzoic acid in the liver
D. is required for synthesis of a proteins and nucleic acids
E. is the coenzyme of carboxylase
F. forms some coenzymes
48
6. Which vitamin can accumulate in body and cause toxicity?
A. Cobalamin
B. Thiamine
C. Ascorbic acid
D. Biotin
E. Calciferol
F. None of the above
7. Match the vitamin with its deficiency.
A. Retinol
1. Scurvy
B. Thiamine
2. Megaloblastic anemia
C. Ascorbic acid
3. Dermatitis, glossitis, hallucination
D. Biotin
4. Osteoporosis
E. Calciferol
5. keratinization of epithelial tissues
F. Folic acid
6. Peripheral neuropathy, anorexia
8. The formation of 1,25 dihydroxycholecalciferol takes place in:
A. skin
B. wall of intestine
C. liver
D. kidneys
E. all of the above
Material for self-study: I. 1.pp.113-162
2.pp.481-497; II.
SECTION IV. BIOENERGETICS
Class 9
MITOCHONDRIAL ELECTRON TRANSPORT CHAIN
The aim of the lesson:
49
1. To know biological oxidation as an energy source in alive systems.
2. To know the structure and biological role of high-energy compounds, respiratory enzymes and coenzymes.
Initial level of knowledge:
- the concept of metabolism, and its component parts;
- coupling of metabolism with the energy exchange;
- the structure of the cell, the role of mitochondrion;
- chemical nature of enzymes;
- coenzyme function of vitamins.
Main topics.
I.2. The main stages of metabolism.
Concepts of metabolism, anabolism, catabolism.
General concepts of metabolic pathways and energy metabolism.
Biological oxidation and tissue respiration.
Dehydrogenation of substrates as an energy source for synthesis of
ATP.
The composition of the electron transport chain.
Role of vitamins in tissue respiration.
Structure and function of coenzymes of dehydrogenases.
ATP and creatine phosphate are the main high-energy compounds of a muscular tissue. ATP is synthesized mainly in the process of oxidative phosphorylation, and creatine phosphate - with the
participation of ATP in the state of muscular rest. When muscle is
active creatine phosphate provides energy for synthesis of ATP from
ADP.
III.2. Control questions.
What are high-energy compounds? Give examples.
What ways to form ATP do you know?
Where and when is creatine phosphate formed?
What is the biological role of creatine phosphate?
III.3. Home assignment.
50
1. Complete the table
Enzyme
Coenzyme
(formula)
Active site of the enzyme
(formula)
oxidized
reduced
NADdependent dehydrogenase
FADdependent dehydrogenase
Coenzyme Q
Cytochrome
oxidase
2. Choose the correct answer. Energy of ATP is utilized to:
A. transportation
B. biosynthesis
C. activation of substances
D. muscle contraction
E. all of the above
3. The electron transport chain consists of proteins except:
A. NAD+
B. FAD
C. Q
D. cyt b
E. cyt a3
4. What electron carrier isn’t the component of mitochondrial ETC?
A. NAD+
B. FAD
C. Q
D. NADP+
E. FMN
51
5. Choose the correct variant. Transportation of electrons to oxygen
is a common catabolic pathway for many substances because of:
A. all substrates are oxidized by a same enzymes
B. these substrates are oxidized only in mitochondrion
C. protons from all substrates are used to water production
D. many substrates are oxidized by NAD+- or FAD-dependent enzymes
6. Choose the correct answer. The only one irreversible reaction in
ETC is:
A. formation of NADH
B. formation of FADH2
C. formation of QH2
D. reaction controlled by cytochrom oxydase
Material for self-study: I. 1.pp.227-235, 247-249
2.pp.86, 92-93; II.
Class 10
OXIDATIVE PHOSPHORYLATION.
REGULATION OF THE ELECTRON TRANSPORT CHAIN
(RESPIRATORY CONTROL)
The aim of the lesson:
to know the pairing of biological oxidation with catabolic and anabolic processes in the cell, stages of synthesis of high-energy compounds in the respiratory chain, the value and mechanism of regulation of oxidative phosphorylation.
Initial level of knowledge:
- anabolism and catabolism as two sides of a single process metabolism;
- the composition of the respiratory chain;
- structure and function of dehydrogenases.
52
Main topics.
I.2. The structure of mitochondrion.
The stages of ATP synthesis in electron transport chain in oxidative
phosphorylation. The ratio Р/О.
Substrate phosphorylation, its energy value.
The uncoupling of tissue respiration and phosphorylation.
Regulation of tissue respiration.
The tricarbonic acid cycle (Krebs cycle) is "focus" in which all metabolic pathways converge.
Succinate dehydrogenase catalyzes the oxidation (dehydrogenation) of succinic acid to fumarate. Its coenzyme is flavin adenine dinucleotide (FAD). As acceptor of hydrogen the oxidized form
of 2, 6- dichlorphenolindophenol (sodium salt) is used, the water solution of which is coloured in blue. The restored form of this compound is colorless. Malonic acid (НООС-СН2-СООН) as a structural
analogue of succinic acid causes competitive inhibition of succinate
dehydrogenase.
III.2. Control questions.
What reaction does succinate dehydrogenase catalyze?
What is the coenzyme of succinate dehydrogenase?
What is a structural analogue of succinic acid causing competitive
inhibition of succinate dehydrogenase?
III.3. Home assignment.
1. Choose the correct answer. The transfer of electrons from
NADH to oxygen gives …. moles of ATP.
A. 1
B. 2
C. 3
D. 6
E. 12
53
2. Match the inhibitors of oxidative phosphorylation with locus of
their action.
A. antimycin A
1. FMN → Q
B. barbiturates
2. cyt b → cyt c
C. rotenone
3. cytochrom oxidase
D. carbon monoxyde
4. phosphorilation
E. cyanide
F. oligomycin
3. Choose the correct variants. Sites in ETC where is enough energy
released for ATP synthesis are:
A. substrate → NAD+
B. NADH → Q
C. FAD → Q
D. Q → cyt b
E. cyt b, c1 → cyt c
F. cyt a3 → oxygen
4. Choose the wrong variant. The chemiosmotic theory postulates
that:
A. main task in oxidative phosphorylation is the translocation of protons to the exterior of a coupling membrane
B. protons accumulate outside the membrane creating an electrochemical gradient
C. Fi-ATP-ase catalyzes the synthesis of ATP
D. protons pass through Fo-Fi complex leading to the formation of
ATP from ADP and Pi
E. the carriers of the respiratory chain form four oxidation-reduction
loops
5. Choose the correct answer. Thermogenin is the uncoupling protein as it:
A. inhibits ATP-ase
B. prevent formation of an electrochemical gradient
C. creates a “proton leak”
54
D. inhibits cytochrom oxidase
E. forms the complex with Q
6. Choose the correct answer. The main function of TCA cycle is to:
A. generate energy
B. produce carbon dioxide
C. decompose the excess of the pyruvate and fatty acids
D. produce GTP
Material for self-study: I. 1.pp.235-240
2.pp.86-88, 93-98; II.
Class 11
MICROSOMAL OXIDATION.
COMPONENTS OF ANTIOXIDANT PROTECTION. DETFRMINATION OF THE ACTIVITY OF RESPIRATORY ENZYMES
The aim of the lesson:
1. To know features of microsomal oxidation, components of
monooxigenase system, biological significance of hydroxylation of
substrates and antioxidants.
2. To master methods of determination of catalase.
Initial level of knowledge:
- anabolism and catabolism as two sides of a single process - metabolism;
- concept of oxidative phosphorylation;
- the structure of the cell, microsome;
- oxidoreductases.
Main topics.
I.2. Structural features of monooxygenases.
Role of microsomal oxidation in the metabolism of regulatory molecules, xenobiotics.
The antioxidant role of vitamin C and Е.
55
The concept of peroxisomal oxidation.
The role of peroxidase, catalase, superoxide dismutase.
II.1. Work 1. QUALITATIVE REACTION FOR CATALASE
The enzyme catalase belongs to the class of oxidoreductases.
This enzyme is found in all tissues and liquids of the body, but for
the most part it is in erythrocytes and liver. The biological role of
catalase is the destruction of harmful hydrogen peroxide to molecular oxygen and water.
2H2O2 → 2H2O + O2
catalase
Principle of the method.
The enzyme catalase contains heme. In the process of destruction of hydrogen peroxide iron which is the part of the enzyme, is
oxidized and restored.
Practical procedure.
Reagents
Control test tube
Sample test tube
Н2О (ml)
1
1
Blood (dr.)
2
2
Boil for 2-3 minutes, then cool
3% solution Н2О2 (dr.)
5-10
5-10
Shake
RESULTS:
Work 2. QUANTITATIVE ASSAY OF CATALASE
Principle of the method and chemism. The basis for the quantitative determination of catalase is the amount of hydrogen peroxide decomposed by an enzyme for a certain period of time, according
to the following equation.
2KMnO4+5H2O2+4H2SO4 → 2MnSO4+8H2O+5O2+2KHSO4
56
Catalase activity is expressed by using catalase number. Catalase number (CN) is called the amount of milligrams of Н2О2 that
decomposed in 1 μL of blood. The number of split Н2О2 is determined by the difference in the volume of KMnO4 solution consumed
before and after the action of catalase.
Practical procedure.
Preparation of blood solution. Pour about 10 ml of distilled
Н2О in a measuring flask (volume 100 ml), add 0.1 ml of blood.
Rinse the micropipette three times with water from a flask. Add water to the mark, to dilute the blood in 1000 times (1 ml of solution
contains 1 μL of blood)
Reagents (ml)
Sample
Control
Н2О
7
7
Blood diluted 1:1000
1
1
1% solution Н2О2
2
10% solution Н2SO4
5
Note. The action of catalase is terminated in the sour
environment, since it operates at рН 7,4
Incubation for 30 minutes at room temperature
10% solution H2SO4
5
1% solution Н2О2
2
Titration with 0.1N KMnO4 solution until a pink colour.
CN = (А - В) ∙ 1.7
Where A - is the number of ml of 0.1N solution KMnO4 used
in the titration of the control;
B - the number of ml of 0.1N solution KMnO4 in ml, used
in the titration of the sample.
The obtained difference is multiplied by 1.7 (1 ml of 0.1N solution KMnO4 is equivalent to 1 ml 0.1N Н2О2, which contains 1.7
mmol/L (1.7 mg) Н2О2 since 1 mol/L Н2О2 equal to 1.7 g)
Clinical-diagnostic value.
Normal catalase number is 10-15 units. In the case of cancer,
anemia, tuberculosis content of catalase in the blood is reduced.
57
RESULTS:
А=
В=
CN =
CONCLUSION:
III.2. Control questions.
What class of enzymes do catalase and peroxidase belong to?
What cells contain the greatest amount of catalase?
What is the specific substrate for catalase?
What is the catalase number?
What is a reference value of catalase number?
Specify the diagnostic value of quantitative assay of catalase in the
blood.
III.3. Home assignment.
1. Choose the correct answers. Microsomal hydroxylase cycle in-
cludes such enzymes as:
E.
F.
G.
H.
monooxygenases
oxidases
dehydrogenases
hydroperoxidases
2. Choose the correct answer. Formation of oxaloacetate from succinate in Krebs cycle:
A. includes reaction of condensation
B. requires CoA
C. requires both NAD+ and FAD
D. produces three ATP
E. produces one mol of GTP
3. Match the letter with number.
A. α-ketoglutarate dehydrogenase
58
1. depends only on FAD
B. succinate dehydrogenase
C. malate dehydrogenase
D. all of the above
4. Complete the table
Enzyme of TAC
2. needs in CoA
3. depends only on NAD+
4. is regulated by NADH+
Activators
Inhibitors
Material for self-study: I. 1.pp.240-244, 260-265
2.pp.88-91, 130-135; II.
Class 12
COLLOQUIUM ON SECTION "BIOENERGETICS"
Questions for colloquium
1. Sources and ways of energy usage for various organisms. Give a
concept of an energy metabolism, biological oxidation, tissue respiration.
2. Metabolism of substances and energy as the basis of a life. Concept of metabolic pathways.
3. Endergonic and exergonic reactions in the alive cell.
4. High energy compounds are universal accumulators of energy during metabolism. Cycle ADP↔ATP. Kinds of phosphorylation as reactions of ATP generation.
5. Modern concept of biological oxidation.
6. The structure of mitochondria. Localization of the components of
oxidative phosphorylation system in it.
59
7. Respiratory chain is a key component of the mitochondrial oxidative phosphorylation system. The structural organization of the electron transport chain.
8. The place and the value of NAD+-dependent dehydrogenases in
the respiratory chain. Structure of the oxidized and reduced forms.
The major substrata of NAD+-dependent dehydrogenases.
9. The place and the value of FAD-dependent dehydrogenases in the
respiratory chain. Structure of the oxidized and reduced forms. The
major substrata of flavin-dependent dehydrogenases.
10. The place and the value of ubiquinone (CoQ) in the respiratory
chain, chemical nature and structure of the oxidized and reduced
forms.
11. Cytochromes, cytochrome oxidase; their chemical nature, place
and role in electron transport chain.
12. Free oxidation, oxidative and substrate phosphorylation.
13. Stages of high energy compound production in the electron
transport chain. The ratio Р/О.
14. Regulation of oxidative phosphorylation. The respiratory control.
The role of some biologically active and medicinal substances.
15. Uncoupling of ATP synthesis from electron transport. Thermoregulative function of tissue respiration.
16. The structure of mitochondria. Selective permeability of the mitochondrial membrane for substrata, ADP and ATP.
17. Infringements of an energetic metabolism as result of hypoxia.
Hypoxic conditions.
18. Vitamins В2 and niacin, ubiquinone. Their role in biological oxidation.
19. Cytochrome P450 monooxygenase system. Oxidation in the peroxisome. Tocopherol, ascorbic acid and β-carotene as antioxidant
chemicals.
20. The tricarboxylic acid cycle: biological role, sequence of reactions.
21 The allosteric mechanisms of the TCA cycle regulation.
22. Integration of common pathways of catabolism with the electron
transport chain (on example of TCA cycle).
60
III.3. Home assignment.
Material for self-study: I. Chapter “Bioenergetics”; II.
SECTION V. BIOSYNTHESIS OF NUCLEIC ACIDS AND
PROTEINS.
MOLECULAR MECHANISMS OF VARIABILITY
Class 13
THE STRUCTURE OF NUCLEIC ACIDS
The aim of the lesson:
1. To revise the structure of nucleotides.
2. To know structural features of DNA and RNA, types of chemical bonds in these molecules.
Initial level of knowledge:
- nitrogenous bases: kinds, structure;
- the structure of mononucleotides;
- the biological role of nucleic acids.
Main topics.
I.2. The structure of DNA.
Kinds of RNA, their quantitative ratio and localization in the cell.
Ribosomes and r-RNA.
t-RNA: uniqueness of structure, the specificity of action.
II.1. Work 1. HYDROLYSIS OF YEAST NUCLEOPROTEINS
To study the chemical composition of nucleoproteins acid hydrolysis of yeast is carried out since they are rich in nucleoproteins.
Specific reactions to each substance discover products of hydrolysis
- polypeptides, purine bases, carbohydrates and phosphoric acid.
а) Biuret reaction of the polypeptides
Chemism and practical procedure see in work 1 (class 1)
(pages 8-10).
RESULT:
61
b) Silver test for purine bases
Chemism of reactions:
NH2
⌡
N
N
+ AgNO3 + NH4OH
N
NH2
⌡
→
NH
N
N
+ NH4NO3 + H2O
N
N-Ag
Practical procedure.
To 10 dr. of hydrolysate add by drops strong solution of ammonia (approximately 10 dr.) before alkaline reaction is determined
by litmus paper dipped in the test tube. Then add 10 dr. of 2% ammonia solution of silver nitrate. Leave the test tube for 3-5 mines
without mixing.
RESULT:
c) Qualitative reaction for pentose
3Н2О
H2COH-(CHOH)3-COH →
strong
H2SO4
ribose
CH3
+
COH
O
furfural
62
→
―ОН
│
H3C-CH-CH3
thymol
→
the condensation products of red
Principle of the method and chemism.
In the interaction of concentrated sulfuric acid and pentoses
the dehydratation occurs and furfural, giving with thymol condensation products of red, is produced.
Practical procedure.
To 10 dr. of yeast hydrolizate add 3 dr. of 1% alcoholic solution of thymol, and stir. Then add 20-30 dr. of concentrated sulfuric acid carefully (by the wall of the test tube) and shake.
RESULT:
d) molybdenum test for phosphoric acid
Chemism of the reaction.
H3PO4+12(NH4)MoO4+21HNO3
→
→
(NH4)3P ∙12MoO3 + 21NH4NO3 + 12H2O
Ammonium phosphomolybdate
(yellow precipitate)
Practical procedure.
Pour 20 dr. of molybdenum reagent to the 10 dr. of hydrolizate and boil for a few minutes. The liquid (not precipitate) is painted in lemon yellow color. Cool the test tube in cold running tap water.
RESULT:
CONCLUSIONS:
III.2. Control questions.
63
What are the nucleoproteins?
What tissues contain a lot of nucleoproteins?
Specify the products of hydrolysis of nucleoproteins.
What proteins are the parts of nucleoproteins? Specify their features.
What qualitative reactions to the products of nucleoprotein hydrolysis do you know?
III.3. Home assignment.
1. Complete the table
Nitrogen base
Nucleoside
Nucleotide
2. Choose one correct answer. Nucleoside is:
А. nitrogen base + pentose
B. nitrogen base + phosphate
C. nitrogen base + pentose + phosphate
D. glucose + phosphate
3. Choose the correct answer. Linkage between nitrogen base and
sugar names:
А. N-glycosidic
B. phosphoester
C. phosphodiester
D. peptide
I. ionic
4. Match the letter and number.
А. adenine
1. In DNA
B. thymine
2. In RNA
C. cytosine
3. Is purine
D. uracil
4. Is pyrimidine
5. In DNA and RNA
64
5. Match the letter and number.
А. Deoxyadenosine monophosphate I. Has ribose
B. Timidine monophosphate
2. Contains the purine base
C. Both
3. Contains the pyrimidine base
D. None
4. Has phosphate at the 5'-position
of pentose
6. Match the letter and number.
А. DNA I. The chains in its molecule are antiparallel
B. RNA 2. Hydrogen bonds are formed between nitrogen bases
C. Both
3.Hydrogen bonds are formed between A and G and between C and U
D. None 4. Its molecule consists of one polynucleotide chain
Material for self-study: I. 1.pp.73-83
2.pp.286-292, 303-312; II.
Class 14
BIOSYNTHESIS OF NUCLEIC ACIDS
The aim of the lesson:
to know the stages of synthesis of nucleic acids in alive organisms.
Initial level of knowledge:
- structure and functions of nucleic acids;
- replication and phases of the cellular cycle;
- the concept of transcription.
Main topics.
I.2. Synthesis of DNA: stages, mechanism, value.
Damage and repair of DNA.
The mechanism of transcription.
The concept of the mosaic structure of genes, the primary transcriptonе.
Posttranscriptional completion of RNA.
65
I.3. Reference work.
SUMMARY QUESTIONS FOR A TEST
«THE STRUCTURE AND FUNCTIONS OF NUCLEIC ACIDS»
1. Structure of pyrimidine and purine nucleic bases (lactam and lactim forms). The complementarity of the bases. The minor nitrogenous bases.
2. The structure of N-glycosides (nucleosides) D-ribose and 2deoxy-D- ribose with nucleic bases and reactions of their hydrolysis.
Pseudonucleosides.
3. The structure of nucleotides, constituting the DNA (deoxyadenilate, deoxyguanilate, deoxycytidilate, thymidylate) and in RNA
(adenilate, guanilate, cytidilate, uridilate). Scheme of incomplete and
complete hydrolysis of these mononucleotides.
4. The structure of diphospho- and triphosphonucleotides, scheme of
their partial and complete hydrolysis.
5. Scheme of incomplete and complete hydrolysis of nucleic acids.
6. The primary structure of nucleic acids. The structure of DNA sites
(structure of triplets, for example, TGA, ACG, CTA, etc.).
7. Secondary and tertiary structure of nucleic acids. Distinction of
structure of DNA and RNA.
8. Complementary polynucleotide chains. The model of the double
helix.
9. Functions of DNA and RNA. Types of RNA.
1О. The structure of RNA sites with sequence of the bases: UGA,
AUG, CUG, etc.
11. The structure of mRNA fragments, obtained by transcription
from TCA, GTA, ACT, etc.
12. Features of the structure and specificity of transfer RNA.
13. The structure of anticodons of tRNA, corresponding to the codons AUG, CUA, UAG, etc.
III.3. Home assignment.
1. Choose the wrong answer. Replication:
66
А. begins after turning the cell in the phase of the synthesis
B. expects formation of replicative fork
C. realizes exact reproduction of DNA before each cellular division
D. provides the frequentative reduplication of genome during Sphase
E. is terminated by formation of tetraploidic chromosome set
2. Choose the correct answer. Transcription:
А. occurs in S-phase of the cellular cycle
B. is initiated by formation of primer
C. always begins with codon AUG
D. occurs at participation TATA-factor
E. does not require local unwinding of double helix DNA
3. Choose the correct answer. Exones are:
А. fragments of DNA, carry out regulatory functions
B. fragments of tRNA
C. coding sequences in mRNA
D. sequences in mRNA which don’t code the protein
E. trinucleotides
4. Choose the wrong answer. Posttranscriptional modification includes:
А. capping
B. addition of a tail
C. removal of introns
D. removal of exones
E. splicing
5. Choose the correct answer. While studying the structure of the
small gene that was recently sequenced during the Human Genome
Project, an investigator notices that one strand of the DNA molecule
contains 19 A’s, 24 G’s, 28 C’s and 32 T’s. How many of each base
is found in the complete double-stranded molecule?
А. 45 A’s, 45 G’s, 45 C’s and 45 T’s
67
B. 15 A’s, 45 G’s, 55 C’s and 35 T’s
C. 35 A’s, 55 G’s, 55 C’s and 35 T’s
D. 30 A’s, 40 G’s, 508 C’s and 60 T’s
6. Match the letter and number.
А. the enzyme synthesizing oligoribonucleotide 1. primer
B. present at 5’-end of DNA chains
2. telomerase
C. oligoribonucleotide
3. telomer
D. the enzyme containing fragment RNA
4. primase
as prosthetic group
Material for self-study: I. 1.pp.421-434
2.pp.314-356; II.
Class 15
BIOSYNTHESIS OF PROTEIN. REGULATION OF PROTEIN
SYNTHESIS IN PROKARYOTES AND EUKARYOTES
The aim of the lesson:
1. Consolidate knowledge of the stages of replication and transcription, the role of DNA-and RNA-polymerases.
2. To know the mechanism of translation and regulation of biosynthesis of protein.
Initial level of knowledge:
- types of nucleic acids;
- structure and function of DNA;
- the structure and functions of different types of RNA;
- biosynthesis of nucleic acids.
Main topics.
I.2. Presentations of students in oral form and discussion.
THE TOPICS OF THE PRESENTATIONS
68
TO THE WORKSHOP
1. History of the study and the general scheme of protein synthesis.
2. Genetic code: the code elements, properties.
3. The flow of genetic information. Matrix synthesis of RNA. RNApolymerases. Reverse transcriptases.
4. Activation of amino acids and transportation of them to the place
of protein synthesis. The role of tRNA as an adaptor between the
nucleotides and amino acids.
5. Stages of protein synthesis. The structure of ribosomes, the functioning of polyribosomes.
6. Mechanism of translation and synthesis of polypeptide chain: initiation and elongation orderly arrangement of amino acids.
7. Termination of protein synthesis, posttranslational processing.
8. Regulation of protein synthesis in prokaryotes. The theory of F.
Jacob and J. Monod.
9. Regulation of protein synthesis in eukaryotes.
10. Molecular mutations and hereditary illnesses.
III.3. Home assignment.
1. Choose the correct answer. Anticodon is:
А. the triplet of nucleotides in DNA, encoding one amino cid
B. the place of the joining of amino acid to tRNA
C. the triplet of nucleotides in tRNA, complementary to codon in
mRNA
D. stop-codon in mRNA
E. triplet of nucleotides in mRNA, encoding one amino acid
2. Match the letter and number.
А. antineoplastic preparations 1. are inhibitors of replication
B. antibacterial preparations 2. inhibit virus dublication in host
cells
C. both
3. repress synthesis of protein in prokariots
D. none
4. effectively reduce velocity of the formation of nucleic acids
69
3. Choose the wrong answer. Operon:
А. is a fragment of the molecule RNA
B. are produced at viral infection
C. activate RNAase which splits rRNA
D. stop synthesis of protein in infected cells
E. break structure of small subunit of ribosome
4. Choose the wrong answer. Interferons:
А. are proteins
B. are produced at viral infection
C. activate RNAase which splits rRNA
D. stop synthesis of protein in infected cells
E. break structure of small subunit of ribosome
5. Choose the wrong answer. Many viruses in human organism:
А. link with subunits of the ribosome
B. stop own transcription
C. stop the synthesis of cellular proteins
D. switch the protein-producing system of the cells to synthesis of
virus proteins
E. synthesize the nucleic acids of the virus in infected cells
6. Choose the wrong answer. There are differences in the set of
proteins of human tissues as:
А. different genes are expressed in different tissues
B. the genome of the different cells is distinguished
C. the genes of proteins not characteristic for this tissue, are firmly
repressed
D. the expression of genes is checked with induction and repression
E. firmly repressed genes are not actuated by external and internal
factors
Material for self-study: I. 1.pp.73, 434-445, 571-576
2.pp.36-37, 358-394; II.
70
Class 16
COLLOQUIUM: "MATRIX BIOSYNTHESES.
MOLECULAR MECHANISMS OF VARIABILITY"
Questions for colloquium
1. Nitrogenous bases. Lactam - lactim tautomerism.
2. Minor bases. Pseudonucleosides. Examples.
3. Nucleosides and nucleotides. The structure and nomenclature.
Cyclic AMP.
4. The primary structure of DNA. Chargaff’s Rules.
5. The structure of nucleoside di- and triphosphates, scheme of partial and complete hydrolysis. High-energy phosphate bonds.
6. Secondary and tertiary structure of DNA.
7. Primary structure and types of RNA, their localization in the cell.
8. Secondary and tertiary structure of different types of RNA. Peculiarity of tRNA structure.
9. The complementarity of heterocyclic bases (nucleotides). Role of
the hydrogen bonds in the formation of secondary structure of nucleic acids.
10. Replication of DNA and phases of the cell cycle. DNApolymerases. Damage and repair of DNA.
11. Biosynthesis of RNA (transcription). RNA-polymerases.
12. The nature and principal features of the genetic code.
13. The main components required for translation and the stages of
protein synthesis. Flow of the genetic information.
14. The structure and function of ribosome and polyribosomes.
15. Activation of amino acids (synthesis of aminoacyl-tRNA) and
transportation of them to the place of protein synthesis. tRNA as
adaptor. Substrate specificity of aminoacyl-tRNA -synthetases.
16. Stages of translation. Initiation of this process.
17. Stages of translation. Elongation of this process.
71
18. Stages of translation. Termination of this process. Formation of
secondary, tertiary and quaternary protein structures. Posttranslational processing.
19. Inhibitors of DNA, RNA and protein synthesis: antibiotics, antiviral drugs, viral and bacterial toxins.
20. Regulation of gene expression. F.Jacob and J.Monod’ hypothesis
about operon model. Repression and induction of protein synthesis
in prokaryotic cells.
21. Induction and repression of protein synthesis in eukaryotic cells.
The role of hormones in regulation of gene expression.
22. A cellular differentiation and ontogenesis as result of differential
activity of genes. Change of protein set of cells during differentiation.
23. Genotypic heterogeneity of populations and polymorphism of
proteins on the example of hemoglobin and some enzymes.
24. Biochemical bases of medical genetics. Molecular mutations and
hereditary diseases. Biochemical methods in the diagnosis of congenital diseases.
III.3. Home assignment.
Material for self-study: I. chapter “Matrix biosyntheses”; II.
SECTION VI. BIOCHEMISTRY OF BLOOD
Class 17
PLASMA PROTEINS. METHODS OF
DETERMINATION OF PROTEINS IN BLOOD SERUM
The aim of the lesson:
1. To know the main plasma proteins, their functions and features
of the structure.
2. To muster methods of separation of proteins in clinic.
Initial level of knowledge:
72
- structure and biological role of proteins;
- classification of proteins;
- the difference albumins and globulins;
- enzymodiagnosis.
Main topics.
I.2. The physiological role of blood proteins.
Characteristic of individual proteins. The concept of hypo- and hyperproteinemia.
Disproteinemia, paraproteinemia, defectoproteinemia.
Canceroembrionic proteins.
II.1. Work 1. QUANTITATIVE ASSAY OF PROTEINS BY THE
BIURET METHOD
Principle of the method.
The method is based on the formation of purple complex of
protein peptide bonds with bivalent ions of copper (copper sulfate)
in an alkaline media (the biuret complex). The colour intensity of the
solution is proportional to the concentration of protein in blood serum and is determined by photometry.
Practical procedure.
Reagents (ml)
Biuret reagent
Blood serum (take micropipette)
0.9% solution NaCl
Sample
5
0.1
-
Control
5
0.1
Stir gently and leave for 30 minutes in a rack at room temperature.
Measure optical density of the sample against the control at a
green optical filter (λ=540 nm). Find concentration of protein (g/L)
using schedule.
RESULT:
CONCLUSION:
73
Clinical-diagnostic value.
Reference value: for adults - 65-85 g/L (6.5 – 8.5 g %), for
children under 6 years - 56-85 g/L (5.6-8.5 g %).
The increase of the concentration of protein (hyperproteinemia) is observed in rheumatism and myeloma (plasmocytoma) up to 120 to 150 g/L. Short-term relative hyperproteinemia is marked
by thickening of the blood due to significant losses of liquid, for example, the intense sweating, uncontrollable vomitting, profuse diarrhea, diabetes, cholera, severe burns.
The decrease in protein level (hypoproteinemia) occurs in
nephrites, malignant tumors and malnutrition.
Work 2. THYMOL TEST
Principle of the method.
Pathologicaly increased β-globulins, γ- globulins and lipoproteins are precipitated from blood serum by sated thymol. Measure the intensity of turbidity dependent on concentration of protein
fractions and their quantitative ratio.
Practical procedure.
Reagents (ml)
Sample
Control
Reagent
3.0
3.0
Blood serum (take a micropi0.05
pette)
Physiological solution
0.05
Stir and keep exactly for 30 minutes at room temperature, then
stir again and measure the optical density of the sample against that
of the control in a 1 cm cuvette at wave length of 650 nm (filter No.
4).
RESULT:
74
CONCLUSION:
Clinical-diagnostic value.
Normal values: 0-4 units S-H (units of turbidity ShankHoagland). Increase thymol test is most characteristic for liver disease (even before the appearance of jaundice). Pathological value is
more than 5 units S-H.
The teacher demonstrates all subsequent work.
Work № 4. REACTION OF PRECIPITATION IN AGAR BY
OUCHTERLONI
Fig.4. Semiquantitative estimation of the results of the analysis. 1-antiserum, 2-test-antigen, 3-10 - serial dilutions of the test
sample: 1/1, 1/2, 1/4, etc. up to 1/128.
Principle of the method.
During a counter diffusion in agar molecules of antigen and
of antibody react with each other, forming a precipitate. Line of
precipitation becomes visible in a passing ray of light, if the reacting
components are in optimum proportion. With this method it is possible to analyze the antigenic mix since each pair of antigen-antibody
forms the line of precipitation, and to compare the antigenic components in various systems.
75
Work 5. SEPARATION OF PROTEINS BY METHOD OF ELECTROPHO-RESIS ON PAPER, IN AGAR AND IN POLYACRYLAMIDE GELS
Fig.5. Separation of human blood serum by electrophoresis in
agarose gel and the curves obtained by the densitometry of electrophoregram. Normal serum (А) and serum of patients with multiple
myeloma (Б), hepatocirrhosis (В) and nephrotic syndrome (Г).
Fig.6. Separation of the proteins of the blood serum in polyacrylamide gel with a linear gradient of concentration in the range of 430%.
76
Principle of the method.
Protein molecules possess a free electric charge whose magnitude and sign depend on the proportion of basic and sour ionized
groups in the molecule. Under the influence of an external electric
field the charged protein molecules move in solution to oppositely
charged pole. The separation of the mix of protein in individual fractions is the result of the difference in relative molecular mass and
charge of the protein molecules which move in an electric field unequally. The speed of movement of protein molecules is proportional
to the amount of their free charge and inversely proportional to the
size of the particles and their degree of hydration. Electrophoretic
separation of proteins is widely used both for diagnostics of diseases, and for preparative purposes.
Work 6. SEPARATION OF PROTEINS BY IMMUNOELECTROPHORESIS
1
a)
albumin
b)
globulins
start
Trench with anti-
bodies
2
Fig.7. 1. Scheme of immunoelectrophoresis: а) electrophoresis of
blood serum proteins; b) immunodiffusion with the formation of individual lines of precipitation for each type of antigenes.
77
2. Immunoelectrophoregram of serum proteins (negative).
Principle of the method.
Immunoelectrophoresis was first proposed by Grabar and Williams in 1953. It is the combination of electrophoretic separation of
proteins followed by immunoprecipitation in the gel which significantly increases the sensitivity of the method. Using immunoelectrophoresis is possible to analyze the antigenic mix, qualitative and
quantitative determination of known antigen. For example, in the
blood serum of healthy adults, this method can allocate more than 30
various protein fractions. Depending on your goals you can use different versions of immunoelectrophoresis.
III.2. Control questions.
What is concentration of total protein in blood serum in norm?
What conditions are characterized by hypoproteinemia?
What conditions are characterized by hyperproteinemia?
What is thymol test in norm?
In what diseases does thymol test increase?
Name the main fractions of blood proteins in electrophoregram.
Specify immunochemical methods for the analysis of proteins.
Explain the basic principle of immunochemical methods.
III.3. Home assignment.
1. Choose the correct answer. Plasma fraction of blood in volume
per cent is:
А. 35
B. 45
C. 50
D. 55
E. 60
2. Match of the following. Type of disproteinemia
А. decrease of albumin and increase of γ–globulin fractions
78
B. decrease of albumin, α2-globulins and increase of β– and γ–
globulins
C. increase of albumin and moderate decrease of all globulins
D. increase of α1- and α2-globulins
1. dehydratation
2. hepatitis
3. acute inflammation
4. chronic inflammation
3. Choose the wrong answer. The serum albumin concentration decreases in:
А. starvation
B. insufficiency of liver
C. nephritis
D. all of the above
E. none of the above
4. Match the protein and the fraction of blood proteins.
А. Ig A
1. α1-globulins
B. ceruloplasmin
2. α2-globulins
C. retinol-binding protein
3. β–globulins
D. transferrin
4. γ–globulins
E. transcortin
F. T4-binding globulin
G. haptoglobin
5. Match the correct statements about anemias below.
А. sickle cell anemia 1. replacement “his” in a pocket of heme
B. methemoglibinemia 2. large deletions in the β–globin gene
C. thalassemia
3. replacement “glu” in 6th position of β–
globin
6. Choose the wrong answer. The serum albumin concentration decreases in:
А. starvation
B. insufficiency of liver
C. nephritis
79
D. diarrhea
E. none of the above
Material for self-study: I. 1.pp.185-193
2.pp.580-597; II.
Class 18
CLOTTING OF BLOOD AND FIBRINOLYSIS
The aim of the lesson:
1. To explore the current views on clotting of blood.
2. To know the role of the main factors of the coagulation system.
3. To know the nature of anticoagulative blood system.
4. To master method of determination of fibrinogen.
Initial level of knowledge:
- structure and biological role of proteins;
- classification of proteins;
- catalyitic function of proteins;
- blood: chemical composition, physical and chemical characteristics, functions;
- solid elements of blood;
- the difference between blood plasma and the serum.
Main topics.
I.2. Plasma clotting factors: the place of synthesis, structure, functions.
Platelet coagulation factors: name, function.
"External" and "internal" ways of activation of blood coagulation.
Scheme of fibrinolysis.
Medicines used for the treatment and prevention of thrombosis.
Hemophilias: the nature, the methods of treatment.
II.1 Work 1. DETERMINATION OF THE CONCENTRATION OF
FIBRINOGEN
80
Principle of the method. When calcium chloride solution is
added to plasma the coagulation of fibrinogen occurs and the sediment is formed.
The amount of the precipitated fibrinogen is determined by
weighing clot on the torsion scales.
Practical procedure.
Pour 1 ml of plasma in a test tube and add 0.1 ml of 5% solution of calcium chloride. Stir the test tube and leave it in a thermostat at temperature 370С for 15 minutes. The fibrin clot is formed to,
press on filter paper until the disappearance of traces of moisture.
Weigh pressed clot on the torsion scales with an accuracy of 1 mg.
Obtained result is multiplied by a factor of 0.222. The quantitative
result is expressed in g/L of plasma.
RESULT:
CONCLUSION:
Clinical-diagnostic value.
Reference value: 2-4 g/L.
Significant increase in the concentration of fibrinogen is observed in case of systemic diseases of the connective tissue and
acute inflammatory processes. The decrease is due, primarily, impaired synthesis at pathology of liver and consumption of fibrinogen
in a large amount in case of thrombotic hemorrhagic syndrome.
Work 2. MICROMETHOD FOR DETERMIINATION of FIBRINOGEN “B” IN PLASMA
Principle of the method. Alcoholic solution of  -naphthol precipitates from the plasma the fibrin-monomers and their complexes
with fibrin peptides A and B, and fibrinogen.
Practical procedure.
In the centrifugal test tube add 0.5 ml of plasma and add 2 dr.
of 2% solution of -naphthol. Shake the test tube and leave it in a
81
rack at room temperature for 10 minutes. In the presence of fibrinogen “B”, the clot, strands or pellets are formed.
Clinical-diagnostic value.
In norm fibrinogen “B” in blood is not determined. The appearance of fibrinogen “B” in plasma is due to pathological increase
in the level of thrombin in the blood, therefore a positive reaction to
fibrinogen “B” is of great importance for the diagnostics of pre- and
- thrombotic states, thrombotic hemorrhagic syndrome.
RESULT:
CONCLUSION:
III.2. Control questions.
What is the concentration of fibrinogen in the blood plasma in
norm?
Specify the reasons of hyperfibrinogenemia, hypofibrinogenemia.
In what case can fibrinogen “B” be found in?
III.3. Home assignment.
1. Choose the correct answer. Vitamin K regulates the synthesis of
blood clotting factors:
А. II
B. VII
C. IX
D. X
E. all of the above
2. Match the letter and number.
А. fibrinogen
1. has structure (α, β, γ)
B. fibrin monomer
2. has structure (Aα2, Bβ2, γ2)
C. thrombin
3. stabilizes fibrin clot by covalent crosslinking
D. transglutamidase 4. is the proenzyme
82
E. prothrombin
5. hydrolyses peptide bonds arg-gly
3. Match the letter and number.
А. inactivates thrombin
1. antithrombin III
B. is activated with heparin
2. α2-nacroglobulin
C. inactivates Xa and VII factors 3. inhibitor of tissue factor
D. interacts with tissue factor
E. forms complexes with serine proteases for their inactivation
4. Choose the correct answer. After reception more than 50 g of
butter fibrinolysis:
А. increases
B. decreases
C. doesn’t change
5. Choose the wrong answer. Substrate for thrombin is:
А. factor V
B. factor VIII
C. fibrinogen
D. protein C
E. thrombomodulin
6. Choose the wrong answer. The inhibitor of blood clotting enzymes is:
А. plasmin
B. α2-nacroglobulin
C. α1-antitripsin
D. antithrombin III
E. anticonvertin
Material for self-study: I. 1.pp.193-196
2.pp.598-608; II.
83
LITERATURE
1.
2.
3.
4.
5.
Биохимический практикум. /Под ред. Д.М.Никулиной. Издво АГМА, Астрахань, 2010, 146с.
Practical and laboratory studies on biochemistry. Part 1 /Edited
by prof. O.V.Ostrovskiy.-Volgograd, 2004-120p.
U. Satyanarayana //Biocnemistry. Books and allied (p) Ltd. 8/1
Chintamoni Das Lane, Kolkata, 700009 (India), 2004.
Robert.K.Murray, D.K.Granner, P.A.Mayes, V.W.Rodwell
//Harper’s Illustrated Biochemistry.26-th edition. McGraw-Hill
Companies, Inc., USA, 2003
A. Lehninger, D. Nelson, M. Boyle Cox //Principles of biochemistry (4th ed.), 2005
84
Contents
Rules on safety for students when working in laboratories…..
The content of the class………………………………………
Material for self-study..……………………………...………
Estimation criteria of student knowledge……………….……
Class 1. The biological role and the physical-chemical properties of proteins. Methods of extraction and fractionation of proteins………………………………………………………
Class 2. Structure of protein molecules. Methods of purification
of proteins…………………………………………… …
Class 3. Classification of proteins. Chemistry of simple and
complex proteins. Methods of protein precipitation…… ……
Class 4. Structure and basic properties of enzymes.…………
Class 5. Mechanism of enzyme action. Kinetics of enzyme reactions………………………………………………..…….
Class 6. Regulation of activity of enzymes. Classification, nomenclature and the practical application of enzymes….…..
Class 7. Colloquium on section "Enzymes"……………….…
Class 8. Classification, structure and biologic role of vitamins.
Methods for the detection of vitamins…………………
Class 9. Mitochondrial electron transport chain……………..
Class 10. Oxidative phosphorylation. Regulation of the electron
transport chain (respiratory control……………….…….
Class 11. Microsomal oxidation. Components of antioxidant
protection. Determination of the activity of respiratory enzymes……………………………………………………...…
Class 12. Colloquium on section "Bioenergetics"……………
Class 13. The structure of nucleic acids……………………..
Class 14. Biosynthesis of nucleic acids………………………
Class 15. Biosynthesis of protein. Regulation of protein synthesis in prokaryotes and eukaryotes…………………………
Class 16. Colloquium: "Matrix biosyntheses. Molecular mechanisms of variability"………………………….………..
85
3
5
5
6
7
12
16
24
30
36
40
43
49
52
55
59
61
65
68
71
Class 17. Plasma proteins. Methods of determination of proteins in blood serum………………………………….………
Class 18. Clotting of blood and fibrinolysis………. .……..…
References……………………………………………………
86
72
80
84
Воробьева Татьяна Борисовна
Никулина Дина Максимовна
THE MANUAL ON BIOCHEMISTRY. Part 1
РУКОВОДСТВО К ЗАНЯТИЯМ
ПО БИОХИМИИ. Часть 1
Учебное пособие
Компьютерный набор - авторский
Технический редактор – Нигдыров В.Б.
ISBN 978-5-4424-0140-0 (основной)
ISBN 978-5-4424-0192-9 (часть 1)
Подписано к печати 04.07.2016
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Издательство Астраханский государственный медицинский
университет 414000, г. Астрахань, ул. Бакинская, 121
87
T.B.Vorobyeva, D.M.Nikulina
Т.Б.Воробьева, Д.М.Никулина
THE MANUAL
ON BIOCHEMISTRY
РУКОВОДСТВО К ЗАНЯТИЯМ
ПО БИОХИМИИ
Part 1
Часть 1
ASTRAKHAN - 2016
АСТРАХАНЬ – 2016
88