Interior Textbook Of Practical Biochemistry (bogdan Manolescu) 2015 Opt [627198]
Bogdan Nicolae Manolescu
Carmina Bușu
TEXTBOOK OF
PRACTICAL
BIOCHEMISTRY
Referenți științifici:
Conf. Dr. Ing. Anca Marton
Departamentul de Chimie Organic ă „C. Nenițescu”,
Facultatea de Chimie Aplicat ă și Știința Materialelor, Universitatea POLITEHNICA Bucure ști
ȘL. Dr. Eliza Oprea
Departamentul de Chimie Organic ă, Biochimie și Cataliză,
Facultatea de Chimie, Universitatea din Bucure ști
Descrierea CIP a Bibliotecii Na ționale a României
MANOLESCU, BOGDAN NICOLAE
Textbook of Practical Biochemistry / Bogdan Nicolae Manolescu, Carmina Bu șu. –
București: Editura NICULESCU, 2015
Bibliogr.
ISBN: 978-973-748-949-4
I. Bușu, Carmina
577.1
© Bogdan Nicolae Manolescu, Carmina Bu șu
© Editura NICULESCU, 2015
Adresa: Bd. Regiei 6D
060204 – Bucure ști, România
Comenzi: (+40)21-312.97.82
Fax: (+40)21-312.97.83
E-mail: [anonimizat]
Internet: www.niculescu.ro
Coperta: Carmen Lucaci
Tipărit în România
ISBN 978-973-748-949-4
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5
Foreword
Biochemistry had an impressive development during the 21st century due to the
parallel improvement of different research techniques. As a consequence, new branches
appeared, including molecular biology, biotechnology, genetic engineering etc.
Nowadays, biochemistry has applications in many areas, from clinical chemistry and
human therapy to industrial production of organic precursors or total synthesis of different
drugs. This is mainly the result of understanding the enzymes, the biocatalysts found inside
all cells.
Due to its enormous advance and the subsequent applications, biochemistry is very
useful to chemical engineers. Biochemistry has its own principles and understanding them
is the first step in understanging the usefulness of this scientific field.
The present Textbook of practical biochemistry is intended for the undergraduate
students in the field of Chemical Engineering, who have a Biochemistry course in their
syllabus.
This textbook presents a total of 35 practical demonstrations which are aimed to
help the student: [anonimizat]: carbohydrates, lipids, amino acids and proteins, and enzymes. In the case of
the quantitative assays the authors presented classic methods as well as the modern versions
that use enzymes as analytical reagents. The methods can be used for the analysis of both
animal and vegetal samples.
We are very thankful to all our colleagues for their suggestions that will help us to
improve the present textbook.
Authors
6 Experiment 1
pH and buffer solutions
Background
Water molecule has a very slight tendency to ionize giving rise to a free proton (H+)
and to the hydroxide ion (HO():
H2O H+ + HO(
The free proton associates with a second water molecule to form the so called
hydronium ion (H 3O+). In fact, the proton passes from a water molecule to another one. The
speed of this process is very high as the hydronium ion has a mean life(time of about 10(12 s
at 25°C. It can be said that a proton ” jumps ” from a water molecule to another one.
H
OH
H
O
H H
O
HHO
HH
O
HH
O
HH
O
HH
OH
H
This is also one of the explanations for the fact that the acid(base reactions are
among the fastest reactions that take place in aqueous solutions. The following equilibrium
constant can be written for water ionization:
7 [][]
[ ]O HHO H
2(
eq+
=K
In fact, Keq is the dissociation constant of the water.
At a pressure of 1 atm and a temperature of 25°C this Keq has a value of 1.8 × 10(16
M. At equilibrium a very small percentage of the water molecules will dissociate to form
H+ and HO(. On the other hand, pure water at 25°C has a concentration of approximately
55.5 M.
The Keq can be rewritten as follows:
[][][](
2 eq HO H O H+= K
Substituting into this equation the afforementioned values gives:
[][]( 2 14 16HO H M 10 0 . 1 5 . 55 10 8 . 1+ − −= × = × ×
The value obtained by multiplying the H+ and HO( concentrations is called ion
product of water or ionization constant of water ( Kw) . It has a value of 10(14 M2 at 25°C.
[][](
w HO H+=K
Pure water is electrically neutral, so it contains equal amounts of H+ and HO(:
[][] M 10 HO H7
w( − += = =K
The concentrations of H+ and HO( are reciprocally related through Kw and, as a
consequence, when the concentration of one of the two chemical species increases the other
one’s must decrease.
Aqueous solutions for which [H+] = [HO(] = 10(7 M are said to be neutral, those with
[H+] > 10(7 M are said to be acidic, and those with [H+] < 10(7 M are said to be basic.
8 For example, in physiological conditions, human blood contains a concentration of
H+ of about 4.0 × 10(8 M. This concentration could be written also as 0.00000004 M which
is totally inconveniently. Moreover, it is impractical to compare different solutions using
this way of expressing the concentration of H+.
For example, it would be difficult to compare a solution with a concentration of H+
of 0.0000000001 M with one with a concentration of 0.00000000001 M it would be
difficult. In fact, in the first case the concentration is 10(10 M, while in the second case the
concentration is 10(11 M.
Due to the aforementioned impractical aspects, in 1909, the danish chemist Søren
Sørenson introduced the concept of pH:
[][ ]++= − =
H1log H log pH10 10
As a consequence, acidic solutions have a pH < 7, basic solutions have a pH > 7 and
neutral ones have a pH = 7. Also, the higher the H+ concentration, the lower is the pH.
Biological molecules have many functional groups which are involved in acid(base
reactions. Moreover, the ionization of these groups is influenced by the pH of the aqueous
medium, which influences the structure and reactivity of the biomolecules.
COOH COO +H+
NH2+H+ NH3
N NR
H HN NR
H+H+
9 It can be seen from the previous reactions that there are groups which can participate
in protonation(deprotonation reactions in direct relation with the pH of the aqueous
medium.
According to Svante Arrhenius (1880), acids and bases were defined as compounds
able to donate protons and hydroxide ions, respectively. This definition can not explain that
NH 3 is a base despite it lacks the hydroxile group.
This definition was reformulated by Johannes Brønsted and Thomas Lowry (1923).
Accordingly, an acid is a compound that can donate protons (like in the case of Arrhenius
definition), while a base is a compound that can accept protons.
A much broader definition was given by Gilbert Lewis (1923): a Lewis acid is a
molecule that can accept a pair of electrons, while a Lewis base is a molecule that can
donate a pair of electrons. This definition is too broad for the biochemical reactions.
The Brønsted(Lowry definition is very important in biochemistry. An acid (HA) is
able to donate its proton according to the following equilibrium:
HA A( + H+
The acid HA donates its proton forming the corresponding conjugate base (A(). The
proton reacts with a base, which in the case of an aqueous solution is a water molecule:
H2O + H+ H 3O+
The water molecule acts like a base forming the corresponding conjugate acid
(H3O+).
By summing the previous two equilibria the following equilibrium can be obtained:
HA + H 2O A( + H 3O+
For the previous equilibrium, the following equilibrium constant, in fact the
dissociation constant, may be written:
10 [][]
[ ][ ] O H HAA O H
2(
3+
=K
Due to the fact that the water concentration is basically a constant (55.5 M), the term
[H2O] is combined with the equilibrium constant. So, the previous constant can be rewritten
as follows:
[ ][][]
[ ]HAA HO H(
2 a+
= = K K
For the same reasons like in the case of H+ concentration, Ka was correlated to p Ka:
a 10 a log p K K− =
The Ka and p Ka for different acids are listed in tables like the following one.
Dissociation constants and pK a values of some acids (25°C)
Acid Ka pK a
H3PO 4 7.08 × 10(3 2.15 (p Ka1)
Acetic acid 1.74 × 10(5 4.76
H2CO 3 4.47 × 10(7 6.35 (p Ka1)
H2PO 42( 1.51 × 10(7 6.82 (p Ka2)
Tris(hydroxymethyl)aminomethane (Tris) 8.32 × 10(9 8.08
NH 4+ 5.62 × 10(10 9.25
HCO 3( 4.68 × 10(11 10.33 (p Ka2)
HPO 42( 4.17 × 10(13 12.38 (p Ka3)
The p Ka value is a measure of the acid strength of a particular compound, while pH
is a measure of the acidity of a solution.
All the acids from the previous table are classified as weak acids because they are
only partially ionized in aqueous solution ( Ka < 1). On the other hand, in the case of the so
called strong acids like HCl, HNO 3 and HClO 4 the dissociation is complet ( Ka >> 1):
11 HCl + H2O Cl( + H3O+
Taking into consideration the aforementioned aspects it is obvious the fact that the
strongest acid that can stably exist in an aqueous solution is H 3O+, while the strongest base
is HO(.
The relationship between the concentrations of an acid and its conjugate base, on
one hand, and the pH value of that solution, on the other, cand be very easily obtained.
[][]
[ ][][]
[ ]HAA Hlog logHAA H(
10 a 10log(
a10+
−+
− = − → = K K
The previous relation can be rearranged as follows:
[ ][]
[ ]HAAlog H log log(
10 10 a 10 − − = −+K
But, taking into consideration that p Ka = ( log Ka and pH = ( log [H+], the previous
relation can be rewritten as follows:
[]
[ ]HAAlog p pH(
10 a+ =K
This relationship is known as the Henderson(Hasselbalch equation and it contains
two terms, a constant one (p Ka), and one that depends on the concentrations of the chemical
species (the weak acid and its conjugate base).
A particular situation is the one when the concentrations of the two chemical species
are equal, [A(] = [HA]. In this case, log 1 = 0, and pH = p Ka.
The Henderson(Hasselbalch equation is very useful for calculating the pH of a
solution that contains known quantities of a weak acid and its conjugate base.
12 Maintaining a relative constant pH is very important for all living systems as a very
small change of the pH can affect the structure and the function of the biomolecules. This is
achieved with the help of buffer systems.
A buffer system contains always an acidic (HA) as well as a basic component (A().
A small amount of added H+ reacts with the basic component of the buffer system (A(),
while a small amount of added HO( reacts with the acidic component of the buffer system
(HA).
As a consequence, an acid(base buffer resists to pH changes when adding a small
amount of acid or base.
A very simple buffer system is a mixture of acetic acid and sodium acetate. When
adding a small amount of H+ the same amount of sodium acetate will be consumed giving
rise to the same amout of acetic acid. When adding a small amount of HO( the same amount
of acetic acid will be consumed giving rise to the same amount of sodium acetate.
For example, let us consider the particular case of a buffer systems containing
[CH 3COOH] = [CH 3COONa] = 1 M. The pH of this system can be calculated using the
Henderson(Hasselbalch equation:
76 . 4 0 76 . 411log 76 . 4 pH10 = + = + =
When adding 0.01 mol of HCl they will consume the same amount of sodium
acetate giving rise to the same amount of acetic acid:
HCl + CH3COONa NaCl + CH3COOH
As a consequence, the pH will become:
75 . 401.0101 . 0 1log 76 . 4 pH10=+−+ =
13 When adding 0.05 mol and 0.1 mol of HCl the pH of the buffer system will become
4.72 and 4.67, respectively.
When adding 0.01 mol of NaOH it will consume the same amount of acetic acid
giving rise to the same amount of sodium acetate:
NaOH + CH3COOH CH3COONa + H2O
As a consequence, the pH will become:
77 . 401.0101 . 0 1log 76 . 4 pH10=−++ =
When adding 0.05 mol and 0.1 mol of NaOH the pH of the buffer system will
become 4.80 and 4.85, respectively.
The previous examples indicate that adding a larger amount of acid or base leads to
a greater change of the pH value. But these amounts are still small and there is no
significant change of the pH.
There are also some more complex buffer systems that can be used for a broader pH
range. For example, the Britton(Robinson can be used in a range from 2 up to 12. It consists
of a complex mixture of H 3BO 3, H 3PO 4 and acetic acid titrated with NaOH to a specific pH.
Some of the previously mentioned buffer systems have some major drawbacks. For
example, the citrate containing buffers bind cations and therefore are not suitable for
studies involving enzymes which need such ions for their activity (kinases need Mg2+,
hydroxylases need Cu2+/Fe2+). Moreover, dehydrogenases, kinases and carboxypeptidases
are inhibited by phosphates. Also, boric acid containing buffers react with compounds that
contain vicinal hydroxyl groups (mono( and polysaccharides, NAD(P)+/NAD(P)H etc). Tris
containing buffers react with compounds containing carbonyl groups giving rise to Schiff
bases.
Starting with 1960s the so(called “ biological buffers ” or “ Good buffers ” where
developed. They were named after Norman Good who investigated them.
The Good buffers have some special features:
1. high water solubility
14 2. they do not alter the ionic strength
3. very low capacity to bind metal ions
4. they do not react enzimatically or non(enzimatically with other compounds
5. they have a very low absorbance in the UV range
6. high degree of purity.
These special buffers are used in different biological experiments like enzyme
kinetics, nucleic acids and proteins electrophoresis, cell cultures etc.
Characteristics of some Good buffers
Trivial name Chemical name pKa pH range
MES 2(( ÎglyphHVâ(morpholino)(ethane sulfonic acid 6.15 5.50 – 6.70
PIPES piperazine( ÎglyphHVâ,ÎglyphHVâ’(bis(2(ethanesulfonic acid) 6.80 6.40 – 7.20
MOPS 3(( ÎglyphHVâ(morpholino)(propanesulfonic acid 7.20 6.50 – 7.90
HEPES ÎglyphHVâ((2(hydroxyethyl)(piperazine( ÎglyphHVâ’(ethanesulfonic acid 7.55 6.80 – 8.20
CHES 2(( ÎglyphHVâ(cyclohexylamino)(ethanesulfonic acid 9.50 8.60 – 10.00
CAPS 3((cyclohexylamino)(propanesulfonic acid 10.40 9.70 – 11.10
NNOH
SO3HONSO3H
NNSO3H
SO3H
HEPES MOPS PIPES
In spite of their advantages, the Good buffers have a high price which is a major
inconvenient. Moreover, some of the Good buffers (HEPES, PIPES) can be involved in free
radicals producing reactions and they can not be used in free radicals studies. They can be
replaced by MES and MOPS.
The efficiency of a buffer system when is evaluated by the buffer capacity (β). The
buffer capacity is expressed as the amount of strong acid or base, in gram(equivalents, that
must be added to 1 liter of solution in order to change its pH by one unit. The buffer
capacity depends on two factors:
15 1. the ratio of the two components of the buffer system; the buffer capacity is maximal
when this ratio equals 1, which means that pH = p Ka,
2. total buffer concentration; the buffer capacity of a buffer solution of 0.5 M will be
greater that the buffer capacity of a buffer solution of 0.05 M because it will take more
acid or base to change the pH.
The buffer capacity can be calculated using the Van Slyke relation:
[]
[ ]()2
3 a3 a
O HO HC 3 . 2
++
+=
KKβ
where C is the total buffer concentration.
Regardless of the chemical composition, a buffer system can keep the pH relatively
constant in a range of pH = p Ka ± 1. As a consequence, if an experiment is to be performed
at pH = 3, MES would be useless, as it can only keep the pH in the 5.50 – 6.70 range.
Reagents and instrumentation
1. reagents: sodium acetate, acetic acid 96% (ρ = 1.05 g/mL), ethanol, phenolphthalein,
NaOH.
2. buffer solution: acetic acid/sodium acetate buffer 0.1 M, pH = 4.76.
3. solutions: 0.1% alcoholic solution of phenolphthalein, solution of NaOH 0.1 M.
4. laboratory glassware: 50 mL buret, Erlenmeyer flasks, volumetric flasks, pipettes,
5. instrumentation: analytical balance, pH meter.
Procedure
Rinse the buret with a small volume of the NaOH solution. After rinsing, fill the
buret with the 0.1 M NaOH about 2 mL above the highest mark. Use a clean and dry funnel
for filling. Tilting the buret at 45° angle, slowly turn the stopcock to allow the solution to
fill the tip. Collect the excess solution dripping from the tip into a beaker to be discarded
later. The air bubbles must be completely removed from the tip. Clamp the buret onto a ring
stand.
16 Add 10 mL of the buffer solution into a 100 mL Erlenmeyer flask without touching
the walls of the flask. Add a few drops of the phenophthalein solution and gently mix the
flask.
Hold the neck of the Erlenmeyer flask in the left hand and swirl it. Place the flask
under the tip of the buret. Note in the lab report the mark indicating the level of the solution
in the buret at the beginning of the titration.
17 The mark can be read easily in the case of Schellbach burettes. Such a buret has an
additional thin and coloured line embedded in the glass.
When watched through the meniscus, this line seems to have a hourglass shape. The
thinnest part of this line must be aligned with the calibration mark. In the case of the
previus image, this line indicates 42.25 (42.2 and half of a mark).
Open the stopcock of the buret slightly with the right hand and start the dropwise
addition of the base. Continue the titration until a faint permanent pink coloration appears.
In that moment stop the titration. The completion of the titration is indicated by the change
of the colour from colourless to a pink hue. At the end of the titration note in the laboratory
report the mark indicating the level of the solution in the buret. By difference calculate the
volume of the NaOH solution used for titration.
In a second 100 mL Erlenmeyer flask add 5 mL of the buffer solution and 5 mL of
distilled water without touching the walls of the flask. Add a few drops of the
phenophthalein solution and gently mix the flask. Repeate the same steps as in the previous
titration. Note in the laboratory report the initial and the final mark indicating the level of
the solution in the buret and calculate the volume of base used.
During the titration do not remove the Erlenmeyer flask from below the buret’s tip.
Laboratory report
Note the volumes of NaOH solution used for the titration of the undiluted and
diluted buffer solution. Explain the difference.
Calculate the buffer capacity of the undiluted and diluted buffer solutions. Fill in the
following table:
V6glyphMMI–aOH (mL) Buffer capacity (β) C (mol/L)
Undiluted buffer
Diluted buffer
18 Experiment 2
Preparation of a liver homogenate
Background
In order to study a particular biomolecule or a biochemical reaction it is necessary to
isolate that compound or the enzyme which catalyzes the biochemical reaction under
investigation from a biological source.
The very first step consists in obtaining the total protein extract from the biological
source. It is mandatory to disintegrate the tissues and the cells to release their content,
including the biomolecule or the enzyme of interest in the extraction medium.
The biological source selected for the extraction must be accessible and must
contain a significant amount of the biomolecule of interest. For example, the liver is a good
source of enzymes involved in glycogen metabolism, the pancreas contains large amounts
of digestive enzymes, while the mammary gland during lactation has large amounts of the
enzymes involved in lactose synthesis.
The liver is a very good biological source for different types of biomolecules:
• monosaccharides and polysaccharides (glycogen)
• amino acids, peptides (glutathione), and proteins
• lipids
• nucleic acids.
In order to obtain a tissue homogenate and the corresponding total protein extract it
is mandatory to prepare an extraction medium. The chemical composition of this extraction
medium is very complex. It contains a buffer system in order to keep the pH at a specific
value during the process of extraction. Besides the components of the buffer system, the
extraction medium could also contain reducing agents (2(mercaptoethanol, dithiothreitol),
detergents (SDS, sodium dodecyl sulfate), organic compounds able to bind transition
metals cations (EDTA), protease inhibitors (diisopropyl fluorophosphate) etc.
All the aforementioned compounds have specific functions during the procedure of
obtaining the tissue homogenate. For example, the reducing agents prevent the oxidation of
sulfhydryl groups ((SH) of different biomolecules, while detergents enable the
19 solubilization of the membrane(associated proteins. The compounds that bind the cations of
the transition metals prevent the generation of different reactive oxygen and nitrogen
species, while the protease inhibitors block the action of different intracellular proteases,
protecting thus the cellular proteins.
There are physical, chemical and enzymatic methods for the disintegration of
different biological samples. The choice of a specific method depends mainly on the type of
biological sample, but also on the biomolecule of interest which will be further
investigated. Moreover, each method has its advantages and disadvantages.
Another important factor is the extraction ratio. Generally, in order to perform an
extraction experiment, it can be used a volume of 10 mL of extraction medium to 1 g of
biological sample (tissue).
Before homogenization remove with scissors and scalpel the undesirable tissues,
like adipose and connective tissues. Also, prior to homogenization the liver could be
perfused with a saline solution in order to remove the blood traces.
Depending on the aim of experiment which will be performed, the animal from
which the liver will be removed should be subject to a caloric restriction of about 24 hours.
In this way the liver glycogen content will decrease as a consequence of the stimulation of
glycogenolysis.
An important aspect during such an experiment is to keep the temperature as low as
possible (around 4°C) in order to limit the action of intracellular hydrolases (proteases,
glycosidases, lipases, nucleases etc). Usually, these enzymes are confined to specific
cellular organelles (lysosomes), but during the homogenization they are released in the
extraction medium and can act upon the biomolecule of interest.
When using vegetal samples there can be interferences due to the photosynthetic
pigments which strongly absorb in the visible range of the spectrum, cellulose and other
compounds like polyphenols.
Reagents and instrumentation
1. reagents: Na 2HPO 4, NaH 2PO 4.
2. solutions: sodium phosphate buffer 0.1 M, pH = 7.5.
3. laboratory glassware: volumetric flasks, Berzelius flasks, scissors, watch glasses,
pipettes.
20 4. laboratory consumables: Eppendorf tubes (1.5 mL).
5. instrumentation: analytical balance, centrifuge, pH meter, micropipettes.
Procedure
In a watch glass cut with the scissors in small pieces a sample of around 1 g of the
defrozen liver. Mince with the scissors the sample as much as possible. Transfer the minced
biological sample into the homogenization tube and then add 10 mL of the buffer. Add also
5 stainless steel balls in each tube.
In order to homogenize, the device must be set to a speed of 5000 rpm (rotations per
minute) for 2 minutes. A longer homogenization time will lead to a temperature increase
with undesired side effects like protein denaturation and inactivation.
21
In order to remove the cell debris it is necessary to perform a centrifugation. The
content of the tube used for homogenization will be transfered in 4(5 Eppendorf tubes of
1.5 mL each. These tubes will be placed in the centrifuge which will be set at 5000 rpm for
15 minutes at 4°C.
The clear supernatant obtained after this step is the total protein extract which will
be used for further experiments. The supernatant will be removed using a micropipett. Take
care not to disarrange the sediment.
The clear supernatant will be aliquoted into 2 Eppendorf tubes of 1.5 mL which will
be stored at ( 4°C.
Avoid as much as possible repeated freezing – thawing procedure of the samples in
order to limit undesired side effects.
Laboratory report
Note the steps and the operations used in order to obtain the total protein extract.
22 Experiment 3
Molisch test for carbohydrates
Background
Monosaccharides undergo intramolecular dehydration reactions in the presence of
concentrated sulfuric acid giving rise to furfural and its derivatives.
CHO
C
C
C
CH2OHOH H
OH H
OH H
CHO
C
C
C
COH H
H HO
OH H
CH2OHOH HOCHO
OCHO HOCH2D(ribose
D(glucose5((hydroxymethyl)(furfuralfurfuralH2SO4
H2SO4( 3 H2O
( 3 H 2O
Pentoses give rise to furfural, while hexoses give rise to 5((hydroxymethyl)(furfural.
They react with α(naphthol to form a purple coloured compound.
O CHOOH
+ 2O
OHOH
H
H+
( H2O
furfural α(naphtholO
OHO
colouredd compound( H2O
23
In the presence of sulfuric acid the glycosidic bonds of di( and polysaccharides are
hydrolyzed to yield the corresponding monosaccharides.
This reaction has no specificity towards the monosaccharides that are assessed and,
therefore can not be used for the identification of specific monosaccharides.
In the case of tri( and tetroses this test is negative as these carbohydrates do not
undergo intramolcular dehydration to furfural or its derivatives.
Reagents and instrumentation
1. reagents: D(ribose, D(glucose, saccharose, α(naphthol, ethanol, sulfuric acid (95%).
2. solutions: 2% solutions of sugars
3. Molisch reagent (5% α(naphthol in ethanol): disolve 0.5 g α(naphthol in 9.5 mL
ethanol.
4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance.
Procedure
Label three different test tubes for the corresponding sugars to be tested. In each test
tube add 2(3 drops of the Molisch reagent over 2 mL of the sugar solution and then mix its
content. Incline the test tube and add very gently 1 mL concentrated sulfuric acid along the
interior side of the test tube. Do not mix the content of the test tube.
At the interface between the two layers appears a purple coloured ring indicating the
presence of a sugar in the tested solution.
Laboratory report
Note the colour of the ring that you see at the interface between the two layers.
24 Experiment 4
Fehling test for reducing carbohydrates
Background
The aldehyde group of aldoses can be oxidized to a carboxy group giving rise to an
aldonic acid. There are three different reagents that can be use for this purpose: Tollens
reagent, Benedict’s or Fehling’s reagents.
CHO
C
C
C
C
CH2OHH OH
H HO
H OH
OH HCOOH
C
C
C
C
CH2OHH OH
H HO
H OH
OH HTollens reagent
Benedict reagent
Fehling reagent
D(glucose D(gluconic acid
In the case of the Tollens reagent the oxidizing agent is Ag 2O giving rise to Ag (the
silver mirror). In the case of Benedict’s and Fehling’s reagents the oxidizing agent is a Cu2+
salt giving rise to Cu 2O (a brick(red precipitate).
25 Monosaccharides that contain a cyclic hemiacetal in equilibrium with acyclic
aldehyde are reducing sugars and they give a positive reaction with the aforementioned
reagents.
OCH2OH
OHOH
OHOH
α(D(glucopyranosefree glycosidic hydroxil group
If the glycosidic hydroxyl group of a monosaccharide is blocked as an acetal, there
will be no equilibrium with the acyclic aldehyde and there will be no positive reaction with
the aforementioned reagents.
OCH2OH
OHOH
OHOCH3
methyl α(D(glucopyranosideblocked glycosidic hydroxyl group
The same is true for disaccharides. For example, lactose contains a free glycosidic
hydroxyl group and it will react with the aforementioned reagents. On the contrary, sucrose
has no free glycosidic hydroxyl group and it will give no reaction with the aforementioned
reagents.
Carbohydrates that react with the three reagents are called reducing sugars, while
those that do not react are called nonreducing sugars.
In the case of the Fehling test, the working reagent is obtained extemporaneously by
mixing equal volumes of the Fehling A reagent (CuSO 4×7H 2O) and Fehling B reagent
(KOH and sodium potassium tartrate).
26 CuSO4 + 2 KOH Cu(OH)2 + K2SO4
Cu(OH)2 CuO + H2O
R(CHO + 2 CuO R(COOH + Cu2O∆
Tartrate anion acts like a bidentate ligand which binds the Cu2+ ions preventing thus
the precipitation of the Cu(OH) 2.
The reaction with the Benedict’s reagent is a version of the Fehling’s test in which
the ligand that binds the Cu2+ ions is citrate. Also, the Benedict test is more sensitive than
the Fehling test in detecting reducing sugars.
Reagents and instrumentation
1. reagents: D(glucose, D(fructose, lactose, sucrose, CuSO 4×7H 2O, KOH, Seignette salt
(sodium potassium tartrate).
2. solutions: 2% solutions of sugars.
3. Fehling A reagent: disolve 7 g of CuSO 4×7H 2O in a volume of 100 mL of distilled
water.
4. Fehling B reagent: disolve 24 g of KOH and 34,6 g of sodium potassium tartrate in a
volume of 100 mL of distilled water.
5. laboratory glassware: test tubes, graduated cylinders, Berzelius flasks, pipettes.
6. instrumentation: analytival balance, water bath.
Procedure
Label four different test tubes for the corresponding sugars to be tested. The working
reagent is made before the assay by mixing equal volumes of the two Fehling A and B
reagents.
In each test tube add 6 mL of the Fehling working reagent and then 3 mL of the
corresponding sugar solutions. Keep the two test tubes on a boiling water bath for about 5(
10 minutes.
Laboratory report
Note the colour of the precipitate and in which test tube it appears.
27 Experiment 5
Seliwanoff test for ketoses
Background
Ketoses undergo intramolecular dehydration reactions when heated in the presence
of concentrated hydrochloric acid giving rise to furfural and its derivatives.
CH2OH
C
C
C
CH HO
OH H
CH2OHOH HO
OCHO HOCH2
D(fructose5((hydroxymethyl)(furfuralHCl
( 3 H2O
Ketopentoses and ketohexoses give rise to furfural and to 5((hydroxymethyl)(
furfural, respectively. They react with resorcinol to form a red coloured compound.
O CHO HOCH2OH
OH+ 2O HO OH
O
CH2OH( H2O ( 2 H2O
5((hydroxymethyl)(furfural resorcinolO HO
O
CH2OHO
coloured compound
In the presence of hydrochloric acid the glycosidic bonds of di( and polysaccharides
are hydrolyzed to yield the corresponding monosaccharides. The released ketoses will be
converted into furfural or into 5((hydroxymethyl)(furfural, giving rise to the coloured
product.
28 In the case of ketotetroses, this reaction will be negative, as these sugars can not
undergo intramolcular dehydration reactions.
Reagents and instrumentation
1. reagents: D(glucose, D(fructose, D(saccharose, hydrochloric acid (37%), resorcinol.
2. solutions: 2% solutions of sugars
3. Seliwanoff reagent: disolve 0.05 g resorcinol in 100 mL of hydrochloric acid 3M.
4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance, water bath.
Procedure
Label three different test tubes for the corresponding sugars to be tested. In each test
tube add 2 mL of the Seliwanoff reagent and 2 mL of the corresponding sugar solution.
Keep the three test tubes on a boiling water bath for about 1(2 minutes.
Laboratory report
Note the colours that appear in each test tube.
29 Experiment 6
Bial test for pentoses
Background
Pentoses undergo intramolecular dehydration reactions when heated in the presence
of hydrochloric acid giving rise to furfural.
CHO
C
C
C
CH2OHOH H
OH H
OH HOCHO
D(ribosefurfuralHCl
( 3 H2O
Furfural reacts with orcinol in the presence of FeCl 3 giving rise to a blue(green
coloured compound.
OCHOOH
CH3HO HO
CH3OH
( 2 H 2OO HO
CH3 CH3OH
O
A further dehydration gives rise to the coloured compound.
30 O HO
CH3 CH3OH
OO HO
CH3 CH3OO
( H2O
coloured compound
Reagents and instrumentation
1. reagents: D(glucose, D(ribose, hydrochloric acid (37%), orcinol, FeCl 3.
2. solutions: 2% solutions of sugars.
3. Bial reagent: disolve 0.3 g orcinol, 0.05 g FeCl 3 in 100 mL of HCl 12M.
4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance, water bath.
Procedure
Label two different test tubes for the corresponding sugars to be tested. In each test
tube add 2.5 mL of the Bial reagent and 0.5 mL of the corresponding sugar solutions. Keep
the two test tubes on a boiling water bath for about 5 minutes.
Laboratory report
Note the colour that appears in one of the test tubes.
31 Experiment 7
The reaction with phenylhydrazine
Background
Aldoses and ketoses react with phenylhydrazine to form osazones. One mol of an
aldose or a ketose reacts with tree mol of phenylhydrazine, regardless of the amount of
phenylhydrazine present in the reaction system.
One mol of phenylhydrazine acts as an oxidizing agent and it is reduced to aniline
and ammonia. The other two mol form imines with the carbonyl groups.
CHO
C
C
C
C
CH2OHOH H
HO H
OH H
OH HNH 2HN
+HC
C
C
C
C
CH2OHOH H
HO H
OH H
OH HN NH C6H5
D(glucose phenylhydrazine phenylhydrazone osazone+ 2 C6H5(NH(NH2
( C6H5(NH2
( NH3
( H2OHC
C
C
C
C
CH2OHHO H
OH H
OH HN NH C6H5
N NH C 6H5
During this reaction the configuration at the C 2 is modified and, as a consequence,
the monosaccharides which are C 2 epimers form the same osazone. For example, D(idose
and D(gulose both form the same osazone.
CHO
C
C
C
C
CH2OHHO H
OH H
HO H
OH HC
C
C
C
C
CH2OHOH H
HO H
OH HN NH C6H5N NH C6H5 CHO
C
C
C
C
CH2OHH OH
OH H
HO H
OH H
D(idose osazone D(gulose
Due to the fact that in the case of ketoses both C 1 and C 2 are involved in the reaction
with phenylhydrazine the aldoses which are epimers at C 2 and the corresponding ketose
32 form the same osazone. For example, D(glucose, D(manose and D(fructose form the same
osazone.
CHO
C
C
C
C
CH2OHH OH
H HO
H OH
OH HC
C
C
C
C
CH2OHOH H
H OH
OH HN NH C6H5N NH C 6H5 CH2OH
C
C
C
C
CH2OHH HO
H OH
OH HO
D(glucose osazone D(fructose
Osazones have crystals with characteristic shapes.
Reducing disaccharides give the test with phenylhydrazine, while nonreducing ones
fail this test as they have no free carbonyl group.
Reagents and instrumentation
1. reagents: D(glucose, D(fructose, sucrose, phenylhydrazine hydrochloride, acetic acid,
sodium acetate.
2. solutions: 2% solutions of sugars.
3. laboratory glassware: test tubes, Berzelius flasks, pipettes.
4. instrumentation: analytical balance, water bath.
Procedure
Label three different test tubes for the corresponding sugars to be tested. In each test
tube add 0.125 g phenylhydrazine hydrochloride, 0.05 sodium acetate, and 5 drops of acetic
acid. Then add more 5 mL of the sugar solutions in the corresponding test tubes.
Mix carefully the test tubes in order to ensure the solubilization of the content and
then keep the test tubes on a boiling water for 30(45 minutes. In the end cool the test tubes.
The shape of the crystals can be observed using a light microscope having a 10X
magnification.
33 Laboratory report
Note the colour of the crystals as well as their shape. Note in which test tube crystals
do not appear.
34 Experiment 8
Assay of D-glucose with 3,5-dinitrosalicylic acid
Background
Reducing monosaccharides are oxidized to the corresponding aldonic acids in the
presence of the 3,5(dinitrosalicylic acid (DNSA). Simultaneously, this reagent is reduced
under alkaline conditions to 3(amino(5(nitrosalicylic acid, which has a maximum
absorbance at 540 nm.
COOH
OH
NO 2 O2NCHO
C
C
C
C
CH2OHOH H
H HO
OH H
OH H+alkaliCOONa
ONa
NH 2 O2NCOOH
C
C
C
C
CH2OHOH H
H HO
OH H
OH H+
3,5(dinitrosalicylic acid D(glucose 3(amino(5(nitrosalicylic acid D(gluconic acid
Using a series of D(glucose solutions with known concentrations a calibration curve
can be obtained and used to assess the D(glucose concentration from a sample.
Reagents and instrumentation
1. reagents: anhydrous D(glucose, benzoic acid, 3,5(dinitrosalicilyc acid, Seignette salt
(sodium potassium tartrate), NaOH.
2. standard D(glucose solution (1 mg/mL): disolve 50 mg of anhydrous D(glucose and
0.05 g of benzoic acid in a volume of 50 mL of distilled water. This stock solution must
be stored at 2(8°C in refrigerator. Benzoic acid acts as a stabilizer for the standard D(
glucose solution.
3. DNSA reagent A: disolve 30 g of sodium potassium tartrate in a volume of 50 mL of
distilled water.
4. DNSA reagent B: disolve 1 g of DNSA in a volume of 20 mL of a 2M NaOH solution.
5. laboratory glassware: test tubes, graduated cylinders, Berzelius flasks, pipettes.
35 6. instrumentation: analytical balance, spectrophotometer, water bath.
The DNSA working reagent is prepared by mixing the reagents A and B and adding
distilled water to a final volume of 100 mL.
Procedure
Pipette into a series of six clean test tubes as indicated in the following table:
Test tube
number Standard D4glucose
solution (mL) Distilled water
(mL) Concentration
(mg/mL)
1 0 1
2 0.2 0.8
3 0.4 0.6
4 0.6 0.4
5 0.8 0.2
6 1.0 0
In each test tube add 2 mL of the DNSA working reagent. Shake the test tubes and
place them in a boiling water bath for 5 minutes.
Cool the test tubes thoroughly and then add 7 mL of distilled water to each test tube.
Read the optical densities of the solutions from all the test tubes at 540 nm. Then,
substract the value obtained in the case of the first test tube (blank) from all the other test
tubes.
The sample with unknown D(glucose concentration is prepared in the same way.
Laboratory report
In order to make the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (mg/mL), while on the Oy axis plot the optical density.
Use the equation of the curve in order to calculate the D(glucose concentration in the
sample.
36 Experiment 9
Assay of D-glucose with o-toluidine
Background
Monosaccharides react with o(toluidine (2(methyl aniline) in the presence of
glacial acetic acid leading to a Schiff base.
CHO
C
C
C
C
CH2OHOH H
HO H
OH H
OH HNH 2
CH3
+HC
C
C
C
C
CH2OHOH H
HO H
OH H
OH HNH3C
( H2O
D(glucose o(toluidine Schiff base
The reaction is quite selective for D(glucose leading to a green coloured Schiff
base with a maximum absorbance at 635 nm.
Using a series of D(glucose solutions with known concentration a calibration
curve can be obtained and used to assess the D(glucose concentration from a sample.
Reagents and instrumentation
1. reagents: anhydrous D(glucose, benzoic acid, o(toluidine, glacial acetic acid.
2. standard D(glucose solution (1 mg/mL): prepared as described in experiment 8.
3. o(toluidine reagent (6% o(toluidine in glacial acetic acid): disolve 6 mL of o(toluidine
in a volume of 94 mL of glacial acetic acid. The reagent must be allowed to age for one
week, during which time the colour will change from yellow to amber. The reagent can
be stored at room temperature but protected form direct action of the light.
37 4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance, spectrophotometer, water bath.
Procedure
Pipette into a series of seven clean test tubes as indicated in the following table:
Test tube Standard D4glucose
solution (mL) Distilled water
(mL) Concentration
(mg/mL) Optical
density (OD)
1 0 4
2 0,5 3,5
3 1 3
4 1,5 2,5
5 2 2
6 2,5 1,5
7 3 1
In each test tube add 4 mL of the o(toluidine reagent. Shake the test tubes and
place them in a boiling water bath for 10 minutes.
Cool the test tubes throughly and read the optical densities of the solutions from
all the test tubes at 635 nm. Then, substract the value obtained in the case of the first test
tube (blank) from all the other test tubes.
The sample with unknown D(glucose concentration is prepared in the same way.
Laboratory report
In order to make the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (mg/mL), while on the Oy axis plot the optical density (OD).
Use the equation of the curve in order to calculate the D(glucose concentration in the
sample.
38 Experiment 10
Assay of D-glucose with anthrone
Background
Under acidic conditions (H 2SO 4) and on heating, monosaccharides undergo
intramolecular dehydration. Hexoses give rise to 5((hydroxymethyl)(furfural and pentoses
give rise to furfural. Both compounds undergo condensation reactions with anthrone.
O CHO HOCH 2O
+O
C
H
O
CH2OH5((hydroxymethyl)(furfural anthrone
coloured condensation product
The products of the condensation reaction have a deep(green colour and a maximum
absorbance at 630 nm.
Using a series of D(glucose solutions with known concentrations a calibration curve
can be obtained and used to assess the D(glucose concentration from the sample
Reagents and instrumentation
1. reagents: anhydrous D(glucose, benzoic acid, sulfuric acid (95%), anthrone.
2. standard D(glucose solution (1 mg/mL): prepared as described in experiment 8.
3. anthrone reagent (0.2% anthrone in concentrated sulfuric acid): disolve 0.2 g of
anthrone in a volume of 100 mL of cool sulfuric acid 95%. Cool the solution throughly.
4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance, spectrophotometer, water bath.
39 Procedure
Pipette into a series of five clean test tubes as indicated in the following table:
Test tube Standard D4glucose
solution (mL) Distilled water
(mL) Concentration
(mg/mL) Optical
density (OD)
1 0 1
2 0,2 0,8
3 0,4 0,6
4 0,6 0,4
5 1 0
In each test tube add 4 mL of the anthrone reagent. Shake the test tubes and place
them in a boiling water bath for 10 minutes.
Cool the test tubes throughly and read the optical densities of the solutions from
all the test tubes at 630 nm. Then, substract the value obtained in the case of the first test
tube (blank) from all the other test tubes.
The sample with unknown D(glucose concentration is prepared in the same way.
Laboratory report
In order to obtain the calibration curve use the optical density values obtained
previously and the corresponding concentration values and make a chart in Excell. On the
Ox axis plot the concentration (mg/mL), while on the Oy axis plot the optical density.
Use the equation of the curve in order to calculate the D(glucose concentration in the
sample.
40 Experiment 11
Assay of D-glucose with glucose oxidase
Background
Glucose oxidase (EC 1.1.3.4) catalyzes the oxidation of D(glucose to δ(D(
gluconolactone.
OCH2OH
OHOH
OHOH
Dăglucose oxidase
β(D(glucopyranose δ(D(gluconolactoneOCH2OH
OHOH
OHOO2H2O2
Glucose oxidase uses FAD as an oxidizing agent, which is transformed into the
corresponding reduced form (FADH 2). The reduced form is reoxidized by transfering the
reducing equivalents to O 2 giving rise to H 2O2.
enzyme FADβ(D(glucopyranoseδ(D(gluconolactone
enzyme FADH 2
enzyme FADH 2enzyme FADO2H2O2
There is a second enzymatic reaction catalyzed by the horseradish peroxidase which
produces a coloured compound with a maximum absorbance at 514 nm.
41 In this reaction H 2O2 is used as an oxidizing agent by the peroxidase in order to
generate a pink dye from the reaction between 3,5(dichloro(2(hydroxybenzensulfonic acid
and 4(aminophenazone.
SO3
HO
Cl ClNNO
CH3H3CH2N
+ O N
ClO3S
NN
CH3CH3O
2 H 2O24 H 2O + HCl
horseradish
peroxidase
The optical density at 514 nm is proportional to the D(glucose concentration in the
sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 514 nm.
42 The D(glucose standard solution provided by the producer has a concentration of
100 mg/dL.
Laboratory report
The concentration of D(glucose in the sample is calculated using the following
formula:
100OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
43 Experiment 12
Assay of D-glucose with hexokinase
Background
Hexokinase (EC 2.7.1.1) catalyzes the phosphorylation of D(glucose giving rise to
D(glucose(6(phosphate.
OCH2OH
O
HOH
OHOHOCH2OPO 32(
OHO
H
OHOH ATPADP
hexokinase
β(D(glucopyranose D(glucose(6(phosphate
In the second enzymatic reaction D(glucose(6(phosphate is oxidized to 6(phospho(
D(gluconate by D(glucose(6(phosphate dehydrogenase (EC 1.1.1.49).
OCH2OPO 32(
OHO
H
OHOHCOO
C
C
C
C
CH2OPO32(OH H
H
O H
OH H
OH HNAD+NADH + H+
Dăglucoseă6ăphosphate
dehydrogenase
D(glucose(6(phosphate 6(phospho(D(gluconate
This is a redox reaction with NAD+ as oxidizing agent. NAD+ is reduced to NADH
which has a maximum absorbance at 340 nm.
44 As a consequence, during the experiment there is an increase in the optical density at
340 nm which is directly proportional with the concentration of D(glucose.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 340 nm.
The D(glucose standard solution provided by the producer has a concentration of
100 mg/dL.
Laboratory report
The concentration of D(glucose in the sample is calculated using the following
formula:
100OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
45 Experiment 13
Liebermann-Burchard test for cholesterol
Background
Under acidic conditions (H 2SO 4), in the presence of acetic acid and acetic anhydride,
cholesterol is gradually oxidized to a compound that has a maximum absorbance at 620 nm.
HOCH3CH3H3C
CH3
H3C
In the first step, the hydroxyl group is protonated and eliminated as water giving rise
to a dienylic carbocation.
HOCH3CH3R
H2SO4/CH 3COOH CH3CH3R
cholesterol dienylic carbocation
The dienylic carbocation is oxidized to a pentaenylic carbocation which is stabilized
by conjugation.
46 CH3CH3R
(CH 3CO) 2O/SO 3CH3CH3R
dienylic carbocation pentaenylic carbocation
Besides the coloured pentaenylic carbocation, there are small amounts of products of
the acetylation and sulfuration of cholesterol.
Because the reaction is quite complex with different intermediate compounds during
the experiment a change in colour from blue to green may be seen.
In specific conditions this reaction can be used for quantitative analysis.
Reagents and instrumentation
1. reagents: acetic acid, acetic anhydride, chloroform, cholesterol, sulfuric acid (98%).
2. solutions: 0.5% cholesterol in chloroform.
3. Liebermann(Burchard reagent: mix carefully in a Berzelius flask acetic acid, acetic
anhydride and H 2SO 4 in a ratio of 7:7:1 (vol:vol:vol). Keep the flask in an ice bucket to
avoid overheating. The reagent must be colourless. If a blue tinge appears the solution
must be discarded and the reagent must be prepared once again.
4. laboratory glassware: test tubes, Berzelius flasks, graduated cylinders, pipettes.
5. instrumentation: analytical balance.
Procedure
Add into a perfectly dried test tube 1 mL of the standard cholesterol solution and
then 1 mL of the Liebermann(Burchard reagent. Shake gently the test tube and then let the
test tube to stand.
Laboratory report
Note the initial colour that appears in the test tube and how it changes.
47 Experiment 14
Assay of total lipids with the phospho-vanillin reagent
Background
In acidic medium, the proton adds to the double bonds of the unsaturated lipids
leading to the formation of a carbocation. This one acts as an electrophilic reagent and
reacts with the carbonyl group of the phosphovanillin leading to a product that has
maximum absorbance at 525 nm.
The reaction take place in three different steps:
• formation of the carbocation (the reaction is favored by heating):
C CH H
CH
CH
HH+
• formation of the phosphoric ester of vanilline:
OH
OCH3
C
OHH3PO4OPO3H2
OCH 3
C
OH
vanillin phosphovanillin
48 OPO 3H2
OCH 3
C
OHOPO 3H2
OCH3
C
H OOPO 3H2
OCH3
C
H O
• electrophilic attack of the carbocation on the carbonyl group of phosphovanilline:
OPO3H2
OCH3
C
OHCH
CH
HOPO3H2
OCH3
C
H O C CH+
Saturated lipids do not undergo this colour reaction. In the case of triacylglycerols
and glycerophospholipids with polyunsaturated fatty acids residues only one carbocation
can result for each fatty acid residue. This could be explained by the lack of stability of
such a chemical structure, as well as by steric hindrance.
Reagents and instrumentation
1. reagents: vanillin, phosphoric acid (85%), sulfuric acid (98%), ethanol.
2. solutions: vanillin solution (disolve 0.6 g of vanillin in a volume of 8(10 mL ethanol
and bring to the mark with distilled water in a 100 mL volumetric flask).
3. phosphovanillin reagent: the vanillin solution is mixed with 400 mL of H 3PO 4. This
reagent must be kept at room temperature in dark.
4. laboratory glassware: volumetric flask, Berzelius flask, graduated cylinder, test tubes,
pippetes.
5. laboratory consumables: Eppendorf tubes (1.5 mL).
49 6. instrumentation: analytical balance, spectrophotometer, water bath.
Procedure
Pipette into a series of three clean test tubes as indicated in the following table:
Sample Standard Blank
Sample (µL) 100 – –
Standard (µL) – 100 –
Distilled water (µL) – – 100
H2SO 4 (µL) 2500 2500 2500
Shake the test tubes and place them into a boiling water bath for 10 minutes. After
that cool the test tubes.
Take three Eppendorf tubes and mark them accordingly to the previous series. In
each Eppendorf tube add 1000 µL of the phosphovanillin reagent. Take from each test tube
belonging to the first series 50 µL and add into the corresponding Eppendorf tube. Shake
the Eppendorf tubes and keep them at room temperature for 30 minutes.
Read the optical density of these solutions at 525 nm.
The total lipids standard solution used has a concentration of 750 mg/dL.
Laboratory report
The concentration of total lipids in the sample is calculated using the following
formula:
750OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
50 Experiment 15
Assay of cholesterol with cholesterol oxidase
Background
Cholesterol esterase (EC 3.1.1.13) hydrolyses different cholesterol esters giving rise
to cholesterol and free fatty acids.
RCOOCH3CH3R
HOCH3CH3R
H2ORCOOH
cholesterol esterase
esterified cholesterol cholesterol
Cholesterol oxidase (EC 1.1.3.6) oxidizes cholesterol to 4(cholesten(3(one giving
rise to H 2O2.
HOCH3CH3R
CH3CH3R
Ocholesterol oxidase
cholesterol 4(cholesten(3(oneO2H2O2
Finally, H 2O2 is used in the reaction catalyzed by the horseradish peroxidase which
produces a coloured compound with a maximum absorbance at 505 nm.
In this reaction H 2O2 is used as an oxidizing agent by the peroxidase in order to
generate a pink dye from the reaction between phenol and 4(aminophenazone.
51 OH
NNO
NC6H5
H3C
H3C O2 H2O24 H2O
peroxidaseNNO
NH2C6H5
H3C
H3C+
4(aminophenazone phenol quinonimine
The optical density at 505 nm is proportional to the cholesterol concentration in the
sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 505 nm.
The cholesterol standard solution provided by the producer has a concentration of
200 mg/dL.
52 Laboratory report
The concentration of cholesterol in the sample is calculated using the following
formula:
200OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
53 Experiment 16
Assay of triacylglycerols with glycerol-3-phosphate dehydrogenase
Background
Lipoprotein lipase (EC 3.1.1.34) hydrolyses triacylglycerols giving rise to glycerol
and fatty acids.
H2C
C
H2CO HO
OCOR 1
COR 3R2CO
lipoprotein lipaseH2C
C
H2CHO HOH
OH+R1COOH
R2COOH
R3COOH+ 3 H 2O
triacylglycerol glycerol free fatty acids
Glycerol kinase (EC 2.7.1.30) catalyzes the phosphorylation of glycerol to sn(
glycerol(3(phosphate.
H2C
C
H2CHO HOH
OHglicerol kinaseATPADP
H2C
C
H2CHO HOH
OPO 32(
glycerol sn(glycerol(3(phosphate
Glycerol(3(phosphate dehydrogenase (EC 1.1.3.21) catalyzes the oxidation of sn(
glycerol(3(phosphate to dihydroxyaceton monophosphate (DHAP) giving rise to H 2O2.
H2C
C
H2CHO HOH
OPO32(snăglicerolă3ăphosphate
dehydrogenaseH2C
C
H2COH
OPO32(O
sn(glycerol(3(phosphate dihydroxyacetone monophosphateO2H2O2
54
Finally, H 2O2 is used in the reaction catalyzed by the horseradish peroxidase which
produces a coloured compound with a maximum absorbance at 505 nm.
In this reaction H 2O2 is used as an oxidizing agent by the peroxidase in order to
generate a pink dye from the reaction between 4(chlorophenol and 4(aminophenazone.
NNO
NH 2C6H5
H3C
H3C+OH
ClNNO
NC6H5
H3C
H3C
OCl H2O2H2O
peroxidase
4(aminophenazone 4(chlorophenol quinonimine
The optical density at 505 nm is proportional to the cholesterol concentration in the
sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
55 Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 505 nm.
The cholesterol standard solution provided by the producer has a concentration of
200 mg/dL.
Laboratory report
The concentration of triacylglycerols in the sample is calculated using the following
formula:
200OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
56 Experiment 17
Assay of amino acids with ninhydrin
Background
Amino acids react, when heated, with ninhydrin giving rise to a purple compound
(Ruhemann’s purple).
Ninhydrin reacts also with ammonia, primary and secondary amines, as well as with
peptides, polypeptides and proteins.
Besides qualitative analysis of amino acids, the reaction with ninhydrin can be used
for quantitative analysis. Under appropriate conditions, the colour intensity is proportional
with the amino acid amount from a sample.
The mechanism of this reaction is not very well understood, but a reasonable one is
the following.
O
OOH
OH ( H2OO
OO+H2N C COO
RHO
ON C
RCH O
O( H2O
O
ON C
RCH O
O( CO 2
OO
N CH R
OO
N CH R
H O
H( HO
ON CH RO
+ H2O
ONH 2O
R CO
H
57 ONH 2O
+O
O
O ONO O
O
ONO O
OH HO
( H2O
ONO
OO
Ruhemann purple
In the same conditions proline and hydroxyproline give rise to a yellow coloured
compound.
NO
O
Due to the fact that this reaction is very sensitive it is used even in forensic studies
to detect fingerprints.
Reagents and instrumentation
1. reagents: glycine, L(alanine, L(proline, ninhydrin, n(butanol.
2. solutions: 0.1% solutions of amino acids.
3. ninhydrin reagent: disolve 0.2 g of ninhydrine in a volume of 100 mL of n(butanol.
4. laboratory glassware: test tubes, graduated cylinders, pipettes.
5. instrumentation: analytical balance, water bath.
58 Procedure
Label three different test tubes for the corresponding amino acids to be tested. In
each test tube add 1 mL of the corresponding amino acid and then 1 mL of the ninhydrin
solution. Keep the two test tubes on a boiling water bath for about 5 minutes.
Laboratory report
Note the colour of the solutions in each of the three test tubes
59 Experiment 18
Xanthoproteic test for aromatic amino acids
Background
Aromatic amino acids (L(tryptophan, L(tyrosine) react with nitric acid giving rise to
the corresponding yellow coloured nitroderivatives.
N
HCOOH
NH2
HOCOOH
NH2
L(tryptophan L(tyrosine
CH2CCOOH
H H2N
OH+ HNO3
( H2OCH2CCOOH
H H2N
OHNO 2
L(tyrosine L(nitrotyrosine
In the conditions of this test L(phenylalanine does not react with nitric acid.
In the presence of alkali nitroderivatives give rise to the corresponding aci(nitro
forms which are orange coloured.
The nitroderivatives and the corresponding aci(nitro derivatives are tautomers.
60
L(tyrosine L(nitrotyrosine sodium salt of the aci(nitro derivative+ NaOH
( H2O
NaONO
OCH2CCOOH
H H2N
CH2CCOOH
H H2N
OHNO2CH2CCOOH
H H2N
N
OO
ONa
Reagents and instrumentation
1. reagents: L(alanine, L(tyrosine, L(phenylalanine, nitric acid (70%), NaOH.
2. solutions: 0.1% solutions of amino acids, solution of NaOH 10 M.
3. laboratory glassware: test tubes, volumetric flask, pipettes.
4. instrumentation: analytical balance, water bath.
Procedure
Label three different test tubes for the corresponding amino acids to be tested. In
each test tube add 1 mL of the corresponding amino acid and then 1 mL of HNO 3. In the
case that the colour does not develop, keep the test tubes on a boiling water bath for about
5(10 minutes. Cool the test tubes and then add a few drops of the NaOH solution in each
test tube.
Laboratory report
Note the colour that appears in each test tube and the change in colour after adding
the NaOH solution.
61 Experiment 19
Separation of amino acids by thin layer chromatography (TLC)
Background
Resolution of the compounds from a sample by thin layer chromatography (TLC) is
based on the differential affinity of the compounds for the two phases: the mobile phase
and the stationary phase. Thus, there will be multiple partition equilibria.
The partition of the compounds between the two phases is a consequence of their
chemical structure, which influences the interactions that the respective compounds can
establish with the components of the two phases. These interactions include ionic bonds,
hydrogen bonds, and hydrophobic interactions. For example, a chromatographic system can
be designed to have a nonpolar mobile phase and a polar stationary phase. When a mixture
of compounds with different polarities is subject to resolution by TLC, the more polar
compounds of the mixture will not travel as quickly as they have a higher affinity for the
stationary phase. As a consequence the concentration of these compounds will be higher in
the stationary phase. On the contrary, the less polar compounds will travel faster in the
same time period as they have a higher affinity for the mobile phase.
TLC uses as a stationary phase a thin layer of absorbent such as silica gel, alumina
or cellulose placed on a glass, aluminium or plastic support.
The properties of the sample to be analyzed must be taken into consideration prior to
the selection of the stationary phase. For example, silica gel can be used for amino acids
and hydrocarbons, while alumina can be used in the case of amines, steroids, lipids,
aflatoxins, alkaloids etc.
Stationary phases and chromatographic mechanisms in TLC
Stationary phase Chromatographic mechanism Compounds
Silica gel adsorption steroids, amino acids, alcohols, bile
acids, aflatoxins
Silica gel RP* reversed phase fatty acids, vitamins, carotenoids,
steroids
Cellulose partition carbohydrates, fatty acids
Alumina adsorption amines, alcohols, steroids, lipids
* RP, reversed phase
62
O Si O Si O Si OOH OH OH
Si OOH
O O O O
CENTRAL PART OF SILICA
As it can be seen from the previous schema, the stationary phase has hydroxyl
groups which act as interacting groups. They can be involved in hydrogen bond formation.
Moreover, this type of stationary phase can be involved in dipole(dipole interactions as well
as van der Waals forces.
H
OHH
OHH
OH
O Si O Si O Si OO O O
Si OO
O O O OH H H HH
OH
CENTRAL PART OF SILICA
The face of a TLC plate which is covered by the stationary phase should not be
touched with the fingers as contamination from the skin oils and amino acids and proteins
can occur. As a consequence, the plates should be handled by keeping from edges or using
a forceps.
The mobile phase can move either upward or downward relative to the stationary
phase. In the case of ascendent TLC, the compounds of the sample with higher affinity for
the mobile phase will migrate over a longer distance from the start.
Selection of the mobile phase (solvent or mixture of solvents) is an important step in
a chromatographic separation. The optimum composition of the mobile phase is determined
after several trials.
63 In order to perform a TLC experiment a developing chamber or vessel is required
(chromatographic tank). The TLC chamber should contain enough solvent to cover the
bottom of the chromatographic plate. Also, the TLC chamber should contain a piece of
filter paper to saturate the atmosphere with solvent vapors.
In the end of a TLC experiment, the chromatographic plate must be developed in
order to highlight the separated compounds as spots. In the simplest case of compounds that
absorb in the visible range of the electromagnetic field the chromatographic plate will
contain a number of coloured spots. In the case of compounds that absorb in the UV range
of the electromagnetic field there must be a suplimentary step in which the compounds
from the sample chemically react with a specific reagent giving rise to coloured spots. The
widely used reagents are iodine, ninhydrin, potassium permanganate, ceric ammonium
molybdate etc. The choice of such a developing reagent depends on the chemical
composition of the sample to be analyzed.
In the end of the experiment there will be a number of coloured spots located at a
specific distance from the starting line where the sample and the standards were applied.
For each spot a retention factor (R f) will be calculated according to the following formula:
64 solvent by travelled distancesample by travelled distanceR=f
In the previous example, the R f of the black spot is 4.5/9.0 = 0.50. R f values range
from 0 to 1 and have no unit.
The numerical value of R f depends on the chemical composition of the mobile
phase, as well as on the polarity of the stationary phase. In other words, if a sample is
separated by TLC using the same stationary phase, but two different mobile phases,
different R f values are observed for all the components of the sample. This implies that, in
order to ensure the reproductibility of the results, the working conditions must be kept
unchanged and reported as well.
TLC can be used to identify the compounds from a given sample (qualitative
analysis). In this case, it is mandatory to use standards of the substances that are supposed
to be found in the sample.
Reagents and instrumentation
1. reagents: K 2HPO 4, KH 2PO 4, amino acids, n(propanol, ammonia, ethanol, 96% acetic
acid (ρ = 1.05 g/mL), ninhydrin.
2. buffer solution: K 2HPO 4/KH 2PO 4 0.02 M, pH = 6.5.
3. solutions: 0.05 M amino acids in the previous buffer solution.
4. mobile phase A: n(propanol:ammonia in ratio 7:3 (vol:vol).
65 5. mobile phase B: ethanol:acetic acid in ratio 9:1 (vol:vol).
6. laboratory glassware: volumetric flasks, pipettes, Berzelius flasks, Erlenmeyer flasks,
chromatographic tanks, watch dishes, chromatographic plates, capillary tips.
7. instrumentation: analytical balance, pH meter, micropipettes.
Procedure
Two TLC experiments will be performed in the same time using the same sample
and the same standard solutions.
Take two chromatographic tanks and mark them as A and B. In each tank introduce
the corresponding mobile phase. Cover the tanks with a watch dish and introduce also a
suitable filter paper along the walls of the tank in order to ensure the saturation of the
interior atmosphere with the vapors of the mobile phase.
Two chromatographic plates must be prepared each with 10 × 7 (cm). Do not touch
with the fingers the surface of the plates in order to avoid the contamination with your own
amino acids and proteins from the skin.
At one of the endges of the chromatographic plate draw a light line (start line) using
a pencil (not a pen). This line must be placed at a distance of about 1.5(2 cm from the
plate’s edge (see below).
66 When drawing the line do not skratch the silica gel. Evenly space four marks on this
line. There must be a distance of about 1.5(2 cm between two following marks. Label each
mark so you can identify correctly the three standard solutions and the sample (see below).
With the help of a capillary tip place on each mark the corresponding standard
solution and the sample. Avoid to have large spots as this is a source of errors in the
determination of the R f values. Do not use the same tip for all the four solutions.
Let the chromatographic plates to stand in the air in order to enable the evaporation
of the solvents. After that, place each plate in a chromatographic tank. The level of the
mobile phase in the tank must be below the start line.
The chromatographic plates are left inside the tanks until the mobile phase arrives at
a distance of about 1 cm to the upper edge of the plate. Remove carrefully the plates from
the two tanks and draw with a pencil the front solvent line.
Place the plates into the oven heated at 110°C in order to ensure the evaporation of
the excess mobile phase.
Take the plates into a hood and spray lightly and evenly a ninhydrin solution. Place
once again the plates into the oven until the colours develop.
Draw a circle arround the outside edge of each spot and mark the center of it.
Laboratory report
For each spot calculate the corresponding R f value like in the following example:
67 Use the R f values to fill in the following table:
Mobile phase A Mobile phase B
Amino acid 1
Amino acid 2
Amino acid 3
Comment the observed differences.
68 Experiment 20
Assay of amino acids using the extinction coefficient
Background
Amino acids do not absorb the radiation in the visible region of the electromagnetic
spectrum and, as a consequence, these compounds appear colourless. On the other hand,
aromatic amino acids (tyrosine, tryptophan, phenylalanine) absorb in the ultraviolet region
of the electromagnetic spectrum.
Tyrosine and tryptophan absorb more than does phenylalanine. All these aromatic
amino acids have maximal optical density at 280 nm.
The concentration of an aromatic amino acid can be assessed from a sample if the
molar extinction coefficient (ε) at 280 nm of that amino acid is known.
Reagents and instrumentation
1. reagents: tryptophan, K 2HPO 4, KH 2PO 4.
2. solutions: tryptophan 10 mM solution (disolve 0.204 g of tryptophan in a volume of
100 mL of potassium phosphate buffer); potassium phosphate buffer 0.02 M, pH = 6.5.
3. laboratory glassware: volumetric flasks, pipettes.
4. instrumentation: analytical balance, spectrophotometer, pH meter.
Procedure
Make a 100 fold dilution of the tryptophan stock solution. Read the optical density
of the diluted solution at 280 nm against the potassium phosphate buffer solution. Use
quartz cuvettes.
Laboratory report
Note the optical density value. In the aforementioned condition the molar extinction
coefficient of tryptophan is 5690 M(1cm(1. The tryptophan concentration is calculated with
the Lambert(Beer formula:
ε l c OD××=
69 Experiment 21
Assay of total protein content through UV absorption spectroscopy
Background
Protein concentration must be assessed during different kinds of experiments.
The protein content from a sample can be assessed using two types of methods: (i)
spectrophotometric assays and (ii) dye(binding assays. The detection can be colorimetric or
by fluorescence.
Proteins have a specific ultraviolet absorption at 280 nm due to the presence of the
aromatic amino acids (tyrosine, tryptophan, phenylalanine). At this wavelength, a 1 mg/mL
protein solution has an optical density of arround 1. When working with samples with a
higher protein content the sample dilution is mandatory prior to the assay. On the other
hand, dilution leads to a decrease of the sensitivity of the test.
Cellular lysates and tissues homogenates contain also nucleosides, nucleotides as
well as nucleic acids which have strong absorption at 260 nm. In the case of samples with a
content of nucleic acids of less than 20% the optical densities at 260 and 280 nm against a
blank must be assessed. The two optical density values can be used to calculate the protein
concentration (mg/mL) according to the following formula:
260 280 OD 76 . 0 OD 55 . 1× − ×
In the case of samples with a content of nucleic acids higher than 20% the previous
equation can not be used. The sample must be processed in order to eliminate a part of the
compounds that contain nucleic bases. For example, this can be achieved through dialysis
against a specific buffer solution.
Another compound that can interfere with this assay is the Trition X(100 detergent.
The ultraviolet absorption spectroscopy can be used to obtain an estimation of the
protein content of a given sample. The major advantage of this assay is the fact that it is not
time consuming. On the other hand, if the molar extinction coefficient of the protein from
70 the sample is known, then the Lambert(Beer equation can be used to accurately assess the
protein amount from the sample.
Despite the fact that the high(molcular mass proteins have a similar tryptophan
content there are also differences regarding this aspect which represents a limitation of this
assay. For this reason, this assay can be used to make a rough estimation of the protein
content of a sample.
Reagents and instrumentation
1. stock bovine serum albumin (BSA) solution 1 mg/mL.
2. laboratory glassware: test tubes, pipettes.
3. instrumentation: analytical balance, spectrophotometer, micropipettes.
Procedure
Pipette into a series of four clean test tubes as indicated in the following table:
Test tube Stock BSA solution
(mL) Distilled water
(mL) Concentration
(µg/mL) Optical
density (OD)
1 10 0
2 5 5
3 2 8
4 1 9
Read the optical densities for all four solutions at 260 and 280 nm against a blank of
distilled water.
Laboratory report
The optical density values will be used to calculate the protein concentration using
the previous formula. Compare the values obtained through this calculation with the ones
obtained through dilution.
71 Experiment 22
Assay of total protein content through biuret assay (version I)
Background
Peptides with more than three peptide bonds react in an alkaline medium (pH ≈ 12)
with Cu2+ ions giving rise to a purple coloured complex. The biuret reactions can be used
for both qualitative and quantitative analysis of peptides, polipeptides and proteins.
The name comes from the name of the compound obtained by heating urea, which
gives a similar coloured complex in the same conditions.
H2NC
NH2O
H2NC
NC
NH2O O
H
H2N
NN
NN COOH
RO
HR
OH
RO
HR
OH
Rurea biuret
peptide
The reaction does not occur with amino acids due to the absence of the peptide
bonds. Also, dipeptides do not react as they contain only one peptide bond.
The biuret reagent contains Cu2+ ions, which are stabilized by the presence of the
sodium potassium tartrate. Under alkaline conditions, the hydrogen atoms of the peptide
bonds are displaced. Each Cu2+ ion is involved in the formation of a tri( or tetradentate
complex leading to the formation of the coloured complex. The complex has a maximum
absorbance at 540 nm.
72 C N C C N
O OC
Rx RyCO
N C C NO
CRx Ry
Cu2+
Due to the fact that this test involves Cu2+ ions it is very sensitive to the presence of
different compounds that can bind these ions (EDTA).
Using a series of proteic solutions with known concentration a calibration curve can
be obtained and used to assess the protein concentration from a sample.
Reagents and instrumentation
1. reagents: bovine serum albumine (BSA), CuSO 4×7H 2O, NaOH, KI, Seignette salt
(sodium potassium tartrate).
2. stock bovine serum albumin (BSA) solution 5 mg/mL.
3. biuret reagent: disolve 0.3 g of CuSO 4×7H 2O and 0.9 g of sodium potassium tartrate in
a volume of 50 mL of NaOH 0.2M; add 0.5 g of KI and distille water to a final volume
of 100 mL.
4. laboratory glassware: test tubes, graduated cylinders, volumetric flasks, pipettes.
5. instrumentation: analytical balance, spectrophotometer, micropipettes.
Procedure
Pipette into a series of seven clean test tubes as indicated in the following table:
Test tube BSA stock solution
(mL) Distilled water
(mL) Concentration
(µg/mL) Optical
density (OD)
1 0 4.0
2 0.1 3.9
3 0.2 3.8
4 0.4 3.6
5 0.6 3.4
6 0.8 3.2
7 1.0 3.0
73 In each test tube add 6 mL of the biuret reagent. Shake the test tubes and keep
them at room temperature for 10 minutes.
Read the optical densities of the solutions from all the test tubes at 540 nm. Then,
substract the value obtained in the case of the first test tube (blank) from all the other test
tubes.
The sample with unknown protein concentration is prepared in the same way.
Laboratory report
In order to obtain the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (µg/mL), while on the Oy axis plot the optical density (OD).
Use the equation of the curve in order to calculate the protein concentration in the
sample.
74 Experiment 23
Assay of total protein content through biuret assay (version II)
Background
The same principle as described in the previous experiment. In this version, the
biuret assay is performed using a commercially available kit.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 20 –
Sample (µL) 20 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 540 nm.
The protein standard solution provided by the producer has a concentration of 5
g/dL.
Laboratory report
The total protein concentration in the sample is calculated using the following
formula:
75 5OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in g/dL.
76 Experiment 24
Assay of total protein content through Lowry assay
Background
This assay was developed by Oliver H. Lowry (1951) and it is a two(step procedure.
In the first step the Cu2+ ions are complexed by the nitrogen atoms of the peptide
bonds of the peptides, polypeptides and proteins and are reduced to Cu+ in the alkaline
medium. This chromophore is stabilized by the presence of the tartrate anion.
In the second step there is the reduction of the Mo6+/W6+ ions of the Folin(Ciocalteu
reagent leading to the formation of a blue coloured complex. In this step are also involved
the tyrosine residues of the peptides, polypeptides and proteins. This complex has a
maximum absorbance at 650 – 750 nm. Usually, the assay is performed at 750 nm due to
the fact that at this wavelength the interferences caused by the presence of different
biomolecules are minimal.
This test is sensitive to the presence of different compounds. Thus, all the reagents
that bind Cu2+, like EDTA can block the reaction. Also, reducing agents and sulfhydryl ((
SH) containing compounds can reduce Mo6+/W6+ ions. Some detergents as well as phenolic
compounds interfere with this assay.
The Lowry assay has been modified to reduce its susceptibility to interferieng
agents, to increase the speed of the reaction and the stability of the coloured complex.
Using a series of proteic solutions with known concentration a calibration curve can
be obtained and used to assess the protein concentration from a sample.
Reagents and instrumentation
1. reagents: bovine serum albumine (BSA), NaOH, Na 2CO 3, Seignette salt (sodium
potassium tartrate), sodium dodecyl sulfate (SDS), CuSO 4×7H 2O, Folin(Ciocalteu
reagent.
2. stock BSA solution 1 mg/mL.
3. Lowry reagent A: disolve 0.2 g of NaOH, 1 g Na 2CO 3, 0.25 g of sodium potassium
tartrate, and 1.25 g of SDS in a volume of 20 mL of distilled water in a volumetric
flask of 50 mL. The mixture is heated on a water bath for SDS solubilization. Cool
77 gradually in order to prevent SDS precipitation. Next day add distilled water into the
volumetric flask up to the mark. This period is necessary for disappearance of the foam
caused by SDS. The reagent is stable for 3 months at room temperature. Before use
check if the solution is clear. If it is not clear make a fresh reagent.
4. Lowry reagent B: disolve 0.1 g of CuSO 4×7H 2O in a volume of 10 mL distilled water.
The reagent is stable 1 year at room temperature.
5. laboratory glassware: test tubes, volumetric flasks, Berzelius flasks, graduated
cylinders, pipettes.
6. instrumentation: analytical balance, spectrophotometer, water bath, micropipettes.
The Lowry working reagent is prepared by mixing the reagents A and B in a ratio of
0.01:1 (vol/vol). The Folin(Ciocalteu is diluted 1:1 with distilled water befor use.
Procedure
Pipette into a series of nine clean test tubes as indicated in the following table:
Test tube BSA stock solution
(mL) Distilled water
(mL) Concentration
(µg/mL) Optical
density (OD)
1 0 1
2 0.10 0.90
3 0.15 0.85
4 0.20 0.80
5 0.25 0.75
6 0.30 0.70
7 0.35 0.65
8 0.40 0.60
9 0.45 0.55
In each test tube add 5 mL of the Lowry working reagent. Shake the test tubes and
keep them at room temperature for 10 minutes. Add 0.5 mL of the diluted Folin(Ciocalteu
reagent. Shake the test tubes and keep them for 30 minutes at room temperature in the dark.
Read the optical densities of the solutions from all the test tubes at 660 or 750 nm.
Then, substract the value obtained in the case of the first test tube (blank) from all the other
test tubes.
The sample of unknown protein concentration is prepared in the same way.
78 Laboratory report
In order to obtain the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (µg/mL), while on the Oy axis plot the optical density (OD).
Use the equation of the curve in order to calculate the protein concentration in the
sample.
79 Experiment 25
Assay of L-lactate
Background
Glycolysis is the metabolic pathway through which one molecule of D(glucose is
converted into two molecules of pyruvic acid. Due to the fact that at physiological pH
pyruvic acid is ionized, it is correct to use „ pyruvate ” instead of „ pyruvic acid ”. The free
energy released in this metabolic pathway is used to synthesize ATP and NADH.
CHO
C
C
C
C
CH2OHH OH
HO H
OH H
OH HCOO
C
CH3O2 NAD+2 NADH + 2 H+
2 ADP + 2 Pa
2 ATP
2 H 2O + 4 H+
Pyruvate has two alternative fates depending on the presence or the absence of
oxygen (aerobic versus anaerobic conditions):
• aerobic conditions – pyruvate enters the mitochondria where it undergoes oxidative
decarboxylation, a process catalyzed by the complex of pyruvate decarboxylase:
COO
C
CH3O CH3COSCoANAD+NADH + H+
CoASH
CO2
80 • anaerobic conditions – pyruvate is reduced in the cytosol of the cells to L(lactate by
lactate dehydrogenase:
COO
C
CH3OCOO
C
CH3H HO
pyruvate L(lactateNADH + H+NAD+
lactate dehydrogenase
The reduction of pyruvate to L(lactate enables the reoxidation of NADH produced in
glycolysis during anaerobic conditions. Without this reaction, due to the use of NAD+ in the
reaction catalyzed by glyceraldehyde(3(phosphate dehydrogenase, all NAD+ would end up
as NADH and glycolysis would stop. Thus, the reduction of pyruvate to L(lactate enables to
keep the flow in glycolysis.
The same reaction takes place also in cells devoided of mitochondria which are
unable to convert pyruvate to acetyl~CoA: (i) red blood cells, (ii) cells located in specific
regions of the liver, (iii) cells located in the fovea centralis of the retina etc. Also, muscles
cells, under hypoxic (during intense physical effort), convert pyruvate to L(lactate.
The aforementioned cells can not use L(lactate. As a consequence, it is released into
the circulation and taken by the liver, and to a lesser extent by the renal cortex. In these
organs, L(lactate is converted back to pyruvate by lactate dehydrogenase. Pyruvate enters
the gluconeogenetic pathway to produce D(glucose.
Lactic acidosis is the consequence of the accumulation of lactic acid in the body.
The most common causes of lactic acidosis include (i) intense physical effort, (ii) thiamine
(vitamin B 1) deficiency, and (iii) pathological conditions (sepsis, respiratory failure, cancer,
diabetes mellitus, renal failure).
The overall process of conversion of D(glucose to L(lactate is called lactic acid
fermentation. It occurs also in bacteria involved in making yogurt.
L(Lactate from the sample is converted to pyruvate by lactate oxidase (EC
1.13.12.4) giving rise to H 2O2.
81 COO
C
CH3H HOCOO
C
CH3O
L(lactate pyruvateO2H2O2
lactate oxidase
Finally, H 2O2 is used in the reaction catalyzed by the horseradish peroxidase which
produces a coloured compound with a maximum absorbance at 505 nm.
In this reaction H 2O2 is used as an oxidizing agent by the peroxidase in order to
generate a red dye from the reaction between 4(aminophenazone and 4(chlorophenol.
NNO
NH 2C6H5
H3C
H3C+OH
ClNNO
NC6H5
H3C
H3C
OCl H2O2H2O
peroxidase
4(aminophenazone 4(chlorophenol quinonimine
The optical density at 505 nm is proportional to the concentration of L(lactic from
the sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
82
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 505 nm.
The L(lactate standard solution provided by the producer has a concentration of 10
mg/dL.
Laboratory report
The concentration of L(lactate in the sample is calculated using the following
formula:
10OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
83 Experiment 26
Assay of uric acid
Background
Uric acid in the final product of purine catabolism in humans and other primates. It
is excreted through the urine.
N
NNNO
OH
HHO HN
NNNO
OH
HHOH
N
NNNO
OH
HHO
urate (conjugate base) uric acid (enol tautomer) uric acid (keto tautomer)
The great majority of the ingested foodstuffs of plant and animal origin contains
nucleic acid. The duodenum is the main site of the dietary nucleic acids digestion. This
process requires the action of pancreatic nucleases and intestinal phosphodiesterases and
gives rise to NMPs ( nucleoside monophosphates):
• from DNA: dAMP, dGMP, dTMP, dCMP,
• from RNA: AMP, GMP, UMP, CMP
N
NNNNH2
O
OH OHCH2 O P OO
O
adenosine monophosphate (AMP)
84
The phosphate group of the NMPs is removed by the action of either a nucleotidase
or a nonspecific phosphatase resulting in the release of the corresponding nucleosides.
The nucleosides are converted to the corresponding free bases through the action of
either a nucleosidase or a nucleoside phosphorylase.
nucleoside + H2O base + ribose ( nucleosidase )
nucleoside + Pi base + ribose(1(phosphate ( nucleoside phosphorylase )
In the case of AMP the reactions catalyzed by the two aforementioned enzymes are
as follows:
AMP + H2O adenosine + D(ribose ( nucleosidase )
AMP + Pi adenosine + D(ribose(1(phosphate ( nucleoside phosphorylase )
Finally, the free bases are converted to uric acid through the action of the xanthine
oxidase, an enzyme which requires Mo ions for its activity.
N
NNNOH
HN
NNNOH
HHON
NNNOH
OHHOH2O + O 2 H2O + O2H2O2H2O2
xanthine
oxidasexanthine
oxidase
hypoxanthine xanthine uric acid
The bases of the cellular nucleic acids are subject to the same reactions as part of the
continual turnover of these biomolecules.
The following schema summarizes the biochemical modifications of the NMPs,
where AMP (adenosine monophosphate), IMP (inosine monophosphate), XMP (xanthosine
monophosphate), GMP (guanosine monophosphate), Rib(1(P (ribose(1(phosphate), and
PNP (purine nucleoside phosphorylase):
85
AMP IMP XMP GMP
adenosine inosine xanthosine guanosine
hypoxanthine xanthine guanine
uric acidnucleotidase nucleotidase nucleotidase nucleotidaseH2O
PiH2O
PiH2O
PiH2O
PiAMP deaminaseH2ONH 4+
adenosine
deaminase
H2ONH 4+PiPi Pi
Rib(1(P Rib(1(P Rib(1(PPÎglyphHVâP PÎglyphHVâP PÎglyphHVâP
xanthine
oxidase xanthine
oxidaseguanine
desaminase
Birds, terestrial reptiles and some insects convert the amino acid nitrogen to uric
acid through purine base biosynthesis as these organisms do not excrete urea. Unlike uric
acid which has a low water solubility, urea is highly water soluble and, as a consequence,
its elimination requires a large amount of water. This could be a problem for some animals
like birds and reptiles.
N
NNNO
OOH
HH
H
NH2
NNN
OO
HH
HO
H2 H2O + O2
CO2 + H2O2urate oxidaseuric acid
allantoin
86 In mammals, other than primates, and different other organisms, uric acid is further
metabolized before excretion.
NH2
NNN
OO
HH
HO
HNH2
NCOOH
NO
HHHNH 2
O H2NH2N
O 4 NH4+2+ H2O
allantoinase allantoicaseH2OCOOH
CHO
urease2 H2O2 CO2
allantoin allantoic acid ureaglyoxylic acid
The final product of this catabolic pathway is species specific and it can be allantoic
acid, urea or ammonia.
Final product Group of organisms
uric acid primates, birds, reptiles, insects
allantoin other mammals (excluding primates)
allantoic acid teleost fish
urea cartilaginous fish, amphibia
ammonia marine organisms, invertebrates
Uric acid from the sample is oxidized to allantoin by uricase (EC 1.7.3.3) giving rise
to H 2O2.
N
NNNO
OOH
HH
HNH2
NNN
OO
HH
HO
H
uric acid allantoinurate oxidase2 H 2O + O 2CO2 + 2 H2O2
Finally, H 2O2 is used in the reaction catalyzed by the horseradish peroxidase which
produces a coloured compound with a maximum absorbance at 520 nm.
87 In this reaction H 2O2 is used as an oxidizing agent by the peroxidase in order to
generate a red dye from the reaction between 2,4(dichlorophenol(6(sulfonic acid and 4(
aminophenazone.
The optical density at 520 nm is proportional to the uric acid concentration in the
sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 25 –
Sample (µL) 25 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 520 nm.
The uric acid standard solution provided by the producer has a concentration of 6
mg/dL.
Laboratory report
The concentration of uric acid in the sample is calculated using the following
formula:
88 6OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
89 Experiment 27
Assay of urea
Background
The excess amino acid nitrogen is eliminated as urea in humans and other species.
Urea is synthesized only in the liver and it is excreted through urine.
The amino acid nitrogen can be excreted as ammonia (ammonotelic organisms),
urea (ureotelic organisms) or uric acid (uricotelic organisms), depending on the availability
of water. Thus, many aquatic organisms simply excrete ammonia. On the other hand, most
of the terestrial vertebrates convert ammonia to the less toxic urea. Terestrial reptiles and
birds use amino acid nitrogen for uric acid biosynthesis. This is due to the fact that water
supply is restricted in the habitats where live different species of reptiles. Also, uric acid
has a limited water solubility and its elimination requires a small amount of water.
The biosynthesis of urea requires the completion of the following steps:
1. transamination:
• removal of amino groups from different amino acids
• transfer of the amino groups to α(ketoglutarate giving rise to L(glutamate
2. deamination:
• release of the amino group from L(glutamate and other amino acids
3. urea cycle:
• urea biosynthesis.
Transamination is the process through which the amino group of an amino acid is
transferred to an α(keto acid giving rise to a new α(keto acid and a new amino acid.
Transamination is a reversible process and it is catalyzed by transaminases, enzymes
that require pyridoxal(5’(phosphate as a coenzyme. Because of the fact that these enzymes
transfer an amino group they are also called aminotransferases.
R1 C COOH
OR2CH COOH
NH2+R1CH COOH
NH2R2 C COOH
O+
keto acid 1 amino acid 2 amino acid 1 keto acid 2
90
Most transaminases accept α(ketoglutarate and to a lesser extent oxaloacetate.
R CH COO
NH3COO
C
CH2
CH2
COOO
+COO
C
CH2
CH2
COOH H3N
R C COO
O+
amino acid α(ketoglutarate keto acid L(glutamate
There are two major transaminases aspartate aminotransferase (AST) and alanine
aminotransferase (ALT), formerly called glutamate oxaloacetate transaminase (GOT) and
glutamate pyruvate transaminase (GPT), respectively.
COO
C
CH2
COOH3N HCOO
C
CH2
CH2
COOO
+COO
C
CH2
COOO
+COO
C
CH2
CH2
COOH H3N
L(aspartate α(ketoglutarate oxaloacetate L(glutamate
COO
C
CH3H3N HCOO
C
CH2
CH2
COOO
+COO
C
CH3O +COO
C
CH2
CH2
COOH H3N
L(alanine α(ketoglutarate pyruvate L(glutamate
91 Transamination does not lead to a net deamination. It leads instead to collection of
the amino groups from different amino acids to α(ketoglutarate resulting in L(glutamate.
Deamination can take place through several reactions, of which the most important
is the oxidative deamination of L(glutamate. This reaction takes place in the mitochondrial
matrix of all the cells of the human body.
COO
C
CH2
CH2
COOH H3NCOO
C
CH2
CH2
COOH2NCOO
C
CH2
CH2
COOONAD(P)+NAD(P)H,H+
H2ONH 4+
Glutamate dehydrogenase is one of the few examples of oxidoreductases which can
use both NAD+ or NADP+ as coenzymes.
The NH 4+ released in the previous reaction is used to synthesize urea throug the urea
cycle pathway.
The urea cycle was the first cyclic metabolic pathway to be discovered in 1932, five
years before the discovery of the Krebs cycle. In the case of mammals, the urea cycle is
confined only to the liver cells.
Urea cycle comprises a sequence of five reactions that are catalyzed by the
following enzymes:
1. carbamoylphosphate synthetase I (EC 6.3.4.16)
2. ornithine transcarbamoylase (EC 2.1.3.3)
3. argininosuccinate synthetase (EC 6.3.4.5)
4. argininosuccinase (EC 4.3.2.1)
5. arginase (EC 3.5.3.1)
The first one takes place in the same cellular compartment as the oxidative
deamination do. The remaining four reactions take place in the cytosol. As it can be seen
from the following schema, one nitrogen atom comes from NH 4+, while the second one
comes from an L(aspartate molecule.
92
H2N
CN
NH 3COO
OHCN
NH 3COO H
NHO
OOC
COONH 2
C
NH
H3NCOOH2N
H3N
NH 3COO
H2N
CN
NH 3COO
OH
H3N
NH 3COOCYTOSOL
MITOCHONDRIAL MATRIX
O
C
NH 2P
OO
OO
2 ATP + NH 3 + HCO 3(2 ADP + P a + H+ornithine transcarbamoylase
carbamoylphosphate
synthetase I
carbamoylphosphateL(citrulline L(ornithineargininosucinateL(arginine
argininosuccinate
synthetaseargininosuccinase
arginase
Urea form the sample is hydrolized by urease (EC 3.5.1.5) giving rise to ammonia
and CO 2.
CNH 2
NH 2OH2O2 NH 3
CO2
Ammonia is used in the reaction catalyzed by L(glutamate dehydrogenase (EC
1.4.1.2). In this reaction NADH, which has a maximum absorbance at 340 nm is oxidized
to NAD+.
As a consequence, during the experiment there is a decrease in the optical density at
340 nm which is directly proportional with the concentration of urea.
93 This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Sample Standard Blank
Working reagent (µL) 1000 1000 1000
Standard (µL) – 10 –
Sample (µL) 10 – –
Shake the Eppendorf tubes and place them into a thermostat at 37°C for a time
period indicated by the manufacturer.
Read the optical densities of the three solutions at 340 nm.
The urea standard solution provided by the producer has a concentration of 50
mg/dL.
Laboratory report
The concentration of urea in the sample is calculated using the following formula:
50OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
94 Experiment 28
Assay of total bilirubin
Background
Hemoproteins are a group of proteins that contain heme as a prosthetic group. Heme
is the complex combination between a heterocycle called porphyrin and an iron ion (Fe2+).
Porphyrin consists of four pyrrolic groups joined together by methenyl bridges. As a
consequence, hemoproteins belong to the larger group of metalloproteins.
NN NN
H
H
There are several types of heme groups which differ by the binding way of some
functional groups to specific carbon atoms of the porphyrin ring. The most important heme
groups are heme A, B, C and D.
Heme B is frequently encountered in biological systems and it consists of a pophyrin
ring which is substituted with four methyl groups, two vinyl groups and two propionyl
groups. These eight substituents can be arranged in different positions giving rise to
different isomers. In biological systems only the isomer IX, called protoporphyrin IX, is
found. As a consequence, heme B is the complex combination of protoporhyrin IX with
Fe2+.
The iron ion has six possible coordination position of which four are occupied by the
four nitrogen atoms of the porphyrin ring. The other two are available for specific amino
acids belonging to the proteic part as well as for different ligands (O 2, CO, NO).
95 The heme groups are bound to the proteic part of hemoproteins either by covalent or
noncovalent bonds. For example, heme C is attached covalently, while heme B is attached
noncovalently.
NN NNH3C
CH3
CH3HOOC
HOOCH3CCH2
CH2Fe2+
heme B
The biological properties of the hemoproteins are a consequence of the presence of
the iron ion. There are many hemoproteins some of which are enzymes.
Hemoglobin transports O 2 from the lungs to the tissues and CO 2 from tissues back to
the lungs. Myoglobin stores O 2 in the skeletal muscles. Hemoglobin is the most abundant
hemoprotein in the human body.
Examples of hemoproteins with catalytic activity are catalase, nitric oxide synthase
(NOS) and cyclooxygenase (COX). Catalase is an enzyme involved in the inactivation of
hydrogen peroxide (H 2O2), a major reactive oxygen species.
2 H2O2 2 H2O + O2
Nitric oxide synthase (NOS) is responsible for the synthesis of nitric oxide which
induces the dilatation of blood vessels. Cyclooxygenase (COX) is an enzyme involved in
the synthesis of different eicosanoids. Aspirin acetylates a serin residue from the catalytic
96 center of COX leading to inactivation of this enzyme. A special case of hemoproteins is
represented by the cytochromes which have a variety of biological functions, including the
detoxication of the exogenous chemicals.
COOH
OCOCH3
+Ser OHCOOH
OH
Ser OCOCH3 +
acetylsalicylic acid active enzyme salicylic acid inactive enzyme
The iron ion is essential for the biological activity of all the hemoproteins. In the
case of cytochromes the biological activity relies on the ability of iron ion to alternate
between the two oxidation states, Fe2+ and Fe3+. On the contrary, in the case of hemoglobin
and myoglobin oxidation of Fe2+ to Fe3+ leads to the loss of the biological activity.
Degradation of heme is catalyzed by heme oxygenase giving rise to biliverdin, CO
and Fe2+. This is the only reaction that takes place in the human body and gives rise to CO.
NN NNM V
M
M PPM
VFe2+released as CO
heme
oxygenaseO2
Fe2+ + CO
NOM V
HNM P
HNP M
NOM V
H
biliverdinheme
97
In the previous reaction the letters M, P and V denote the methyl, propionyl and
vinyl groups, respectively.
A special situation is that of red blood cells which contain large amounts of
hemoglobin. After a period of 120 days, the senescent red blood cells are removed from the
circulation and their components degraded. Senescent red blood cells are removed by
special cells found in the bone marrow, spleen as well as in the liver.
The carbon atom eliminated as CO comes from the carbon atom that joined the A
and B rings of the protoporphyrin IX. Biliverdin is a green pigment.
The reduction of the biliverdin’s central methenyl bridge by the biliverdin reductase
gives rise to bilirubin which is an yellow pigment.
NOM V
HNM P
HNP M
NOM V
H
biliverdin
NOM V
HNM P
HNP M
NOM V
H H
bilirubinNADPH + H+
NADP+biliverdin
reductase
In the previous reaction the letters M, P and V denote the methyl, propionyl and
vinyl groups, respectively.
The conversion of heme to bilirubin can be seen in the case of a bruise that heals.
The colour of the bruise turns from red (due to heme) at the beginning, to green (due to
biliverdin) and finally to yellow (due to bilirubin).
98 Bilirubin is a lipophilic compound which is transported to the liver in complex with
serum albumin. In the liver, each molecule of bilirubin is conjugated with two molecules of
D(glucuronic acid. The conjugated bilirubin is then excreted in the bile and enters the small
intestine. The two residues of D(glucuronic acid are removed and bilirubin is modified to
give rise to urobilinogen. Some of the urobilinogen is reabsorbed and transported via the
bloodstream to the kidney where it is converted to the yellow urobilin which is excreted
giving the urine its specific colour.
Most of the urobilinogen is converted by the gut bacteria to the stercobilin which is
a deeply red(brown pigment which gives the feces the specific colour.
The blood contains two main fractions of bilirubin, the conjugated one and the
unconjugated one. The former has two D(glucuronic acid residues attached, while the later
is noncovalently bound to serum albumin.
Bilirubin from the sample is assessed through the van den Bergh reaction. Bilirubin
reacts with the diazotized sulphanilic acid to produce the purple coloured azobilirubin.
NOM V
HNM P
HNP M
NOM V
H H
1. HCl
2. NaNO2
3. sulphanilic acidbilirubin
NOM V
HNM P
HN N SO3H
NP M
NOM V
H HN N HO3S
The conjugated bilirubin reacts directly in aqueous solution to give the azobilirubin
and it is called the direct bilirubin. On the other hand, the unconjugated bilirubin which is
99 noncovalently attached to serum albumin requires prior solubilization with DMSO in order
to react with the diazotized sulphanilic acid. This fraction is called the indirect bilirubin.
The sum of the two fractions represents the total bilirubin.
The intensity of the colour formed is proportional to the bilirubin concentration from
the sample.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions. Pipette into three Eppendorf
tubes as indicated in the following table:
Blank Total bilirubin
Reagent R 1 (µL) 1500 1500
Reagent R 2 (µL) – 50
Sample/Standard (µL) 100 100
Reagent R 1 contains sulphanilic acid (30 mM), hydrochloric acid (50 mM) and
DMSO (dimethylsulphoxide) (7 M). Reagent R 2 contains sodium nitrite (29 mM).
Shake the Eppendorf tubes and incubate them at room temperature for 5 minutes.
Read the optical densities at 555 nm.
Laboratory report
The concentration of urea in the sample is calculated using the following formula:
50OD ODOD OD C
blank standardblank sample×−−=
The results are expressed in mg/dL.
100 Experiment 29
Assay of aspartate aminotransferase (AST/GOT)
Background
Aspartate aminotransferase (EC 2.6.1.1) is an aminotransferase involved in the
reversible transfer of an amino group from L(aspartate to α(ketoglutarate:
COO
C
CH2
COOH3N HCOO
C
CH2
CH2
COOO
+COO
C
CH2
COOO
+COO
C
CH2
CH2
COOH H3N
L(aspartate α(ketoglutarate oxaloacetate L(glutamate
Aspartate aminotransferase is found in high concentration in heart muscle, skeletal
muscle and liver cells, as well as in other cells of the body.
Aspartate aminotransferase and alanine aminotransferase are routinely used in the
clinical laboratory to evaluate the liver function.
The AST from a sample catalyzes the formation of oxaloacetate. This is reduced in
the second enzymatic reaction to L(malate by L(malate dehydrogenase (EC 1.1.1.37).
COO
C
CH2
COOOCOO
C
CH2
COOH3N HNADH,H+NAD+
Lămalate dehydrogenase
oxaloacetate L(malate
101 This is a redox reaction with NADH as a reducing agent. NADH, which has a
maximum absorbance at 340 nm is oxidized to NAD+.
As a consequence, during the experiment there is a decrease in the optical density at
340 nm which is directly proportional with the aspartate aminotransferase activity from the
sample.
This is an example of the technique of coupled reactions.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions.
Pipette into quartz cuvette 1000 µL of the working reagent and then 100 µL of the
sample. Mix and incubate for 1 minute. Read the initial optical density (OD i) and then the
final optical density (OD f) after 3 minutes.
Read the optical densities at 340 nm.
Laboratory report
The catalytic concentration of aspartate aminotransferase in the sample is calculated
according to the following formula:
(U/L) 1750 )/min OD ( (ODi f×
The catalytic concentration is expressed as units per liter (U/L).
One international unit (IU) of aspartate aminotransferase is defined as the amount of
enzyme that transforms 1 µmol of substrate per minute, in standard conditions.
102 Experiment 30
Assay of glutathione S-transferase
Background
All organisms are exposed to different kind of foreign chemicals which are called
xenobiotics. These include both natural and synthetic chemical compounds like secondary
plant metabolites and toxins produced by molds, plants, and animals, drugs, pesticides,
industrial chemicals, pyrolysis products in cooked food etc. In the case of the human body,
many of the aforementioned compounds can be absorbed through different epithelia (skin,
lungs, gastrointestinal tract). Because most of them are highly lipophilic compounds, they
tend to accumulate in the human body with potential side effects.
Most of the xenobiotics are excreted from the human body after biotransformation.
Biotransformation is a term that indicates a number of biochemical reactions through which
the lipophilic compounds are transformed into more water soluble products which are
excreted in urine or/and feces.
There are phase I and phase II biotransformation reactions. The phase I reactions
include oxidation, reduction and hydrolysis leading to the exposure or addition of specific
functional grups. The phase II reactions include conjugation with different compounds like
glutathione, amino acids, D(glucuronic acid etc, leading to highly hydrophilic products.
The enzymes involved in the biotransformation reactions are present in all tissues of
the human body, with higher concentrations in liver, intestine, kidney, testis lung etc.
Glutathione S(transferases (EC 2.5.1.18) are a family of enzymes which catalyzes
the conjugation of different xenobiotics with reduced glutathione (GSH).
These enzymes deprotonate the –SH group of GSH leading to a thiolate anion. This
acts like a nucleophilic reagent and attacks an electrophilic C, O, ÎglyphHVâ or S atom, which
belongs to the compound to be conjugated. The name of these enzymes stands for the fact
that the nucleophilic agent is the thiolate anion.
There are two types are glutathione conjugation reactions: displacement reactions
and addition reactions. In the first case, GSH (in fact the GS( anion) displaces an electron(
withdrawing group. In the second case, GSH is added to a double bond or to an ring.
103 H2N C
NN COOH
COOHO
H OHSH
reduced glutathione
(γ(glutamylcysteinylglycinã)isopeptidic bond
The products of the glutathione conjugation reaction formed in the liver are excreted
through the bile and then through feces. In kidney, these conjugation products are subject to
the sequential action of three enzymes: γ(glutamyltranspeptidase, aminopeptidase M and ÎglyphHVâ(
acetyl transferase. The final products are the so(called mercapturic acids which are excreted
through urine.
R S CysGlu
GlyR S CysNH 2
GlyR S CysNH 2
COOHH2OGluH2OGly
γăglutamyl
transpeptidaseaminopeptidase M
R S CysNH 2
COOHR S CysHN
COOHCOCH 3acetyl(CoACoASH
mercapturic acidÎglyphHVâăacetyl transferase
The displacement of an electron(withdrawing group is facilitated by the presence of
one or more electron(withdrawing groups (F(, Cl(, Br(, (NO 2, (CN, (CHO). This is the
reason for which 1(chloro(2,4(dinitrobenzene is the most frequently used substrate in order
to assess the glutathione S(transferases activity.
A simple procedure for the assay of glutathione S(transferase consists of following
the increase of optical density at 340 nm due to the formation of the conjugation product
between reduced glutathione and 1(chloro(2,4(dinitrobenzene.
104 Cl
NO2
NO 2S
NO2
NO2G
HCl
glutathion Sătransferase
1(chloro(2,4(dinitrobenzene conjugation productGSH
The increase in the optical density is proportional with the glutathione S(transferase
activity from the sample.
Reagents and instrumentation
1. reagents: Na 2HPO 4, NaH 2PO 4, reduced glutathione (GSH) (307.22 g/mol), 1(chloro(
2,4(dinitrobenzene (202.55 g/mol), ethanol.
2. solutions: GSH 20 mM solution (disolve 0.1536 g of GSH in a volume of 25 mL of
distilled water); 1(chloro(2,4(dinitrobenzene 20 mM solution (disolve 0.2026 g of 1(
chloro(2,4(dinitrobenzene in a volume of 50 mL of ethanol 95%); sodium phosphate
buffer 0.2 M, pH = 6.5.
3. laboratory glassware: volumetric flasks, Berzelius flasks, graduated cylinder, pipettes.
4. laboratory consumables: Eppendorf tubes (1.5 mL).
5. instrumentation: analytical balance, spectrophotometer, pH meter, micropipettes.
Procedure
Make a series of four dilutions (1:9, 1:19, 1:29, 1:39) from the liver homogenate.
For dilution use the sodium phosphate buffer.
Pipette into a quartz cuvette 800 µL of the buffer solution, 50 µL of the GSH
solution and 100 µL of diluted sample. Start the enzymatic reaction with 50 µL of the 1(
chloro(2,4(dinitrobenzene solution. When adding the sample start the stopwatch. Mix with
the tip of the micropipettes the solution from the cuvette for about 10 seconds and read the
initial optical density. Record the increase of optical density for 5 minutes at intervals of 1
minute.
105 Laboratory report
Plot the graph OD = f(t) and use the linear portion. The time range during which the
previous function is linear is used to calculate the enzymatic activity. Keep in mind the
dilution factor of the four dilutions used.
The catalytic concentration is expressed as µmol/min/mL as well as units/mL. One
unit of glutathione S(transferase is defined as the amount of enzyme catalyzing the
conjugation of 1 µmol of 1(chloro(2,4(dinitrobenzene during 1 minute.
Calculate the specific activity by dividing the enzymatic activity by the protein
concentration in the sample.
Calculate the GSH and 1(chloro(2,4(dinitrobenzene concentrations in the final assay
volume.
106 Experiment 31
Assay of gamma-glutamyl transferase ( γ-GT)
Background
Gamma(glutamyl transferase (EC 2.3.2.2) is an enzyme involved in the transfer of
the γ(glutamyl residues from different peptides to acceptor molecules like water, amino
acids and other peptides. The most important donor of γ(glutamyl residues is the tripeptide
glutathione (γ(glutamyl(cysteinyl(glycine).
H2N C
NN COOH
COOHO
H OHSH
γ(glutamyl(cysteinyl(glycinepeptide COO H3N +
H3N C
COOO
Npeptide
HCOOH3NN COO
OHSH
cysteinyl(glycine
γ(glutamyl(peptideisopeptidic bond
This enzyme is present in the cellular membrane of cells from different tissues
including kidney, bile duct, pancreas, gallblader, spleen, heart, brain etc.
107 It is involved in the transfer of amino acids across the cellular membrane from the
extracellular space to the intracellular one. Also, this enzyme is involved in the biosynthesis
of leukotrienes and catabolism of glutathione (γ(glutamyl cycle).
Gamma(glutamyl transpeptidase from the sample catalyzes the transfer of the γ(
glutamyl residue from γ(glutamyl(3(carboxy(4(nitroanilide to glycylglycine giving rise to
γ(glutamyl(glycylglycine and 2(nitro(5(amino(benzoic acid.
H3N C CH2CH2CO H
COON
HNO2
COOH3N NO2
COOglycylglycineγ(glutamyl(glycylglycine
γ(glutamyl(3(carboxy(4(nitroanilide 5(amino(2(nitro(benzoic acidgammaăglutamyl
transpeptidase
The rate of formation of the 5(amino(2(nitro(benzoic acid is directly proportional to
the catalytic concentration of gamma(glutamyl transpeptidase present in the sample.
Reagents and instrumentation
1. use commercially available kits (must be stored at 2(8°C).
2. laboratory glassware: pipettes.
3. laboratory consumables: Eppendorf tubes (1.5 mL).
4. instrumentation: spectrophotometer, thermostat, micropipettes.
Procedure
Work accordingly to the manufacturer’s instructions.
Pipette into quartz cuvette 1000 µL of the working reagent and then 100 µL of the
sample. Mix and incubate for 1 minute. Read the initial optical density (OD i) and then the
final optical density (OD f) after 3 minutes.
Read the optical densities at 405 nm.
108 Laboratory report
The catalytic concentration of gamma(glutamyl transpeptidase in the sample is
calculated according to the following formula:
(U/L) 1190 )/min OD ( (ODi f×
The catalytic concentration is expressed as units per liter (U/L).
One international unit (IU) of gamma(glutamyl transpeptidase is defined as the
amount of enzyme that transforms 1 µmol of substrate per minute, in standard conditions.
109 Experiment 32
Assay of xanthine oxidase
Background
Xanthine oxidase (EC 1.17.3.2), known as well as xanthine oxidoreductase, is an
enzymatic activity of a protein involved in the last steps of purines catabolism. In fact, the
same protein can have two different enzymatic activities: xanthine oxidase and xanthine
dehydrogenase (EC 1.17.1.4). This protein is very abundant in the mammalian liver.
Oxidation of specific sulfhydryl groups or irreversible proteolytic modification of
the protein changes the activity from xanthine dehydrogenase to xanthine oxidase.
NADH + H+
N
NNN
HOH
HON
NNN
HOH
HOOHH2O + NAD+
xanthine dehydrogenase
xanthine uric acid
N
NNN
HOH
N
NNN
HOH
HOH2O + O2H2O2
xanthine oxidase
hypoxanthine xanthine
N
NNN
HOH
HON
NNN
HOH
HOOHH2O + O 2H2O2
xanthine oxidase
xanthine uric acid
110
As it can be seen from the previous reactions, the xanthine oxidase activity of the
protein catalyzes a sequence of two reactions: conversion of hypoxanthine to xanthine and
xanthine to uric acid, respectively.
Moreover, the product of the two enzymatic reactions is uric acid, which in
mammals is the final product of purines catabolism and it is excreted through urine.
A simple procedure for the assay of xanthine oxidase consists in following the
increase of optical density at 290 nm due to uric acid formation. The increase in the optical
density is proportional with the xanthine oxidase activity from the sample.
Reagents and instrumentation
1. reagents: K 2HPO 4, KH 2PO 4, xanthine (152,11 g/mol).
2. solutions: xanthine 0.15 mM solution (disolve 0.0023 g of xanthine in a volume of 100
mL of distilled water); potassium phosphate buffer 0.05 M, pH = 7.5.
3. laboratory glassware: volumetric flasks, Berzelius flask, pipettes.
4. laboratory consumables: Eppendorf tubes (1.5 mL).
5. instrumentation: analytical balance, spectrophotometer, pH meter, micropipettes.
Procedure
Make a series of four dilutions (1:9, 1:19, 1:29, 1:39) from the liver homogenate.
For dilution use the potassium phosphate buffer.
Pipette into a quartz cuvette 633 µL of the buffer solution and 333 µL of the
xanthine solution. Start the enzymatic reaction with 34 µL of diluted sample. When adding
the sample start the stopwatch. Mix with the tip of the micropipettes the solution from the
cuvette for about 10 seconds and read the initial optical density. Record the increase of
optical density for 5 minutes at intervals of 1 minute.
Laboratory report
Plot the graph OD = f(t) and use the linear portion. The time range during which the
previous function is linear is used to calculate the enzymatic activity. Keep in mind the
dilution factor of the four dilutions used.
111 The catalytic concentration is expressed as µmol/min/mL as well as units/mL. One
unit of xanthine oxidase is defined as the amount of enzyme catalyzing the oxidation of 1
µmol of xanthine to uric acid during 1 minute.
Calculate the specific activity by dividing the enzymatic activity with the protein
concentration from the sample.
Calculate the xanthine concentration in the final assay volume.
112 Experiment 33
Assay of total antioxidant capacity (TEAC assay)
Background
Reactive oxygen and nitrogen species are involved in food deterioration as well as in
the pathogenesis of some human diseases (atherosclerosis, diabetes mellitus, cancer,
chronic inflammation etc).
In the biochemical papers and textbooks the terms „ proăoxidant” and „ antioxidant ”
are very often used. Pro(oxidant is any compound that can induce an oxidative modification
to a target biomolecule, while an antioxidant is any compound that can reduce or inactivate
a pro(oxidant. An antioxidant can be defined as „ any substance that when present at low
concentrations compared to those of an oxidizable substance significantly delay or prevents
oxidantion of that substrate ”.
Reactive oxygen and nitrogen species act as pro(oxidants agents and can be
classified as either radical or non(radical species. The first group includes chemical species
like the hydroxyl radical (HO•) and nitric oxide radical (NO•). The second group includes
chemical species like hydrogen peroxyde (H 2O2), hydroxide and peroxynitrite anions (HO(,
OONO(, O 2•−).
All living systems developed a plethora of mechanisms for the protection against the
action of the reactive oxygen and nitrogen species with the help of the antioxidants. There
can be enzymatic and nonenzymatic antioxidants.
The first group includes enzymes like catalase, superoxide dismutase, glutathione
peroxidase, glutathione reductase, while the second group includes different low and high
molcular(mass antioxidants.
Plants contain many secondary metabolites which can act ast low molecular(mass
antioxidants: phenolic acids, flavonoids, carotenoids, tocopherols, ascorbic acid. On the
other hand, animal tissues contain both low molecular(mass (glutathione, uric acid, ascorbic
acid) and high molecular(mass (albumin) antioxidants.
113 H2O2
O2
O222 H+superoxide dismutaseH2OcatalaseH2OGSHGSSGNADPH, H+ NADP+
glutathione peroxidaseglutathione reductase
COOH
OHOH HO
OHHOCOOH
O
OOHHO
OHOH
OH
gallic acid caffeic acid quercetin (flavonol)
O
HOCH3
H3C
CH3CH3CH3
CH3 CH3 CH3
alpha tocopherol
The protection against the deleterious effects of the pro(oxidants can be achieved by
several mechanisms: inhibition of generation and/or inactivation of the reactive oxygen and
nitrogen species, transitional metal ions chelation, the activity of the antioxidant enzymes,
and inhibition of the enzymes that produce reactive oxygen and nitrogen species.
The in vitro analytical methods used to assess the antioxidant capacity of a specific
sample can be classified as competitive and non(competitive assays.
114 In the case of the competitive assays there is a competition between a biomolecule
and an antioxidant for the reactive oxygen or nitrogen species. The antioxidant capacity is
assessed by the quantification of an analytical signal which depends on the biomolecule or
its oxidized form. The antioxidant capacity can be also assessed after the aforementioned
reaction took place and in the system is added a chemical compound to react with the
remaining reactive oxygen or nitrogen species or with the biomolecule. There are many
factors that influence these assays like the rate of the reaction between biomolecules and
the reactive species etc.
reactive oxygen or nitrogen
speciesreduced reactive oxygen or nitrogen
speciesbiomoleculeoxidized biomolecule
antioxidant
oxidized antioxidant
In the case of the non(competitive assays the potential antioxidant compound reacts
directly with the reactive oxygen or nitrogen species without requiring the presence of any
other chemical compound. Similarly, after the reaction took place, the remaining reactive
oxygen or nitrogen species can react with a specific chemical to give rise to an analytical
signal.
reactive oxygen or nitrogen
speciesreduced reactive oxygen or nitrogen
species
antioxidant
oxidized antioxidant
There are different assays in use to measure the total antioxidant capacity of pure
compounds or biological samples (plasma, urine, tissue homogenates, food extracts,
beverages etc). Each method is based on generation of a radical through a specific method
and the measurement of a signal at a fixed moment or during a time period.
One of the simplest methods is the so(called TEAC (Trolox equivalent antioxidant
capacity) assay. It is based on the generation of the ABTS•+ radical cation through the
115 reaction between ABTS [2,2’(azinobis((3(ethylbenzothiazoline(6(sulfonic acid] and
potassium persulfate (K 2S2O8).
K2S2O8S
N
SN
NNSO3H
HO 3SH3C
CH3
S
N
SN
NNSO3H
HO 3SH3C
CH3ABTS
ABTS radical cation
The ABTS•+ radical cation has two resonance structures which are depicted below:
S
N
SN
NNSO3H
HO 3SH3C
CH3
SN
HO 3SH3C
S
N NNSO3H
CH3
116 The blue(green ABTS•+ radical cation has maximum absorbance at wavelenghts 645
nm, 734 nm, and 815 nm. Usually, the TEAC assay is performed at 734 nm. After the
addition of the solution to be tested, the antioxidants found in the sample reduce the
ABTS•+ radical cation to the colourless ABTS.
SN
HO3SH3C
S
N NNSO3H
CH3
R OH
R O
SN
HO3SH3C
S
N NNSO3H
CH3H
The extent of this reduction depends on the concentration of antioxidants in the
sample as well on the duration of the reaction. The extent of decolorization is expressed as
percentage inhibition (I%) of the ABTS•+ radical cation optical density at 734 nm. Usually,
this test is performed for exactly one minute. In the end, the results are calculated with a
calibration curve made with solutions of Trolox with concentrations ranging from 0.5 to 2.5
mM. Trolox (6(hydroxy(2,5,7,8(tetramethylchroman(2(carboxylic acid) is an water(soluble
analogue of vitamin E and it is used as an antioxidant standard.
O
HO
CH3H3CCH3
CH3
COOH
Trolox
117 Reagents and instrumentation
1. reagents: ABTS, K 2S2O8, Trolox, ethanol, K 2HPO 4, KH 2PO 4.
2. solutions: ABTS 7 mM solution (disolve 0.0384 g of ABTS in a volume of 10 mL of
distilled water); K 2S2O8 140 mM (disolve 0.3784 g of K 2S2O8 in a volume of 10 mL of
distilled water); postassium phosphate buffer 5 mM, pH = 7.4; Trolox 2.5 mM solution
(disolve 0.0156 g of Trolox in a volume of 25 mL of potassium phosphate buffer or
ethanol).
3. samples: beverages (red and white wines, apple juice), liver homogenate prepared in
the experiment 2.
4. laboratory glassware: volumetric flasks.
5. laboratory consumables: Eppendorf tubes (1.5 mL and 10 mL).
6. instrumentation: analytical balance, spectrophotometer, pH meter, micropipettes.
Procedure
ABTS•+ radical cation is generated by adding 176 µL of the K 2S2O8 solution to 10
mL of the ABTS solution. The formation of the ABTS•+ radical cation begins immediately,
and the solution turns greener. The mixture must be allowed to stand in the dark at room
temperature for 12(16 hours prior to use.
The oxidation of ABTS by K 2S2O8 is incomplete as these two compounds react in a
ratio of 1:0.5. The optical density of the ABTS•+ radical cation stock solution is not
maximal and stable before 6 hours from the mixing of the two solutions. This stock solution
can be kept for about two days in the dark at room temperature.
For the study of beverages and other food extracts the ABTS•+ radical cation stock
solution is diluted with ethanol to an optical density of 0.68 ± 0.02 at 734 nm. Also, the
Trolox solution is made using ethanol. In the case of the study of biological samples like
plasma or tissue homogenates the ABTS•+ radical cation stock solution is diluted to the
same optical density with the phosphate buffer. Also, in this case the Trolox solution is
made using phosphate buffer.
In a 1 cm cuvette add 990 µL of the diluted ABTS•+ solution and record the initial
optical density (OD i). Add 10 µL of the sample to be tested and start the stopwatch. After
exactly 1 minute read the final optical density (OD f). Addition of the sample leads to the
118 decolorization of the diluted ABTS•+. Using the two values of optical densities calculated
the percentage inhibition (I%) is calculated according to the following formula:
100ODOD ODI %
if i×−=
This percentage inhibition (I%) is a function of concentration and it is calculated
using a calibration curve made with solutions of Trolox with concentrations ranging from
0.5 to 2.5 mM. Each solution of Trolox must be used as in the case of the sample.
Bellow it is presented a table containing the data obtained in the case of a typical
calibration curve with Trolox:
OD i OD f OD i – OD f %I Trolox concentration (mM)
0.713 0.397 0.316 44.32 2.5
0.711 0.463 0.248 34.88 2
0.711 0.557 0.154 21.66 1.5
0.712 0.628 0.084 11.80 1
0.713 0.675 0.038 5.33 0.5
One can see that increasing Trolox concentrations leads to increased consumption of
the ABTS•+ radical cation, leading to lower OD f values. Moreover, the percentage
inhibition (%I) is greater as the Trolox concentration increases.
Using the data from the last two columns of the previous table the following
calibration curve is made:
Calibration curve y = 0,0489x + 0,3462
R2 = 0,9883
00,511,522,53
0,00 10,00 20,00 30,00 40,00 50,00
% IConcentration (mM)
119
The equation is y = 0.0489 x + 0.3462, where y is the concentration of Trolox and x
is the corresponding percentage inhibition (%I).
In the case of a specific sample, the two optical densities (OD i and OD f) will be used
to calculate the percentage inhibition (%I). This value will be introduced in the equation of
the calibration curve in order to obtain the results.
The results are expressed as Trolox equivalent antioxidant capacity (TEAC), which
is defined as the concentration (mM) of Trolox having the equivalent antioxidant activity to
a 1 mM solution of the compound or sample under investigation.
Laboratory report
Create your own calibration curve with the help of a similar table presented above.
Use the equation of the calibration curve to calculate the antioxidant capacities of the
samples taken into analysis. Comment the differences obtained.
120 Experiment 34
Assay of total phenol content with the Folin-Ciocalteu reagent
Background
Plant phenols are some of the most important secondary metabolites in the plant
kingdom. They act as phytoalexins, attractants for pollinators, pigments, antioxidants as
well as protective agents againts UV light.
These compounds can be classified as simple phenols, phenolic acids, coumarins,
flavonoids, stilbenes, hydrolyzable and condensed tannins, lignans and lignin. Of these,
phenolic acids and flavonoids are among the best studied compounds.
Flavonoids are a large family of compounds that share the following chemical
structure, of 2(phenylchromane:
O
Flavonoids are classified into subclasses according to some structural peculiarities:
anthocyanidins, flavanols, flavanones, flavonols, flavones, and isoflavones.
O
OOHO
OO
O
flavonol flavone isoflavone
121 O
OO
OH
OHHOOHOH
O HO
OHOHOHR1
R2
flavanone flavanol (catechins) anthocyanidine
Flavonoids are highly investigated natural compounds due to their potential
beneficial effects on the health status. In spite of the huge number of studies involving these
compounds there is no meta(analysis to clearly demonstrate that the consumption of a
flavonoid(rich diet can prevent some pathological conditions, like cancer.
Several methods were developed to assess the total phenol content of different plant
extracts. These methods rely on reactions that give rise to specific compounds that absorb
in the UV/Vis. They are easy and rapid to perform, and need low(cost reagents.
One of the most widly spread method involves the use of the Folin(Ciocalteu
reagent to give rise to a blue complex that has a maximum absorbance at 746 nm.
Moreover, this method is described by several pharmacopoeias, including the European and
the Romanian ones. The Folin(Ciocalteu reagent is a mixture of phosphomolybdate and
phosphotungstate and in the presence of phenolic compounds, at alkaline pH, the Mo6+ is
reduced to Mo5+ to form a blue chromophore.
Using a series of gallic acid solution with known concentrations a calibration curve
can be obtained and used to assess the total phenol content from a sample.
COOH
OHOH HO
gallic acid
122 The method with Folin(Ciocalteu reagent can be applied as a macromethod as well
as a micromethod.
The results are expressed as mM of gallic acid equivalents.
Reagents and instrumentation
1. reagents: gallic acid, Folin(Ciocalteu reagent, Na 2CO 3, ethanol.
2. standard gallic acid solution (2.5 mM): disolve 0.0425 g of gallic acid in a volume 100
mL of distilled water.
3. Na2CO 3 0.7M solution: disolve 3.71 g of Na 2CO 3 in a volume of 50 mL of distilled
water.
4. laboratory glassware: volumetric flask, Berzelius flask.
5. laboratory consumables: Eppendorf tubes (1.5 mL).
6. instrumentation: analytical balance, spectrophotometer, micropipettes.
Procedure
Pipette into a series of clean Eppendorf tubes as indicated in the following table:
Test tube Standard gallic acid
solution (µL) Distilled water (µL) Concentration (mM)
1 100 900
2 50 950
3 10 990
4 40 (from test tube 3) 960
Mix well the four test tubes and prepare a second series of four test tubes and
number as the previous ones. In each test tube from the second series add a volume of 100
µL solution from the corresponding test tube from the first series. Then add in each test
tube of the second series 200 µL of a 10% (vol/vol) of Folin(Ciocalteu reagent solution and
mix. In the end add in each test tube 800 µL of the Na 2CO 3 solution and mix.
Incubate the assay test tubes at room temperature in dark for 2 hours.
For the blank use 100 µL of distilled water and add the same volumes of reagents.
Incubate in the same conditions.
In the case of the sample make a 1:9 dilution using distilled water. Take 100 µL of
the diluted sample and add the same volumes of reagents. Incubate in the same conditions.
123 Read the optical densities of the solutions at 765 nm. Substract the value obtained in
the case of the blank from the optical densities of all the other test tubes.
Laboratory report
In order to obtain the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (mM), while on the Oy axis plot the optical density (OD).
Calibration curve for the total phenol content assay
y = 0,9947x + 0,0108
R2 = 0,995300,050,10,150,20,250,3
0 0,05 0,1 0,15 0,2 0,25 0,3
Concentration (mM)OD
Use the equation of the curve in order to calculate the total phenol content from the
sample. The concentration from the sample is obtained by multiplying with the dilution
factor.
124 Experiment 35
Assay of total free thiols
Background
Free thiols is a general term used to define the biological compounds that have free
–SH groups. These compounds are classified as being low(molecular mass thiols (free
cysteine, glutathione, other peptides) and high(molecular mass thiols (proteins with a high
content of cysteine residues of which –SH groups are free).
These thiols are very important for biological processes including antioxidant
protection, gene regulation etc.
Glutathione is the most abundant non(protein thiol found in the biological systems.
It is present in all cells of the human body in concentrations that range from micromolar to
millimolar. For example, liver cells contain as much as 10 mM of glutathione. This is a
consequence of the fact that liver is involved in the biotransformation of many endo( and
exogenous (xenobiotics) compounds by conjugation with glutathione.
Inside cells, glutathione is found almost in its reduced form (GSH), but the oxidized
form (GSSG) is also present. GSSG is reduced back to GSH through the action of
glutathione reductase which uses NADPH produced by the pentose phosphate pathway.
R(O(OH + 2 GSH R(OH + H2O + GSSG
NADPH + H+
NADP+glutatione
peroxidase
glutathione
reductase
pentose phosphate
pathway
125 The assessment of reduced glutathione and the ratio of reduced to oxidized forms
are useful markers to evaluate the redox status of a tissue.
A simple procedure for the assay of total free thiols (proteic and nonproteic)
consists in the reaction with 5,5’(dithiobis((2(nitrobenzoic acid) (Ellman’s reagent).
O2N
HOOC SS
NO 2COOH R(SHO2N
HOOC SSRHS
NO 2COOH
+
Ellman reagent 2(nitro(5(thiobenzoic acid
The 2(nitro(5(thio(benzoic acid ionizes to a dianion in aqueous solution at alkaline
pH. This dianion has a maximum absorbance at 412 nm. This test is very sensitive to the
pH value.
S
NO2COO
Using a series of glutathione solutions with known concentration a calibration
curve can be obtained and used to assess the total free thiols content from a sample. The
result is expressed as glutathione equivalents.
Reagents and instrumentation
1. reagents: Na 2HPO 4, NaH 2PO 4, Na 2EDTA, reduced glutathione (GSH) (307.22 g/mol),
Ellman’s reagent (396.34 g/mol), Na 3citrate, sodium dodecyl sulfate (SDS).
2. solutions: GSH stock solution (disolve 6.25 mg of GSH in a volume of 25 mL distilled
water); Ellman’s reagent solution (disolve 19.82 mg Ellman’s reagent and 1 g of
Na3citrate in a volume of 25 mL of distilled water); sodium phosphate buffer 5 mM,
pH = 8 with 1 mM Na 2EDTA; SDS 10% solution (disolve slowly and without shaking
5 g of SDS in a volume of 45 mL of distilled water).
126 3. laboratory glassware: volumetric flasks, Berzelius flasks, test tubes, graduated cylinder,
pipettes.
4. instrumentation: analytical balance, spectrophotometer, pH meter, micropipettes.
Procedure
Pipette into a series of six clean test tubes as indicated in the following table:
Test tube GSH stock solution (µL) Buffer (µL) Concentration
(µmol/L) Optical
density (OD)
1 0 300
2 25 275
3 50 250
4 75 225
5 100 200
6 150 150
In each test tube add 2400 µL of the sodium phosphate buffer, 300 µL of the SDS
solution and 300 µL of the Ellman’s reagent solution. Mix gently in order to avoid foam
formation due to the presence of the SDS.
Incubate the test tubes for 1 hour at 37°C. After this time mix gently the test tubes
and read the optical densities of the solutions from all the test tubes at 412 nm. Then,
substract the value obtained in the case of the first test tube (blank) from all the other test
tubes.
The sample with unknown free thiols concentration is prepared in the same way.
Laboratory report
In order to obtain the calibration curve use the optical density values previously
obtained and the corresponding concentration values and make a chart in Excell. On the Ox
axis plot the concentration (µmol/L), while on the Oy axis plot the optical density (OD).
Use the equation of the curve in order to calculate the total free thiols concentration
in the sample. The results are expressed as reduced glutathione equivalents.
127 References
1. Bogdan Nicolae Manolescu. Lucrări practice de biochimie . Editura NICULESCU,
2014 .
2. Angela Popescu, Mihaela Iordache, Virgil Iordache. Biochimie generală. Îndrumar de
lucrări practice . Editura Universității POLITEHNICA București, 1996 .
3. Mihaela Zăuleț, Marieta Costache. Lucrări practice de biochimie și biologie
moleculară. Volumul I – biochimie . Editura Universității din București, 2011 .
4. Arti Nigam, Archana Ayyagari. Lab manual in bicohemistry, immunology, and
biotechnology . Tata McGraw(Hill Publishing Company Limited, 2007 .
5. Kurt M. Dubowski. An oătoluidine method for bodyăfluid glucose determination .
Clinical Chemistry, 8: 215(235, 1962 .
6. Joseph A. Knight, Shauna Anderson, James M. Rawle. Chemical basis of the sulfoă
phosphoăvanilin reaction for estimating total serum lipids . Clinical Chemistry, 18:
199(202, 1972 .
7. R.W. Burke, B.I. Diamondstone, R.A. Velapoldi, O. Menis. Mechanism of the
LiebermannăBurchard and Zak color reactions for cholesterol. Clinical Chemistry, 20:
794(801, 1974 .
8. Harold Edelhoch. Spectroscopic determination of tryptophan and tyrosine in proteins .
Biochemistry, 6: 1948(1951, 1967 .
9. Bradley J.S.C. Olson, John Markwell. Assay for determination of protein
concentration . Current Protocols in Protein Science, 48: 3.4.1(3.4.29, 2007 .
10. James E. Nobel, Marc J.A. Bailey. Quantitation of protein . Methods in Enzymology,
463: 73(95, 2009 .
11. William H. Habig, Michael J. Pabst, William B. Jakoby. Glutathione Sătransferase.
The first enzymatic step in mercapturic acid formation . Journal of Biological
Chemistry, 249: 7130(7139, 1974 .
12. Stefan Marklund, Gudrun Marklund. Involvement of the superoxide anion radical in
the autoxidation of pyrogallol and a consistent assay for superoxide dismutase .
European Journal of Biochemistry, 47: 469(474, 1974 .
128 13. Re R., Pellegrini N., Proeggente A., Pannala A., Yang M., Rice(Evans C. Antioxidant
activity applying an improved ABTS radical cation decolorization assay . Free Radical
Biology and Medicine, 26: 1231(1237, 1999 .
14. Elizabeth A. Ainsworth, Kelly M. Gillespie. Estimation of total phenolic content and
other oxidation substrates in plant tissues using FolinăCiocâlteu reagent . Nature
Protocols, 2: 875(877, 2007 .
15. Carla M. Stinco, Maria V. Baroni, Romina D. Di Paola Naranjo, Daniel A. Wunderlin,
Francisco J. Heredia, Antonio J. Melendez(Martinez, Isabel M. Vicario. Hydrophillic
antioxidant compounds in orange juice from different fruit cultivars: composition and
antioxidant activity evaluated by chemical and cellular based (Saccharomyces
cerevisiae) assays . Journal of Food Composition and Analysis, 37: 1(10, 2015 .
129 Contents
Experiment 1. pH and buffer solutions 6
Experiment 2. Preparation of a liver homogenate 18
Experiment 3. Molisch test for carbohydrates 22
Experiment 4. Fehling test for reducing carbohydrates 24
Experiment 5. Seliwanoff test for ketoses 27
Experiment 6. Bial test for pentoses 29
Experiment 7. The reaction with phenylhydrazine 31
Experiment 8. Assay of D(glucose with 3,5(dinitrosalicylic acid 34
Experiment 9. Assay of D(glucose with o(toluidine 36
Experiment 10. Assay of D(glucose with anthrone 38
Experiment 11. Assay of D(glucose with glucose oxidase 40
Experiment 12. Assay of D(glucose with hexokinase 43
Experiment 13. Liebermann(Burchard test for cholesterol 45
Experiment 14. Assay of total lipids with the phospho(vanillin reagent 47
Experiment 15. Assay of cholesterol with cholesterol oxidase 50
Experiment 16. Assay of triacylglycerols with glycerol(3(phosphate dehydrogenase 53
Experiment 17. Assay of amino acids with ninhydrin 56
Experiment 18. Xanthoproteic test for aromatic amino acids 59
Experiment 19. Separation of amino acids by thin layer chromatography (TLC) 61
Experiment 20. Assay of amino acids using the extinction coefficient 68
Experiment 21. Assay of total protein content through UV absorption spectroscopy 69
Experiment 22. Assay of total protein content through biuret assay (version I) 71
Experiment 23. Assay of total protein content through biuret assay (version II) 74
Experiment 24. Assay of total protein content through Lowry assay 76
Experiment 25. Assay of L(lactate 79
Experiment 26. Assay of uric acid 83
Experiment 27. Assay of urea 89
Experiment 28. Assay of total bilirubin 94
Experiment 29. Assay of aspartate aminotransferase (AST/GOT) 100
130 Experiment 30. Assay of glutathione S(transferase 102
Experiment 31. Assay of gamma(glutamyl transferase (γ(GT) 106
Experiment 32. Assay of xanthine oxidase 109
Experiment 33. Assay of total antioxidant capacity (TEAC assay) 112
Experiment 34. Assay of total phenol content with the Folin(Ciocâlteu reagent 120
Experiment 35. Assay of total free thiols 124
References 127
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