Protocols in Biochemistry and Clinical Biochemistry 1st Edition Buddhi Prakash Jain Shyamal K Goswami Shweta Pandey

Protocols in Biochemistry and Clinical
Biochemistry 1st Edition Buddhi

Prakash Jain Shyamal K Goswami
Shweta Pandey
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Protocols in
Biochemistry and
Clinical
Biochemistry
Protocols in
Biochemistry and
Clinical
Biochemistry
BUDDHI PRAKASH JAIN
Assistant Professor, Department of Zoology, School of Life Sciences, Mahatma Gandhi
Central University Bihar, Motihari, India
SHYAMAL K. GOSWAMI
Professor, School of Life Sciences, Jawaharlal Nehru University, New Delhi, India
SHWETA PANDEY
Govt VYT PG Autonomous College, Durg, Chhattisgarh, India
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Techniques
1. Centrifugation
G ¼ ω2 r
2. Electrophoresis
3. Spectrophotometer or
4. Chromatography
4π 2 s2 r
5. Titration G¼
3600
The unit of G is cm/s.
CENTRIFUGATION Relative centrifugal force is
In biology, centrifugation is a technique that utilizes the G
centrifugal force for the separation of the biological con- RCF ¼
g
tents from the liquid medium. The separation of the
or
contents depends on their shape, size, density, and the
viscosity of the medium. Analytical centrifugation is 4π 2 s2 r
RCF ¼
used for the analysis of the purified macromolecules ð3600Þð981Þ
and was first developed by Svedberg in the late 1920s.
or
Preparative centrifugation deals with the actual separa-
tion of biological contents. RCF ¼ 1:12 105 s2 r
Application of preparative centrifugation: Subcellular
According to the Stokes’ law, when a particle moves
fractionation, affinity purification of membrane
through a viscous medium, it experiences a frictional
vesicles, etc.
force:
Application of analytical centrifugation: Determination
of the purity of macromolecules and their relative F ¼ 6πrp ηv
molecular mass, detection of changes in conforma-
where
tion, and ligand-binding studies.
η ¼ viscosity of the medium
The process utilizes the principle that a particle, in a
ν ¼ velocity of the particle
liquid medium, will sediment much faster when placed
rp ¼ radius of the particle
under the influence of centrifugal force. While moving
Also, when particle sediment, it feels an upward
through a viscous medium, they experience a frictional
force. This force will be equal to the weight of water
force. This force acts in the opposite direction to its
displaced.
sedimentation.
4
Relative centrifugal force: F ¼ πrp3 ρp ρm ω2 r
3
F ¼ Mω2 r
where
where ρp ¼density of the particle
M ¼mass of particle ρm ¼density of the medium
r ¼ radius of the rotor r ¼ distance of the particle from the medium
ω ¼ angular velocity rp ¼ radius of the particle
Angular velocity in rad/s: ω ¼ angular velocity
2π s A nonhydrated spherical particle will accelerate
ω¼ under centrifugal force till the net force of sedimenta-
60
tion becomes equal to the frictional force.
where
Or
s ¼ speed of the rotor (revolution/min)
ω ¼ angular velocity (rad/s) 4 3
πrp ρp ρm ω2 r ¼ 6πrp ηυ
Applied centrifugal force (G) is 3
vii
viii TECHNIQUES
Thereafter, the particle will sediment at a constant Types of Centrifuges

velocity, There are various types of centrifuges, which differ from
each other in the following ways:
dr 2 2 ρp ρm ω r
2
1. The maximum speed that can be attained.
v¼ ¼ rp 2. The maximum volume of samples can be
dt 9 η
centrifuged.
or the time of sedimentation is
3. The presence or the absence of a vacuum.
9 η r 4. The presence or the absence of a refrigeration
t¼ ln b
2 ω2 r 2 ρ ρ
p p m
rt system, etc.
Some commercially available centrifuges are
where 1. Large capacity low-speed preparative centrifuges.
t ¼sedimentation time 2. Refrigerated high-speed preparative centrifuges.
η ¼viscosity of the medium 3. Small-scale laboratory microfuges.
rp ¼ radius of the particle 4. Large-scale clinical centrifuges.
rb ¼ radial distance from the center of rotation to the bot- 5. Analytical ultracentrifuges.
tom of the tube 6. Preparative ultracentrifuges.
rt ¼ radial distance from the center of rotation to the liq-
uid meniscus Types of Rotor
ρp ¼ density of the particle 1. Fixed angle rotor.
ρm ¼ density of the medium 2. Vertical rotor.
ω ¼angular velocity 3. Swinging bucket rotor.
Sedimentation rate of a particle is expressed as its
sedimentation coefficient, s, Types of Centrifugation
v 1. Differential centrifugation: In this method, the com-
s¼ ponents of various size, shape, and density are sepa-
ω2 r
rated according to the difference in their
sedimentation rate. In this method, the centrifugal
The sedimentation rate per unit centrifugal field can
field is gradually increased by increasing the speed
be obtained in various mediums as well as various
of the rotor. The medium of separation is homoge-
temperature, therefore the experimental values of
nous in this case.
the sedimentation coefficient corrected and expressed
2. Density gradient centrifugation: In this method, the
as sedimentation constant in water as media at 20°C
particles are separated according to their density in
and are denoted as S20,w. For biological particles, it
a medium that is nonhomogenous in its density.
is very low and can be expressed as the Svedberg unit,
The density of the medium increases toward the bot-
S. One Svedberg unit is equal to 1013 s (Fig. 1).
tom of the centrifuge tube. These can be of two types
Before centrifugation After centrifugation
Buoyant force and frictional force
Centrifugal force
Particles of various sizes Particles sedimented according to

are evenly distributed the net force applied on them.
FIG. 1 Pictorial representation of forces involved in centrifugation.
TECHNIQUES ix
of gradients—continuous and discontinuous. It has Ff ¼ frictional force

two types of variations: r ¼ radius of the particle
a. Rate zonal centrifugation: The gradient used has a η ¼ viscosity of the medium
maximum density less than that of the least dense ν ¼ velocity of the particle
particle. The density gradient is reasonably shal- Remember, it is an oversimplified situation, where
low. Centrifugation is done at a comparatively the pore size of the medium is not considered, and
low speed for a short time. After centrifugation, the particle is assumed to be spherical.
the particles form discrete zones depending on Therefore, the velocity of the particle in an electric
their sedimenting rate. The centrifugation is termi- field is
nated before any of the zones reach the bottom of 1. Directly proportional to the electric field applied and
the tube. It is used to separate particles that differ charge of the molecule.
in their size but not density. 2. Inversely proportional to the viscosity and its size.
b. Isopycnic centrifugation: In this method, the gra- Factors affecting the electrophoretic mobility:
dient used has a maximum density greater than 1. Size of the sample: Bigger the sample size greater is
the densest particle. The particles travel various the frictional force.
distances and become stationary when their den- 2. Charge of the sample: Higher the net charge of the
sity becomes less than that of the region below it. particle greater is the mobility.
It is used to separate the particle, which differs in 3. The shape of the particle: A globular protein (com-
their density. pact shape) will migrate faster than a fibrous protein.
4. Electric field: The rate of migration under unit poten-
tial gradient is known as mobility of ion. The current
ELECTROPHORESIS in the solution is conducted by ions of buffer and can
The migration of charged particles under the influence be expressed as
of the electric field is termed as electrophoresis. It was
V ¼ IR
first performed in 1861 by Quincke. There are various
biological macromolecules, for example, nucleotides, It shows that if the potential gradient increases, it will
nucleic acids, amino acids, peptides, proteins, etc., that increase the current and, therefore, the mobility of an
possess ionizable groups, which get ionized at a pH and ion. It seems the mobility of an ion increases with the
migrate when placed in an electric field. Certain cells increase in potential gradient. However, during electro-
such as bacteria and red blood cells also migrate in phoresis, power is generated in the medium of electro-
the electric field. The migration toward anode or cath- phoresis and is dissipated in the form of heat. This heat
ode depends on the kind of net charge on the molecule. generation will have the following effects:
When a spherical molecule, possessing a charge q, is o Increased rate of diffusion of buffer ions and sample,
under the influence of the electric field, then the force which leads to the broadening of the samples.
acting on it can be described as follows: o Convection current will form, which will mix the sep-
arated samples.
△Eq
F¼ o The decrease in viscosity and hence the resistance of
d
the medium.
where o Heat-sensitive materials may get affected, e.g., dena-
ΔE¼potential difference between the two electrodes turation of proteins can occur.
q ¼net charge of the molecule Therefore, it is a better option to experiment with a
d ¼ distance between the electrodes constant power supply.
While migrating, the particle will also experience a 5. The medium used: It should not lead to the adsorp-
frictional force, opposite to the direction of its migration of the sample. Similarly, molecular sieving of
tion. This frictional force depends on the size and shape the medium affects the migrations of the molecules.
of the molecule, pore size of the medium in which elec- The medium generally used are agarose and
trophoresis occurs and the viscosity of the medium. polyacrylamide.
Neglecting the pore size of the medium, it can be 6. Composition of the buffer: The choice of the buffer
expressed as depends on the sample to be separated. For example,
Ff ¼ 6πrηϑ carbohydrates that are uncharged can be separated
using borate buffers, which interact with carbohy-
where drates to form charged complexes.
x TECHNIQUES
Direction of migration of nucleic acids
Negative Positive
electrode − +
electrode
Buffer
Gel
Wells for Electrophoresis

samples tank
FIG. 2 Electrophoresis apparatus.
7. The pH of buffers: pH affects the ionization of the

sample molecules and, therefore, their rate and I0
ln ¼ kb
sometimes direction of the sample molecules (in I
case of ampholytes). 2:303 log 10 ¼ kb
8. Electroendosmosis due to the presence of charged According to the Beer’s law, the amount of light
groups on the surface of the support medium too absorbed by a material is directly proportional to the
can affect the rate of migration of the particles. Aga- number of molecules of the absorbent, or in other
rose can contain sulfate groups; paper has carboxyl words, the concentration of the absorbing solution. This
groups and in capillary electrophoresis, the surface can be represented as
of glass comprises silanol (Si-OH) groups (Fig. 2).
I 0
¼ ek c
I0
SPECTROPHOTOMETER or
It is an instrument that measures the amount of light
2:303 log 10 ¼ k0 C
absorbed by a sample. It is mainly used to measure
the concentration of a sample. Both equations can be combined as
Light has dual characteristics, particle as well as wave I0
nature. Any object that absorbs light, obeys two laws— log 10 ¼ abC
I
Bouguer-Lambert law and Beer’s law.
where
Bouguer-Lambert law—the amount of absorbed by a
a ¼combined constant
material is directly proportional to the thickness of the
b¼ path length
material. It is independent of the intensity of the inci-
C¼ concentration of the absorbing material
dent light. It can be represented as
This equation is Beer-Lambert law, and it states that
I the amount of the light absorbed is directly propor-
¼ ekb
I0 tional to the concentration as well as path length of
where the absorbing material. I/I0 is known as transmittance,
I ¼intensity of the transmitted light it is the amount of light that escape absorption by the
I0 ¼ intensity of the incident light material. I0/I is known as absorbance or optical
k¼ linear absorption coefficient of the absorbing density (OD).
material It is easier to use absorbance as an index of concen-
b¼ thickness of the absorbing material, also known as tration as it varies linearly with concentration, while
path length transmittance varies with a concentration in a
The above equation can also be written as nonlinear manner.
Absorbance (A) can also be written as
I
ln ¼ kb I0
I0 A ¼ log 10 ¼ abC
I
or
TECHNIQUES xi
where a is the absorptivity or absorption coefficient or 5. Fluorescence: Some solutes, especially drugs or their
extinction coefficient. intermediates fluoresce, and their fluorescent inten-
Specific absorption coefficient—when path length is sity are also detected by the spectrophotometer.
expressed in cm and concentration in g/L, then the Absorption spectrum—it is the pattern of energy
absorption coefficient is termed as a specific absorp- absorption by a substance, when, the light of varying
tion coefficient. wavelength passes through it. It is a unique characteristic
Molar absorption coefficient—when path length is of the substance as every substance is made up of mol-
expressed in cm and concentration in mol/L, then ecules and each element/ion of the molecule has unique
the absorption coefficient is termed as the molar arrangements of an electron in their orbits/orbitals.
absorption coefficient and is denoted as am. When light is passed through them, they absorb energy,
according to their electronic configuration and their
am ¼ as Mw ðmolecular weightÞ electrons get excited. These electrons get promoted to
a higher energy level. According to the quantum theory,
For a given compound, if the solvent and wavelength
the electron gets excited only after accepting the radia-
are defined, then molar absorptivity is a physical con-
tion, which has exact quantized energy that can push
stant. Molar absorptivity of some compounds are listed
the electron to a permitted energy level. The wavelength
as below:
at which it absorbs the maximum amount of light is
Compound Solvent λmax am × 1023 λmax. When the concentration of any substance is
Adenine Water 260 13.3 increased, the absorbance at all wavelength increases,
NADH Water 340 6.22 although the change in absorbance per unit change in
ATP Water 260 15.4 concentration is maximum at λmax. Therefore, during a
FAD Water 445 11.3 quantitative experiment, absorbance of a compound
is measured at λmax.
When a sample is placed in a cuvette and light is passed Factors that can affect the absorption spectra are
through it, we can calculate the concentration of the polarity (affect the transition states of the electrons)
sample. When path length is constant, then the optical and pH of the solvent (it can change the ionization state
density is directly proportional to the concentration of of the molecules) as well as the relative orientation of
the sample, it can be written as follows: the neighboring absorbing groups.
Some isolated covalently bonded groups characteris-
OD1 OD2
¼ tically absorb the light in esters, carbonyl, and nitrile
C1 C2
group of ethylenic or acetylenic groups. Similarly, there
where are auxochromes that themselves don’t act as chromo-
OD1 and OD2 ¼ optical density of sample 1 and 2, phores, but their absence or presence shifts the absorp-
respectively tion spectrum toward longer wavelengths. They are also
C1 and C2 ¼ concentration of samples 1 and 2, known as color enhancers, e.g., OH, OR, SH, NH2
respectively groups. They can share the nonbonding electrons by
Deviations from Beer-Lambert law: extending the conjugation.
1. When the concentration of the reagent is high: It may There are different conditions like change in the
lead to dimerization or polymerization of the reagent. polarity of the solvent or the presence of chromophore,
OD of monomer differs from polymers. High concen- etc., which can shift the absorption spectrum in four dif-
tration may also lead to aggregation leading to the for- ferent directions and result in the following shifts:
mation of aggregates, which scatter light. 1. Bathochromic shift: Due to the presence of auxo-
2. Temperature: On heating, the solvents may expand. chrome, the shift is toward the higher wavelength.
The change in the degree of solubility, dissociation/ This is also known as the redshift.
association of solutes, and hydration of solutes may 2. Hypochromic shift: The shift is toward shorter wave-
vary according to the temperature. These changes can length, due to the removal of conjugation or the
lead to variation in absorbance. change in the polarity of the solvent. This is also
3. Turbidity: A turbid solution absorbs more light. known as a blueshift.
4. Sample instability: Some colored complexes are 3. Hyperchromic shift: This results in an increase in the
unstable and their intensity can increase or decrease intensity of the absorption maximum. It results in a
with time. For example, the ANSA method of phos- higher extinction coefficient. This occurs mostly due
phate determination. to the presence of auxochrome.
xii TECHNIQUES
Body of
spectrophotometer
1 2 3 4
Lid
Slot for
Display unit cuvette
1 2 3
Absorbance of a solution depends on its concentration and path length: (a) As the absorbance depends on the
concentration, the absorbance of the tube 4 will be highest and that of the tube 1 will be lowest. (b) (a) As the
absorbance depends on the pathlength, the absorbance of the tube 3 will be highest and that of the tube 1 will be
lowest.
FIG. 3 Spectrophotometer.
4. Hypochromic shift: The shift of absorption maxi- The term partition coefficient or distribution coeffi-
mum to a lower value. It occurs due to the introduc- cient (Kd) is used to describe the mode of distribution
tion of the groups that is appropriate result in the of a compound between two immiscible phases. At a
distortion of the geometry of the absorbing molecule given temperature, when a sample is allowed to equili-
(Fig. 3). brate itself between two different liquids that are immis-
cible and are present in equal volumes, the ration in
which it does the distribution is known as partition coef-
ficient. It is the concentration of a component present in
CHROMATOGRAPHY one phase divided by that in another phase. It can be
The first account of chromatography was given by expressed as
Michael Tswett in 1906. In the experiment, chlorophyll
was separated from a mixture of plant pigments, the col- Concentration in phase A
Kd ¼
ored pigments and chlorophyll get separated forming Concentration in phase B
distinct colored bands. Hence, the term chromatogra- The effective distribution coefficient is the total amount
phy was given to the separation technique. It is a of the analyte in a phase divided its total amount in
method of separation of an analyte between two immis- another phase. It also considers the volume of the
cible phases. One of the phases remains stationary, phases, for example, if the distribution coefficient of
while the other is mobile. The mobile phase usually an analyte between A and B phases is 1, but the total vol-
moves over the surface of the stationary phase. It can ume of phase A is 5 times more than that of B, then the
also percolate through the interstices of the stationary effective distribution coefficient is of the analyte is 5. Its
phase. A sample is introduced in the system with the amount in phase A is 5 times more in A than B. In the
mobile phase and the components of the sample inter- technique, the stationary phase (immobilized) can be
act with two phases in a differential manner, depending solid, gel, liquid, or solid/liquid mixture, while the
on their properties. Each component migrates at a differ- mobile phase is liquid or gas (Fig. 4).
ent rate based on the rate of interactions between the
components of the sample and each of the phases. Types of Chromatographic Techniques
The component showing the least interaction with the Based on the nature of the support that is used to hold
mobile phase but strong interaction with the mobile the stationary phase, the techniques can be classified as
phase will migrate slowly. The separation is done by follows:
the virtue of the differential movement of the compo- 1. Plane chromatography: The following are the types
nents of the sample. of plane chromatography:
TECHNIQUES xiii
Support Support
Pin Pin
Paper for Paper for

chromatography chromatography
Beaker Beaker
Spot of sample/ Resolved

Direction origin line components
Direction of
of flow Solvent flow Solvent
FIG. 4 Chromatography.
a. Paper chromatography: In this technique, the sup- In the case of an anion exchanger, the negatively
port for the stationary phase is paper, i.e., cellu- charged sample will interact with the stationary
lose. The separation is done by adsorption or phase, while the positively charged and neutral mol-
partition principle. ecules will elute first.
b. Thin-layer chromatography: In this technique, the 4. Molecular sieve chromatography: In this chromatog-
glass plate coated with a layer of silica (partition raphy, a gel consisting of porous beads is the support
principle), Kieselguhr, and alumina (adsorption medium. The molecules smaller than the pore size
principle). will get entrapped in the pore of the gel, while the
2. Column chromatography: In this, the stationary larger molecules will not and thus get eluted first.
phase is packed in a column of glass or metal. 5. Affinity chromatography: In this chromatography,
The mixture of analytes is applied to the column the specificity of an enzyme toward its substrate or
and the mobile phase passes through it. The substrate analogs. The technique utilizes the specific-
mobile phase is termed as eluent. The stationary ity of an enzyme for its substrate (also receptor for its
phase is coated on the matrix and packed in the agonist, the antibody for an antigen) or substrate
column or is applied on the walls of the column. analog for the enzyme’s (other proteins with biolog-
The eluent flows through the column and separates ical specificity) separation. A substrate analog is
the sample according to the partition coefficient of coupled to the gel matrix and the cellular suspension
the components in the sample. The components can percolate through. The enzyme which is specific
that leave the column individually are termed as for the substrate analog binds to the gel becoming
an eluate. immobile, while all other components move down
Based on the interaction phenomenon between sta- and out. The technique has a very high-
tionary and mobile phase, the technique can be classi- resolution power.
fied in the following types:
1. Partition chromatography: In this, the separation
involves a large number of partition steps on the TITRATION
granules of an insoluble hydrated inert substance, It is a method to determine the concentration of an
e.g., silica gel or starch. The granules are hydrated identified analyte. It is a volumetric analysis, i.e., a
and hold water in a very tightly bound manner. method in which the amount of a solution is deter-
Hence, water is the stationary phase. mined by measuring its volume. In this process, a vol-
2. Adsorption chromatography: In this chromatogra- ume of a solution of known strength is added to
phy, the components are absorbed with the help of another solution to complete the reaction. There are a
weak electrostatic forces or hydrogen bonding on few important terms used in the procedure, which can
the surface of the stationary phase. be listed as below:
3. Ion exchange chromatography: In this chromatogra- 1. Titer: It is the weight of solution in 1 mL of solution
phy, an ion exchanger consisting of inert support or weight of a substance that will react with 1 mL of
covalently coupled with negative (cation exchanger) solution or is equivalent to 1 mL of solution.
or the positive (anion exchanger) functional group. 2. Standard solution: A solution of an accurately known
The support can exchange the charged functional concentration is known as the standard solution.
group with that of the components of the sample. They can be of two types—primary standard
xiv TECHNIQUES
solutions—these solutions are prepared directly by Types of reaction in titrimetry:

mixing the salt and solvent, e.g., N/10 oxalic acid. 1. Neutralization reaction: The reaction between acid
Secondary standard solutions—these are prepared and base.
after standardization through the reaction with pri- 2. Acidimetry: The reaction in which the amount of
mary standard, e.g., N/7 NaOH. base is determined by titrating it against a standard
3. End point: The point at which titration is stopped, it acid solution.
is indicated by the change in color of an indicator. 3. Alkalimetry: The reaction in which the amount of
The indicator that is used depends on the pH at acid is determined by titrating it against a standard
which the indicator changes the color. base solution.
4. Indicator: A high molecular weight substance that 4. Precipitation and complexation reactions: A reaction
indicates the physical and chemical status of a reac- which forms a precipitate, or a complex approaches
tion. When dissolved in water, they act as a weak acid the completeness. It occurs due to the removal of
or base. The acidic indicator comprises a colored ions from the solution.
anion, while the basic indicator comprises a colored 5. Redox reactions: Reactions involving the transfer of
cation. electrons from one molecule to another.
Common indicators used are Common equations useful in titration are
Volume Normality ¼ Gram equivalent weight
and
Color in Color in
The volume of solution 1 ðV1 Þ Normality of solution 1 ðN1 Þ
pH Acidic Basic Concentration
¼ Volume of solution 2 ðV2 Þ Normality of solution 2 ðN2 Þ
Indicator Range Medium Medium Used
Methyl orange 3.1–4.4 Orange- Orange- 0.1% In the titration process, the volume of a solution of
red yellow unknown strength is taken in a burette and this is
Methyl red 4.2–6.2 Red Yellow 0.2% in 95% titrated against a known volume of known strength in
alcohol
a flask. The flask also contains a few drops of indicator.
Phenolphthalein 8.2–10.0 Colorless Pink 1 g in 110 mL
95% alcohol
As the reaction reaches the end point, the color of the
and 90 mL solution in the flask changes, and the volume of the
of water solution of unknown strength (solution in the burette)
Bromocresol 3.6–5.2 Yellow Blue 0.04% in is noted. By applying the above formula, the strength of
green alcohol the solution can be obtained.
CHAPTER 1
Solutions and Acids and Bases
The solution is a homogenous mixture of two or more But it is true if the conc. HCl is 100%, but in the bot-
components. Solutions that are made up of two compo- tle, the conc. HCl is 37.23%, so the volume needed is
nents are binary. The component in which something is 82.38 mL in 1 L of solution.
dissolved or, in other words, the component which is in
larger quantity is known as a solvent; the component
which is dissolved or, in other words, which is present MOLALITY (M)
in smaller quantity is the solute. A solution containing 1 mol of solute in 1 kg of a solvent
is known as 1 molal solution.
Solute + Solvent ¼ Solution
No:of moles of solute
Each component may be in any state, i.e., solid, liquid, Molality ¼
Weight of solvent in kg
or gaseous state. The strength of any solution is very
important in doing any experiment and it has to be For example,
made cautiously. The strength of a solution can be To prepare 1 m solution of NaOH in water, we need
defined as the amount of solute dissolved per unit solu- 1 mol of NaOH in 1000 g of water.
tion or solvent. It can be expressed in various terms. As specific gravity of water is 1,
1 g water ¼ 1 mL of water.
1 mol of NaOH ¼ 40 g of NaOH.
MOLARITY (M) Now, if we add 40 g of NaOH in 1000 mL of water, it
It is the most common way to represent the strength of will result in a 1 M solution of NaOH in water.
any solution. It is defined as the no. of moles of a solute Note that molarity refers the total volume of solution
per liter of the solution. (solvent + solute) whereas molality refers the weight of
the solvent.
No:of moles of solute
Molarity ¼
Volume of the solution in liters
For example, NORMALITY
To prepare a solution of 2 M NaOH in 100 mL of It is the number of gram equivalents of the solute that
solution, we need 2 M of NaOH dissolved in 100 mL has been dissolved in 1 L of solution. Gram equivalent
of the solution. or equivalent weight of a given substance is the mass
1 mol NaOH ¼40 g of NaOH of a substance that combines with or displaces a fixed
2 mol NaOH ¼80 g of NaOH quantity of other substances. This can be defined as
Now, if in 1000 mL of solution, 80 g of NaOH is dis- the following:
solved then 2 M solution could be obtained. 1. For an element the mass of the element that com-
For 100 mL of solution, 8 g of NaOH is needed. bines with/displaces 1.008 g of hydrogen, 8.0 g of
But in case of liquid it differs as here density comes oxygen, or 35.5 g of chlorine. It can be obtained by
into play: dividing the atomic weight of an element by its
Preparation of 1 M HCl: valency.
MW of HCl is 36.5 g. 2. For acids and bases, it is the mass of acid or base that
So, if 36.5 g of HCl is dissolved in 1 L of solution then provides or reacts with 1 mol of H+.
1 M HCl could be obtained. The unit of equivalent weight is g.
The specific gravity of HCl is 1.19, which means, For example, in the case of NaCl: anion is Na+ and
1 mL of HCl weigh 1.19 g. cation is Cl. The total charge on anion or cation is 1.
So, the volume needed is mass/specific gravity or So, the equivalent weight is (23 +35.5)/1 ¼ 58.5.
36.5/1.19. In the case of Al2(SO4)3: anion is two molecules of
The volume of HCl needed is 30.67 mL. Al3+ and cation is three molecules of (SO4)2.
Protocols in Biochemistry and Clinical Biochemistry. https://doi.org/10.1016/B978-0-12-822007-8.00009-X

© 2021 Elsevier Inc. All rights reserved. 1
2 Protocols in Biochemistry and Clinical Biochemistry
The total charge on either of them is 3 2.

So, the equivalent weight is 342.15/6 ¼ 57. HCl!H + + Cl
Gram equivalent of solute

Normality ¼ HNO2 Ð H + + NO
2
Volume of the solution in a liter
Base: The compound that can produce hydroxide ion
Mass Concentration (OH) when dissolved in water, e.g., NaOH.
Mostly, the macromolecules like nucleic acids, proteins
NaOH ! Na + + OH
do not have a uniformly defined composition. In that
case, their concentration is defined as weight per unit 2. Bronsted-Lowry theory
volume or in terms of percentage. According to this theory, any compound that yields
proton (H+) is acid and any compound that accepts
the proton is base. So, according to this theory:
PERCENT
Part of solute dissolved to form a hundred parts of the Acid Base
solution. HCl H+ Cl
For example, 2% NaOH (w/v) means, 2 g of NaOH H2SO4 H+ SO4 2
in 100 mL of solution. HNO3 H+ NO3
Similarly, 2% HCl means,
2 g of HCl in 100 mL of solution (w/v). Acid and its base are said to be conjugated. A strong acid
2 g of HCl in 100 g of solution (w/w). has a weak conjugate base, and weak acid has a strong
2 mL of HCl in 100 mL of solution (v/v). conjugate base. Similarly, the weak base has a strong
conjugate acid and a strong base has a weak conjugate
acid.
MASS FRACTION 3. Lewis theory: According to this theory, acids are the
Mass fraction of a component in the solution means the compounds that accept a pair of electrons while
mass of the component per unit mass of the solution. bases donate a pair of electrons, e.g., NH3 is a base
Let component A of the solution has mass Wa and while BF3 is an acid.
component B has Wb.
Then, Electrolytes
Electrolytes are the compounds that form ions when dis-
Wa
Mass fraction of A ¼ solved in water or any polar liquid, e.g., NaCl. Major
Wa + Wb
biomolecules in our cells, e.g., peptides, amino acids,
Similarly, nucleotides, nucleosides, and nucleic acids, are weak
Wb
electrolytes and dissociates partially in aqueous solu-
Mass fraction of B ¼ tions. The biological function of most of these biomol-
Wa + Wb
ecules depends on the pH. The pH of a solution is the
measure of the alkalinity or acidity. It is the negative
If a substance is present in a very low amount the term
of the logarithm of the molar hydronium (or hydrogen)
parts per million is used. It is the gram of solute per mil-
ion concentration.
lion grams of the solution.
g or mL of solute pH ¼ log ½H3 O + ¼ log ½H +
Conc of solute in ppm ¼ 106
g or mL of solution
Water is the most important weak electrolyte and par-
tially exists in its ionized form:
Acids and Bases
There are three major theories describing acids and H2 O Ð H + + OH
bases. +
The H ion again reacts with water to form hydronium
1. Arrhenius theory ions (H3O+):
Acid: According to Arrhenius, the compound that
H + + H2 O Ð H3 O +
ionizes to produce a proton (H+) when dissolved in
water, e.g., HCl. The equilibrium constant of water at 25°C is 1.81016.
HCl is strong acid while HNO2 is a weak acid. So,
CHAPTER 1 Solutions and Acids and Bases 3
½H + ½OH Hasselbalch equation is very important. According to

Keq ¼ ¼ 1:8 1016
½H2 O the equation, for weak acid:
The molarity of pure water is 55.6 M. So, ½conjugated base
pH ¼ pKa + log
½ weak acid
½H + ½OH
1:8 1016 ¼
55:6 or
or ½ionized form
pH ¼ pKa + log
16 14 ½unionized form
½H ½OH ¼ 1:8 10
+
55:6 ¼ 1 10 ¼ Kw
Similarly, for a weak base:
Kw is also known as the autoprotolysis constant of ½weak base
water. It is specific for 25°C and depends on pH ¼ pKa + log
½conjugated acid
temperature.
In the case of acids and bases, or
Ka + Kb ¼ Kw ½unionized form
pH ¼ pKa + log
½ionized form
where Ka and Kb are equilibrium constant of acid and
base, respectively. Buffering capacity is the ability of a buffer to resist a
For example, in the case of acetic acid (CH3COOH) change in pH when an acid or base is added to it. It is
denoted as β and can be defined as the moles of acid
CH3 COOH Ð CH3 COO + H + or base which is required to change the pH of the buffer
and by one unit.
½CH3 COO ½H + ½conjugated base½H db da
Ka ¼ ¼ β¼ ¼
½CH3 COOH ½weak acid dpH dpH
5
¼ 1:75 10 and its pKa is 4:75 where db and da are the amount (moles) of base and
Further, acid, respectively, and dpH is the change in pH.
CH3 COO + H2 O Ð CH3 COOH + OH Problem

and Calculate the pH of 0.01 M of a weak acid with Ka ¼ 1.75
105.
½CH3 COOH½OH ½weak acid½OH
Kb ¼ ¼
½CH3 COO ½conjugated base
Solution
¼ 1:77 1010 and its pKb is 9:25 Let the formula of weak acid be HA.
On multiplying both we get Then,
Ka Kb ¼ ½H + ½OH ¼ Kw ¼ 1 1014 HA Ð H + + A
and ½H + ½Cl
Ka ¼ ¼ 1:75 105
pKa + pKb ¼ pKw ½HA
Stronger acid will have a smaller numerical value of pKa HA will dissociate to form equal amounts of ions H+
(and larger Ka) than a weaker acid. The smaller the pKa and Cl.
value, the more it will be in ionized form. Let x be the amount of H+ forms. Then Cl will be x
and HA will be 0.01 x M.
Buffer ½x½x
1:75 105 ¼
A solution that resists change in its pH, upon the addi- ½0:01 x
tion of acid or base, is known as a buffer solution.
Mostly they are weak electrolytes, due to this their ionic 1:75 107 1:75 105 x ¼ x2
status varies with pH. In an aqueous solution, they are In this equation we can neglect x as its value is very
either a mixture of a weak acid with its conjugate base small, so the above equation can be written as
or a weak base with its conjugate acid. For the prepara-
tion of buffer solution or to estimate its pH Henderson- 1:75 107 ¼ x2
or get precipitated with ethanol, so cannot be used in

DNA and RNA precipitation process.
x ¼ 4:18 104
2. Tris buffer: It is the most used buffer in biochemical
pH is log [H+], so, experiments. pKa of tris base is 8, therefore it has a
high buffering capacity between 7.5 and 8.5. It does
pH ¼ log 4:18 104
not interfere with most of the biochemical reactions.
It has very low toxicity. It is highly polar and thus sol-
pH ¼ ð4 + log 4:18Þ
uble in aqueous solution; therefore, it cannot diffuse
or across the biological membrane and cannot affect
pH ¼ 3:38
the intracellular pH. It is not susceptible to salts. It
does not absorb any light in the visible or ultraviolet
region. However, the dissociation of tris is affected by
Question 2 temperature and concentration of tris, e.g., with a 1°
The pH of a solution, which is a mixture of acetic acid C increase in the temperature, the pH of the solution
and sodium acetate, is 5.06. The concentration of free decreases by 0.03 units and 10 mM and 100-mM tris
acetic acid in it is 0.1 N and that of sodium acetate is solution differ in pH by 0.1 unit of pH. The higher
0.2 M. Calculate pKa. the concentration of tris the higher the pH. It reacts
with some metal ions like Ca2+, Cu+, Ni2+, etc. (like
Answer 2 phosphate buffers). It also reacts with glass electrode
According to Henderson-Hasselbalch equation hence can lead to error in pH reading. Tris reacts with
½conjugated base various fixatives like glutaraldehyde, formaldehyde,
pH ¼ pKa + log and glyoxal.
½weak acid
3. Carbonated buffer: The disadvantage is that most
In this case, metal carbonates are insoluble in water. Its pH is sen-
½acetate sitive to temperature. It works in the range 10–10.8.
pH ¼ pKa + log 4. Glycylglycine buffer: It is often used in enzymologi-
½acetic acid
cal experiments. Its best working range is 7.5–8.0. It
½0:2 shows no affinity toward divalent cations like Ca2+
5:06 ¼ pKa + log
½0:1 and Mg2+. It also has very low UV absorbance. How-
ever, being a peptide, it can be cleaved by protease
pKa ¼ 5:06 log ½0:2 and hence cannot be used in reactions involving pro-
teases and crude protein preparation.
pKa ¼ 5:06 0:301 5. Triethanolamine buffer: This too is majorly used in
enzymological experiments. Like the above buffer,
pKa ¼ 4:76 its best working range is 7.5–8.0 and shows no affin-
In an experiment, the type of the buffer used depends on ity toward divalent cations like Ca2+ and Mg2+. It too
the pKa value. In addition to this, the properties of the has very low UV absorbance and unlike glycylglycine
buffers are also considered in the selection of buffers buffer, it is not cleaved by proteases. However, the
for experiments. The properties of a few buffers are dis- buffer is volatile and is therefore suitable for experi-
cussed in the following: ments in which buffer is ultimately removed, e.g.,
1. Phosphate buffers: They have high buffering capac- purification experiments.
ity, both Na+ and K+ are highly soluble in water so 6. Good buffers: Discovered by Norman Good. These
any ratio of Na+ and K+ can be used to prepare the are the best suitable buffers for a variety of molecular
buffer. Even with low molarity, a solution of high biology work. They are not toxic, do not precipitate
ionic strength can be obtained. However, it is not divalent ions, do not absorb in the UV range, and
possible to form a phosphate buffer solution of high their pH is not sensitive to the change in temperature.
buffer capacity with low ionic strength. Another dis- Their name is long so is written in abbreviations, e.g.,
advantage of using phosphate buffers is that they MOPS [3-(N-morpholino) propane sulfonic acid]—
may bind polyvalent cations, especially Ca2+ ions. working range 6.5–7.9 and PIPES
They are known to be toxic to mammalian cells. They (1,4-piperazinediethanesulfonic acid)—working
lack buffering capacity in the range of 7.5–8.0, they range 6.4–7.2.
CHAPTER 2
Carbohydrate
A. Qualitative test for carbohydrates 19. Extraction and analysis of soluble carbohydrates
1. Determination of the presence of carbohydrates from plants.
in the given sample by Molisch’s test.
2. Determination of the presence of reducing carbo- QUALITATIVE TEST FOR CARBOHYDRATES
hydrates in the given sample by picric acid test. Definition
3. Determination of the presence of reducing car- Determination of the presence of carbohydrates in the
bohydrates in the given sample by Fehling’s test. given sample by Molisch’s test.
4. Determination of the presence of reducing carbo-
hydrates in the given sample by Benedict’s test.
Rationale
5. Determination of the presence of reducing carbo-
All types of carbohydrates—monosaccharides (except
hydrates in the given sample by Tommer’s test.
triose and tetrose), disaccharides, and
6. Determination of the presence of reducing car-
polysaccharides—get dehydrated upon treatment with
bohydrates in the given sample by
conc. sulfuric acid or hydrochloric acid and produce
Nylander’s test.
an aldehyde. This aldehyde condenses with two mole-
7. Distinguish between monosaccharides and
cules of α-naphthol to form a purple-colored ring.
reducing disaccharides by Barfoed’s test.
8. Distinguish between aldose and ketose by conc:H2 SO4 or HCl
Pentose sugar ! Furfural
Seliwanoff’s test.
conc:H2 SO4 or HCl
9. Determination of pentose sugar by Bial’s test. Hexose sugar ! 5 Hydroxymethylfurfural
10. Determination of the presence of galactose in a
Both aldehydes (furfural and 5-hydroxymethylfurfural)
test sample by the Mucic acid test.
condense with Molisch’s reagent (10% α-naphthol in
11. Differentiation between the presence of keto-
ethanol). Other phenols such as thymol and resorcinol
hexose and aldohexose in the test sample by
also give a colored product. This test results in the
Foulger’s test.
purple-colored product by all carbohydrates that are
12. Determination of the presence of reducing car-
larger than tetrose. Nucleic acid and glycoproteins too
bohydrates in the given sample by Fearon’s
give a positive test. In this test, oligosaccharides or poly-
methylamine test.
saccharides hydrolyzed to form monosaccharides,
13. Confirmation of the presence of reducing carbo-
which ultimately react as above.
hydrates in the given sample by osazone
formation test. Molisch’s reagent
Furfural=5 hydroxymethylfurfural ! Purple
14. Determination of the presence of polysaccha- colored dye
rides in the given sample by iodine test.
B. Quantitative test for carbohydrates
15. Quantitative estimation of carbohydrates in the Materials, equipment, and reagents
given sample by anthrone test. A. Reagents: Molisch’s reagent (10% α-naphthol in eth-
16. Quantitative estimation of reducing carbohy- anol), test sample (sugar solution), concentrated
drates in the given sample by 3,5-dinitro salicylic H2SO4.
acid (DNSA) test. B. Glassware: Test tube, test tube holder, dropper.
C. Other tests for carbohydrates
17. Extraction and estimation of glycogen from the Protocol
liver and muscle of the well-fed and starved rat. 1. Take 2 mL of the sample solution in a test tube.
18. Separation and identification of sugars present 2. To this add 2–3 drops of Molisch’s solution.
in fruit juices using thin-layer chromatography 3. Add 1 mL of conc. H2SO4 or HCl along the sides of
(TLC). the test tube, so that two layers are formed.
Protocols in Biochemistry and Clinical Biochemistry. https://doi.org/10.1016/B978-0-12-822007-8.00007-6

Analysis and statistics Pros and cons

The purple ring indicates the presence of sugar/ Pros Cons
carbohydrates. The easy and primary Trioses and tetrose do not give
method to test the positive result of this test
presence of carbohydrates
Precursor techniques
1. Molisch’s reagent: Add 10 g of α-naphthol to 95% Alternative methods and protocols
ethanol. DNS method, anthrone method.
2. Sugar solution: If sugar is in powdered form add 1 g
of sugar to 100 mL of water. Summary
1. The carbohydrates get dehydrated to form an alde-
hyde when treated with concentrated acid.
Safety considerations and standards 2. The aldehyde formed by dehydration is condensed
1. Concentrated acids should be handled very carefully, with Molisch’s reagent to form a purple-colored ring.
one must wear gloves while handling it. 3. Nucleic acids and other biomolecules that contain
2. Always add acid to water. sugars too yield color in the test.
MOLISCH TEST
Purple ring
Acid
along the
Molisch side of Positive
solution test tube Molisch test
Sample
+
NaOH
Sample +
Sample + Acid
Molisch
Negative
Molisch test
CHAPTER 2 Carbohydrate 7
Definition Analysis and statistics

Determination of the presence of reducing carbohy- Reducing sugar is present in the solution.
drates in the given sample by picric acid test.
Rationale 1. Sugar solution: If sugar is in powdered form add 1 g
Reducing sugars react with picric acid, i.e., 2,4,6-
of sugar to 100 mL of water.
trinitrophenol (yellow-colored solution) in alkaline
2. Saturated picric acid solution: 1.3 g of picric acid dis-
solution and reduce it to picramic acid, i.e.,
solved in 100 mL of water.
2-amino-4,6-dinitrophenol (red-colored solution.
3. 10% sodium carbonate solution: Add 10 g of
Reducing sugars alkaline solutions Na2CO3 to 100 mL of distilled water.
Picric acid ðyellowÞ
! Picramic acid ðredÞ
The sugars which have a free aldehyde or ketone group, Safety considerations and standards
in other words if the aldose C1 and ketose C2 are not 1. Picric acid is a highly acidic solution with pH 2 and
involved in any bond formation, are known as reducing hence should be handled very carefully.
sugars. The free aldehyde or ketone group enables them
to work as reducing agents. All monosaccharides are
reducing sugars while disaccharides can be classified Pros and cons
as reducing or nonreducing.
Picric acid is reduced under the alkaline condition in Pros Cons
the presence of reducing sugar. This results in the change Easy and sensitive Indicates the presence of reducing
of color of the solution from yellow to red. experiment sugars only
Materials, equipment, and reagents

A. Reagents: Test sample (sugar solution), sodium car-
bonate (10%) solution, saturated picric acid Alternative methods/procedures
solution. Fehling’s test, Benedict’s test, Tommer’s test.
B. Glassware: Test tube, test tube holder, dropper.
C. Instrument: Burner.
Summary
Protocols 1. In the test, reducing sugars reduces picric acid (yel-
1. Take 1 mL of test solution in a test tube. low) to picramic acid (red) in the presence of alka-
2. To this add 1 mL of saturated picric acid followed by line conditions.
the addition of 0.5 mL of sodium carbonate solution. 2. Both monosaccharides and reducing sugars yield a
3. Mix the above gently and heat for 10–20 s. positive result in the test.
PICRIC ACID TEST
r
olo
dc
Sodium Mix and Re Positive Picric
Picric acid carbonate heat acid test
No
Sample co
lo
+ rc
ha
Picric ng
Sample acid e
+ +
Sample
Picric Sodium
acid carbonate
Negative Picric
acid test
Definition The aldehyde/ketone group is oxidized to acid and

Determination of the presence of reducing carbohy- Cu2+, which is bound to copper tartrate complex,
drates in the given sample by Fehling’s test. reduces to copper(I) hydroxide.
Rationale Reduction
Reducing sugars get oxidized to acids and the cupric ion 4
+2 +1
gets reduced to cuprous ions, resulting in red- colored Cu ðC4 H3 O6 Þ2 + 2OH + 2e ! 2 Cu OH + 4C4 H3 O6 2
precipitate.
The copper ions of Fehling’s I solution result in its This is followed by the dehydration of copper(I) hydrox-
blue color as they exist in [Cu(H2O)4]2+ or [Cu ide and red precipitate of copper(I) oxide is formed.
(H2O)6]2+ in the aqueous solution. Upon the addition +1
2 Cu OH ! Cu2+1 O # + H2 O
of Fehling’s solution II, copper reacts with tartrate ions
and the solution becomes dark blue. The aldehyde/ketone group is oxidized to acid:
2 +
CuðH2 OÞ4 + 2C4 H4 O6 2 + 2OH RCHO + 2OH ! RCOOH + H2 O + 2e
CopperðaqÞ ðTartrate ionÞ
4 This is followed by deprotonation of carboxylic acid:
! CuðC4 H3 O6 Þ2 + 6H2 O
Copper complexed with tartrate ions RCOOH + OH ! RCOO + H2 O
Therefore, the net reaction is Precursor techniques

4 1. Sugar solution: If sugar is in powdered form add 1 g
2 CuðC4 H3 O6 Þ2 + 5OH ! Cu2 O # of sugar to 100 mL of water.
4
+ 3H2 O + CuðC4 H3 O6 Þ2 + RCOO 2. Fehling’s solution A: Dissolve 3.5 g of cupric sulfate
pentahydrate (CuSO45H2O) in water to make a
In aqueous solution, only less than 1% sugars exist in final solution of 50 mL.
open-chained form. However, the reaction products, 3. Fehling’s solution B: Dissolve 10 g of sodium hydrox-
i.e., copper(I) oxide and gluconic acid, are continu- ide (NaOH) and 17.5 g of potassium-sodium tartrate
ously removed as precipitate and gluconic ion in the (Rochelle salt; KNaC4H4O64H2O) in 50 mL of
reaction, therefore the ring form of sugar is converted water.
to the open-chain form till the completion of the 4. Before the test, both Fehling’s solutions A and
reaction. B should be added in equal amount, as per the
required amount.
A. Reagents: Test sample (sugar solution), Fehling’s Safety considerations and standards
solution A, Fehling’s solution B. 1. Fehling’s solution B can cause serious eye damage
B. Glassware: Test tube, test tube holder, dropper. and skin irritation, so it must be handled carefully
C. Instrument: Water bath. and one must wear gloves while handling it.
Pros and cons

Protocols Pros Cons
1. Take 1 mL of test solution in a test tube.
An easy and quick Determines the presence of reducing
2. To this add 1 mL of Fehling’s solution (A+ B, 0.5 mL
method sugars only
each).
3. Mix the above gently and keep in a water bath at 80° Alternative methods/procedures
C for 3–5 min. Benedict’s test and Tommer’s test.
Analysis and statistics Summary

Mixing of Fehling’s solution, A (blue), and B (colorless) 1. Fehling’s solution A is blue and becomes dark blue
results in a darker blue-colored solution. When this when mixed with Fehling’s solution B.
mixture is added to a sugar solution and heated in a 2. When the mixture of the above solution is mixed
water bath and the red precipitate is formed, then it with reducing sugars, they get oxidized to acids
indicates the presence of reducing sugar in the test and the cupric ion gets reduced to cuprous ions,
solution. resulting in red-colored precipitate.
FEHLING TEST
Red ppt
Fehling Fehling Mix and Positive

Solution A Solution B heat Fehling test
Sample
+
Fehling
hling
Sample Solution A
+ +
Sample hling
Fehling
F
Fehling
ehling
hling
Solution A Solution B
Negative
Fehling test
Definition and sodium citrate complexes with cupric ions to pre-

Determination of the presence of reducing carbohy- vent them from deteriorating to cuprous ions on
drates in the given sample by Benedict’s test. storage.
Rationale
Benedict’s solution contains mild alkali Na2CO3. Upon Materials, equipment, and reagents
heating in the presence of alkali, reducing sugars are A. Reagents: Test sample (sugar solution), Benedict’s
converted into enediol. Enediols are alkenes with a reagent.
hydroxyl group on each carbon of C]C and are very B. Glassware: Test tube, test tube holder, dropper.
powerful reducing agents. The cupric ions (Cu2+) of C. Instrument: Water bath.
Benedict’s reagent get reduced to cuprous form (Cu+)
as cuprous oxide (which gets precipitated) by enediols.
Protocols
Reducing sugar + CuðcitrateÞ2 2 ! RCOO + Cu2 O # 1. Take 1 mL of test solution in a test tube.
Carboxylate ion Cuprous oxide
2. To this add 2 mL of Benedict’s solution.
In Benedict’s solution, copper sulfate furnishes cupric 3. Mix the above gently and keep in a water bath at 80°
ions, sodium carbonate makes the medium alkaline C for 5 min.
Analysis and statistics Materials, equipment, and reagents

If the solution turns red, it indicates the presence of A. Reagents: Test sample (sugar solution), Tommer’s
reducing sugar in the test solution. reagent.
Precursor techniques C. Instrument: Water bath.
of sugar to 100 mL of water. Protocols
2. Benedict’s solution: Dissolve 17.3 g of sodium citrate 1. Take 1 mL of test solution in a test tube.
and 10 g of sodium carbonate in 80 mL water. In a 2. To this add 1 mL of Tommer’s reagent.
separate beaker, add 1.73 g of copper sulfate to 3. Mix the above gently and keep in a water bath at 80°
10 mL of water. Now, mix both the above solutions C for 2–3 min.
and make up the volume up to 100 mL.
Analysis and statistics
Safety considerations and standards
If a red-colored precipitate is formed, then it indicates
1. Sodium carbonate is a skin irritant, so it must be han-
the presence of reducing sugar in the test solution.
dled carefully and one must wear gloves while
handling it.
Pros and cons 1. Sugar solution: If sugar is in powdered form add 1 g
Pros Cons of sugar to 100 mL of water.
2. Tommer’s solution: Dissolve 5 g of sodium hydrox-
Easy and quick method Determines the presence
ide (NaOH) in 90 mL water and 5 g copper sulfate
of reducing sugars only
(CuSO4) in 90 mL water separately. Mix both and
Benedict modified the Fehling’s
test, to by producing a single make the solution to a final volume of 200 mL.
solution, which is more stable
and convenient Safety considerations and standards
1. Sodium hydroxide is a skin irritant, so it must be
Alternative methods/procedures
handled carefully and one must wear gloves while
Fehling’s test, Tommer’s test.
handling it.
Summary
Pros and cons
1. The reducing sugar gets oxidized and reduces the
Pros Cons
cupric ions present in Benedict’s solution to yield
red color under alkaline conditions. Easy and quick Determines the presence of reducing
2. The alkaline condition is provided by sodium method sugars only
carbonate. Alternative methods/procedures
Fehling’s test and Benedict’s test
Definition
Determination of the presence of reducing carbohy-
Summary
drates in the given sample by Tommer’s test.
1. The reducing sugars reduce the cupric ions to form
red precipitate under alkaline conditions provided
Rationale
by sodium hydroxide.
Reducing sugars get oxidized to acids and the cupric ion
gets reduced to cuprous ions, resulting in the red-
colored precipitate. Definition
Determination of the presence of reducing carbohy-
Alkali
Reducing sugar + Cu2 + ! RCOO + Cu2 O # drates in the given sample by Nylander’s test.
Carboxylate ion Cuprous oxide
In Tommer’s test, copper sulfate furnishes cupric ions; Rationale

sodium hydroxide makes the medium alkaline. In the Reducing sugars get oxidized to acids and reduce bis-
presence of alkali, the cupric ions get reduced to cuprous muth subnitrate to black precipitate of metallic bismuth
ions forming a red-colored precipitate. under alkaline condition.
Bismuth subnitrate + alkali!Bismuth hydroxide

Rationale
Reducing sugar, heat
Barfoed’s reagent contains cupric acetate, which gets
! BismuthðsÞ reduced in the presence of monosaccharides and reduc-
In Nylander’s test, the bismuth subnitrate is reduced to ing disaccharides to cuprous oxide (red precipitate).
black-colored metallic bismuth in the presence of alkali Reducing disaccharides take a much longer time than
and reducing sugar. The reducing sugar gets oxidized the monosaccharides as the reducing disaccharides
to acid. hydrolyze first (in acidic medium) and then react with
cupric acetate.
Materials, equipment, and reagents RCHO + 2Cu2 + + 2H2 O ! RCOOH + 4H + + Cu2 O #
A. Reagents: Test sample (sugar solution), Nylander’s
solution. Materials, equipment, and reagents
B. Glassware: Test tube, test tube holder, dropper. A. Reagents: Test sample (sugar solution), Barfoed’s
C. Instrument: Water bath. reagent.
Protocols C. Instrument: Water bath.
2. To this add 1 mL of Nylander’s solution. Protocols
3. Mix the above solution gently and keep it in a water 1. Take 1 mL of test solution in a test tube.
bath at 80°C, for 3 min. 2. To this add 3 mL of Barfoed’s solution.
3. Mix the above gently and keep in a water bath at 80°
Analysis and statistics C for 2–5 min.
If the solution turns black, it indicates the presence of
reducing sugar in the test solution. Analysis and statistics
The formation of red precipitate within 1–5 min indi-
Precursor techniques cates the presence of monosaccharides. If it takes
1. Sugar solution: If sugar is in powdered form add 1 g 7–12 min for the red precipitate to form then it indicates
of sugar to 100 mL of water. the presence of reducing disaccharides.
2. Nylander’s reagent: Dissolve 20 g of potassium
hydroxide (KOH) in 150 mL of water. To it dissolve Precursor techniques
4 g of bismuth subnitrate [Bi5O(OH)9(NO3)4] and 1. Sugar solution: If sugar is in powdered form add 1 g
8 g of potassium sodium tartrate (Rochelle salt, KNa- of sugar to 100 mL of water.
C4H4O64H2O) and make the final volume to 2. Barfoed’s solution: Dissolve 0.33 mol [60 g of
200 mL. Heat the mixture at 50°C till all the salts Cu(CH3COO)2 or 65.9 g of Cu(CH3COO)2H2O]
get dissolved. Cool and then filter it. of cupric acetate in 1% acetic acid solution.
Pros and cons Safety considerations and standards
Pros Cons
1. Acetic acid should be handled carefully, and one
Easy and quick Determines the presence of reducing must not inhale its fumes.
method sugars only 2. It is very necessary to keep track of time.
Pros and cons
Fehling’s test, Tommer’s test.
Pros Cons
Summary Easy and quick method Nonreducing sugars cannot be

1. Reducing sugars get oxidized to acids and reduce bis- detected
Distinguish between Barfoed’s solution is required
muth subnitrate that is present in the reagent to a
monosaccharides and in comparatively in more
black precipitate of metallic bismuth under alkaline reducing disaccharides quantity than the test
condition. sample
Definition Summary
Distinguish between monosaccharides and reducing 1. The monosaccharides and reducing disaccharides
disaccharides by Barfoed’s test. reduce the cupric ions to red precipitate of copper.
2. This test can even distinguish between monosaccha- Pros and cons
rides and reducing disaccharides as they take time to Pros Cons
produce a red color of copper. Easy and quick method
3. Disaccharides first get hydrolyzed in the acidic It can detect ketose sugars specifically
medium followed by their reaction with cupric
acetate. Summary
1. The test is for the differentiation of aldose from
Definition ketose sugars.
Distinguish between aldose and ketose by 2. Only ketose sugars get dehydrated in the presence of
Seliwanoff’s test. conc. acid.
3. The dehydrated ketose reacts with resorcinol to pro-
Rationale duce a deep cherry red color.
The test is named after the scientist, Theodor Seliwanoff,
who devised it. Polysaccharides and oligosaccharides Definition
ketose get hydrolyzed in the presence of concentrated Determination of pentose sugar by Bial’s test.
acids to yield simple sugar followed by furfural. The
dehydrated ketose reacts with resorcinol to produce a Rationale
deep cherry red-colored product. Aldose may react (at The test is named after the scientist, Manfred Bial. Pen-
a much slower rate) to form a faint pink-colored tose form furfural in acidic medium. This furfural con-
product. denses with orcinol in the presence of ferric ions to
form a blue-green-colored complex.
½H + 2 resorcinal
Ketose ! Furfural ! Deep cherry red color ½H + Orcinol, Fe3 +
Pentose sugar ! Furfural ! Blue green product
½H + Orcinol, Fe3 +
Materials, equipment, and reagents Hexose sugar ! Hydroxymethyl furfural ! Brown
A. Reagents: Test sample (sugar solution), Seliwanoff’s grey product
reagent.
In Bial’s reagent, hydrochloric acid provides the acidic
medium, ferric chloride provides the Fe3+ ions, while
C. Instrument: Water bath.
orcinol condenses with furfural to form the resultant
blue-green-colored product. Hexose generally reacts to
Protocols
form brown-gray color product.
2. To this add 2 mL of Seliwanoff’s solution.
3. Mix the above gently and keep in a water bath at
A. Reagents: Test sample (sugar solution), Bial’s
80–100°C for 1–3 min.
reagent.
The formation of deep cherry red-colored precipitate
within 1–2 min indicates the presence of ketose. If it
takes a longer time for a faint pink-colored precipitate
Protocols
to form, then it indicates the presence of aldose.
2. To this add 2 mL of Bial’s solution.
3. Mix the above gently and keep it in a water bath at
Precursor techniques 80–100°C for 3–5 min.
1. Sugar solution: If sugar is in a powdered form add 1 g
of sugar to 100 mL of water. Analysis and statistics
2. Seliwanoff’s solution: Dissolve 0.05 mg of resorcinol The appearance of blue-green color indicates the pres-
in 100 mL of 3 N HCl. ence of pentose and that of brown-grey color indicates
the presence of hexose sugar.
1. Hydrochloric acid is corrosive and must be handled Precursor techniques
carefully. 1. Sugar solution: If sugar is in powdered form add 1 g
2. It is very necessary to keep track of time. of sugar to 100 mL of water.
2. Bial’s reagent: Dissolve 0.4 g orcinol in 200 mL Analysis and statistics

hydrochloric acid and 0.5 mL of 10% ferric chloride The formation of crystals indicates the presence of galac-
solution. tose in the test solution.
Safety considerations and standards Precursor techniques

1. Hydrochloric acid is corrosive and must be handled 1. Sugar solution: If sugar is in powdered form add 1 g
carefully. of sugar to 100 mL of water.
Pros and cons Safety considerations and standards

Pros Cons 1. Hydrochloric acid is corrosive and must be handled
carefully.
Pentose and hexose sugars can be distinguished
by this easy method
Various versions of this test are used for the detection of Pros and cons
RNA Pros Cons
The presence of galactose can be It is a time taking
Alternative methods/procedures detected process
Tauber’s benzidine test.
Summary
Summary 1. The test is specific for galactose and galactose con-
1. The test is specific for pentose sugars. taining sugars like lactose.
2. Only pentose sugars get converted to furfural in the 2. Lactose forms an insoluble mucic acid when it is
presence of an acid, which in turn condenses with heated with conc. nitric acid.
orcinol and ferric ions to form the colored complex.
Definition
Definition Differentiation between the presence of ketohexose and
Determination of the presence of galactose in the test aldohexose in the test sample by Foulger’s test.
sample by the Mucic acid test.
Rationale
Rationale The dehydration of carbohydrates in the presence of
The oxidation of monosaccharides by nitric acid yields conc. H2SO4 yields furfural. The furfural reacts with
soluble dicarboxylic acid. On the other hand, oxidation stannous chloride in the presence of urea and gives a col-
of galactose as well as lactose yields insoluble mucic acid ored complex.
(also called galactaric acid). Lactose is hydrolyzed to glu-
cose and galactose; it is galactose which gives a positive ½H + SnCl2 , urea
Aldohexose ! Hydroxy methyl furfural ! Green
reaction to the test. blue product

½ H + SnCl2 , urea
A. Reagents: Test sample (sugar solution), conc. Ketohexose ! Hydroxy methyl furfural !
nitric acid.
Yellow or olive green color
Protocols A. Reagents: Test sample (sugar solution), Foulger’s
1. Take 2 mL of test solution in a test tube. reagent.
2. To this add 5 mL of conc. nitric acid. B. Glassware: Test tube, test tube holder, dropper.
3. Mix the above gently and keep in a water bath at C. Instrument: Bunsen burner.
80–100°C for 1 h or till the solution is reduced to
1 mL. Protocols
4. Scratch the inner wall of the test tube. 1. Take 0.5 mL of test solution in a test tube.
5. Let it cool. 2. To this add 3 mL of Foulger’s reagent.
6. Observe the crystals formed under the microscope. 3. Mix the above gently and boil for 45 s.
Analysis and statistics B. Glassware: Test tube, test tube holder, dropper.
The presence of ketohexose in the test solution is C. Instrument: Bunsen burner.
indicated by green-blue color, while the presence of
aldohexose results in yellow to olive green color. Protocols
(Although this test is used to distinguish between keto- 1. Take 1 mL of test solution in a test tube.
hexose and aldohexose; pentoses too react with the 2. To this add 2 mL of methylamine hydrochloride, fol-
reagent and yield yellow color.) lowed by the addition of 0.2 mL of sodium hydrox-
ide solution.
Precursor techniques 3. Mix the above gently and heat at 56°C for 30 min or
1. Sugar solution: If sugar is in powdered form add 1 g at 100°C for 5 min.
of sugar to 100 mL of water.
2. Foulger’s reagent: Add 40 g urea to 80 mL of 40% Analysis and statistics
H2SO4. To this add 2 g of stannous chloride and boil The appearance of red color in the solution indicates the
it vigorously till it becomes clear. Cool it and then presence of reducing sugar in the test sample.
make up the volume up to 100 mL by adding 40%
H2SO4. Precursor techniques
Safety considerations and standards of sugar to 100 mL of water.
1. Sulfuric acid is corrosive and must be handled 2. Methylamine hydrochloride solution: Add 0.2 g of
carefully. methylamine hydrochloride (CH3NH2HCl) to
2. The process of boiling should be done carefully. 100 mL of water.
3. 10% sodium hydroxide solution: Add 10 g of NaOH
Pros and cons to 100 mL of distilled water.
Pros Cons
Easily distinguish between Other carbohydrates too Safety considerations and standards
aldohexose and ketohexose give similar color Methylamine hydrochloride can cause skin and eye irri-
tation; when inhaled can irritate the lungs, so it should
Summary be handled carefully.
1. The test is to distinguish between ketohexose and
aldohexose. Pros and cons
2. Both get dehydrated in the presence of conc. sulfuric Pros Cons
acid to hydroxymethylfurfural. But react with stan-
Easy and sensitive Indicates the presence of reducing
nous chloride to give different colored complexes.
experiment sugars only
Definition Alternative methods/procedures
Determination of the presence of reducing carbohy- Fehling’s test, Benedict’s test, Tommer’s test.
drates in the given sample by Fearon’s methylamine test.
Summary
Rationale 1. In this test, reducing sugars are treated with methyl-
Reducing carbohydrates undergo hydrolysis in alkaline amine hydrochloride in alkaline conditions.
conditions to form enediol. The enediol forms a red- 2. The sugars get hydrolyzed in the presence of alkali
colored product after reacting with methylamine hydro- and form enediol.
chloride. Enediols are alkenes with a hydroxyl group on 3. This enediol reacts with methylamine hydrochloride
each carbon of C]C and are very powerful reducing to form a colored product.
agents.
Alkaline solutions CH3 NH2 HCl Definition
Reducing sugar
! Enediol ! Red
Confirmation of the presence of reducing carbohydrates
colored product
in the given sample by osazone formation test.
Materials, equipment, and reagents Rationale

A. Reagents: Test sample (sugar solution), sodium Osazones are carbohydrate derivatives that are formed
hydroxide (10%) solution, 0.2% methylamine when sugars react with an excess of phenylhydrazine.
hydrochloride. They are colored crystalline compounds and can be
detected under the microscope. Each sugar forms a char- Safety considerations and standards
acteristic crystal. Reducing sugars react with one molecule Be careful while handling the glacial acetic acid.
of phenylhydrazine hydrochloride to form phenylhydra-
zone hydrochloride, this again reacts with another mole- Pros and cons
cule of phenylhydrazine hydrochloride to give a keto Pros Cons
derivate. Finally, the keto derivative reacts with the third It is a confirmatory test for respective sugars None
molecule of phenylhydrazine hydrochloride.
Summary
Sugar + Phenylhydrazine hydrochloride 1. This is a confirmatory test for various reducing
! Sugar phenylhydrazone + H2 O sugars.
Sugar phenylhydrazone + 2ðphenylhydrazine hydrochlorideÞ 2. Each sugar reacts with an excess of phenylhydrazine
! Osazone + C6 H5 NH2 + NH3 + H2 O to form a characteristic crystal, which can be
observed under the microscope.
Materials, equipment, and reagents Definition

A. Reagents: Test sample (sugar solution), sodium ace- Determination of the presence of polysaccharides in the
tate, phenylhydrazine, glacial acetic acid. given sample by an iodine test.
C. Instrument: Water bath. Rationale
Polysaccharides, especially starch, on reaction with an
Protocols aqueous solution of iodine, form colored complexes.
1. Take 5 mL of test solution (2%) in a test tube. When dissolved in an aqueous solution of potassium
2. To 5 mL of the test solution add 1 g of phenylhydra- iodide, the elemental iodine yields a triiodide anion
zine mixture and 2 drops of glacial acetic acid. ðI3 Þ. The triiodide anion form complexes with polysac-
3. Mix the above gently and heat in a boiling water bath charides as the result of an intermolecular charge-
for approximately 30 min or till the formation of osa- transfer complex. The intensity of the color decreases
zone crystals, whichever is earlier. with the increase in temperature and organic solvents.
4. Cool it down and take 2 drops of sugar crystals on a
I3
glass slide, put a coverslip on it, and examine it under Starch ! Blue black color
a microscope. I3
Glycogen ! Reddish brown color
Osazone crystals of various shapes can be observed Materials, equipment, and reagents
under a microscope. The characteristics of the crystals A. Reagents: Test sample (sugar solution), iodine
are summarized as follows (Shah, 2016): solution.
Name of sugar Shape of crystal
Glucose Needle Protocols
Fructose Needle 1. Take 1 mL of test solution in a test tube.
Mannose Needle 2. To this add five drops of iodine solution.
Galactose Balls with a thorny edge
3. Mix the above and observe the color obtained.
Arabinose Dense ball needle
Xylose Fine long needle
Maltose Sunflower Analysis and statistics
Lactose Cotton ball The appearance of blue-black color in the solution indi-
cates the presence of starch and an intermediate reddish-
Precursor techniques brown color indicates the presence of glycogen in the
1. 2% Sugar solution: If sugar is in powdered form add test sample.
2 g of sugar to 100 mL of water.
2. Phenyl hydrazine mixture: It is prepared by adding 1 Precursor techniques
part of phenylhydrazine hydrochloride to 2 parts of 1. Sugar solution: If sugar is in powdered form add 1 g
sodium acetate. of sugar to 100 mL of water.
2. Iodine solution: Prepare 0.005 N iodine solution in Materials, equipment, and reagents
3% potassium iodide solution. To make this add A. Reagents: Anthrone reagent (0.2% anthrone in conc.
3 g of KI to 100 mL of the final solution in water fol- sulfuric acid), test sample (sugar solution), standard
lowed by the addition of 0.063 g of iodine to it. solutions.
Safety considerations and standards C. Instruments: Water bath, colorimeter.
As glycogen and partially hydrolyzed starch give inter-
mediate color (reddish-brown), it should not be con- Protocol
fused with the yellowish-brown color of iodine- Preparation of standard curve:
potassium iodide. (A) Stock standard (1 mg/mL of glucose):
Weigh 100 mg of glucose and add 100 mL of
Pros and cons distilled water to it.
Pros Cons (B) Working standard (0.1 mg/mL of glucose):
Dilute 10 mL of stock solution to a final volume of
Used for distinguishing the It cannot be performed at a
presence of very low pH, as it results in 100 mL by adding distilled water to it.
polysaccharides hydrolysis of 1. Take six test tubes and label them.
polysaccharides 2. To these add 0–1mL of working standard solution
Increased temperature and and add water to make the final volume to 1 mL.
presence of organic 3. Simultaneously, take 1 mL of the sample solu-
solvents decrease the tion in another test tube.
intensity of the resultant 4. Add 4 mL of anthrone reagent, mix well, and
color cover each of the test tubes.
5. Put the tubes in a boiling water bath for 10 min.
Summary
6. Cool the test tubes to room temperature.
1. This test can distinguish between starch and
7. Measure the optical density at 620 nm.
glycogen.
8. Subtract the value of the absorbance of blank
2. Polysaccharides react with an iodine solution to
from the absorbance value for each of the test
form a unique color.
tubes.
9. Plot the standard calibration curve with concen-
QUANTITATIVE TEST FOR
tration plotted on the X-axis and optical density
CARBOHYDRATES
on the Y-axis.
Definition
Quantitative estimation of carbohydrates in the given
Test Tube Test
sample by anthrone test.
No. Blank 1 2 3 4 5 Sample
Working 0 0.2 0.4 0.6 0.8 1.0 1.0
Rationale solution
All types of carbohydrates, monosaccharides (except tri- (mL)
ose and tetrose), disaccharides, and polysaccharides, get Water added 1 0.8 0.6 0.4 0.2 0 0
dehydrated upon treatment with conc. sulfuric acid or (mL)
hydrochloric acid and produce aldehydes. This alde- Anthrone 4 4 4 4 4 4 4
reagent
hyde reacts with anthrone to yield a bluish-green- Heat in boiling water bath for 10 min. Cool it to room temperature
colored complex. OD at A0 A1 A2 A3 A4 A5 AT
620 nm
conc:H2 SO4 or HCl
Pentose sugar ! Furfural Final OD A0 A1–A0 A2–A0 A3–A0 A4–A0 A5–A0 AT–A0
conc:H2 SO4 or HCl
Hexose sugar ! 5 Hydroxymethylfurfural
Calculate the concentration of the test sample corre-
Both the aldehydes (furfural and 5-hydroxymethylfurfural)
sponding to the optical density obtained, using the lin-
condense with anthrone reagent (10% α-naphthol in etha-
ear standard graph.
nol). This test results in a bluish-green-colored product by
all carbohydrates, which are larger than tetrose.
Anthrone reagent
Furfural=5 hydroxymethylfurfural
! Bluish 1. Anthrone reagent: Add 0.2 g of anthrone reagent to
green colored product 100 mL of conc. sulfuric acid (H2SO4).
2. Test sample solution: Dissolve the test sample in yellow-colored DNSA is reduced to orange-red
10mL. The solution should be diluted accordingly so 3-amino-5-nitrosalisyclic acid (ANSA).
that the optical density of the test sample does not Alkali
exceed that of test tube no. 5, i.e., 0.1 mg/mL of sugar. Reducing sugar + 3, 5 dinitrosalicyclic acid ! RCOO
ðyellowÞ
+ 3 amino 5 nitrosalicylic acid
Calculation ðorangeredÞ
The concentration can be obtained using a standard
graph. The dilution (if made) of the test sample should
be multiplied to obtain the final concentration of the
A. Reagents: Dinitro salicylic reagent, sodium hydrox-
test sample solution.
ide solution (2 mol/L), test sample (sugar solution),
standard solutions.
1. Concentrated acids should be handled very carefully,
C. Instruments: Water bath, spectrophotometer.
one must wear gloves while handling it.
2. Always add acid to water.
Protocol
Preparation of standard curve:
Pros and cons
A. Stock standard (1 mg/mL of glucose):
Pros Cons
Weigh 100 mg of glucose and add 100 mL of distilled
An easy method to test the Trioses and tetrose do not give water to it.
quantity of positive result of this test B. Working standard (0.25 mg/mL of glucose):
carbohydrates Not suitable when proteins
Dilute 10 mL of stock solution to a final volume of
having a high amount of
tryptophan is present, as it
40 mL by adding distilled water to it.
too contributes to the 1. Take six test tubes and label them.
production of red color 2. To these add 0–1 mL of working standard solu-
tion and add water to make the final volume to
Alternative methods and protocols 1 mL.
Molisch’s method. 3. Simultaneously, take 1 mL of the sample solution
in another test tube.
Summary 4. Add 1 mL of DNS reagent, mix well, and cover
1. It is a quantitative test for carbohydrates, except tri- each of the test tubes.
ose and tetrose. 5. Put the tubes in a boiling water bath for 5 min.
2. In this test, the aldehyde produced by the treatment 6. Cool the test tubes to room temperature.
of carbohydrates with conc. acid reacts with anthrone 7. Measure the optical density at 540 nm.
reagent and yields colored product based on the ini- 8. Subtract the value of the absorbance of blank
tial concentration of the sugars. from the absorbance value for each of the test
3. All the tubes containing carbohydrates are treated tubes.
simultaneously so that the test sample can be com- 9. Plot the standard calibration curve with concen-
pared with the standards. tration plotted on the X-axis and optical density
4. The reaction is sensitive to temperature; therefore, all on the Y-axis.
tubes should be cooled to room temperature before
reading the absorbance.
Test Tube Test
No. Blank 1 2 3 4 5 Sample
Definition
Working 0 0.2 0.4 0.6 0.8 1.0 1.0
Quantitative estimation of reducing carbohydrates in the solution
given sample by 3,5-dinitro salicylic acid (DNSA) test. (mL)
Water added 1 0.8 0.6 0.4 0.2 0 0
Rationale (mL)
DNS reagent 1 1 1 1 1 1 1
All types of reducing sugars undergo oxidation to form Heat in boiling water bath for 5 min. Cool it to room temperature
acids and in turn reduce many reagents like DNSA. OD at A0 A1 A2 A3 A4 A5 AT
Under alkaline conditions, the free aldehyde or ketone 620 nm
group of reducing sugar is oxidized to acids and Final OD A0 A1A0 A2A0 A3A0 A4A0 A5 A0 AT A0
Analysis and statistics Rationale

Calculate the concentration of the test sample corre- Glycogen is a storage polysaccharide in animals. It is pri-
sponding to the optical density obtained, using the lin- marily synthesized and stored in the liver and muscles.
ear standard graph. In the liver, the level of glycogen is affected by the state
of diet. In the normal rat, glucose is converted into gly-
Precursor techniques cogen by the process of glycogenesis. This process
1. Sodium potassium tartrate: Add 30 g of the salt to involves many enzymes like phosphorylase, alpha-1,6-
100 mL of the final solution in water. glucosidase, phosphoglucomutase, and glucose-6-phos-
2. 3,5-Dinitrosalicylic acid: Dissolve 5 g of this reagent phatase. During starvation, this glycogen which is stored
in 100 mL of water. in the liver is broken down in a process termed as glyco-
3. Dinitro salicylic acid (DNS) reagent: Add 50 mL of genolysis. This process lasts for 1–2 days; after this, the
(1) to 20 mL of (2) and make the solution to a to formation of blood glucose occurs from noncarbohy-
a final volume of 1 L. drate sources in the process of gluconeogenesis. While
4. Test sample solution: Dissolve the test sample in in muscles, its level remains almost constant and is
10mL. The solution should be diluted accordingly so not dependent on the state of diet. As muscles lack
that the optical density of the test sample does not the enzyme glucose-6-phosphatase, which aid in the
exceed that of test tube no. 5, i.e., 0.25mg/mL of sugar. conversion of glycogen to glucose, the glycogen in mus-
cles cannot get converted to blood glucose directly.
Calculation The glycogen is released from tissues by homogeni-
The concentration can be obtained using a standard zation with TCA or by boiling with 30% KOH. It is then
graph. The dilution (if made) of the test sample should get precipitated using 95% ethanol. Sodium sulfate
be multiplied to obtain the final concentration of the increases the yield of glycogen as it acts as a
test sample solution. coprecipitate.

A. Reagents: Tissue (liver, muscle), 5% trichloro acetic
1. The test is sensitive to the temperature, so cool down
acid, 95% ethanol.
all the samples to room temperature before reading.
2. Sodium hydroxide solution is corrosive and irritates
C. Instruments: Centrifuge (table top), mortar-pestle.
skin and eyes, so it must be handled carefully.
Protocols
Pros and cons
1. Weigh 5 g of tissue, cut into small pieces, now add
Pros Cons
an appropriate amount of 5% TCA to it, and grind it
An easy and quick method for Not suitable for the in mortar and pestle.
determining the amount of determination of 2. Pour the homogenate in a centrifuge tube.
reducing sugars concentration in a
3. Centrifuge it at 3000 rpm for 5 min at 4°C.
complex mixture of sugars
4. Transfer the supernatant to a graduated tube, again
Summary add 5% TCA and repeat the process of homogeniza-
1. The test is specific for the reducing sugars, which react tion, and centrifuge as above.
with DNS reagent in alkaline conditions to form a 5. Collect the supernatant to the same graduated tube.
colored product, which depend on the concentration 6. This can be repeated two more times.
of the sugar solution. 7. Take 1 mL of above and to this add 5 mL of 95%
2. All the tubes containing reducing carbohydrates are ethanol with blowing to effect proper mixing.
treated simultaneously so that the test sample can 8. Now cap the tubes and allow it to stand overnight at
be compared with the standards. room temperature. Alternatively, it can be kept in a
water bath at 37°C for 3 h.
9. After the precipitation is completed, centrifuge the
OTHER TESTS FOR CARBOHYDRATES tubes at 3000 rpm for 15 min.
Definition 10. Discard the supernatant and dissolve the pellet in
Extraction and estimation of glycogen from the liver and 5 mL water and reprecipitate by adding 10 mL of
muscle of the well-fed and starved rat. 95% ethanol.
11. Recentrifuge the tubes at 3000 rpm for 15 min. Dis- 4. 0.5 M NaOH: 2 g NaOH in a final volume of 100 mL
card supernatant. of solution in distilled water.
12. Dissolve pellets in 2 mL of water. 5. 5. 30% KOH: 30 g KOH in a final volume of 100 mL
of solution in distilled water.
Alternative method: Using KOH
Materials, equipment, and reagents Safety considerations and standards
A. Reagents: Tissue (liver, muscle), 30% KOH solution, 1. Handle all the reagents very carefully.
saturated Na2SO4, 95% ethanol, 1.2 M HCl,
0.5 M NaOH. Pros and cons
B. Glassware: Test tube, test tube holder, dropper. Pros Cons
C. Instrument: Centrifuge (table top). An easy method to isolate glycogen Time
Isolation of glycogen: from tissues taking
1. Accurately weigh 1.5 g of the liver and skeletal
muscle. Summary
2. In a centrifuge tube take 2 mL of KOH, place the tis- 1. This is a method to isolate glycogen from mice by
sue in it. homogenization of the tissue with TCA or boiling
3. Heat in a boiling water bath for 20 min, with inter- with KOH.
mittent shaking. 2. The glycogen is obtained from the liver; muscles do not
4. Let the tube cool in ice. yield it as they lack the enzyme glucose-6-phosphatase.
5. Add 0.2 mL of saturated Na2SO4 and mix it.
6. Add 5 mL of 95% ethanol and allow the tube stand Definition
on ice for 5 min. The glycogen gets precipitated. Separation and identification of sugars present in fruit
7. Centrifuge the tube at 3000 rpm for 5 min. juices using thin-layer chromatography (TLC).
8. Discard the supernatant and add 5 mL of water, mix
thoroughly (warm it in case it does not dissolve). Rationale
9. Add double distilled water to it to make a solution Fruit juices contain a wide variety of sugars. These sugars
up to 10 mL. This is glycogen solution in water. can be identified accurately by thin layered chromatogra-
10. If the rat is NOT starved and is well fed, then dilute phy. In the experiment, various sugars have varying sol-
the solution to 100 mL with water. ubility in a solvent and this property is utilized in TLC.
Hydrolysis of glycogen: Depending on the solubility, the sugars travel the dis-
11. Take 1 mL of glycogen solution in a test tube and tance (upward) with the solvent. As the experiment pro-
add 1 mL of 1.2 M HCl to it. Heat in boiling water ceeds, the most soluble sugar will travel the longest
bath for 2 h. distance with solvent and reaches the highest point,
12. To this add 1 drop of phenolphthalein and then while the least soluble sugar will travel the shortest dis-
neutralize with 0.5 M NaOH, till the pink color of tance. The sugars can be detected after the spraying the
the indicator turns yellow. solution of aniline-diphenylamine followed by heating.
13. Then dilute it to 5 mL.
14. This results in the hydrolysis of glycogen to glucose, Materials, equipment, and reagents
which can be determined by the glucose oxidase A. Reagents: Thin layered plates of silica gel G, solvent,
method (specific for glucose). fruit juices, absolute ethanol, standard sugar solu-
tion, aniline-diphenylamine location reagent.
Analysis and statistics B. Glassware: Glass plates, separation chambers.
The glycogen extracted from tissues is hydrolyzed and glu- C. Instruments: Oven, spray gun.
cose concentration, which is an indication of glycogen
content in this context, should be higher in well-fed mice. Protocols
1. Take 1 mL fruit juice; add 3 mL of ethanol to remove
Precursor techniques denatured protein.
1. 5% TCA: Weigh 5 g of TCA and add to distilled water 2. Spot the supernatant in a thin layered plate with
to a final volume of 100 mL. standard sugar solutions.
2. 95% ethanol: To 95 mL of ethanol add 5 mL of dis- 3. Irrigate the plate with the solvent system in ascending
tilled water. direction in a chamber.
3. 1.2 M HCl: Add 10.6 mL of HCl to a final volume of 4. Let it develop until the solvent front reaches the top
100 mL of solution in distilled water. of the plate.
5. Draw a line across it and remove the plate. Definition

6. Let the plate dry in the stream of air. Extraction and analysis of soluble carbohydrates from
7. Spray the aniline-diphenylamine solution in a fume plants.
chamber and heat the plates at 100°C.
8. Note the color developed for each sugar and calculate Rationale
the Rf value. Soluble carbohydrates are the carbohydrates that dis-
9. Compare the color and Rf values of each sugar sam- solve in the aqueous environment of the cells. Under
ple with that of the standards. The same sugar will stress and various stimulations, the respective quantities
have the same color and Rf value. of the soluble sugars and starch get altered. The pool of
soluble carbohydrates varies under various environ-
Analysis and statistics mental conditions. The soluble carbohydrates get
A comparison of color and Rf value indicates the sugars extract with ethanol or water. After the removal of solu-
present in the solution, i.e., fruit juices. ble sugars, the starch is hydrolyzed by enzymes and then
extracted with water.
1. Thin layered plate of silica gel G: Prepare 0.02 M Materials, equipment, and reagents
sodium acetate solution by adding 0.164 g of A. Reagents: 2% amyloglycosidase and 0.5% α amylase
CH3COONa to a final volume of 100 mL in water. in 0.1 M sodium acetate buffer (pH 4.5), and 95%
To this solution add silica gel to make a slurry. Pour ethanol.
it in plates approximately 0.25 mm thick, let it dry. It B. Apparatus: Volumetric flask, centrifuge tubes, and
should be activated before the experiment. mortar pestle.
2. Solvent: Mix ethyl acetate, isopropanol, water, and C. Instrument: Water bath.
pyridine in the ratio 26:14:7:2.
3. Standard sugar solution (1% in sugar in 10% isopro- Protocols
panol): Add 10 mL of isopropanol to make a final 1. Excise the desired part of the plant and cut it into
volume of 100 mL, to this add 1 g of desired sugar. small pieces with the help of a sharp, clean scalpel.
4. Location reagent: It could be prepared freshly by mix- 2. This can be frozen or heat inactivated to inactivate
ing 1% aniline, 1% diphenylamine in acetone, and the enzymes and slow down the metabolic rate
85% phosphoric acid in the ratio of 5:5:1. thereby preventing the conversion of sugars, or it
can proceed for the extraction of sugars.
Safety considerations and standards 3. Rapid freezing in liquid nitrogen.
1. Phosphoric acid should be handled carefully. 4. Heat inactivation is done at 90°C for 90 min (min-
2. Do not inhale the solvent. imum for 60 min) followed by grinding in mortar
3. The Rf value should be calculated precisely. and pestle.
5. For the extraction of sugar, weigh 100–400 mg of
Pros and cons dry ground or frozen wet sample in a round bottom
Pros Cons plastic 50 mL centrifuge tube.
6. Add 15–20 mL of 95% ethanol and cap it with a
The identification of individual sugar in a mix is
possible rubber stopper having a hole, it should be equipped
with glass reflux.
Alternative methods 7. Mix and place the mixture in a water bath at 85°C.
Other chromatographic techniques. Note the boiling of ethanol and keep it in boiling
ethanol for 20 min.
Summary 8. Uncap the tubes and centrifuge it at 10,000 g for
1. The fruit juice along with the standard sugar solution 10 min.
is spotted on the silica plate and allowed to travel 9. Decant the supernatant in a volumetric flask of suit-
with the solvent in an upward direction. able capacity and repeat the extraction for three
2. As the solvent front reaches the highest point, the more times. Make up the volume with 95% etha-
plate is dried, and spots are detected. nol. This is an ethanolic extract.
3. The distance traveled by a sugar is the same in a par- 10. Note that ethanol extract can be held for 2 weeks at
ticular solvent and the sugars can be detected by 4°C. Similar to ethanol, as used in the above extrac-
comparing the spots developed by spraying the tion, water too can be used at 60°C followed by cen-
detection reagent. trifugation at 20,000–30,000 g for 10 min. The water
extract can be kept for 5 days at 4°C. Since the starch Alternative methods
is partially soluble in water, the water extract is not Method of Ichimura and Hisamatsu (1999).
suitable for subsequent starch determination.
11. For starch extraction, take the pellet of ethanolic Summary
extract, let the ethanol evaporate completely. 1. The plant material is ground using liquid nitrogen
12. Add 10 mL of water to sample (ethanolic extract in a and the soluble carbohydrates are extracted using
50-mL centrifuge tube) and mix. water or ethanol.
13. To gelatinize the starch, incubate the samples at 90° 2. The starch is extracted using sodium acetate and
C with intermittent mixing for 30 min. enzymatic solution.
14. Let the samples cool at RT.
15. Add 0.2 M of sodium acetate, pH 4.5, 1 mL amylo- Summary of the sugar test.
glycosidase, and α amylase to it.
16. Incubate at RT for 1–2 h for α amylase to gelatinize Sl
No. Test Substrate Result
the starch, followed by incubation at 55°C for
16–24 h. 1 Molisch’s All The purple ring is
17. Let the tubes cool at RT then centrifuge it at 30,000 g test carbohydrates formed
for 20 min. 2 Picric acid Reducing sugars The yellow color
test turns red
18. Pour the supernatant in a volumetric flask; add
3 Fehling’s test Reducing sugars The blue solution to
10 mL water to the pellet, mix it well, and incubate red ppt
it for 10 min at 60°C. 4 Benedict’s Reducing sugars The blue solution to
19. Centrifuge it at 30,000 g for 20 min. test red ppt
20. Keep the supernatant in the same volumetric flask. 5 Tommer’s Reducing sugars The blue solution to
21. Repeat the extraction with water twice. test red ppt
22. The starch extract can be stored at 4°C for up to 6 Nylander’s Reducing sugars Black ppt
5 days. test
23. The total sugar content can be determined by the 7 Fearon’s test Reducing sugars The blue solution to
anthrone method. red ppt
8 Barfoed test Mono and Monosaccharides
disaccharides turn blue to red in
Analysis and statistics a short time
The soluble sugar extracts can be obtained as both etha- Disaccharides turn
nolic extract and water extract. The starch extract is blue to red take a
obtained after removing soluble sugars from the ethano- long time
lic extract. 9 Seliwanoff’s Ketose and Ketose give deep
test aldose cherry red; aldose
turns slowly to a
Precursor techniques faint pink
1. 95% ethanol: Add 95 mL of ethanol to make a final 10 Bial’s test Pentose and Pentose give blue-
volume of 100 mL in water. hexose green color;
2. 0.2 M sodium acetate solution (pH 4.5): Mix 2.72 g hexose give
of sodium acetate with 80 mL water, maintain the brown-gray color
pH of 4.5, and adjust the volume to 100 mL with 11 Foulger’s Ketohexose and Ketohexose gives
water. test aldohexose green-blue color;
aldohexose gives
yellow or olive
Safety considerations and standards green color
1. The blank and sample should be processed carefully 12 Mucic acid Galactose Crystals of mucic acid
in the same manner. test are formed
13 Osazone test Various sugars Different types of
crystals formed
Pros and cons
14 Iodine Polysaccharides Starch gives a blue-
Pros Cons
black color;
An easy method for determination of sugars Time taking glycogen gives a
and starch process red-brown color
CHAPTER 3
Lipid
1. Solubility test for lipids. 3. Shake well and check their solubility.
2. Emulsion test for lipids. 4. Heat it to 50°C for 5 min and again check their
3. Determination of the degree of unsaturation of fatty solubility.
acids. 5. Repeat the test using other solvents and record the
4. Determination of the saponification value of the solubility.
given fat/oil sample.
5. Determination of the fatty acid value of fat. Analysis and Statistics
6. Acrolein test for the presence of glycerol. The samples that are more soluble in polar solvent have
7. Libermann-Buchard test for the detection of small fatty acids, while the samples having larger fatty
cholesterol. acids are less soluble in polar solvents even after heating.
8. Extraction of lipid from leaves (Bligh and Dyer’s
method and single-extraction method). Safety Considerations and Standards
9. Extraction of lipid from egg yolk and estimation of 1. Alcohol is highly inflammable so utmost precaution
phosphorous content in it. should be applied while heating.
10. Extraction of lipid from tissues (Folch method).
Pros and Cons
SOLUBILITY TEST FOR LIPIDS Pros Cons
Rationale A preliminary idea about the lipid can be This test is not
Fats are esters of fatty acids. They are amphoteric in obtained specific
nature and harbor a hydrophilic head of glycerol and
Alternative Methods/Procedures
hydrophobic tail of fatty acids. Other lipids too are
hydrophobic. They get dissolved in organic solvents. Tri- Emulsion test, grease spot test, Sudan test.
glycerides with a small-chain fatty acids are slightly sol-
uble in polar solvents as water, while those having long- EMULSION TEST FOR LIPIDS
chain fatty acids are insoluble in it and form emulsions. Rationale
All triglycerides are soluble in diethyl ether, benzene, The emulsion is a mixture of two immiscible liquids.
and chloroform like nonpolar solvents. Their solubility When the sample containing lipids is suspended in eth-
in polar solvents like methanol, ethanol, acetone, etc. anol, they get partially solubilize. When the above
increases on heating. Understanding their solubility solution is mixed with water, the formation of cloudy
characteristics aids in the extraction process. emulsion occurs due to the insolubility of lipids in
water. Emulsification agents like soap, bile salts, pro-
Materials, Equipment, and Reagents teins, etc. aid in the formation of a permanent
A. Reagents: Fatty acids (butyric, palmitic acid), fats, emulsion.
and oils (butter, vegetable oil), solvents. Solvent
(water, acetone, ethanol, diethyl ether, chloroform, Materials, Equipment, and Reagents
etc.). A. Reagents: Food sample, ethanol, distilled water.
B. Glassware: Test tubes, dropper. B. Glassware: Test tubes, dropper.
C. Instrument: Burner.
Protocol
Protocol 1. Crush the food sample (only in case of solid food
1. Label various test tubes and put 1 g of different lipids sample) and place it in a dry test tube. In case of liq-
in it. uid sample, add a few drops of the sample in a dry
2. To these add 5 mL water. test tube.
2. Add 2 mL of ethanol to the samples.

3. Shake it thoroughly. Precursor Techniques

4. In case of solid sample, allow the sample to settle 1. Huble’s reagent: (a mixture of 7% mercury chloride
down (5 min) to allow lipid to be extracted and in alcohol and 5% iodine in 96% alcohol is added
transfer the ethanol to another test tube. in equal proportion). Add 3.5 mL of mercury chlo-
5. Add 2 mL of distilled water to it. ride (HgCl2) to a final solution of 50 mL in ethanol.
6. Observe the formation of a white-colored emulsion. In a separate beaker, add 2.5 g of iodine in 48 mL of
ethanol to a final volume of 50 mL. Now mix both
Analysis and Statistics the solutions in equal ratio to a final volume of
The samples containing lipids will form a white-colored 100 mL.
emulsion.
Analysis and Statistics
The number of drops of lipid samples required for the
Safety Considerations and Standards
disappearance of pink color is more for a lipid having
1. In case of solid samples, the crushing of the samples
an unsaturated fatty acid with less number of unsatu-
should be done properly.
rated bond than that having more number of saturated
bonds. In other words, the more the unsaturation the
Pros and Cons less the number of lipid drops required.
Pros Cons
A preliminary idea about the lipid can be This test is not Safety Considerations and Standards
obtained specific 1. Glassware should be washed before and after use.
2. Carefully note down the mL required in the process.
Alternative Methods/Procedures 3. The change in the color of the solution should be
Solubility test, grease spot test. monitored carefully.
DETERMINATION OF DEGREE OF Pros and Cons

UNSATURATION OF FATTY ACIDS
Pros Cons
Rationale
The degree of unsaturation of The exact number of double
Unsaturated fats are mostly obtained from plants, while
fats can be easily determined bonds cannot be predicted
animals are the common source of saturated fats. The
unsaturated fats contain a double or triple bond in Summary
the hydrocarbon chain of their fatty acids. The halogens 1. The degree of unsaturation is indicated by the
such as iodine can be accommodated in these unsatu- amount of iodine solution accommodated in a lipid
rated bonds. In this test, the Huble’s reagent comprising solution.
iodine is added to the lipid and the degree of the unsa- 2. The pink color of the Huble’s reagent with chloro-
turation can be determined by the decolorizing of the form disappears with the addition of fatty acids to it.
reagent. 3. The more the number of drops of fatty acids required,
the less is the degree of unsaturation of fatty acid.
Materials, Equipment, and Reagents
A. Reagents: Test samples, Huble’s reagent, DETERMINATION OF THE SAPONIFICATION
chloroform. VALUE OF THE GIVEN FAT/OIL SAMPLE
B. Glassware: Test tubes, dropper. Rationale
Saponification is a process involving the hydrolysis of
Protocols fats on its reaction with alkali, thereby leading to the for-
1. Take 3 mL of chloroform in the test tubes and mation of salts of fatty acids and glycerol. The salts of
label them. fatty acids are known as soap. The amount of potassium
2. Now add 3 mL of Huble’s reagent to each test tube. hydroxide required in the hydrolysis indicates the sapon-
3. Shake well. ification value of fat. It can be simply described as the
4. The solution turns pink because of the presence of amount of alkali (e.g., potassium hydroxide) required
free iodine. to saponify 1 g of fat. Each molecule of triacylglycerol
5. Now add the test samples in a dropwise manner. requires three molecules of KOH for saponification. Also
6. Note down the number of drops required for the 1 g of triacylglycerol with shorter length of fatty acids has
pink color to disappear from the solution. more number of molecules of triglycerides as compared
CHAPTER 3 Lipid 25
to the same with a longer length of fatty acids; therefore, Then, the titer value for the sample is (Y X) mL ¼
the former will require more amount of KOH. In this mL of KOH required to saponify 1 g of fat.
way, the saponification value depicts the average molec- It is also known as the saponification value of the fat.
ular weight of a triglyceride. In this process, excess of
28:05 titre value in mL
alkali is used and the remaining KOH is determined by Saponification value ¼
ðmg=gÞ Weight of sample in gram
titrating it with 0.5 N HCl.
C H2 OR C H2 OH
Since 1 mL of 0.5 N KOH contains 28.05 mg of KOH, the
| | multiplication factor of 28.05 is included in the above
C HOR + 3KOH ! C HOH + 3RCOO K + equation.
| |
CH2 OR CH2 OH The molecular weight of KOH is 56 and a triglyceride
Fat Alkali Glycerol Soap releases three molecules of fatty acids.
Therefore,
Materials, Equipment, and Reagents 3 56 1000
Saponification value ¼
A. Reagents: Fat solvent (mixture of 95% ethanol and ðmg=gÞ Average molecular weight of fat
ether in 1:1 v/v), 0.5 N alcoholic KOH, 1% phenol-
phthalein solution in 95% alcohol, 0.5 N HCl, test OR
sample. 3 56 1000
Average molecular weight of fat ¼
B. Glassware: Burette, conical flask. Saponification value ðmg=gÞ
C. Instruments: Water bath, reflux condenser.
Protocol Analysis and Statistics

1. Weigh 1 g of fat sample in a conical flask and dissolve The saponification value in mg/g and the average molec-
it in 3 mL of the solvent of fat. ular weight of fat in g can be calculated.
2. To this add 25 mL of 0.5 N of alcoholic KOH, attach a
reflux condenser to it. Safety Considerations and Standards
3. Parallelly, set up another reflux condenser with 3 mL 1. Alcohol is highly inflammable so utmost precaution
fat solvent and 25 mL of 0.5 N of alcoholic KOH as a should be applied while heating.
blank as it is devoid of fat. 2. There should be effective cooling of the condenser to
4. Heat both condensers on boiling water bath for prevent the evaporation of alcohol.
30 min.
5. Let both the flasks to cool to the room temperature. Pros and Cons
6. Add a few drops of phenolphthalein to both the Pros Cons
flasks.
The average molecular weight of fatty acids can be
7. Titrate both with 0.5 N HCl, till pink color
determined
disappears.
8. The difference between the blank and test sample gives Summary
the volume of KOH required to saponify 1 g of fat. 1. It is a method to determine the average molecular
weight of the fatty acids.
Precursor Techniques
2. Fat is added to a fixed amount of alkali and attached
1. Fat solvent: Mix 47.5 mL of ethanol and 2.5 mL of to a reflux condenser. The amount of alkali required
water to prepare 95% ethanol, mix 95 mL ethanol to saponify 1 g of fat is determined by titrating it
and 5 mL water. Take 25 ml of 95% ethanol and against HCl.
mix 25 mL ether in it.
2. 1% phenolphthalein solution in 95% alcohol: Mix
95 mL of ethanol and 5 mL of water, to this add 1 g
of phenolphthalein. DETERMINATION OF THE FATTY ACID
3. 0.5 N HCl: Add 4.44 mL of conc. HCl to water to VALUE OF FAT
make a final solution of 100 mL. Rationale
The fats become rancid on storage. The rancidity is
Calculation caused due to the formation of peroxide by atmo-
Suppose, spheric oxygen at the double bonds and also because
0.5 N KOH in the test sample is X mL. of microorganisms as they cause its hydrolysis to liber-
0.5 N KOH in blank is Y mL. ate fatty acids and glycerol. Therefore, the age and
quality of fats can be determined by the acid value of Safety Considerations and Standards
fat or the amount of free fatty acids present in it. Their 1. Glassware should be washed before and after use.
presence is determined by the titration of the sample 2. Carefully note down the mL required in the process
with KOH. The acid value is defined as milligrams of of titration.
KOH required to neutralize the free fatty acids in 1 g 3. The change in the color of the solution should be
of fat or oil. monitored carefully.
Pros and Cons

A. Reagents: 1% phenolphthalein solution in 95% alco- Pros Cons
hol, fat solvent (95% ethanol: ether in 1:1 v/v), 0.1 N The quality of fat can be easily determined
potassium hydroxide.
B. Glassware: Burettes, conical flask. Summary
1. The fatty acid value of fat is determined by titrating it
against an alkali.
Protocols 2. It depicts the age and quality of the fat.
1. Take 10 g of fat in a conical flask and add 50 mL of fat
solvent to it. ACROLEIN TEST FOR THE PRESENCE OF
2. Shake the above and add a few drops of phenol- GLYCEROL
phthalein to it. Rationale
3. Titrate it with KOH till a faint pink color solution When glycerol is heated with potassium hydrogen sulfate,
appears for 15 s (approximately). it gets dehydrated to acrolein, which is an unsaturated
4. Note down the volume of KOH used. aldehyde. This glycerol can be in its free form or as alcohol
5. Repeat the above for a blank solution that is devoid in fats. Acrolein has a characteristic pungent smell.
of fat.
6. Again, note down the volume of KOH used. C H2 OH C HO
| |
heat
C HOH + KHSO4 ! C H + 2H2 O
| |
Precursor Techniques CH2 OH CH2
1. Fat solvent: Mix 47.5 mL of ethanol and 2.5 mL of Glycerol Acrolein
water to prepare 95% ethanol, mix 95 mL ethanol
and 5 mL water. Take 25 mL of 95% ethanol and Materials, Equipment, and Reagents
mix 25 mL ether in it.. A. Reagents: Test samples, anhydrous potassium
2. 1% phenolphthalein solution in 95% alcohol: Mix hydrogen sulfate.
95 mL of ethanol and 5 mL of water, to this add 1 g B. Glassware: Test tubes, test tube holders.
of phenolphthalein. C. Instrument: Bunsen burner.
3. 0.1 N potassium hydroxide: Add 2.8 g of KOH in
500 mL of solution in water. Protocol
1. Take 1.5 g of potassium hydrogen sulfate in the
test tube.
Calculation 2. Add five drops of sample to it, in case of solid sample
The volume of KOH used for blank ¼ X mL. take the equivalent weight of the sample.
The volume of KOH used for sample ¼ Y mL. 3. Cover the sample by adding a few more crystals of
Therefore, the titer value of fat ¼ Y X mL. potassium hydrogen sulfate to it.
5:6 titre value in mL 4. Heat the test tube on a burner and note the odor of
Acid value ¼
ðmg=gÞ Weight of sample in gram the fumes coming out of the tube.
Since, 1 mL of 0.1 N KOH contains 5.6 mg of KOH, the Precursor Techniques
multiplication factor of 5.6 is included in the above
None
equation.
CHAPTER 3 Lipid 27
Analysis and Statistics Test Tube Test

The pungent smell of the fumes indicates the presence of No. Blank 1 2 3 4 Sample
glycerol in the test sample. Working 0 0.5 1.0 1.5 2 2.0
solution
Safety Considerations and Standards (mL)
1. Be careful while heating. Do not inhale the fumes. Chloroform 2 1.5 1.0 0.5 0 0
added (mL)
Alternative Methods/Procedures Liberman- 2 2 2 2 2 2
Buchard
Dichromate test for glycerol.
reagent
OD at A0 A1 A2 A3 A4 AT
Summary 680 nm
1. The test is to detect the glycerol in a sample. Final OD A0 A1 A0 A2 A0 A3 A0 A4 A0 AT–A0
2. The pungent smell on heating the sample with
potassium hydrogen sulfate is the indicator of the
glycerol. Precursor Techniques
1. Libermann-Buchard reagent: 50 mL of acetic anhy-
dride is pipetted in a glass vial, keep it in ice. Also
LIBERMANN-BUCHARD TEST FOR THE 5 mL of concentrated sulfuric acid is added to
DETECTION OF CHOLESTEROL the vial.
Rationale 2. Cholesterol standard:
Libermann-Buchard test is a colorimetric test for choles- Stock standard (2 mg/mL of cholesterol): Weigh
terol, which yields a deep green color. Although the 200 mg of cholesterol and add 100 mL of chloroform
exact nature of chromophore is not known, it is sug- to it.
gested that in this the hydroxyl group of cholesterol Working standard (0.4 mg/mL of cholesterol):
reacts with acetic anhydride and conc. sulfuric acid, this Dilute 2 mL of stock solution to a final volume of
increases the conjugation of the unsaturation of the 10 mL by adding chloroform to it.
adjacent fused ring.
Analysis and Statistics
Materials, Equipment, and Reagents The concentration of cholesterol is obtained by
A. Reagents: Stock solution, working solution, chloro- spectrophotometer.
form, Libermann-Buchard reagent.
B. Glassware: Test tubes, test tube holders. Safety Considerations and Standards
C. Instrument: Spectrophotometer. 1. Be careful while heating. Do not inhale the fumes.
Protocol Pros and Cons

1. Take six test tubes and label them. Pros Cons
2. To these add 0–2 mL of working standard solution
and add chloroform to make the final volume to The widely accepted method The reagents like acetic
for the determination of anhydride, sulfuric acid,
2 mL.
cholesterol in the blood and chloroform are
3. Simultaneously, take 2 mL of the sample solution in corrosive and toxic
another test tube.
4. Add 2 mL of Libermann-Buchard reagent, mix well, Alternative Method
and keep in dark for 90 min. Salkowski method.
5. Measure the optical density at 680 nm.
6. Subtract the value of the absorbance of blank from Summary
the absorbance value for each of the test tubes. 1. It is a quantitative test for cholesterol.
7. Plot the standard calibration curve with concentra- 2. The reaction between the Libermann-Buchard
tion plotted on the X-axis and optical density on reagent and the sample yields a colored product that
the Y-axis. can be estimated using the standard graph.
EXTRACTION OF LIPID FROM LEAVES 6. Shake the extracts at 100 rpm for 24 h in an orbital
(BLIGH AND DYER’S METHOD AND SINGLE- shaker.
EXTRACTION METHOD) 7. Transfer the intact, extracted leaf materials to a new
Rationale vial using forceps.
One phase system is used in Bligh and Dyer’s method, 8. Let it dry overnight at 105°C.
with chloroform:methanol:water (1:2:0.8), with the tis- 9. Add 280 μL of 600 mM ammonium acetate to the
sue water included in the water amount. The extraction is sample before lipid analysis.
followed by the addition of chloroform and methanol to
make two phases. Lipid is found in the chloroform phase. Precursor Techniques
BHT acts as an antioxidant. KCl form cations, which aid 1. 0.01% BHT: Weigh 1 mg of BHT and add it to 10 mL
in shifting of lipid separation to the organic phase. of water.
2. 1 M KCl: Add 7.45 g of KCl to make a 100-mL solu-
Materials, Equipment, and Reagents tion with water.
A. Reagents: Test samples, anhydrous potassium hydro-
gen sulfate, 0.01% butylated hydroxytoluene (BHT). Analysis and Statistics
B. Glassware: Test tubes, test tube holders. The pungent smell of the fumes indicates the presence of
C. Instrument: Bunsen burner. glycerol in the test sample.
Protocol Safety Considerations and Standards

Bligh and Dyer’s method 1. Chloroform is toxic.
1. Heat 4 mL isopropanol with 0.01% BHT at 75°C in
a 50 mL tube. Pros and Cons
2. Harvest three leaves of Arabidopsis and add it to Pros Cons
the tube. One of the most practiced Less efficient as compared to
3. Heat the samples at 75°C for 15 min. methods for extraction of a other methods of extraction
4. Let the samples cool down to RT. broad range of lipids for specific types of lipids
5. Add 1.5 mL of chloroform and 0.6 mL of water to Usage of chloroform is
the samples. inappropriate for large-scale
6. Shake the samples for 1 h. production, due to its toxic
7. Transfer the extracts to a new tube and extract again and carcinogenic nature
the leaf materials four more times with 4 mL chloro-
form:methanol (2:1) with 0.01% BHT with 30 min
Floch’s method.
of shaking.
8. Combine all the extracts.
Summary
9. Wash the extracts with 1 mL of 1 M KCl and then
1. The method utilizes the phase separation method to
with 2 mL of water.
isolate the lipids from the samples. Lipids get sepa-
10. Evaporate the combined extract and dissolve it
rate out with the chloroform.
again in 1 mL chloroform.
11. Transfer the intact, extracted leaf materials to a new
vial using forceps and weigh it.
EXTRACTION OF LIPID FROM EGG YOLK
Single-step extraction method
AND ESTIMATION OF PHOSPHOROUS
1. Heat 4 mL isopropanol with 0.01% BHT at 75°C in a
CONTENT IN IT
20 mL tube.
2. Harvest three leaves of Arabidopsis and add it to
Rationale
the tube. The extraction of lipid content from the yolk is desirable
3. Heat the samples at 75°C for 15 min. due to its high nutritional value. The lipids are extracted
4. Let the samples cool down to RT. based on different solubilities of neutral lipids and polar
5. Add 12 mL of a mixture of chloroform:methanol: lipids in various organic solvents. The experiment also
water in a ratio of 30:41.5:3.5 so that the final ratio includes β-CD. The removal of cholesterol by adsorp-
of the components is 30:41.5:3.5:25 for chloroform: tion by β-CD is a better protocol as it is nontoxic, non-
methanol:water:isopropanol. hygroscopic, and chemically stable.
CHAPTER 3 Lipid 29
Materials, Equipment, and Reagents 9. Take 1, 2, and 3 mL of phosphate lipid obtained

A. Reagents: Eggs, β-CD solution 95% ethanol, H2O2, above in the test tube nos. 5, 6, and 7, respectively,
5 M H2SO4, KH2PO4, and 0.2 mL of 5% and 1, 2, and 3 mL of nonphosphorylated lipid in
(NH4)2MoO4. test tube nos. 8, 9, and 10, respectively.
B. Glassware: Test tubes, test tube holders, round bot- 10. Take 4 mL of water in the test tube no. 11, this will
tom flask, Kjeldahl flask. serve as blank.
C. Instruments: Homothermic reactor with a magnetic 11. Make up the volume of test tubes to 4 mL by adding
stirrer, centrifuge. water.
12. Add 5 mL of 5 M H2SO4 and 0.2 mL of 5%
Protocol (NH4)2MoO4 solution.
1. Crack the egg and place it in a homothermic reactor 13. To this add 0.3 mL of reducing agent.
with a magnetic stirrer. 14. Shake all the tubes vigorously and heat for 10 min
2. Add 95% ethanol to it. in a boiling water bath.
3. Stir it for 1 h at 65°C. 15. Cool down the tubes and make the volume up to
4. Filter the ethanol extracted fraction and collect it in 50 mL with distilled water.
a beaker. 16. Read the absorbance at 630 nm with reference to the
5. Repeat the extraction process twice. blank.
6. Collect all the extracts in a beaker. The solid egg res-
idue is egg yolk protein free from lipids.
Precursor Techniques
7. Crystalize the extracted fraction at 4°C for 8 h.
1. 1% NaCl: Add 1 g of NaCl to a final volume of
8. Centrifuge the above at 2800 g, 10 min at 4°C.
100 mL in water.
Remove the solidified triacylglycerol.
2. Standard phosphate solution: Add 4.392 g of
9. Collect the supernatant (lipid fraction devoid of
KH2PO4 in 1 L water. Take 10 mL of it and make final
triacylglycerol) and place it in a homothermic reac-
volume 1 L by water.
tor with a magnetic stirrer.
3. 5% (NH4)2MoO4: Add 5 g of (NH4)2MoO4 to a final
10. Add an aqueous solution of β-CD dropwise.
volume of 100 mL.
11. The molar ratio of β-CD: cholesterol is set to
4. Reducing agent: Dissolve 3.75 g of sodium metabi-
be 5:1.
sulfite Na2S2O5, 125 mg of Na2SO3, and 65 mg of
12. Incubate it for at 30 min.
1-amino-2-naphthol-4-sulfonic acid in 25 mL water.
13. Centrifuge the above at 2800 g for10 min.
5. 95% ethanol: Add 95 mL of ethanol to make a final
14. The precipitate is cholesterol.
volume of 100 mL with water.
15. After cholesterol removal, transfer the supernatant
6. β-CD solution: Add 15 mL of water to per gram of
to a previously weighed round bottom flask.
β-CD.
16. Remove ethanol by rotatory evaporation.
17. Collect the dried phospholipids.
Determination of phosphate present in the extracted Analysis and Statistics
lipid The phospholipid and nonphosphorylated lipids were
1. Dry the lipid solution on a water bath. extracted from the egg yolk. The determination of phos-
2. Take 40 mg of phosphorylated and 100 mg of non- phorous content in phospholipid and nonphospholi-
phosphorylated lipids. pids was done.
3. Digest both in a micro-Kjeldahl flask with 1 and
2 mL of conc. H2SO4.
4. Add H2O2 in drops at interval till the mixture
becomes colorless, follow it with incubation for 1. Handle the egg yolk properly.
30 min for the decomposition of H2O2. 2. Care should be taken while working with H2SO4.
5. Transfer the solution to a volumetric flask and make
up to the volume. Summary
6. Now label 11 series of the test tube. 1. The extraction of phospholipids depends on the fact
7. Make a standard phosphate solution with that it is more polar in nature.
10 μg P/mL. 2. The triglyceride is separated from the mixture by
8. Take 1, 2, 3, and 4 mL of the standard solution in the solidifying it by cooling and cholesterol is removed
tubes labeled as 1, 2, 3, and 4, respectively. using the β-CD solution.
EXTRACTION OF LIPID FROM TISSUES Precursor Techniques

(FOLCH METHOD) 1. Chloroform-methanol mixture: Mix 20 mL of chloro-
Rationale form and 10 mL of methanol.
2. Pure solvent upper phase: Mix chloroform, metha-
This protocol includes a step of homogenization of a tis-
nol, and water in a separatory funnel in 8:4:3 by vol-
sue sample in a mixture of chloroform and methanol.
ume. Let it stand, this will result in a biphasic system.
Methanol serves as a polar component, which increases
The upper part of it is a pure solvent upper phase,
the solubilization of lipids present in the cells. Water
while the lower is a pure solvent lower phase. In
and methanol, being polar, comprises salt while the
the upper phase, the proportion of chloroform,
chloroform phase contains lipids.
methanol, and water is 3:48:47 and in the lower
phase, it is 86:14:1.
A. Reagents: Tissue, chloroform-methanol mix, dis- Analysis and Statistics
tilled water, pure solvent upper phase. The lower phase containing the lipid portion of tissue is
B. Glassware: Test tubes, glass-stoppered vessel. obtained.
C. Instrument: Centrifuge.
Protocol 1. Handle the upper and lower phase carefully, do not
1. Weigh 1 g of tissue and homogenize it with 20 mL of mix it.
2:1 chloroform-methanol mixture.
2. Filter the homogenate in a glass-stoppered vessel. Pros and Cons
3. Mix the crude extract with 0.2 of its volume of water Pros Cons
(i.e., 4 mL). The most suitable method for Less efficient as compared to
4. Let it stand (or centrifuge it at 2400 rpm for 20 min) the extraction of a broad the hexane-isopropanol
to separate in two phases (the volume of upper and range of lipids method) for nonpolar
lower phases are 40% and 60%, respectively). lipids
5. Remove the upper phase, as much as possible, by
siphoning.
6. Add 1.5 mL of pure solvent upper phase gently, rotate Bligh and Dyer’s method.
the tube gently to allow the mixing of rinsing fluid
with remaining upper phase. Now remove it. Summary
7. Rinse like above twice. 1. The lipid, being nonpolar in nature, is extracted with
8. Take out the upper phase. a nonpolar phase of the extraction solution.
9. The lower phase consists of lipids.
CHAPTER 4
Protein
A. Isolation (extraction) of protein Materials, equipment, and reagents

1. Isolation of protein from animal cells/tissue. A. Reagents: Tris base, HCl, NaCl, EDTA, NP-40, sodium
B. Qualitative test for proteins and amino acids deoxycholate, SDS, KCl, Na2HPO4, KH2PO4.
1. Biuret test. B. Glassware: Microcentrifuge tubes, pipette, tips, cell
2. Xanthoproteic test. scraper.
3. Millions test. C. Instruments: Tissue homogenizer, refrigerated
C. Quantitative test (estimation) for proteins microcentrifuge.
1. Biuret test.
2. Bradford method. Protocols
3. Bicinchoninic acid (BCA) method. A. From animal tissue:
4. Folin-Lowry method. 1. Dissect the animal, remove the required organ,
5. UV method. and cut into small pieces.
D. Separation of proteins/amino acid 2. Wash the tissue in PBS buffer and transfer it to a
1. Protein separation by gel electrophoresis under microcentrifuge tube. (Store in liquid nitrogen or
denaturing condition (SDS-PAGE). 80°C for later use.)
2. Separation of amino acids by paper 3. Use 1 mL of ice-cold RIPA buffer for 100 mg tissue
chromatography. (this volume of tissue weigh can vary).
4. Homogenize the tissue with the help of a homog-
enizer (with an interval of 30 s).
ISOLATION OF PROTEINS FROM ANIMAL 5. After homogenization keep the tube in ice for
CELLS/TISSUE (USING 10 min.
RADIOIMMUNOPRECIPITATION ASSAY 6. Centrifuge the tube at 13,000 rpm in a refrigerated
BUFFER) microcentrifuge for 15 min.
Definition 7. Transfer the supernatant layer to a separate tube.
Protein extraction or isolation from tissues/cultured 8. Aliquot and store at 20°C or 80°C.
cells or any other sample is the first step for any bio- B. From cultured cells:
chemical as well as molecular analysis (protein purifica- 1. Aspirate the culture media form the adherent cells.
tion, PAGE, western blotting, crystallization, 2. Wash the cells twice with ice-cold PBS buffer.
proteomics, etc.). For protein extraction, cells are first 3. Scrap the cells in 1 mL PBS buffer (in 100 mm
lysed using appropriate lysis buffer, then proteins are dish) with help of scrapper.
isolated. Different lysis buffers are used for extracting 4. Centrifuge to pellet down the cells at 5000 rpm
proteins from the tissues or cultured cells depending for 10 min at 4°C.
upon the cellular fractions (location) needed for pro- 5. Discard the supernatant PBS buffer and wash the
teins. Radioimmunoprecipitation Assay (RIPA) buffer pellet again in ice-cold PBS (optional).
is used for extracting whole-cell proteins as well as 6. Add 1 mL of ice-cold RIPA buffer and lyse the cells.
membrane-bound and nuclear fraction. 7. Tap the tube for some time and keep in ice for
5–10 min. Repeat this procedure for 30–40 min
Rationale till the cell pellet lysed completely.
RIPA lysis buffer is used for rapid lysis of cells and effi- 8. Centrifuge the tube for 15min at 13,000rpm at 4°C.
cient solubilization of proteins from membranes, 9. Carefully transfer the supernatant in the separate
nuclear as well as cytoplasmic fractions. The RIPA buffer tube and discard the debris at the bottom.
contains ionic [sodium dodecyl sulfate (SDS)] as well as
nonionic (Triton X-100) detergents that disrupt the Precursor techniques
membrane and separate the membrane-bound, cyto- 1. Phosphate buffer saline (PBS): Dissolve 8 g NaCl,
plasmic proteins in the buffer. 0.2 g KCl, 1.44 g Na2HPO4, and 0.24 g KH2PO4 in

800 mL distilled water. Adjust the pH 7.4 with HCl. Pros and cons
Make the final volume of 1000 mL. Pros Cons
2. RIPA buffer: Extracted proteins are compatible Not suitable for study
with various assays protein-protein
Volume for 10 mL
interaction
Reagents Stock Working Buffer
The lysis buffer does not interfere
Tris-HCl pH 7.4 1M 50 mM 500 μL with the immunoreactivity
NaCl 4M 150 mM 375 μL hence suitable for western
EDTA (pH 8) 0.5 M 1 mM 20 μL assay, immunoassays
SDS 10% 0.1% 100 μL
Triton X-100 100% 1% 100 μL
Na-deoxycholate 10 mg Alternative methods/procedures
Distilled water Remaining to make NP-40 buffer, Laemmli sample buffer (LSB).
the volume 10 mL
Protease 3–4 μL
inhibitors Troubleshooting and optimization
Phosphatase 3–4 μL Problem Solution
inhibitors or Low protein Use less RIPA buffer
sodium concentration
orthovanadate Protein Use protease inhibitors in the buffer just
degradation before use
• 1 M Tris HCl (pH 7.4): Dissolve 121.1 g Tris base in
Low protein yield Lyse the cells completely in the buffer
800 mL distilled water. Adjust the pH 7.4 using HCl. and incubate in the buffer for a
Make the final volume of 1000 mL and autoclave. longer time
• 4 M NaCl: Dissolve 58.44 g NaCl in 250 mL of dis-
tilled water.
• 0.5 M EDTA (pH 8): Dissolve 18.61g disodium ethyl- Summary
enediaminetetraacetate. 2H2O into 80mL of distilled 1. The cultured cells are first harvested, then lysed in
water. Mix and shake vigorously using a magnetic stir- RIPA buffer. The lysed cells are centrifuged at
rer. Adjust the pH 8 using NaOH pellets. Once the pH 13,000 rpm for 15 min at 4°C. The solubilized
adjusted and salt dissolves, make the final volume proteins in the supernatant is then transferred to
100 mL by distilled water and autoclave. another tube.
• 10% SDS: Dissolve 10g SDS in 100 mL distilled water. 2. The animal tissue is dissected and cut into pieces.
• Amount of RIPA buffer used for cultured cells (grown The tissues are homogenized in RIPA buffer, centri-
in). fuged, and the supernatant is transferred to another
100 mm dish 1 mL tube.
60 mm dish 0.5–0.6 mL
35 mm dish or 6 well plate 0.3 mL Definition
12 well plate 0.2 mL
To detect the presence of protein in the given sample by
24 well plate 0.1 mL
the Biuret test.
1. Protease and phosphatase inhibitors must be added Rationale
just before use. Biuret test is used for the detection of peptide bonds in
2. All the steps for cell lysis and protein extraction must the protein. This test is not suitable for amino acid only.
be carried out at 4°C or ice. The protein solution when reacts with CuSO4 in alkali
3. Autoclave all the reagents and solutions after condition (NaOH/KOH), purple to violet-colored cop-
preparations. per (II) cation complex is formed. The intensity of the
complex is directly proportional to the number of the
Analysis and statistics peptide bond in the protein sample.
The isolated proteins are quantitated by an appropriate
NaOH
method and stored at 20°C. Protein + CuSO4 ! Copper cation complex ðpurple to violetÞ
CHAPTER 4 Protein 33
Materials, equipment, and reagents Analysis and statistics

A. Reagents: Sodium hydroxide, copper sulfate, Color changes to violet indicate the presence of proteins
sample. in the sample. No change in color means the absence of
B. Glassware: Test tube, test tube holder, dropper, bea- protein.
ker, spirit lamp.
Pros and cons
Pros Cons
Protocols
1. Take 2 mL of samples in different test tubes. Very easy and quick assay Less sensitive
The assay does not depend on Detection interfered by the
2. Add 2 mL of 40% NaOH solution in each of the test
the amount and presence of other
tubes.
composition of amino contaminants like lipid,
3. Add four to five drops of 1% CuSO4 in the solution. acids in the protein nucleic acids in the sample
4. Warm the mixture for about 5 min (optional).
5. Observe the color change.
Millions test, Xanthoproteic test, Ninhydrin test.
1. 40% NaOH—dissolve 40 g NaOH in 100 mL of
water. Summary
2. 1% CuSO4—dissolve 1 g CuSO4 in 1 mL of water. 1. This is a widely used method for the identification of
proteins.
2. The reaction of the peptide bond in the protein with
Safety considerations and standards CuSO4 in alkali conditions forms a purple-violet
1. Glassware should be washed before and after use. complex.
2. Handle the chemicals carefully. 3. This method can be used for the detection of proteins
3. Keep the test tube away from the body while heating. in the biological fluid as urine, plasma, blood, etc.
BIURET TEST
Negative
r
Colo Biuret test
ue
Bl
NaOH CuSO4 Heat
Vi
ol
et
Sample Co
lo
+ rc
NaOH ha
Sample ng
+ e
Sample + CuSO4
NaOH
Positive Biuret
test
Definition
To detect the presence of protein in the given sample by Heating NaOH
Protein + conc:HNO3
! Yellow nitro compound ! Orange color
the Xanthoproteic test.
Rationale Materials, equipment, and reagents

Biuret test is used for the detection of proteins with tyro- A. Reagents: Sodium hydroxide, conc. HNO3, sample,
sine and tryptophan amino acid in their composition. tyrosine, tryptophan powder.
Phenylalanine does not give a positive result. B. Glassware: Test tube, test tube holder, dropper, bea-
Proteins having these amino acids give yellow color ker, spirit lamp.
when heated with conc. HNO3. The aromatic benzene
ring undergoes nitration. Upon adding the alkali solu- Protocols
tion (NaOH), the yellow color changes to orange. Thus, 1. Take 1 mL of the sample in a test tube.
the appearance of the orange color in a solution with 2. Take 1 mL of distilled water in another test tube as a
HNO3 and NaOH indicates the presence of protein with negative control.
tyrosine and tryptophan amino acids. This test also gives 3. Take 1 mL of 1% tryptophan or 1% tyrosine solution
positive results with a solution that contains only tryp- in another test tube as a positive control (optional).
tophan or tyrosine. 4. Add 1 mL of conc. HNO3 in each of the test tubes.
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people according to their capacity, and the more gifted according to
their abilities. Certain fundamental truths are therefore set forth in the
Law and the Prophets in an authoritative, poetical, or dialectical
style; but the higher order of intellects are encouraged to search for
proper proofs. Thus the whole nation is addressed by Isaiah: ‘Lift up
your eyes on high and see who hath created these,’ and the like.
Also the chief of the Prophets tells the Israelites: ‘Hear, O Israel, the
Lord is our God; the Lord is One.’ Those who are more highly
endowed than their fellow-men are exhorted, either directly or
indirectly, to follow the course which is suitable to them. The direct
exhortation to philosophical research is contained in the words:
‘Know then this day, and take it to thy heart, that the Lord He is God,’
&c.; and indirectly it is contained in the commandment [16]to love and
to fear God, as has been explained by R. Moses Maimonides.—The
study of science will certainly be of use to the scholar; it leads to a
knowledge of the created things, and through these to a knowledge
of the Creator. Such study may even be considered as necessary to
the Jewish scholar, though not to the ordinary Jew. The scholar
must, however, not entirely rely on his research, but on that which is
taught in the Law. In this the scholar and the ordinary man are equal,
that both accept the teaching of the Torah as infallible; only with this
difference, that the scholar can in addition satisfy his thirst for
knowledge and confirm by scientific proof what he has already
accepted as true on the authority of the Bible.”
Of modern scholars I only quote Moses Mendelssohn’s theory. He

accepts unconditionally the teaching of the Bible; all its truths are
absolute and perfect; no reasoning whatever can refute them; but
difficulties may sometimes present themselves to us in reconciling
the teaching of the Bible with that of our reason. What have we then
to do? The philosopher declares: “If I were to find my reason in
contradiction to the Word of God, I could command reason to be
silent; but the arguments, so long as they have not been refuted, will
nevertheless assert themselves in the innermost recesses of my
heart; the arguments will assume the form of disquieting doubts,
which will resolve themselves into childlike prayers, earnest
supplication for enlightenment. I should utter the words of the
Psalmist: ‘Lord, send me Thy light, Thy truth, that they may [17]guide
me, and bring me to Thy holy mount, to Thy dwelling-place!’ ”
The conception which Moses Mendelssohn had of Jewish belief and

its relation to reason we learn from the following passage:—“I
recognise no other eternal truths than those which are not only
comprehensible to the human mind, but also demonstrable by
human powers. This principle by no means brings me into conflict
with my own religion; on the contrary, I consider it an essential
element in Judaism, and the characteristic difference between
Judaism and Christianity. Judaism has no revealed religion in the
sense in which Christianity has. The Jews have a revealed
legislation which instructs them in the divinely ordained means by
which they may attain the eternal bliss. Laws and rules for conduct in
life were revealed to Moses in a supernatural way, but no doctrines,
no saving truth, and no general laws of logic. The latter the Eternal
reveals to us, as to all men, through nature and through the things
themselves; never through words and letters. The divine book
revealed to Moses, though a book of laws, includes an inexhaustible
treasure of truths and doctrines.… The more we study it the more we
wonder at the depth of the knowledge contained in it. But these
truths are taught, and not forced upon us as dogmas. Belief does not
allow itself to be commanded; it is based upon conviction. In the
Hebrew language, the very word which is generally translated ‘faith,’
viz., ‫‏אמונה‬‎denotes originally confidence, trust that the promise made
will also be fulfilled, and not what we understand by ‘religious faith.’ ”
[18]
These words of Mendelssohn show how greatly those err who quote
his opinions in support of the dictum that Judaism recognises no
dogmas. According to Mendelssohn, Judaism does not consist
entirely of laws; it teaches also certain truths. We have certain
dogmas without which the laws can have no meaning, yet there is no
precept, “Thou shalt believe.” Nowhere in our Law, whether written
or oral, is a solemn declaration of our creed demanded. In so far
Mendelssohn’s view is correct; but when he believes that all the
truths we are taught in Scripture can be made evident by logical
demonstration he is mistaken. As to the meaning of ‫‏אמונה‬‎comp.
supra, p. 4. [19]
[Contents]
THE THIRTEEN PRINCIPLES
OF
OUR CREED.
The main source of our creed is the Bible, and among the Biblical
books, chiefly the Pentateuch (‫‏תורה‬‎). In these books we find many
truths taught by God Himself, or by His inspired messengers, and
they form the substance of our creed. It matters little how we arrange
them, how we collect them into groups, and subdivide these again,
provided we believe in them implicitly. In the Bible they are not
arranged systematically; they are intermingled with, and are
contained implicitly in, the history and the laws that form the subject-
matter of the Scriptures; it is the observance of those laws which
constitutes the best evidence of the belief seated in the heart. No
declaration or recital of a creed is commanded in the Pentateuch; no
tribunal is appointed for inquiring whether the belief of a man is right
or wrong; no punishment is inflicted or threatened for want of belief.
It became, however, necessary to formulate the truths taught in the
Bible, when disputes arose as to their meaning and to their validity.
The Mishnah, therefore, declares certain opinions as un-Jewish and
contrary to the teaching of the Divine Word. Later on, when
controversies [20]multiplied between the various sections of the
Jewish nation, as well as between Jews and Christians and Jews
and Mohammedans, it was found most important to settle the form
and arrangement of our beliefs. Moses Maimonides, the great
religious philosopher, taught, in his Commentary on the Mishnah,
thirteen principles of faith, which found general acceptance among
the Jews, and are known as the Thirteen Principles. They have
found their way into the Prayer-book in two different forms, one in
prose and one in poetry. Maimonides, in commending them to the
reader, says: “Read them again and again and study them well, and
let not your heart entice you to believe that you have comprehended
their full meaning after having read them a few times; you would then
be in a great error, for I have not written down what occurred to my
mind at first thought. I first thoroughly studied and examined what I
was going to write, compared the various doctrines, the correct ones
and the incorrect ones, and when I arrived at what we ought to
accept as our creed, I was able to prove it by arguments and
reasoning.” The thirteen articles as put forth by Maimonides, and
called by him principles and foundations of our religion, are the
following:—
1. The first principle: The belief in the existence of the Creator; that
is, the belief that there exists a Being who requires no other cause
for His existence, but is Himself the cause of all beings.
2. The second principle: The belief in the Unity of God; that is, the
belief that the Being who is the cause of everything in existence is
One; not like the unity of a group or class, composed of a certain
[21]number of individuals, or the unity of one individual consisting of
various constituent elements, or the unity of one simple thing which
is divisible ad infinitum, but as a unity the like of which does not
exist.
3. The third principle: The belief in the Incorporeality of God; that is,
the belief that this One Creator has neither bodily form nor
substance, that He is not a force contained in a body, and that no
corporeal quality or action can be attributed to Him.
4. The fourth principle: The belief in the Eternity of God; that is, the
belief that God alone is without a beginning, whilst no other being is
without a beginning.
5. The fifth principle: The belief that the Creator alone is to be

worshipped, and no other being, whether angel, star, or ought else,
all these being themselves creatures.
6. The sixth principle: The belief in Prophecy; that is, the belief that
there have been men endowed with extraordinary moral and
intellectual powers, by which they were enabled to reach a degree
and kind of knowledge unattainable to others.
7. The seventh principle: The belief that our teacher Moses was the
greatest of all prophets, both those before him and those after him.
8. The eighth principle: The belief in the Divine origin of the Law; the
belief that the whole Pentateuch was communicated to Moses by
God, both the precepts and the historical accounts contained therein.
9. The ninth principle: The belief in the integrity of the Law; that both
the written and the oral Law are of Divine origin, and that nothing
may be added to it or taken from it. [22]
10. The tenth principle: The belief that God knows and notices the
deeds and thoughts of man.
11. The eleventh principle: The belief that God rewards those who
perform the commandments of His Law, and punishes those who
transgress them.
12. The twelfth principle: The belief that Messiah will come at some
future time, which it is impossible for us to determine; that he will be
of the house of David, and will be endowed with extraordinary
wisdom and power.
13. The thirteenth principle: The belief in the revival of the dead, or
the immortality of the soul.
These thirteen principles (‫‏שלשה עשר עקרים‬‎) may be divided into three
groups, according to their relation to the three principles:—1.
Existence of God. 2. Revelation. 3. Reward and punishment. The
first group includes the first five principles, the second the next four,
and the third the remaining four. In this order they will now be
considered.
[Contents]
1. Existence of God ‫‏מציאות הבורא‬‎.
The notion of the existence of God, of an invisible power which

exercises its influence in everything that is going on in nature, is
widespread, and common to almost the whole human race. It is
found among all civilised nations and many uncivilised tribes. The
existence of God may be regarded as an innate idea, which we
possess from our earliest days. This is the origin of Natural Religion.
Thinkers of all ages and nations have attempted to confirm this
innate idea by convincing arguments. Prophets and divine poets
[23]have frequently directed the attention of those whom they
addressed to the marvels of nature in order to inspire them with the
idea of an All-wise and All-powerful Creator.
“Lift up your eyes on high, and behold who hath created these? Who
is He who bringeth them forth by number? All of them He calleth by
name, by the greatness of His might, and for that He is strong in
power, not one is lacking” (Isa. xl. 26). “The heavens declare the
glory of God, and the firmament sheweth His handywork” (Ps. xix. 2).
The regularity in the rising and setting of the heavenly bodies, which
enables us to foretell the exact time and duration of an eclipse of the
sun or the moon, is certainly a strong argument for the belief that
there is a mighty and wise Creator who fixed the laws in accordance
with which these luminaries move.
“Beautiful are the luminaries which our God has created. He has
formed them with knowledge, reason, and understanding; He
endowed them with power and strength to rule in the midst of the
world. Full of splendour and beaming with light, they illumine the
whole world; they rejoice when they rise, they are glad when they
set, doing in reverence the will of their Master” (Sabbath Morning
Service).
A similar regularity we notice when looking on the face of the earth.

The various seasons of the year, each with its peculiar aspect and
influence, the sequence of day and night at regular intervals, the
gradual and systematic development of vegetable and animal life—
all point forcibly to the fact that these [24]things do not owe their
existence to chance, but to the will of an Almighty and All-wise
Creator.
Again, if we consider the structure of a single plant, or of a single

animal, we find that every one of the members and parts of which it
is composed has its peculiar function or purpose in the economy of
the whole plant or the whole animal. Let one of these component
parts refuse its function or cease to fulfil its purpose and the whole is
disorganised. Certainly there must be a Being who makes the
different members of an organism co-operate for the development
and advantage of the whole. The idea of purpose which regulates
this co-operation cannot have originated in the parts nor in the
whole, but in the conception of Him by whose Will these were
created.
“The finger of God” is further recognised in the important events of

the life of the individual as well as in the history of whole nations. We
are frequently reminded of the lesson, “The heart of man deviseth
his way, but the Lord directeth his step” (Prov. xvi. 9). “Salvation is
the Lord’s, and on Thy people it is incumbent to bless Thee” (Ps. iii.
9).
Another argument in support of the belief in the existence of God is

taken from the moral consciousness which every human being
possesses. This points to the existence of a higher Being, perfect in
goodness, as the origin and cause of the moral consciousness in our
own heart.
These and similar arguments are employed to strengthen and purify

our belief in God. The question, however, arises, are these
arguments alone sufficient [25]to convince us? Are they strong
enough to resist the attacks of scepticism?
On examining them thoroughly we shall find them of excellent

service to the believer. His belief is strengthened against many
doubts by which he may be assailed; and scepticism will be kept at
bay by these arguments. But of themselves and unsupported they
may not always suffice to establish belief in God; and if they carry
conviction with them for the moment, we are not sure whether fresh
arguments of opponents might not again unsettle the mind. Another
method was therefore chosen by the Almighty, by which certainty is
attained, and a sure guide is given for our moral and religious life. It
is Revelation. Of this we shall speak later on.
The principal forms of religion or worship that sprang from the

natural belief in God are Polytheism, Pantheism, Atheism, Theism,
and Deism.
1. The first form of Divine worship of which history and archæology

give us information is Polytheism. The creating and ruling power of
some invisible Being was noticed everywhere. Every manifestation
of such influence was ascribed to its peculiar deity, which was
worshipped according to the peculiar conception of the deity in the
mind of the individual person, family, or nation. This is chiefly the
kind of idolatry mentioned in the Bible and combated by the
prophets.
A very general object of worship were the stars. Rabbi Jehudah ha-
Levi, in Kuzari iv. 1, in trying to explain the origin of this practice,
says as follows:—“The spheres of the sun and the moon do not
move in the same way. A separate cause or god was therefore
[26]assumed for each, and people did not think that there was a
higher force on which all these causes depended.” The ancient
monuments and the treasures stored up in our museums show how
great was the variety of forms which idolatry took, and to how great
an extent people adhered, and still adhere, to this kind of worship.
But there have been thinkers and philosophers even among the
idolatrous nations who sought a unity in the construction and working
of the universe, and early arrived at the idea of a First Cause as the
sole source of all that exists.
2. The fact that the influence of the Divine power makes itself
perceptible to the observing eye of man everywhere produced
another kind of human error: Pantheism (All-God). Modern
Pantheism dates from Spinoza; but long before Spinoza, when the
secret forces at work in the changes noticed by us in all material
objects were recognised as properties inherent in the substance of
things, these forces were considered as the sole independent
causes of the existing universe, and the combination of these forces,
called Nature, was considered to be the First Cause, or God. A
modification of this theory is contained in the philosophy of Spinoza.
According to this great philosopher’s system, the universe in its
entirety has the attributes of the Deity: there exists nothing but the
Substance (God), its attributes, and the various ways in which these
attributes become perceptible to man. Spinoza tried to defend
himself from the reproach of describing God as corporeal, but he did
not succeed. The attribute of extension or space which God
possesses, according to Spinoza, is only conceivable [27]in relation
to corporeal things. The philosophy of Spinoza is in this dilemma:
either God is corporeal, or the corporeal world does not exist. Both
assumptions are equally absurd. It is true, in one of his letters he
complains that he has been misrepresented, as if he believed God to
consist of a certain corporeal mass. But we cannot help assuming
the existence of a certain corporeal mass, and if this is not God, we
must distinguish in our mind God and something that is not God,
contrary to the fundamental doctrine of Pantheism. Besides, there
are many incongruities and improbabilities involved in this theory. It
has no foundation for a moral consciousness. The wicked and the
good are alike inseparable from God. They both result with necessity
from the attributes of God, and they cannot be otherwise than they
actually are. If we, by the consideration that injury done to us by our
fellow-man was not done by that person alone, but by a series of
predetermined necessary causes, may be induced to conquer hatred
against the apparent cause of our injury, we may equally be induced
by the same reasoning to consider the kindness and benefits of our
friends not worthy of gratitude, believing that they were compelled to
act in this manner, and could not act otherwise.
3. Pantheism, by teaching All in One and One in All, is opposed to

the theory of man’s responsibility to a higher Being, denies the
existence of God in the ordinary sense of the word, and is, in its
relation to true religion, equal to atheism.
In the Bible atheism is stigmatised as the source of all evils. Thus the
patriarch Abraham suspected the [28]people of Gerar, that there was
“no fear of God” in the place, and was afraid “they might slay him”
(Gen. xx. 11); whilst Joseph persuaded his brothers to have
confidence in him by the assertion, “I fear God” (Ib. xlii. 18). The first
instance of an atheist we meet in Pharaoh, king of Egypt, when he
defiantly said, “I know not the Lord, neither will I let Israel go” (Exod.
v. 2). Another form of atheism is warned against in the words of
Moses: “Lest thou sayest in thine heart, My strength and the power
of my hand has got for me all this wealth” (Deut. viii. 17); and “Lest
they say, Our hand is high, and it is not the Lord that hath done all
this” (Ib. xxxii. 27). The prophets likewise rebuke the people for want
of belief in God. In the Psalms, the crimes and evil designs of
oppressors are traced to godlessness. “The wicked says in his heart,
There is no God” (Ps. xiv. 1). But this atheism of the Bible is not a
theoretical or dogmatic one; it is not the result of thought, or of deep
inquiry into the causes of things, but merely the voice of an evil
inclination which tempts man to act contrary to the command of God,
and assures him of immunity, under the impression that his actions
are not watched by a higher authority. In post-Biblical literature we
meet with the phrase, ‫‏לית ִּד ין ולית ַּד ָּין‬‎“There is no judgment, and there
is no judge,” as the basis of atheism.
4. Although the conviction of man’s responsibility to a higher

authority is the essential element in the belief in God, yet the notion
of godlessness was so intimately connected with crime and
wickedness, that those who rejected the authority and mastership of
the [29]Deity refused to be called godless or atheists. Many
philosophers retained the name “God” (theos, deus) for their “First
Cause” of the universe, although it is deprived of the chief attributes
of God. Thus we have as the principal religious theories resulting
from philosophical investigations, Theism and Deism. Literally these
two terms denote, Theory of God, or Belief in God; the one word
being derived from the Greek theos, the other from the Latin deus,
both meaning “God.”
There is, however, an essential difference between the two theories.

Theism and Deism have this in common, that both assume a
spiritual power, a divine being, as the cause and source of
everything that exists. They differ in this: to Theism this power is
immanent in us and the things round us; Deism considers this power
as separate from the things. Revelation or prophecy is altogether
denied by the Deists, whilst the Theists would accept it after their
own fashion and rationalise it.
All these various systems of religion have this in common, that they
attempt to remove from religion everything that cannot be
comprehended by human reason. But all attempts to substitute
human reason for Divine authority have failed. A limit has been set to
human reason, and that cannot be overcome. In every system of
religion—the natural and the rational included—there is a mystic
element, which may be enveloped in a mist of phrases, but remains
unexplained. Whether we call the Creator and Ruler of the universe
God, Deus, or Theos, His relation to the universe, and to man in
particular, cannot be [30]determined by the laws which determine the
natural phenomena in the universe, created by His Will.
What is our conception of the Deity? The fundamental idea, from

which all our notions concerning God are derived, and which we
have in common with all other believers in God, is that He is the First
Cause, the Creator of the universe. This idea expressed in the term
‫‏הבורא יתברך שמו‬‎forms the basis of our creed. It is the Creator that is
described in it. Seven of the articles begin, “I believe with a perfect
faith that the Creator, blessed be His name,” &c.
We do not use the term “First Cause,” because it is too narrow; it

only expresses part of the truth, not the whole of it. By “First Cause”
some understand the cause of the gradual development of the
primitive matter into the innumerable variety of things contained in
the universe; the development of the original chaos into system and
order. It is true that the Creator is the cause of all this; but He is more
than this: He is the cause of the primitive matter, and of the original
chaos. For He has created the world out of nothing. The first verse of
the Bible teaches us creation from nothing (creatio ex nihilo): “In the
beginning God created the heaven and the earth” (Gen. i. 1); that is,
the whole universe. It is true that there were men who explained the
meaning of the Hebrew root ‫‏ברא‬‎in a different manner, and desired to
assign to it the meaning: cutting out, forming out of a given material.
But they certainly misunderstood the spirit of the Scriptures. The
eternal coexistence of God and matter would imply a dualism utterly
incompatible with the teaching of the Bible. The frequently repeated
declaration, “He is our God; there [31]is none besides” (‫‏אין עוד‬‎),
clearly excludes every form of dualism. Those who assert that the
universe could not come from nothing belong to the class of people
of whom the Psalmist says, “And they returned and tempted God,
and set limits to the Holy One of Israel” (Ps. lxxviii. 41).
If we cannot understand the act of the Creation, it is our own intellect

that is limited; and if we were to persuade ourselves that we
understand better the eternity of matter, we should deceive
ourselves. We cannot conceive matter without form as existing in
reality, nor can we have a clear notion of anything infinite. We are
human beings, endowed by the will and wisdom of the Creator with
limited physical and intellectual faculties, and in things that surpass
our powers we cannot do better than follow the guidance of the
Divine Word. If we do so we may be sure that we shall be on the
right way to truth.
The first principle declared in our creed is this: God is not only the
Creator of the heavens and the earth, with all their hosts; He is also
the constant ruler of all created beings; He is ‫‏בורא ומנהיג‬‎. We
therefore praise Him in our daily Morning prayer as “Doing wonders;
renewing in His goodness the work of the creation every day.” When
we observe the ordinary phenomena in nature, occurring in
accordance with certain fixed laws which have been discovered and
described by man, we see in them the greatness of the Creator by
whose will these laws are still in force, and by whose will any or all of
these laws may one day cease to continue.
It has been asserted that any interruption or change [32]of these fixed
laws would indicate a weakness and want of foresight on the part of
the Creator, and a fault in the plan of the Creation. This notion has
led people either to deny the truth of the Biblical accounts
concerning the miracles wrought by the Almighty, or to admit the
correctness of the facts while denying their miraculous character, or
to consider the fixed laws of nature, together with their exceptions,
as designed in the original plan of the Creation. How short-sighted is
man! He cannot even fully comprehend his own short-sightedness!
God made him ruler over the works of His hands, and he presumes
to be the ruler of God Himself! When we learn from numerous
observations and experiments the law that seems to regulate certain
recurring phenomena, have we then fathomed the infinite wisdom of
God in the Creation? Do we know the reason which led Him to
produce certain things according to certain laws, and not otherwise?
Have we in discovering a law of nature obtained the power of
prescribing the same law to God, and disallowing Him to deviate
therefrom? Far be it from us human beings, dust and ashes, to
arrogate to ourselves such a right! It may even be one of the objects
with which miracles were wrought to teach us that we do not yet
know all things, that events may happen which we are unable to
foresee, that phenomena may appear which we are unable to
explain according to the laws hitherto discovered; in short, that our
knowledge and wisdom are limited.
The fact that God has created the universe ex nihilo has been
expressed by Jewish philosophers as follows:—God is the only
Being who demands no cause for His existence; the very idea of
God implies existence, [33]and cannot be conceived without it. All
other beings owe their existence to certain causes, in the absence of
which they would not exist. God alone is therefore only active,
without ever being passive, only cause without ever being effect,
whilst every other being is both active and passive, cause and effect;
it has been produced by certain causes, and is in its turn the cause
of the existence of other beings. In the first article a phrase
expressing this idea has been added: “And He alone is the active
cause of all things, whether past, present, or future.” By the addition
of this sentence it was intended to deny the Eternity of matter (‫‏קדמות‬
‫העולם‬‎). The reference to past, present, and future is to emphasise
the constant action of the Creator, and the dependence of the natural
forces on His Will. The first principle has, therefore, the following
form:—
“I firmly believe that the Creator, blessed be His name, is both

Creator and Ruler of all created beings, and that He alone is the
active cause of ALL things, whether past, present, or future.” 1
Before passing on to the second principle concerning God, let us

briefly answer a question that has frequently been asked: What is
the relation between the theory of evolution, or in general the results
of modern science, and the history of the creation as related in the
Bible? In the Biblical account of the creation the various kinds of
plants and animals are described as the result or different and
distinct acts of the Creator, whilst according to the theory of
Evolution one creative act sufficed, and the great variety of creatures
is the result [34]of gradual development according to certain laws
inherent in the things created. The Bible tells us of six days of the
creation, whilst according to the theory of evolution it must have
taken millions of years before the various species could have
developed the one from the other. Whilst the Biblical account
describes the earth as the centre of the universe, astronomy shows
that the earth is one of the most insignificant of the bodies that fill the
infinite space of the universe. According to astronomy and geology,
the age of the earth numbers millions of years; from the Biblical
account we infer that the earth is comparatively young. In the Bible
man is described as the aim and end of the whole creation; natural
history and the theory of evolution consider man simply as one of the
forms resulting from a natural development of the animal world.
What shall be our decision in this discrepancy? Shall we shut our
eyes to the results of modern science in our firm belief in the truth of
the Bible? Or shall we accept the former and abandon the latter?
We should adopt neither of these alternatives. We have great

confidence in our reasoning power, and in the results of science
based on reason, but we have still greater confidence in the
truthfulness of Divine teaching. The conflict is not a modern product;
it existed in former times as well. When the Jews first became
acquainted with Greek literature and philosophy, faith was shaken in
the heart of many a Jew that was led away by the attractive
language and the persuasive arguments of the Greek. Such was the
case with the Jews in Alexandria, who were almost [35]more Greek
than Jewish. Feeling that their faith in their old traditions was
beginning to give way, they looked about them for the means of
reconciling faith and philosophy. Where the literal sense of Holy Writ
was awkward, the allegorical interpretation was substituted for it; but
the authority of the Bible was recognised. Later on, in the Middle
Ages, when Aristotle, as understood and interpreted in the Arabic
schools, was infallible, perplexity again became general, among the
educated and learned, as to the course to be pursued in case of a
conflict; whether to remain true to the Bible or to join the banner of
Aristotle. The most prominent amongst the Jewish theologians who
sought the way of reconciliation was Moses Maimonides. This
philosopher wrote his famous work, “Guide of the Perplexed,”
expressly for those scholars who, whilst firmly adhering to the
inherited faith, had been trained in the study of philosophy, and were

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Protocols in Biochemistry and Clinical

Biochemistry 1st Edition Buddhi

Concepts and Cases in Biochemistry As per the revised

Clinical Case Discussion in Biochemistry 1st Edition

Marks Basic Medical Biochemistry A Clinical Approach

Practical Biochemistry with Clinical Correlation for

Marks Basic Medical Biochemistry A Clinical Approach

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Biochemistry and Cell Biology of Ageing Part IV

Plant Biochemistry Caroline Bowsher

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Thereafter, the particle will sediment at a constant Types of Centrifuges

Before centrifugation After centrifugation

Buoyant force and frictional force

Particles of various sizes Particles sedimented according to

of gradients—continuous and discontinuous. It has Ff ¼ frictional force

Direction of migration of nucleic acids

Wells for Electrophoresis

FIG. 2 Electrophoresis apparatus.

7. The pH of buffers: pH affects the ionization of the

Paper for Paper for

Spot of sample/ Resolved

solutions—these solutions are prepared directly by Types of reaction in titrimetry:

Solutions and Acids and Bases

Protocols in Biochemistry and Clinical Biochemistry. https://doi.org/10.1016/B978-0-12-822007-8.00009-X

The total charge on either of them is 3 2.

Gram equivalent of solute

½H + ½OH Hasselbalch equation is very important. According to

CH3 COO + H2 O Ð CH3 COOH + OH Problem

or get precipitated with ethanol, so cannot be used in

Protocols in Biochemistry and Clinical Biochemistry. https://doi.org/10.1016/B978-0-12-822007-8.00007-6

Analysis and statistics Pros and cons

Definition Analysis and statistics

Materials, equipment, and reagents

PICRIC ACID TEST

Definition The aldehyde/ketone group is oxidized to acid and

Therefore, the net reaction is Precursor techniques

Pros and cons

Analysis and statistics Summary

Fehling Fehling Mix and Positive

Definition and sodium citrate complexes with cupric ions to pre-

Analysis and statistics Materials, equipment, and reagents

In Tommer’s test, copper sulfate furnishes cupric ions; Rationale

Bismuth subnitrate + alkali!Bismuth hydroxide

Summary Easy and quick method Nonreducing sugars cannot be

2. Bial’s reagent: Dissolve 0.4 g orcinol in 200 mL Analysis and statistics

Safety considerations and standards Precursor techniques

Pros and cons Safety considerations and standards

Materials, equipment, and reagents

Materials, equipment, and reagents Rationale

Materials, equipment, and reagents Definition

Analysis and statistics Rationale

Materials, equipment, and reagents

5. Draw a line across it and remove the plate. Definition

Protocols in Biochemistry and Clinical Biochemistry. https://doi.org/10.1016/B978-0-12-822007-8.00006-4

3. Shake it thoroughly. Precursor Techniques

DETERMINATION OF DEGREE OF Pros and Cons

Bismuth subnitrate + alkali!Bismuth hydroxide