Centrifugation techniques such as velocity sedimentation centrifugation and fractional centrifugation are used to separate cellular components by size and weight. Velocity sedimentation centrifugation involves spinning a sample at low or high speeds to separate heavier precipitates from lighter organelles. Fractional centrifugation uses successive centrifugations at increasing speeds to separate cellular organelles into pellet and supernatant fractions. Spectrophotometry measures the absorption or emission of light by molecules. Beer's law states that absorbance is directly proportional to concentration, with absorbance measured using spectrophotometers and concentration determined using calibration curves.

1. 1
Application of centrifugation
◘ 1- Velocity Sedimentation Centrifugation :-
♦ Centrifuges are used most often For Preparative-Scale Separation
procedures .
♦ this technique consist of :-
- Placing the sample in a tube or similar container
- Inserting the tube in the rotor
- Spinning the sample for a fixed period .
- The sample is removed and the two phases ( Pellet and supernatant )
may be separated by careful decantation .
♦ It separate particles ranging in size :
- From Coarse precipitates to cellular organelles .
- Heavy precipitates are sediment in Low-Speed Centrifuges .
- Lighter Organelles such as ribosomes require the High centrifugal
forces of an ultracentrifuge
♦ Further Characterization or analysis is usually carried out in the
individual phases .

2. 2
◘ 1- Fractional Centrifugation :
♦ Much of our current understanding of cell structure and function depends
on separation of subcellular components by centrifugation .
♦ This specific method of separation consists of successive centrifugation at
increasing rotor speeds .
♦ the rotor chamber must be kept at low temperature to maintain the native
structure and function of each cellular organelle ant its component
biomolecules .
♦ The Operation :-
- A high-speed centrifuge equipped with a
fixed angle rotor is the most appropriate
for the first two centrifugations at 600 xg
and 20,000 xg .
- After each centrifuge run , the supernatant
is poured into another centrifuge tube ,
which is then rotated at the next higher
speed .
- The final centrifugation at 100,000 xg to
sediment microsomes and ribosomes must be done in an ultracentrifuge .
- The 100,000 xg supernatant , the cytosol , is the soluble portion of the cell and
consists of soluble proteins and smaller molecules.

3. 3
Spectrophotometry
◘ The instruments that measure electromagnetic radiation have several
concepts and components in common .
◘ Photometric instruments measure light intensity without consideration of
wavelength .
◘ Most instruments today use :-
- filter (photometers)
- prisms , gratings ( spectrometers )
♠ to isolate or select a narrow range of the incident wavelength .
◘ Radiant energy that passes through an object will be partially reflected ,
absorbed , and transmitted .

4. 4
◘ Electromagnetic radiation : is described as photons of energy travelling
in waves .
◘ The relationship between wavelength and energy E
is described by Planck’s formula :
- Because the frequency of a wave is inversely
proportional to the wavelength .., the energy of
electromagnetic radiation is inversely
proportional to wavelength .
◘ Electromagnetic radiation includes a spectrum of energy from ( short-
wavelength , high energetic x-rays & γ-rays ) to ( long-wavelength Radio-
frequencies )
◘ Visible light falls in between , the color violet at 400 nm and red at 700
nm wavelengths being the approximate limits of the visible spectrum .

5. 5
♣ Absorption & Emission Spectra :
• Instruments measure either absorption or
emission of radiant energy to determine
concentration of atoms or molecules .
• The two phenomena , absorption and
emission , are closely related .
• For a ray of electromagnetic radiation to be absorbed , it must have the
same frequency as a rotational or vibrational frequency in the atom or
molecule that it strikes ( hits ) .
• levels of energy that are absorbed move in discrete steps and any
particular types of molecule or atom will absorb only certain energies and
not others .
• when energy is absorbed valence electrons move to an orbital with a
higher energy level .
• Following energy absorption , the
excited electron will fall back to the
ground state by emitting a discrete
amount of energy in the form of a
characteristic wavelength of radiant
energy .
• Absorption or emission of energy by atoms
results in a line spectrum .

6. 6
• Because of the relative complexity of molecules , they absorb or emit a
bank of energy over a large region .
• light emitted by incandescent solids ( tungsten or deuterium ) is in
continuum .

7. 7
◘ Principle :-
○ Beer’s Law :-
♦ Definition :
- The relationship
between absorption of light by a solution and the concentration of
that solution .
- States that the conc. Of the a substance is directly proportional to
the amount of light absorbed or inversely proportional to the
logarithm of the transmitted light .
♦ when a beam of monochromatic light entering a solution , some of the
light is absorbed and the remainder passes through , strikes a light detector .
- The detector convert the light to an electric signal .

8. 8
♦ Percent Transmittance ( % T ) :-
- Is the ratio of the radiant energy transmitted ( I ) divided by the
radiant energy incident on the sample ( Io )
- All light absorbed or blocked results in 0% (T) transmittance .
- A level of 100% T is obtained if no light is absorbed .
- In practice , the solvent without the constituent of interest is placed
in the light path , most of the light is transmitted , but a small
amount is absorbed by the solvent and cuvet or is reflected away
from the detector .
- The electrical readout of the instrument is set arbitrarily at 100% T,
while the light is passing through a “blank” or reference .
- The sample containing absorbing molecules to be measured is
placed in the light path .
- The difference in amount of light transmitted by the blank and that
transmitted by the sample is due only to the presence of the
compound being measured .

9. 9
- The % T measured as : the ratio of
the sample transmitted beam
divided by the blank transmitted
beam .
- %T = sample beam signal / blank
beam signal ×100
♦ Absorbance :-
• Is the amount of light absorbed .
• It cannot be measured directly by a spectrophotometer .
• It is mathematically derived from %T as follow :-
%T = ( I/Io) ×100
A = - log ( I/Io)
A = log ( 100%) – log %T
A = 2-log %T
- Where : ( Io) is incident light & ( I ) is Transmitted light .
• According to beer’s law :- Absorbance is directly proportional to
concentration .
A = ɛ*l*c
- Where ( ɛ ) is the molar absorptivity , the fraction of a specific
wavelength of light absorbed by a given type of molecule .
- ( l ) is the length of light path through the solution .
- ( c) is the concentration of absorbing molecules .

10. 10
♦ Absorptivity :
• Depends on molecular structure and the way in which the absorbing
molecules react with different energies .
• For any particular molecular type , absorptivity changes as wavelength of
radiation changes .
• The amount of light absorbed at a particular wavelength depends on the
molecular and ion types present and may vary with , PH and temperature.
• because the path length and molar absorptivity are constant for a given
wavelength , C .
♦ Unknown concentration are determined
from a calibration curve that plots
absorbance at a specific wavelength
versus concentration for standards of
known concentration .
• For calibration curves that are linear and
have a zero Y-intercept , unknown
concentration can be determined from a
single calibrator .
• Beer’s Law deviations :
- Not all calibration curves result in straight lines .
- Deviation from linearity are typically observed at
high absorbance .
- The stray light within an instrument will
ultimately limit the maximum absorbance that a
spectrophotometer can achieve , typically 2.0 absorbance units .

11. 11
♦ Beer’s law Examples :
%T = ( I/Io) ×100
A = - log ( I/Io)
A = log ( 100%) – log %T
A = 2-log %T
A = ɛ*l*c
(ε ): (Greek letter, epsilon) is the molar absorptivity of the solute with units of M-1
cm-1
(or ( L mol-1
cm-1
or mol-1
dm3
cm-1
)
(l) is the path length of the light through the solution in units of cm.
(C) is the concentration of the solution in mol L-1
(or mol dm-3
or M )
◘ Using a cuvette with a length of 1 cm, you measured the absorbance of a solution with a
concentration of 0.05 mol/L. The absorbance at a wavelength of 280 nm was 1.5. What is
the molar absorptivity of this solution?
 ɛ280 = A/lc = 1.5/(1 x 0.05) = 30 L mol-1
cm-1
◘ A solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm length cuvette.
Given the absorbance coefficient of Tryptophan is 6.4 × 103
LMol-1
cm-1
. What is the
concentration of solution?
Solution:
As : ε = A / l c
l = 0.5 cm
A= 0.54
ε = 6.4 × 103
LMol-1
cm-1
C=?
So c = A/ε l
= 0.54 / 6.4 × 103
× 0.5
Answer = 0.000168 M
◘ A solution of thickness 2 cm transmits 40% incident light. Calculate the concentration of the
solution, given that ε = 6000 dm3
/mol/cm.
Solution:
A = 2 - log %T = 2 – log 40 = 2 – 1.6020 = 0.398
A = ε l c
l= 2cm
ε = 6000 dm3
/mol/cm
A=0.398
c=?
So c = A/ ε l = 0.398/ 6000 × 2
Answer = 3.316 X 10 – 5
mol / dm3

12. 12
◘ A solution shows a transmittance of 20%, when taken in a cell of 2.5 cm
thickness. Calculate its concentration, if the molar absorption coefficient is
12000 dm3
/mol/cm.
Solution:
A = 2 – log %T = 2 - log 20 = 2 – 1.301 = 0.698
A = ε l c
l= 2.5 cm
ε = 12000 dm3
/mol/cm
A=0.698
c=?
So c = A/ ε l
= 0.698/ 12000 × 2.5
Answer = 2.33 X 10 – 5
mol / dm3
◘
Calculate the molar absorptivity of a 1 x 10-4
M solution, which has an
absorbance of 0.20, when the path length is 2.5 cm.
Solution:
A = ε l c
l= 2.5 cm
A= 0.20
C= 1 x 10 – 4
M
ε =?
So ε = A / l c
= 0.20/2.5 ×1×10-4
Answer = 800 dm3
/mol/cm
◘ Calculate the molar absorptivity of a 0.5 x 10-3
M solution, which has an
absorbance of 0.17, when the path length is 1.3 cm.
Solution:
A = ε l c
l= 1.3 cm
A= 0.17
C= 0.5 x 10-3
M
ε =?
So ε = A / l c
= 0.17/ 1.3 × 0.5 x 10-3
Answer = 261.53 dm3/mol/cm.

13. 13
◘ A CaCO3 solution shows a transmittance of 90%, when taken in a cell of 1.9 cm thickness.
Calculate its concentration, if the molar absorption coefficient is 9000 dm3
/mol/cm.
Solution:
A = 2 - log %T = 2 - log 90 = 2 – 1.954 = 0.045
A = ε l c
l= 1.9 cm
ε = 9000 dm3/mol/cm
A=0.045
c=?
So c = A/ ε l
= 0.045/ 9000 × 1.9
Answer = 2.631 × 10 -6
mol / dm3
◘ A 1.00 × 10–4 M solution of an analyte is placed in a sample cell with a path length of 1.00
cm. When measured at a wavelength of 350 nm, the solution’s absorbance is 0.139. What is
the analyte’s molar absorptivity at this wavelength?
l = 1.00 cm
c = 1.00 × 10–4
M
A=0.139
ε =?
So ε = A / l c
= 0.139/ 1.0 × 1.00 x 10-4
Answer = 1390 cm−1
M−1
Spectrophotometry