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Friday, April 1, 2011

ABSORPTION SPECTRA OF NUCLEIC ACIDS IN ULTRA-VIOLET RANGE


  ABSORPTION SPECTRA OF NUCLEIC ACIDS IN ULTRA-VIOLET RANGE
Aim:
To obtain absorption spectra of Nucleic acid in the Ultra-Violet range.
Principle: 
Ultra violet spectroscopy is concerned with the study of absorption of UV radiation which ranges from 200 to 400 nm. Compounds which are colored absorb radiation from 400 to 800 nm. But compounds which are colorless absorb radiation in the UV region. In both UV as well as visible spectroscopy, only the valence electrons absorb the energy, thereby the molecule undergoes transition from ground state to excited state. This absorption is characteristic and depends on the nature of electrons present. The intensity of absorption depends on the concentration and path length as given by Beer-Lambert’s law.
The types of electrons present in any molecule may be conveniently classified as
1.      ‘σ ' electrons: these are the ones present in saturated compounds. Such electrons do not absorb near UV, but absorb vacuum UV radiation (<200nm).
2.      ‘π ‘electrons : These electrons are present in unsaturated compounds (eg) double or triple bonds.
3.      ‘n' electrons: these are non bonded electrons which are not involved in any bonding (eg) lone pair of electrons like in S, O, N and Halogens.

The nucleic acid absorbs strongly in UV region of spectrum due to the conjugated double bond system of the constituents purine and pyrimidine. They show characteristic maxima at 260 nm and minima at 230 nm.
Chemicals required:
1.      DNA Solution
2.      Saline  Citrate solution
3.      Dis.H2O
Apparatus required:
1.      Spectrophotometer
2.      glass cuvettes 
3.      standard flasks, etc


Procedure:
10mg of DNA is weighed accurately and dissolved and made up to 100ml of saline citrate solution. And prepare the following concentrations: 2mg%, 4mg%, 6mg%, 8mg%, and 10mg% as given below:
Test Tube No.
10mg% of DNA Solution
Distilled Water in ml
Concentration in mg%
1 (Blank)
-
10
-
2
2
8
2
3
4
6
4
4
6
4
6
5
8
2
8
6
10
0
10
7( Unknown)
5
5
?
 
Now measure the extinction of all the solutions at 210 nm taking distilled water as blank and adjusting the spectrophotometer to 100% T with blank. Plot a graph of extinction against protein concentration (in mg %). You get a straight line graph passing through the origin. Determine the concentration of protein in the unknown solution by extrapolation.
 Observations:
Test Tube No.
DNA Concentration in mg%
Extinction
1
-

2
2

3
4

4
6

5
8

6
10

7
-


 Calculations:
1.      From Observations:
Extinction ‘A’ = 6 mg % of DNA
Extinction ’B’ =  mg% DNA
1ml .of unknown solution =  mg% of DNA
DNA concentration in the unknown solution is= …….mg%
2.      From Graph:
The concentration of DNA in the unknown solution =…….mg%
Results:
1). From Calculations:
  Amount of DNA in the unknown sample=…….. gm%
2). From Graph:
Amount of DNA in the unknown sample = ……..gm%

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