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Aim and objectives

Discuss about the Partition Technology by Chromatography.

Chromatography consist of the method called Gel filtration, this method helps to separate the molecules according to the molecular size of them. This process is also sometime referred to as “molecular sieving”, “gel permeation chromatography”, and “gel exclusion chromatography” (6). This method mainly depends on the polymer beads that is the gel which have the controlled pore sizes. The molecules are given access through the pores and spaces of the gel matrix depending on their size (2). The molecules which are small have the ability to penetrate completely through the gel matrix, on the other hand the larger molecules are totally excluded from the beads. The molecules which have an intermediate size can move to a certain fraction of the internal volume of the beads just according to their size (4). So a large molecule possess the ability to move through the column in a lesser volume and so moves faster than the small molecules. Therefore the small molecules will move slow through the column and will be collected after large and intermediate sized molecules (8).

To separate the mixture of three compounds including blue Dextran (a polysaccharide of a high molecular weight with a blue dye linked with it covalently), Haemoglobin (lysed horse blood), and bromophenol blue by the use of gel columns and also the determination of the Ve for each of these compounds.

  1. The Sephadex gel
  2. Polypropylene chromatography column,
  3. Separation of the mixtures around 0.3 mL is to be done in an eppendorf microfuge tube.
  4. 5 M NaCl, 20 mM Pi, buffer solution is prepared with a pH 7.4.
  5. Eluate collecting tubes.
  6. For the procedure of determination of the spectral properties the separation of the components of the mixture was done.

On a retort stand the setting of the column was done vertically with a clamp. A flexible tubing piece and a clamp that can be adjusted must be positioned on the outlet so that the effluent flow can be restricted.

The Sephadex suspension should be mixed gently but thoroughly in the column with few mL of the buffer solution and it should be 50 percent Sephadex by volume. Then the buffer is allowed to run through the column unless and until the level of the buffer becomes more than the column’s one third. More gel can be added if necessary. The column should be equilibrated by running one column volume of the buffer through it.

The tubes should be prepared for the collection of the effluent of the column should be done immediately the sample is placed on the column.

Then buffer should be permitted to drain down to the topmost point of the gel. It should keep in mind that the top of the gel bed can never be allowed to dry up.

Materials

At first the sample solution should be given a spin briefly to remove any type of aggregates or insoluble substances and then 0.1mL of the sample should be loaded very carefully on the entire surface of the gel without making any disruption to the column.

The column elution should be started very carefully by adding small volume of the buffer when the sample is absorbed onto the bed of the column.

When the absorption of the sample on the gel bed is complete then extra level of the buffer being further added to the top of the gel and thus can be increased. In this respect it should be noted that the run rate of the column should not be too high to avoid poor form of resolution.

The collection of the eluate in small beakers should be started immediately after the sample has been added on the column.

Once there is the presence of the noticeable colour in the eluate, the collection of the 1ml fraction of the eluate in the tubes should be started

The collection procedure should be continued until all the three components are eluted.

The volumes of all the fraction collected should be measured.

Water about 2mL should be given extra to all the fraction and the and the determination of the absorbance was done at suitable wavelength for those three chromophores.

An absorbance graph was plotted contrary to the volume of elution (Ve) for the three chromophores each.

The Sephadex should not be discarded as it is very costly and placed it in a container provided for recycling.

The total volume (Vt) of the gel column is given by

Vt = Vo + Vi + Vg

Where Vo is the volume of the liquid that came out of  the gel. This is also known as void volume.
Vi is the volume of liquid inside the gel particles.     
Vg is the volume of the insoluble Sephadex matrix.

The dispersal procedure of the material among the external and the internal liquid phases of a Sephadex gel is demarcated as its distribution coefficient (Kd), it is the function of its size of molecule. The value of Kd specifies the portion of internal volume available to the molecules of the specific material (5). The volume present inside the beads that is the molecular species can penetrate is therefore KdVi. For larger molecules which are not able to enter inside the Sephadex particles Kd = 0 and for small molecules which have full access to it, Kd = 1.

Methods

The volume of elution (Ve) for a molecular species is represented as the whole column volume accessible to the molecule and is therefore made up of the volume outside the gel plus the volume of gel inner space the molecule can penetrate.

Ve = Vo + KdVi

The Ve of a molecule is represented experimentally as the volume taken by the molecule to come through the column that is the volume of effluent of the column that is collected between applying the sample to the top of the column to the mid-point of the peak of the molecule as it comes off the other end (9).

Table1. The Absorbance of the Running volume at different wave length

Fraction

Running volume (mL)

Absorbance

Wavelength 630 nm

Wavelength 415 nm

Wavelength 590 nm

1

52

0

0

0

2

54

0.032

0.06

0.233

3

56

0.721

0.018

0.554

4

58

0.112

0.163

0.124

5

60

0.135

0.239

0.145

6

62

0.115

0.163

0.129

7

64

0.114

0.162

0.126

8

66

0.044

0.042

0.359

9

68

0.033

0.046

0.37

10

70

0.042

0.047

0.376

11

72

0.031

0.045

0.364

12

74

0.03

0.044

0.349

13

76

0.029

0.045

0.329

14

78

0.039

0.045

0.315

15

80

0.035

0.045

0.299

16

82

0.031

0.047

0.273

17

84

0.031

0.044

0.233

18

86

0.025

0.041

0.2

19

88

0.021

0.048

0.163

20

90

0.02

0.043

0.134

21

92

0.017

0.04

0.109

22

94

0.018

0.045

0.087

23

96

0.016

0.041

0.09

24

98

0.011

0.044

0.056

25

100

0.009

0.046

0.043

26

102

0.013

0.044

0.035

27

104

0.008

0.041

0.027

28

106

0.005

0.043

0.021

29

108

0.01

0.042

0.016

30

110

0.007

0.046

0.018

Graph1. Showing the graphical representation of the absorbance of the friction volume at different wavelength

absorbance of the friction volume at different wavelength

The definition of void volume is the volume of the liquid or buffer outside the gel and it is represented by Vo. Considering that Blue Dextran is totally left out from the gel. Then the void volume of the column is 66 ml because before 66 ml the absorbance was very high in all the wave lengths. As the blue dextran has the highest molecular weight of greater than 106. Therefore it can be assumed that the Blue Dextran is excluded from the gel in small volume and faster than all other compounds as the compound having greater molecular weight will excluded in the first but in smaller volume because the pores or spaces in the gel matrix allow access to molecules depending on their size. So the by the time when the larger molecule is totally excluded of the gel it can be assured that the small molecules are already present in the gel and will be excluded afterward in a slow process as they are capable to completely penetrate the gel matrix.

The elution volume of a molecule is the whole column volume available to the molecule and is consequently composed of the volume exterior of the gel plus the volume of gel interior space it is able to penetrate. It is represented as Ve.

Therefore,  Ve = Vo + KdVi

It is evident from the table of the result that the void volume for bromophenol blue is 110mL because it is the molecule which has the lowest molecular weight 670 and it will be excluded at the last from the gel because it is able penetrate completely through the matrix of the gel.  The Kd given for bromophenol blue is 1. As there is no presence of the compound after the elution of the void volume. So, here Vi is zero.

Results

Therefore,   Ve = Vo + KdVi

                        Or, Ve = 110+1x0

                        Or, Ve = 110+0

                        Or, Ve = 110. So the elution volume for the case of bromophenol blue is 110mL that is equal to its void volume.

In the case haemoglobin the void volume will be 76ml as estimated from the result of the table and here also the molecular weight is less around 64000 and this compound has also totally came out of the gel but after Blue Dextran, because it is a compound of intermediate molecular weight. So there will be no presence of the compound after the elution of the void volume. So, here Vi is also zero.

Therefore, Ve will be equal to void volume and that will be 76ml so it can be calculated that

Ve = Vo + KdVi

Or, 76 = 76+ Kd x 0

Or, Kd = 76/76

Or, Kd = 1. So the Kd for the haemoglobin is 1.

Conclusion

The compounds having larger molecular weight came out fast because they are excluded from the gel as their molecule due to its larger molecule cannot penetrate in the gel matrix as a result the larger molecular weight compounds pass through the column in smaller volume but faster than the intermediate sized molecule and the smaller molecules. On the other hand the smaller molecules gets complete penetration in the matrix of the gel and so passes through the gel very slow but with a larger volume and the intermediate sized molecule get an penetration in the gel matrix to certain fraction only so they pass through the a little faster than the smaller molecules.

It quite vivid from the absorbance result of the fractions of the running volume at different wavelength  that the efficacy of the  separation of the three components is no doubt efficient. The table of the result and the graph clearly gives the evidence that Blue dextran was excluded first and then it was followed by haemoglobin and the compound excluded at last was bromophenol blue.

Reference

Bodman J. long will the fractions drop from the lower margin of the supporting. Zone Electrophoresis: Chromatographic and Electrophoretic Techniques. 2013 Sep 3:190.

Charkiewicz E. Identification of antioxidant active trace element proteins in animal cells and human cell lines using bioanalytical methods (Doctoral dissertation, Freie Universität Berlin).

Colel GT, Sun SH, Dominguez J, Yuanl L. M. Franco", and TN Kirkland 3. Fungal Antigens: Isolation, Purification, and Detection. 2013 Nov 11:395.

Ire FS. Evaluation of Some Bioprocess Strategies for Control of Proteolyic Degradation of Raw Starch Digesting Amylase (RSD) of Aspergillus Carbonarius (Doctoral dissertation).

Kim KH, Tran JC, Compton PD, Kelleher NL, SKINNER O, inventors; Northwestern University, assignee. System and method for high throughput mass spectrometric analysis of proteome samples. United States patent application US 15/026,655. 2014 Oct 3.

Le Maire M, Chabaud R, Hervé G. Laboratory guide to biochemistry, enzymology, and protein physical chemistry: a study of aspartate transcarbamylase. Springer Science & Business Media; 2012 Dec 6.

Smith I. PAPER ELECTROPHORESIS AT LOW. Zone Electrophoresis: Chromatographic and Electrophoretic Techniques. 2013 Oct 22:17.

Timerman AP. The Isolation of Invertase from Baker's Yeast-An Introduction to Protein Purification Strategies. INTECH Open Access Publisher; 2012.

Van den Berghe G, Bontemps F. incorporation of" C in the end products of purine catabolism. Purine Metabolism in Man, III: Biochemical, Immunological, and Cancer Research. 2012 Dec 6;122:85.

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