Pipette Double Sided Sellotape Hair Dryer A Solvent Butaniol Glacial Ethanoic Acid Distilled H2O With a ratio of 4:1:1 Method I will first need to prepare the egg white by separating it from the egg yolk and then adding the protease mixture to the egg white. This will then be left over night in an incubator so that the enzymes have time to react with the proteins in the egg whites, and break down the bonds between the amino acids, so that they are more accessible and easier to identify.
After the egg/protease mixture has been left over night I will then get a strip of chromatography paper and draw a line approximately 2cm from the end of the strip I will then get the egg/protease mixture and using the pipette slowly and carefully drop the mixture into the middle of the line that has just been drawn, when the paper looks saturated I will take the strip over to the hair dryer which I will be using to accelerate the drying process, once the paper has been fully dried using the hair dryer I will take the strip back over to the bench and repeat the process of saturating the paper and then drying it.
I repeated this process of saturating and drying 4 times in total; I did this because the more of the mixture that was present on the chromatography paper then logically the more possibility of more and different amino acids being present. I then placed the double-sided sellotape onto the Gas Jar Lid and folded the other end of the chromatography paper and stuck it to the sellotape so that the saturated end of the chromatography paper was dangling just above the bottom of the Gas Jar, this is so that when I put the solvent into the Gas Jar it was just submerged into the solvent.
So as I just mentioned I then put the solvent into the bottom of the Gas Jar, so that it could rise up the chromatography paper, put then lid on so that the vapours from the solvent were trapped in the Gas Jar and left it overnight. I then removed the chromatography paper from the Gas Jar and sprayed it with Ninhydrin in a fume cupboard I have to do this because amino acids have no colour; therefore I have to spray on a revealing agent (Ninhydrin) so that we can see them.
I then drew a line on it when it was thoroughly dry to mark where the solvent had run too (called the solvent front). I then measured the distance from the solvent front to the start line (where I had originally dropped the egg/protease mixture) this is the distance moved by the solvent front. I then measured the distance from each spot of amino acid to the start line, this is the distance moved by each by the spot. I will then work out the Rf Factor for each spot of amino acid and compare the Rf factors to the table and work out what amino acid each spot is.
Once the experiment had been carried out on the Chicken egg I then repeated the exact same process for the Duck egg so the that it was a fair test. Prediction My prediction is that the egg whites will contain many different amino acids and my prediction is that they will have over 8 different amino acids present in them; I think this because an egg white contains approximately 40 different proteins so obviously when it contains that many proteins it will contain a lot of amino acids as they are what makes up the proteins but in different formations.
Diagram Gas Jar Lid Gas Jar Chromatography Paper Solvent Results Chicken egg Rf = Retention Factor Rf = x y x = Distance from starting point to spot y = Distance to solvent front Distance to solvent front/y = 166mm Distance to amino acid/x1 = 86mm Distance to amino acid/x2 = 132mm Calculation for the Rf factor of x1 Rf factor of x1 = 0. 158 166mm = 0. 158 86mm x1 = Methionine Calculation for the Rf factor of x2 Rf factor of x2 = 0. 795 166mm = 0. 795 132mm x2 = Leucine Amino Acid Distance Rf Factor Identified Amino Acid x1 86 0. 158 Methionine x2 132.
0. 795 Leucine Duck egg Rf = Retention Factor Rf = x y x = Distance from starting point to spot y = Distance to solvent front Distance to solvent front/y = 220mm Distance to amino acid/x1 = 25mm Distance to amino acid/x2 = 60mm Distance to amino acid/x3 = 97mm Distance to amino acid/x4 = 132mm Calculation for the Rf factor of x1 Rf factor of x1 = 0. 113 220mm = 0. 113 25mm x1 = Lysine Calculation for the Rf factor of x2 Rf factor of x2 = 0. 272 220mm = 0. 272 60mm x2 = Serine Calculation for the Rf factor of x3 Rf factor of x3 = 0. 440 220mm = 0.
440 97mm x3 = Tyrosine Calculation for the Rf factor of x4 Rf factor of x4 = 0. 60 220mm = 0. 60 132mm x4 = Valine Amino Acid Distance (mm) Rf Factor Identified Amino Acid x1 25 0. 113 Lysine x2 60 0. 272 Serine x3 97 0. 440 Tyrosine x4 132 0. 60 Valine Analysis The proteins in an egg white are globular proteins, which means that the long protein molecule is twisted and folded and curled up into a more or less spherical shape. A variety of weak chemical bonds keep the protein curled up tight as it drifts placidly in the water that surrounds it.
When these proteins are broken up into amino acids by the protease the amino acids all become free and so we are able to perform the chromatography test on them. From my results I can gather that only two amino acids showed up on the chromatography paper for the chicken egg, the two amino acids that showed up were Methionine and Leucine the results from my test lead me to deduce that from the two Leucine was the lighter molecule this is because in chromatography the lighter molecules travel further than the heavier molecules.
From the results of the Duck egg though I can clearly see that more amino acids showed up on the chromatography paper these were Lysine, Serine, Tyrosine and Valine out of these Valine had the lighter molecule because it travelled further along the chromatography paper, lysine therefore has the heaviest molecule because it moved the shortest distance along the chromatography paper, serine was the second heaviest molecule followed by tyrosine up o valine the lightest out of the four. Evaluation.
Overall I think my experiment was pretty fair although I think that a couple of things affected the results; I think that my prediction was correct in the fact that although there were only two amino acids in the chicken egg white and four amino acids in the duck egg white that showed on the chromatography paper I think that because the Rf factors of them were so close that they were indistinguishable on the paper and so I think that there were more amino acids present but they could not separate in the time scale that I did the experiment in.
I think to correct this we could do the experiment for longer as this would let the amino acids have longer to separate out into there respective positions on the chromatography paper, I also think that I could have also done a two way paper chromatography where by which a chromatogram is turned by ninety degrees, and placed in a different solvent in the same way as before; some spots separate in the presence of more than one pigment. As before, the value is calculated, and the two pigments are identified.
I could have also used a square of chromatography paper and done the experiment turning it 90 degrees each time until four results are shown on the paper. From research, I have found that chicken eggs contain 17 different amino acids, this is shown in the table below and so in my prediciton I do not think I was entirely wrong as it may not have shown all of these on the chromatography paper but I have already discussed a possible reason for this. Amino Acidsg. Alanine 0. 348 0. 203 0. 143 Arginine 0. 375 0. 191 0. 199 Aspartic acid.
0. 628 0. 358 0. 272 Cystine 0. 145 0. 091 0. 05 Glutamic acid 0. 816 0. 467 0. 353 Glycine 0. 21 0. 123 0. 086 Histidine 0. 148 0. 079 0. 072 Isoleucine 0. 341 0. 199 0. 141 Leucine 0. 534 0. 296 0. 244 Lysine 0. 449 0. 239 0. 221 Methionine 0. 195 0. 121 0. 069 Phenylalanine 0. 332 0. 205 0. 119 Proline 0. 249 0. 137 0. 116 Serine 0. 465 0. 242 0. 238 Threonine 0. 3 0. 16 0. 148 Tryptophan 0. 076 0. 043 0. 033 Tyrosine 0. 255 0. 137 0. 124 Valine 0. 381 0. 224 0. 155.
Table from: http://www. aeb. org/LearnMore/NutrientBreakdown. htm.