You have just calculated the amount of GFPS65T (in mg) recovered by student Alex
ID: 273104 • Letter: Y
Question
You have just calculated the amount of GFPS65T (in mg) recovered by student Alex. Do you expect this to be less, more, or the same as the mg amount of GST-GFPS65T in the initial lysate sample (before purification commenced)? Your answer should include a consideration of milligrams of protein and moles of protein. Answer less, more, or same, then justify your answer briefly. This section should be no more than 1/3 page 3 (1 mark) 4 (1 mark) 6 (1 mark) Dil'n of A for GFPS65T] in sample each Actual [GFPS65T]Vol. of the Total mg of in the undiluted GFPS65T diluted sample GFPS65T in sample (ug/ml). Find this sample (mg/ml) eluate (ml) entire sample by solving the eg'o volume 0.299 250.2 0.2502 0.7506 0.287 238.2 0.2382 0.7146 0.300 251.2 0.2512 0.7536 0.5958 0.5688 0.5778 1/3 0.115 66.2 0.1986 0.112 63.2 0.1896 1/3 1/5 1/5 1/5 Amount of GFPS65T in the sample (mg) (averaged from samples) (1 mark) 0.113 64.2 0.1926 0.075 26.2 0.131 0.393 0.076 27.2 0.136 0.408 0.081 32.2 0.161 0.483 =0.5828Explanation / Answer
It shall be expected to be less as compared to the amount of milligram of protein obtained before purification . This is so because, before purification the protein sample is a very crude one which may contain contamination of other proteins of similar size , molecular weight, etc. Hence as these would be removed during the various steps of purification, the actual pure GFPS65T shall be of lower amount (weight).
On the other hand the moles of the protein shall remain same. Mole is the number of units of an entity at its molecular weight. This number will not change as GFPS65T remains unchanged in the lysate only the unwanted contaminants get removed during purification.
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