Q1. YIELD : How do the total number of mg of lysozyme recovered in Carb #1 compa
ID: 210220 • Letter: Q
Question
Q1. YIELD : How do the total number of mg of lysozyme recovered in Carb #1 compare with the total number in the 6 ml of 8X dil HEW that you originally mixed with the Sephadex? This gives you the yield. For example if you started with 10 mg of enzyme mixed with Sephadex and you recovered 8 mg in Carb #1, than you have an 80% yield of enzyme (an 80% recovery and 20% loss). Ifyou have 80% yield,you would have lost 20% ofthe total enzyme somewhere, but the benefit of doing the procedure is that the purity of the enzyme was been significantly increased. In other words, most of what you separated away from the Carb#1 fraction would be non-lysozyme instead of lysozyme. Q2. SIZES : The length of an amino acid is approximately 10 atoms long, which is equal to nm. The length of lysozyme is approximately 10 amino acids long, which is equal to nm. The width of an E. coli cell is about lysozyme molecules laid end to end. What might be a problem for E. Coli if it was 10 times smaller than your answer, or 10 times larger than your answer? 03. PURITY : How does the % of protein which is lysozyme in Carb #1 compare with the % for 8xDil HEW? For example if 5% of the total mgs of protein in 8xDil HEW is lysozyme but 50% of the total mgs of protein in Carb#1 is lysozyme, then you have had a 10-fold increase purity in /42Explanation / Answer
Answer 1:
The Crab 1 stock concentration would be 1 mg/ml. if it was 8X diluted then, the total concentration would be-
1/8 mg/ml
Now we have 6 ml of 8X diluted HEW, then we would recover the 80% of this solution.
80% of 6 ml = 4.8ml
So 1 ml of 8X diluted HEW has 1/8 mg Crab 1
Now, 4.8 ml would have = 1/8 x 4.8
= 0.6 mg
Answer 2:
1 atom has minimum 0.1nm diameter.
Therefore, 10 atoms would have 1nm size.
Lysozyme would then 10 nm diameter.
Ecoli width is 250 nm. That is equals to 25 lysozymes.
10 times smaller E.coli will digest rapidly by lysosome than 10 times larger E.coli.
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