M Tris-HCl pH 8.0 0.5M EDTA 1M Glucose 10% SDS 10M NaOH Please calculate the vol
ID: 24194 • Letter: M
Question
M Tris-HCl pH 8.00.5M EDTA
1M Glucose
10% SDS
10M NaOH
Please calculate the volume of each of the above stock reagents required to make 1ml of the following solutions. For example, calculate the amount you need for each of the chemicals for GTE. All three volumes that you calculate will be placed into the same tube and the final volume should be brought to 1ml using dH2O.
A. GTE
50mM glucose
25mM Tris-HCl pH8.0
10mM EDTA
B. NaOH/SDS
1%SDS
0.2M NaOH
C. Potassium acetate/acetic acid
600 ?l 5M KoAc
115 ?l glacial acetic acid
For this solution you only need to calculate the amount of water required to
make the final volume of this solution 1ml.
D. T.E.
10mM Tris-HCl pH 8.0
1mM EDTA
Explanation / Answer
1a) You just need to divide the protein concentration by the cell concentration (8e-5 grams/mL divided by 5e8 cells/mL = 1.6e-13 grams/cell). 1b) 2 mg = 2e-3 grams. Divide this by the grams/cell you (I) calculated in part a. 2e-3 grams /1.6e-13 grams/cell = 1.25e10 cells. 1c) Divide what you got in part 1b by the original concentration of bacteria given in 1a. 1.25e10 cells/5e8 cells/mL = 2.5 mL. 1d) You loaded 2mg, 1/5000 of that was protein X. So divide 2mg by 5000. So, 4e-4 mg or 4e-7 grams. 1e) You loaded 2mg, 1/2000 of that is protein Y, so divide 2mg by 2000. 1e-3 mg or 1e-6 grams. 1f) You do not know the stoichiometric relationship of the interaction, so you cannot say which is limiting without that information. Assuming that it's a 1:1 relationship, since they have the same molecular weight, you don't need to take into account their molar concentration, so there's less of protein X (4e-7 grams compared to 1e-6 grams), so protein X is limiting. For the last question, you'd need to know the average length of an ORF that the average protein of 110 kDa is encoded by (you wrote 110 Da, but that's super small; average protein length is between 120-180 kDa). Since I don't really know for sure, I can't answer this question. You'd just take the ratio (35 kDa / 110 kDa = X basepairs / Y basepairs in the average ORF) and solve for X. So, if we say the average ORF size is about 1300 basepairs long, the ORF of protein X would be about 413 basepairs long.
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