wiy Pegarding Lab 3: part 1 from Table 1. A. How much of your 1 M Tris-HCI did y
ID: 703944 • Letter: W
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wiy Pegarding Lab 3: part 1 from Table 1. A. How much of your 1 M Tris-HCI did you use to make the TE8 solution? B. What was the dilution factor for your transform solution? 40ml I0 DF C. How much of your stock solutions of NaOH & SDS did you use to make the Alkaline Lysis solution? 4m? 9. (18 pts) Answer the following regarding Lab 3: part 2, problem sets. A. How much NaCl is required to make the desired volume of a 1M solution? B. How much SDS is required to make the desired volumes of a 20% (w/v) solution? C. How much of a 5 M solution of NaCI stock solution will you need to make the following solutions? D. What are the dilutions factors for the following stock to working solution dilutions? E. What are the dilutions factors for the following stock working solution dilutions? F. You are following a protocol which says you need to add 100 ug of BSA and 100 ng of genomic DNA 1) 50 ml 2) 500 ml 2) 250 mL 2) 500 mL of 1 M NaCI 1) 2.5 L 1) 20 ml of 1 M Nac 1) 10% SDS 0.5% SDS 1) 12 M HCI>0.6 M HCI 2) 20% SDS ? 1% 2) 4 M NaCl ? 200 mM NaCl volume of 50 uL. How much of each will you add to what volume of dH20 given the following stocks 25 ug/uL BSA uL, 10 ng/ul genomic DNA= uL uL, dH20 10. (1 pt) Show the dimensional analysis for converting ug/ml>Explanation / Answer
Ans 9
Part a
Molarity of NaOH solution = 1 mol/L
When 50 mL volume is desired
Moles of NaOH = volume x molarity
= 50 mL x 1L/1000 mL x 1 mol/L
= 0.050 mol
Mass of NaOH = moles x molecular weight
= 0.050 mol x 40g/mol
= 2 g
In 100 mL
Moles of NaOH = volume x molarity
= 100 mL x 1L/1000 mL x 1 mol/L
= 0.100 mol
Mass of NaOH = moles x molecular weight
= 0.100 mol x 40g/mol
= 4 g
Part b
Basis 100 g/mL
Density of solution = 100 x 20 g/mL
= 2000 g/mL
Volume = 2.5 L
Mass of SFS required = density x volume
= 2000 g/mL x 1000mL/L x 2.5 L
= 5 x 10^6 g
Basis 100 g/mL
Density of solution = 100 x 20 g/mL
= 2000 g/mL
Volume = 250 mL
Mass of SFS required = density x volume
= 2000 g/mL x 250 mL
= 5 x 10^5 g
Part c
Molarity of NaCl solution M1 = 5 mol/L
V1 =?
M2 = 1 M
V2 = 20 mL
M1 x V1 = M2 x V2
V1 = 1 x 20/5 = 4 mL
Molarity of NaCl solution M1 = 5 mol/L
V1 =?
M2 = 1 M
V2 = 500 mL
M1 x V1 = M2 x V2
V1 = 1 x 500/5 = 100 mL
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