5:45 PM oooo Verizon K CHEM372 practice quiz for fin... l. A 150 mM glucose stoc

Question

5:45 PM oooo Verizon K CHEM372 practice quiz for fin... l. A 150 mM glucose stock is diluted 10 times in series, mixing 50 HL of glucose and 25 uL of water each time. What are the glucose concentrations in wells 3, 7 and 10? 2. A 14uM stock of protein is measured to have a concentration of 875 ug/mL. What is the molecular weight of the protein? 3. A 10mM stock of a compound is serially diluted by mixing 10 uL of compound with 70 L of diluent, in 6 steps total. What are the concentrations in wells 2, 4 and 6? 4. A DNA stock at 1 mg mL is diluted 36 fold into a final volume of 22 mL. What is the total amount of DNA in the final 22 mL? 5. You have a stock of a protein solution at 12 mg/mL.The molecular weight of the protein is 23 kDa. What is the molarity? 6. A 12 uM protein stock (mol weight 56000) is diluted in series, 2 fold at each step, for a total of times. What is the final concentration in well #7 in mg/mL? 7. A DNA stock solution with A260 3.2 needs to be diluted such that a final 12 mL of l4 ug/mL DNA is obtained. How do you do that? 8. A stock of glycine at 0.2 M is diluted in series, by mixing 5 uL of glycine and 95 L of buffer at each step, for a total of8 steps. What are the concentrations of glycine in wells 2,4,6 and 8? 9. You mixed 120 HL of a DNA stock solution with 8 mL of buffer and measure an A260 of 1.2 for this dilution. What is the concentration of the undilutedDNA stock in ug/mL? 10. You have a stock of a DNA solution at 1.2 mg/mL, the DNA fragment being 980 bp in length. How do you prepare a dilution of this DNA that has a concentration of 0.5 uM Courses Calendar To Do Notifications Messages

Explanation / Answer

1)

Mixing 50 uL stock with 25 uL water produces a 1.5 time dilution.

So, conc in well #3 = 150/(1.5*3) = 33.33 mM

Conc in well #7 = 150/(1.5*7) = 14.28 mM

Conc in well #10 = 150/(1.5*10) = 10 mM

2)

14 uM = 875 ug/mL = 0.875 ug/L

So, 14 uM = 14 umoles/L = 0.875 ug/L

Thus, 14 moles = 0.875 g, so MW = 0.875/14 = 0.0625 g/mol

3)

Similar approach as used in part 1, except that here the solution is diluted by a factor of 8 every time ( because mixing 10 uL with 70 uL gives a final volume of 80 uL, which means 10 uL is converted to 80 uL so dilution factor is 8 )

4)

After dilution, conc = 1/36 mg/mL

Total amount = final vol * final conc = 1/36 * 22 = 0.61 mg

5)

23 kDa = 3.8*10-17 mg

So moles present = 12/(3.8*10-17) = 3.15*1017 in 1 mL

So molarity = 3.15*1014 moles/L

6)

After 7 dilutions of 2 fold, final dilution is 2*7 = 14 fold

So final conc = 12/14 uM = 12/14 umoles/L = 12/14 * 56000 * 10-6 = 0.048 mg/mL

7)

Need the molar absorptivity value of DNA for this

8)

Similar approach as used in part 1, except that here the solution is diluted by a factor of 20 every time ( because mixing 5 uL with 95 uL gives a final volume of 100 uL, which means 5 uL is converted to 100 uL so dilution factor is 20 )

9)

Need the molar absorptivity value of DNA for this

10)

Avg MW of one base pair = 660 g/mol

So MW of DNA = 660*980 = 646800 g/mol

So original stock conc = 1.2 mg/mL = 1.85 uM

This has to be diluted to 0.5 uM

So dilution factor = 1.85/0.5 = 3.7 times

Take 10 uL DNA, add 27.7 uL water , final vol will be 37.7 uL thus giving the desired dilution of 3.7 times.

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