Question

1. calculate the volume in ml of 50x protease required to make 2.0 l of 2x protease solution.
2. calculate the mass in grams of agarose to prepare a 2.5% agarose in 40 ml of 1x tae buffer. report your answer to two decimal places.
3. calculate the volume of 6x loading dye needed to run 15μl of dna sample by gel electrophoresis. note the loading dye is always at 1x concentration. report the answer to one decimal place with the correct units.
4. what is the dilution factor if you add 100μl of protease solution to 100 μl of diluent?

a. 1/1
b. 1/2
c. 1/4

1. if you perform a 5 - step 2 - fold dilution, what is the dilution factor at the last step?

a. 2^(-2)
b. 2^(-3)
c. 2^(-4)
d. 2^(-5)

1. calculate the volume in ml of a 20.5 mg/ml kanamycin needed to make 500 ml of lb with a final concentration of kanamycin at 23μg/ml. report the answer to one decimal place.
2. how will you prepare a 5mm solution of silver nitrate in 100 ml using a stock solution of 0.1m silver nitrate. this solution is used to prepare nanoparticles.
3. calculate the mass in grams to prepare a 10mg/ml solution of albumin in 30 ml using a stock solution of 0.5g/ml of albumin solution. report your answer to 2 decimal places.

Explanation:

Step1: Use dilution formula for protease

\(C_1V_1 = C_2V_2\)

Step2: Solve for \(V_1\) of protease

\(V_1=\frac{C_2V_2}{C_1}\)

Step3: Use percentage formula for agarose

\(\frac{m}{V}=\%\) (w/v)

Step4: Solve for mass of agarose

\(m=\%\times V\)

Step5: Use dilution formula for loading - dye

\(C_1V_1 = C_2V_2\)

Step6: Solve for \(V_1\) of loading - dye

\(V_1=\frac{C_2V_2}{C_1}\)

Step7: Calculate dilution factor for protease

\(DF=\frac{V_{total}}{V_{original}}\)

Step8: Determine dilution factor for multi - step dilution

\(DF = 2^n\)

Step9: Use dilution formula for kanamycin

\(C_1V_1 = C_2V_2\)

Step10: Solve for \(V_1\) of kanamycin

\(V_1=\frac{C_2V_2}{C_1}\)

Step11: Use dilution formula for silver nitrate

\(C_1V_1 = C_2V_2\)

Step12: Solve for \(V_1\) of silver nitrate

\(V_1=\frac{C_2V_2}{C_1}\)

Step13: Use dilution formula for albumin

\(C_1V_1 = C_2V_2\)

Step14: Solve for \(V_1\) of albumin

\(V_1=\frac{C_2V_2}{C_1}\)

Step15: Calculate mass of albumin

\(m = C_2V_2\)

Answer:

1. For the volume of 50X protease to make 2.0 L of 2X protease solution:

• Use the dilution formula \(C_1V_1 = C_2V_2\), where \(C_1 = 50X\), \(C_2=2X\), and \(V_2 = 2.0L = 2000mL\).
• \(V_1=\frac{C_2V_2}{C_1}\)
• \(V_1=\frac{2\times2000}{50}=80mL\)

1. For the mass of agarose to prepare a 2.5% agarose in 40 mL of 1X TAE buffer:

• A 2.5% (w/v) agarose solution means 2.5 g of agarose in 100 mL of solution.
• Let \(m\) be the mass of agarose. Using the proportion \(\frac{m}{40}=\frac{2.5}{100}\)
• \(m=\frac{2.5\times40}{100}=1.00g\)

1. For the volume of 6X loading - dye needed to run 15μL of DNA sample:

• Using the dilution formula \(C_1V_1 = C_2V_2\), where \(C_1 = 6X\), \(C_2 = 1X\), and \(V_2=15\mu L\).
• \(V_1=\frac{C_2V_2}{C_1}=\frac{1\times15}{6}=2.5\mu L\)

1. For the dilution factor when adding 100μL of protease solution to 100μL of diluent:

• The total volume after dilution \(V_{total}=100 + 100=200\mu L\), and the volume of the original solution \(V_{original}=100\mu L\).
• The dilution factor \(DF=\frac{V_{total}}{V_{original}}=\frac{200}{100}=2\), or in fraction form \(\frac{1}{2}\), so the answer is b.

1. For the dilution factor in a 5 - step 2 - fold dilution:

• In a 2 - fold dilution, each step multiplies the dilution factor by 2. After \(n\) steps, the dilution factor is \(2^n\). Here \(n = 5\), so the dilution factor is \(2^5\), and the answer is d.

1. For the volume of 20.5 mg/mL kanamycin needed to make 500 mL of LB with a final concentration of 23μg/mL:

• First, convert the units. \(23\mu g/mL=0.023mg/mL\), \(V_2 = 500mL\), \(C_1 = 20.5mg/mL\), \(C_2=0.023mg/mL\).
• Using \(C_1V_1 = C_2V_2\), \(V_1=\frac{C_2V_2}{C_1}=\frac{0.023\times500}{20.5}\approx0.6mL\)

1. For preparing a 5mM solution of silver nitrate in 100 mL using a 0.1M stock solution:

• First, convert the units. \(5mM = 0.005M\), \(V_2 = 100mL\), \(C_1 = 0.1M\), \(C_2=0.005M\).
• Using \(C_1V_1 = C_2V_2\), \(V_1=\frac{C_2V_2}{C_1}=\frac{0.005\times100}{0.1}=5mL\). So, pipette 5 mL of the 0.1M silver - nitrate stock solution into a 100 - mL volumetric flask and fill it up to the mark with the appropriate solvent.

1. For the mass of albumin to prepare a 10mg/mL solution in 30 mL using a 0.5g/mL stock solution:

• First, convert the stock - solution concentration to mg/mL. \(0.5g/mL = 500mg/mL\).
• Let \(V_1\) be the volume of the stock solution. Using \(C_1V_1 = C_2V_2\), where \(C_1 = 500mg/mL\), \(C_2 = 10mg/mL\), \(V_2 = 30mL\).
• \(V_1=\frac{C_2V_2}{C_1}=\frac{10\times30}{500}=0.6mL\). The mass of albumin in the final solution \(m = C_2V_2=10\times30 = 300mg=0.30g\)