Determine the amounts/volumes of the chemicals needed to prepare the following s
ID: 92963 • Letter: D
Question
Determine the amounts/volumes of the chemicals needed to prepare the following solutions (‘w’ = weight; ‘v’ = volume; ‘mw’ = molecular weight):
1. 50 ml of 0.9% (w/v) saline (NaCl)
2. 30 ml of 50% glycerol (v/v – in distilled water) [you will need this solution as a stock solution when making 10X gel loading dye]
You will need to make 50X TAE so that you can do gel electrophoresis. One of the ingredients is 0.5 M EDTA pH = 8.0. EDTA stands for disodium ethylenediamindetratraacetate•2(H2O). This EDTA solution will be provided to you. However, you should know how it was made, and the following two questions address this.
3. Determine the amount of EDTA needed to prepare 100 ml of 0.5 M EDTA pH = 8.0. The EDTA MW = 372.2 g/mole
4. As EDTA dissolves in water, the [H+] will increase (note that full name of EDTA includes "acetate"). To increase to a pH = 8.0, would you need to add HCl or NaOH
Determine the amounts/volumes of the chemicals needed to prepare the following solutions:
5. 100 ml of 0.1 N HCl from concentrated (12.0 M) HCl [For HCl, 1 N = 1 M]
6. 100 ml of 1.0 M Tris pH = 7.6. MW of Tris base is 121 g/mole [you will be asked to make this solution as practice, including titrating it with the solution in question 7, to obtain pH 7.6]
7. pH 7.6 is close to a neutral pH. When making 1.0 M Tris base, would you expect to need HCl or NaOH to decrease the pH to 7.6?
To do electrophoresis, you will need to make a solution called TAE, which is a specific mixture of Tris base, acetic acid, and EDTA. Assuming no errors and no spills, you will need a minimum of 50 ml 1X TAE to make your agarose gel plus 1 liter of 1X TAE for gel running buffer. TAE is normally made as a 50X concentrated stock.
8. What is the minimum volume of 50X TAE that you will need to make?
You determined in the previous assignment the exact volume of 50X TAE that you will need. I suggest that you make 40 ml of 50X TAE so that you have some extra (just in case you need it).
The recipe for one liter of 50X TAE is as follows:
• 242g Tris base
• 57.1 ml glacial acetic acid
• 100 ml 0.5 M EDTA pH 8.0
9. Provide a recipe for 40 ml of 50X TAE. [You need to make this solution
10. When you make your agarose gel, you will need to make 50 ml of 1% (w/v) agarose in 1X TAE. How much agarose will you require? [you will make this later, not on Solution day]
When adding your DNA samples to the agarose gel, you will need to add loading dye to the samples. Loading dye contains:
• glycerol, which helps the samples sink into the sample wells of the gel; and
• tracking dyes so that you have a way of estimating how far the samples have travelled through the gel during electrophoresis.
The recipe for 6X loading dye is as follows:
• 0.25% bromophenol blue (w/v)
• 0.25% xylene cyanol (w/v)
• 30% glycerol in water (v/v)
11. Provide a recipe for 10 ml of 10X loading dye, using the information above. (Important tip: Remember that you have already made 30ml of a 50% stock solution of glycerol. I recommend determining the concentration of glycerol required for 10X loading dye, then figuring out how much of your 50% glycerol stock will be needed to make 10ml of 10X loading dye.)
Solution to make
Exact pH of final solution (if appropriate)
Final container for solution
Completed (Y or N)
100 ml of 0.9% (w/v) NaCl
100 ml of 1.0M Tris, pH 7.6
40 ml of 50X TAE
Solution to make
Exact pH of final solution (if appropriate)
Final container for solution
Completed (Y or N)
100 ml of 0.9% (w/v) NaCl
100 ml of 1.0M Tris, pH 7.6
40 ml of 50X TAE
Explanation / Answer
1) 0.9% of saline (NaCl) can be prepared by adding 9 grams of sodium chloride (NaCl) in 1 litre of water.
Therefore, 50 ml of 0.9% (w/v) saline (NaCl) can be prepared by adding 0.45 grams of NaCl in 50 ml of water.
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