Answers
 

Explanations

Being able to work out dilutions is a necessary part of your CMR practicals, and we feel that it is important for you to be able to do these calculations confidently. As a consequence, this section if quite a large one.
It can be split into three groups, however.

1)     Q 51 to 56.     Dilution factors.
e.g. 1 ml of enyzme solution added to 9 ml of buffer is a 1:10 dilution of the enzyme solution. One ml enzyme solution in a total of 10 ml.

Q 51.    You have a 20 mM NaOH stock, and want to make a 1 mM NaOH solution.  This 1 mM solution = 1/20^th the concentration of the stock.  ie a 1 in 20 dilution.  This means that to make your 1 mM NaOH solution you need 1 part stock in a total of 20 parts of solution.  The remaining volume of solution has to be something else - in this case water.  So you need 1 ml stock and 19 ml (19 + 1 = 20ml) water.  However, this is not one of the available options.  If you make up 10 ml of 1 mM NaOH instead of 20 ml, you will need 0.5 ml of stock and 9.5 ml of water (both divided by 2).  Answer 3.
 

It is the same principle when diluting to percentage of the orginial solution.
Q 55.     To dilute an aqueous solution of glucose to 40% of its original concentration, take 100 ml of the solution and mix it with the following volume of water (in ml):
So if 100 ml is 40% of the total volume, the remaining 60% is water.  If you can work out how many ml = 1%, you can work out how many ml = 60%.
1% = 100ml/40.  1% also = volume of water (X ml)/60.  Thus 100ml/40=Xml/60.
So X = (100 ml / 40) x 60 = 150 ml (answer 2)

2)     Q 57 to 63.     How you use dilution factors.
These questions are how you would use your knowledge of dilution factors in a practical exercise. You need to be able to pick out the information you require to answer each question. For example:

Q 57.     If 50 ml of a 10 molar solution of NaOH is diluted with water to a total volume of 1 ml, the final concentration of NaOH is:

First, work out the dilution factor. 50 ml in a total of 1000ml (1 ml).
= 50 in 1000
= 1 in 20

This means that the 10 M solution has been diluted by a factor of 20.
10 M / 20 = 0.5 M = 500 mM (answer 2)

Q 59.     If you put 0.5 ml of 3 mmol.litre^-1 4-nitrophenol phosphate and 0.4 ml of buffer in a test tube and start the reaction by adding 100 ml of a 1 mg.ml^-1 protein solution containing alkaline phosphatase, what is the final concentration of protein in your tube?

This question contains a lot of information. It may help you to break it up into smaller pieces.
First, work out the dilution factor. 100 ml has been added to 0.9 ml of reagents, making a total volume of 1000 ml (1 ml)
This is a 1 in 10 dilution. A 1 in 10 dilution will result in a 1 / 10 mg.ml^-1 protein solution = 0.1 mg.ml^-1 (answer 1)

An alternative method is to work out the final volume in the test tube. = 0.5 ml + 0.4 ml + 0.1 ml = 1 ml.
By adding 0.1 ml of a 1 mg.ml^-1protein solution, you have added 0.1 x 1 mg of protein. This 0.1 ml of protein is now in 1 ml of liquid. Hence 0.1 mg.ml^-1.

3)        Dilution factors that are greater than 1.
An example of how to work through these questions is the method to answer Q 65.
The absorbance of 3 ml of a 10^-5 mol.l^-1 solution of 4-nitrophenol, measured at 400 nm, is 0.18. The absorbance at 400 nm of a 1 mol.l^-1 solution of 4-nitrophenol is therefore:
A 1 mol.l^-1 solution is 10⁵ times more concentrated than a 10^-5 mol.l^-1 solution. And remember that the absorbance of light by a solution is directly proportional to the concentration (Beer's Law).
Thus the absorbance of a 1 mol.l^-1 solution would be 0.18 x 10⁵ = 18000 (answer 2)

The volume of the solution (3 ml) is a red herring. The absorbance is proportional to the path length of the cuvette (usually 1 cm), not the volume of liquid in it.