Mixed up mixture
Can you fill in the mixed up numbers in this dilution calculation?
A standard calibration for a spectrophotometric assay for lysozyme was created by diluting some initial concentrations of the reagents Tris HCl buffer, NaCl, Lysozyme and a Micrococcus lysodeikticus suspension.The final volume was 5ml.
The numbers in the calculation have become mixed up. Can you work out how consistently to fill in the table? There will be two numbers left over. What will these represent?
Here is a printable version of the table and cards that you can cut out.

Initial conc. 
Final conc. 
Dilution factor 
Volume added 
Tris 




NaCl 




Lysozyme 




Cells 




171 mM

10^{6} /ml 
10mM 
500 pmol/ml 
0.050 ml

136 
20 
0.250 ml 
68 nmol/ml

50 mM 
100 
0.037 ml 
0.292 ml

10^{8} /ml 
17.1 
0.629 ml 
4.371 ml

1000 mM 


First divide the numbers into groups according to whether they are concentrations, pure numbers, or volumes.
Secondly, group the concentrations into pairs, along with the required dilution factor. Then, divide 5ml by the dilution factor to additionally find the volume added.
The numbers should be in groups of 4 now, with three additional outlying numbers.
Thirdly, think which of the components added would be measured as a number per ml, rather than a number of moles per ml. This will assign one group of 4 numbers.
Fourthly, think which of the remaining items has a large molecular weight, and so will only have a low molar concentration added. Assign this group of 4 numbers.
The final two groups are hard to assign: research what the intracellular concentration of sodium ions is, since the NaCl is added to keep the solution isotonic.
The data should be all assigned now, and the sum of the added volumes should be equal to one of the leftover numbers. What might the final number represent?
Image
The solution is shown above, but is little help without the logical thought process which allowed its deduction.
Firstly, looking at the mass of data, it is notable that of the concentrations, two of them are given solely as a number per ml, rather than in moles per ml or l. These clearly match up with the 'cells' row, as it is common to talk about the number of cells per ml, rather than as a molar concentration. It is immediately clear that the $10^8$ must be the initial concentration and the $10^6$ value the final concentration, since a dilution occurs. This dilution factor is 100, as given by $\frac{10^8}{10^6}$. It is now possible to match these three pieces of data with the necessary 'volume added'. Dividing the total 5ml volume by the dilution factor gives an added volume of 0.050ml.
Matching the rest of the data is somewhat trickier, and it is perhaps useful to organise the data into three piles: concentrations, numbers, and volumes. Dilution factors can be coupled to pairs of concentrations through observation or trial and error. This reveals the two unused volumes, which can be put to one side.
It should hopefully be evident that Lysozyme is an enzyme, which is a protein formed from many amino acids. It thus has an exceptionally large RMM compared to either NaCl or Tris, and so its molar concentration is likely to be much smaller than either of these. Therefore, Lysozyme can be matched with the 68 nmol/ml, 500pmol/ml 136 and 0.037ml category.
Matching the final data to the Tris and NaCl is much trickier. One way of deducing this is to look up the cellular concentration of sodium ions, since the NaCl may be being added to the assay to ensure that the solution is isotonic, and so the cells will not lose or gain water by osmosis. Research reveals that these sodium ion concentrations are between 5 and 15 mMol, and so it is likely that data set with the final concentration of 10 mMol matches with NaCl.
Therefore, the final data set must match with TrisHcl!
The final part of the questions asks what the two unused volumes correspond to. 0.629ml corresponds to the total volume of the 4 solutions added so far to the assay, and 4.371ml corresponds to the amount of water that should be added to make up the solution to 5ml.
Why do this problem?
This problem gives practice in converting between a variety of measures. Don't be put off if your students are not aware of the meaning of the various reagents: the table can be reconstructed using knowledge of units alone and a little common sense. Engaging with a problem containing some unfamiliar elements is a good general aspect of mathemaical training.Possible approach
The most intimidating part of this question is the scientific
names of the reagents. Encourage students to see past this to the
mathematical structure of
the problem. If it helps, the names of the reagents could be
replaced by letters.
Key questions
 What is the question asking mathematically?
 Is it possible easily to rule out certain cards from certain positions? What possibilities remain?
Possible extension
You might naturally try Reallife
equations next.
Possible support
Change the names of the reagents to letters or something
simpler.