50 mM glucose

From 2011.igem.org

Glucose has several forms, and here we are using D-glucose [[1]], also known as anhydrous dextrose. It has a stereoisomer called L-glucose - perhaps you have had organic chemistry, in which case the concept of right and left-handed molecules may still be accessible in your brain's long-term memory. BTW, the D-glucose our team used was donated - it had been collecting dust in a colleague's cabinet.

Making 1M glucose stock solution

D-Glucose is C_6H_{12}O_6, with a molar mass of 180.16 gr/mole. To make 1M d-glucose stock solution one uses

1 liter distilled water
180.16 grams anhydrous dextrose 

Since we did not need an entire liter of the 1M glucose stock, the following recipe sufficed:

50 ml distilled water
(50 ml/1000 ml)(180.16 gr) = 9.0 gr anhydrous dextrose

A beaker, magnetic stir bar, and magnetic stir plate are handy. No heat is needed for this sugar-water to become a 1M solution, but a little warmth won't hurt. The rapid dissolution of the d-glucose at this concentration makes it unlikely that you'd have it on the plate long enough for it to caramelize. In other words, it's hard to screw up.

Using 1M glucose stock to achieve a 50mM final concentration of glucose in 100ml of Solution I

Now, we must add some of the 1M glucose stock to make the final concentration 50mM in 100ml of Solution I. How much should we add? Use the formula

 C1 *  V1 = C2 * V2 , where '*' represents the multiplication operator

C1 is the stock solution's concentration - (1M for this example), V1 is the unknown quantity that we wish to discover, C2 is the final concentration of glucose in Solution I being prepared - (0.050M in this example), and V2 is the final volume of the solution being prepared - in this case, 100ml.

Important Note: The units of measure one selects for volume and concentration must be the same on both sides of the equation. If we use 50mM concentration for C2, then we must use 1000mM in place of 1M for the working stock concentration, C1. Likewise, if we wish to use volume measured in ml for V2, then we will obtain an answer for V1 that is in ml. One does not have to use ml just because one used mM - one could measure volume in liters and concentration in millimoles.

The quantity V1 = 5ml is obtained by the calculation below, which involves cancellation of the mM units top and bottom:

  V1 = (50 mM)(100 ml)/(1000 mM) = (5000 mM * ml)/(1000 mM) = 5 ml

There are other ingredients in Solution I, and for each reagent's stock solution one uses the same type of calculation. To bring the final volume up to the desired level, after all such reagents have been added (final volume of 100 ml is desired for this example) one uses distilled water, or double-distilled water if possible.