Update: I realized that I was off by a factor of 10 in my calculation for the amount of D2O to add to DDW. It turns out that the amount you need to add is 0.14ul of D2O per 1mL of water, which is vastly different from 1.4ul.

I was hoping to start the third attempt at the “Effects of DDW on Tobacco Seeds” experiment but got caught up doing some “trivial” calculations. It’s hard to do trivial things when you haven’t done math in a while. Why was I trying to do these calculations?

Well it occurred to me that perhaps the root hairs we saw weren’t a product of there being no D2O in water, it could have been as simple as thinking there was nothing at all in the DDW besides H2O. Perhaps this water is so pure that the plants are looking for something in the water and the hairs are a byproduct of that.

So I thought, what if I added enough D2O to the DDW so that it has the same amount of D2O as naturally occurring water. This way the only real difference between the ddw sample and the di water sample is how the waters were purified. Maybe the Thermo purifier water we have actually has trace amounts of something in it and this is why the hairs aren’t as prominent.

In order to set that up, I needed to know how much D2O is in natural water. The short answer turned out to be 156ppm (parts per million) D2O in H2O according to some standard from ocean water. And according to my calculations results in a concentration of 7.8mM, which is quite a staggeringly large number coming from a molecular biology background. Steve put it best in this blog post:

0.03% does sound trivial.  But the way I look at it, biology has somehow evolved to make use of different divalent cations in much lower concentration, such as magnesium, zinc, calcium, etc.  And it can distinguish between potassium (K+) and sodium (Na+).  How much more different are K+ and Na+ from each other, compared to the difference between D and H?

So 7.8mM would translate to about .14ul of “pure” D2O in 1mL of “pure” deuterium depleted water. Of course we don’t have pure of either of those. We have 99.9% pure D2O (99.9 atom % D according to the bottle) and < 1ppm D2O in deuterium depleted water.

Now I became stressed! Can I assume that those amounts are so trivial in 1.4ul of D2O that I don’t need to worry about adjusting the calculations? Well I ran some other numbers to make sure.

99.9% pure D2O is roughly 1000ppm H2O (assuming H2O is the contaminate in the bottle), roughly 10x more concentrated than the amount of D2O in natural water. It turned out the concentration of H2O in my D2O was around 55mM. So in 0.2ul of D2O there would be ~2nl of H2O. An extra 2nl of H2O in 1.999ml of D2O is not much change at all.

What about the amount of D2O in the DDW? Well 1ppm turns out to be about 10nM which equates to about 0.9nl in 1mL of DDW. Again less than 1nl won’t affect (too much) the 0.2ul that I’m already adding. Instead of there being 2ul it will be 2.001ul. I think I can live with that!

Unless I come to the realization that all these numbers are important. If that’s the case then I’ll go back through and figure out exact amounts to add to offset the potential impurities in both solutions. Yay for me!