OK, I see from Mike’s post what Randy was after. You want a calculation to determine how much of a neonic is required to alter bee behavior or do worse. So we can work it out, but it will be a ~long~ post…here’s what we need to do:
1. We need to estimate how many neurons we will have to depolarize to debilitate the bees.
2. We need to discover how many AChR have to be bound by persistent agonist at each postsynaptic membrane (or endplate in the case of muscle cells) in order to chronically depolarize each neuron sufficiently to exhaust it (or in order to depolarize the muscle cells enough to damage them.)
3. Multiply the number in #1 by the number we get in #2, that will give us the number of molecules of neonic effecting the result.
4. Calculate how many molecules are in the measured residual in the Suchail paper. We also have to keep in mind….as apparently some are forgetting…that Suchail et al were unable to account for 30% of the dose they gave the bees. This is a lot to lose! Recall that they said the discrepancy was probably due to IMI breaking down into metabolites that they weren’t tracking. But the fact is, they don’t know.
So I will now work through the above steps. If this bores you, stop reading. If you are good at math and interested, please correct any errors you see. This is “just a draft” of effort.
1. How many neurons do we need to debilitate in order to debilitate the bee? The ones involved are many of the brain neurons and all of the motoneurons. (Let’s leave out the motor end plates for this although they too are susceptible.) Rather than trying to figure out how many nerve cells are in the whole bee, let’s consider that if we just knock out a large part of the brain that would “do the trick”. We know that in a realistic situation, not all of the brain neurons need to be damaged before the bee is damaged, and the same is true of the motoneuron system. So for the purpose of this calculation we will use the published number of neurons in the mushroom bodies and assume that every single one has to be damaged before an effect is seen. This will more than allow for the actual reality….that only a small percentage of the mushroom body neurons and a small percentage of the other neurons and motoneurons need to become damaged when the bee visibly suffers. OK? This is a very conservative approach that should satisfy any who wish to see extreme rigor in any assumptions made here.
From Mobbs, P.G. The brain of the honeybee Apis mellifera. I. The connections and spatial organization of the mushroom bodies. Philosophical transactions of the Royal Society of London. Vol 298, no. 1091 “The mushroom bodies of the bee are paired neuropils in the dorsal part of the brain. Each is composed of the arborizations of over 17 10^4 small interneurons of similar architecture called Kenyon cells.”
Here is our number: 170,000 neurons.
2. How many ACh receptors do we need to damage in order to damage each neuron? The number of these receptors varies among neurons. Let’s use the number required to generate an end plate potential. That is another estimate, but it is a good one because it, too, is highly conservative. If that number of AChR are bound to a neonic, it means the current leak is high enough to functionally depolarize the cell to such an extent that it cannot engage in normal activity. Again, the REAL number required to incapacitate said neuron would only be a fraction of this.
From “IS the MEPP due to the Release of One Vesicle or to the Simultaneous Release of Several Vesicles at One Active Zone?” by Tremblay et al, 1983, Brain Research Reviews, vol 6, p 299-314, the number of ACh molecules required to produce one endplate potential in vertebrate systems is 6000. There are a minimum of two binding sites/receptor in vertebrates, but this is not true of insects, which have only one binding site/receptor. So this would translate to 3000 molecules of acetylcholine required to produce an endplate potential in the insect. Endplate potentials produce depolarizations that are larger than we need to functionally impair a neuron, so if we use this number we are being cautious on the side of “more needed to cause dysfunction”.
3. So the total number of neurons in the mushroom bodies X number of molecules needed to depolarize each one = 170,000 X 3,000 or 510M molecules are needed to disable every neuron in the mushroom bodies of the bee brain.
4. Imidacloprid molecular weight http://en.wikipedia.org/wiki/Molar_mass
255.661
One mole = 255.661 grams
Avagadro’s number = 6.02214×10^23
0.5 nanograms = 1.96 X 10^-12 moles of imidacloprid
1.96 X 10^-12 X 6.02214×10^23 = 1.18033944 X 10^12 molecules in 0.5 nanograms
This is half of the chronic dose needed to kill bees over a week to ten days.
The number of molecules we need to disable the mushroom bodies divided by the number in 0.5 ng of IMI is 5.1 x 10^8 / 1.18 x 10^12 = 4.3. This means that there are 4.3 times as many molecules in 0.5 ng of IMI as needed to kill bees after 10 days.
Do you want to repeat all this for the acute LD50 dose? It’s not really necessary but I will if you like.
Christina
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