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From:
Bill Hesbach <[log in to unmask]>
Reply To:
Informed Discussion of Beekeeping Issues and Bee Biology <[log in to unmask]>
Date:
Tue, 6 Nov 2018 19:45:26 -0500
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John Oliver >If one estimates the colonies strength by multiplying the number of frames of bees by 1500 per deep frame or 1000 bees per medium frame >one can estimate the number of bees in the colony, and if this number is divided by the mite fall after a treatment and then multiplied by 100 we will get a >figure of the number of phoretic mites per 100....


I can't see how this formula works. Let's say I estimate the colony population at 30,000 bees. Also, let's say 300 mites drop after the treatment.  Then if I'm reading this right I would divide 30,000 by 300 (mite drop) which is equal to 100, and then multiply by 100, I would get 10,000 mites - what I'm I missing?  


Another possibility would be to divide the estimated population by 100 first to get how many groups of 100 bees there are in the population. That then becomes the denominator in a fraction where the numerator is the drop number, which then gets you to how many mites per 100 bees. Example  = 30,000/100= 300 groups of 100. If the drop is 300 mites then it would be 300(mite drop) /300(groups of 100) = 1 mite per 100 bees.  As the drop increases, let's say to 400, the mite drop per hundred bees increases.    30,000/100=300  then 400/300= 1.3 mites per hundred and so on. 

Having said that, it's only useful in a totally phoretic situation. Otherwise, if you want to use this method as an equivalent to a mite wash on nurse it would require knowing the population of nurse bees because that's when the correlations are meaningful. 




Bill Hesbach
Cheshire CT 

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