Update: Altering Photopreference of Drosophila with Diazepam

In my last post I presented results of testing wing-clipped flies in the Benzer Paradigm, where half of them were fed with Diazepam. I repeated this experiment, but added 40 normal flies to 40 wing-clipped flies. However the following figure doesn’t show any increase of photopreference in Diazepam treated flies.

To complement the individual T-Maze experiment, that I started recently with wing-clipped flies, I performed it with normal flies as well (n=20). The procedure was repeated 10 times. While the first figure in the following shows all 10 choices, the second one only refers to the first 6 choices.

Update: Altering Photopreference of Drosophila with Diazepam

After testing wing-clipped individual flies in the Single Fly T-Maze, that is described in my last post, I reran the experiment changing two things in the setup: Firstly the procedure on each fly was repeated 10 times instead of only 6 and secondly each fly was tested twice, which results in a direct comparison of the same organism with and without the drug. On the first day of the experiment (yellow bar) all of them were fed with the yeast-solution containing 10% ethanol and on the second day (green bar) the same flies were tested again, but half of them got the 5 mM Diazepam treatment. Looking at the following figure (n=23) the control group (nD>nD) shows a similar behavior in both days, but compared to the Diazepam treated flies (nD>D) there is a perceivable increase in photopreference.

Besides the T-maze I started testing flies using the Benzer Paradigm. In this experiment they don’t have the options of light and darkness, but they can choose between light and less light. They can either go towards the light and transfer to another vial or just stay where they are. I started testing only wing-clipped flies (n = 10) with a groupsize of about 50 flies, but I will test groups of wing-clipped and normal flies taken together in the same vial as well.

The Preference Index is calculated by

PI = ((#F5×5)+(#F4×4)+(#F3×3)+(#F2×2)+(#F1×1)+(#F0×0)) / (#FT )