This is an image of a typical male C. elegans worm. 1mm is the length if it is s

Question

This is an image of a typical male C. elegans worm. 1mm is the length if it is stretched straight. Using the image estimate the volume in its digestive tract, and how many E. coli could line up end to end from the mouth to the cloacal opening. You may approximate the GI tract as a cylinder. The volume of a cylinder is Product r^2 I_ R is the radius (i/2 diameter) and L is the length. If the experiment in question 1 showed that the food passed through in 5.5 minutes, then what the dilution rate for the contents of the C. elegans?

Explanation / Answer

volume of a cylinder = pi x r2 x h, where pi= 3.14, r = radius of the cylindrical object, and h= length/height of object.

E. coli is 2 micron in length.

so number of E coli x 2 micron = 1 mm.

E coli = 1 x 10-3 / 2 x 10-6

= 1 x 103 / 2.

=1000 / 2

= 500 E. coli.

you haven't provided the diameter of the worm. so I am unable to provide the volume.

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