# Calculating the number of isoprenes in Lanosterol

Hey guys I'm having trouble understanding the structure of lanosterol regarding the isoprenes in the structure shown below:

The first isoprene I think is where the 3rd carbon would be at the OH. But I thought that carbon 3 of isoprene doesn't have any groups sticking out- only carbon 2 with the methyl group sticking out. Can isoprene units have hydroxyl groups too?

Also from looking at the image above, there also appears to be another isoprene unit right in the middle of lanosterol in blue- but it looks like a straight 5 carbon chain unlike your typical isoprene unit with a methyl group sticking out somewhere. Why is this? Thanks!

Squalene 1 is produced in vivo by reductive coupling of two farnesyl pyrophosphate (FPP) units. FPP is ultimately formed from acetate through the agency of the mevalonic acid pathway. Squalene is enantioselectively epoxidized to form squalene epoxide 2. In vivo acid-promoted cyclization leads to carbocation 3a with stereocontrol at the newly formed centers. Note that the red mechanism arrow in structure 2 leads to an anti-Markovnikov addition creating vicinal quaternary carbons at $$\ce{C8 and C14}$$. At this juncture all of the isoprene units are intact.
The conformation of the chain at $$\ce{C20}$$ in 3a is critical because it determines the stereochemistry at this center in lanosterol 4. The hydride at $$\ce{C17}$$ in structure 3b migrates to $$\ce{C20}$$ while the $$\ce{C13}$$ hydride shifts to $$\ce{C17}$$. The continuity of the original isoprene units is now broken as the $$\ce{C14}$$ methyl (methide) group migrates to $$\ce{C13}$$ and the $$\ce{C8}$$ methyl group moves to $$\ce{C14}$$. Finally, a proton is lost from $$\ce{C9}$$ to form the tetrasubstituted double bond of lanosterol. The facial integrity of the four migrating groups is maintained throughout the rearrangements. The migration of the two methyl groups destroys the integrity of the isoprene rule within lanosterol itself.