2006-11-15 08:53:28 · 9 answers · asked by Frank m 2 in Science & Mathematics ➔ Biology
the rough endoplasmic reticule is called so because of the ribossomes attached to it. both the parts of the endoplasmic reticulum are responsible for membrane synthesis. As you might know, cell membranes are composed of lipids and proteins. Now, the smooth e.r. can only synthesize the lipidic portion, while the rough e.r., since it has ribosomes (responsible for the translation of RNA into proteins), can synthesize the membrane proteins, as well.
2006-11-15 08:57:40 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Rough ER contains ribosomes on it's surface which give it the "rough" appearance unlike the Smooth ER that has no ribosomes on it.
2016-03-28 21:42:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
protein synthesis takes place on the rough ER (because of ribosomes). The smooth ER is more like a bunch of tubes that synthesize lipids.
2006-11-15 09:02:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
funny you would ask that....i just took a test on cells in biology today!!!! the difference is that the smooth ER has NO ribosomes on it....and the rough ER does.
2006-11-15 09:02:03 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Rough ER-saccules with ribosomes attached, proteins made and transported into saccule.
Smooth ER-tubules carry material through the cell, modifications made as proteins and other material travel through system
2006-11-15 09:01:55 · answer #5 · answered by scook0424 1 · 0⤊ 0⤋
smooth er does not have any ribosomes associated with it. For a really good explanation go to wikipedia and look up Endoplasmic Reticulum.
2006-11-15 09:00:24 · answer #6 · answered by biosciguy 3 · 0⤊ 0⤋
rough Er has ribosomes on them and smooth ER don't have any on them
2006-11-15 10:26:51 · answer #7 · answered by amipatel333 1 · 0⤊ 0⤋
rough er has ribosomes attached but it is basically that they look different.
2006-11-15 09:01:09 · answer #8 · answered by Clark H 1 · 0⤊ 0⤋
texture
2006-11-15 08:54:40 · answer #9 · answered by Anonymous · 0⤊ 1⤋