How much do membrane invaginations (folding) increase surface area?

Reply Wed 10 Oct, 2012 12:03 am
There are two cells cell A and Cell B. Cell A is a rectangular prism and Cell B is the same thing but with invaginations in the membrane. Dimensions for cell A are: height = 3 um width = 6 um and depth = 2 um

i) calculate SA for cell A : I got 72 um^2
ii) calculate V for cell A : I got 36 um^3
iii) calculate SA/V for cell A : I got 2

Now they say that the volume of Cell B is 10 um^3 less than Cell A because of the folding and the SA/V is 48 , so what is cell B's SA

I did this:

SA/26 = 48
SA =1248 um^2

doesn't this seem ridiculous ?? Is this correct and it's just crazy or did I go wrong somewhere?
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Reply Wed 10 Oct, 2012 03:38 am
Cell A

SA=(3x6+6x2+2x3)x2=(18+12+6)x2=36x2=72 um^2

V=3x2x6=36 um^3

SA/V=72 um^2/36 um^3=2 um^-1

Cell B

Vb=Va-10 =26 um^3

SAb/Vb= 48 um^-1

SAb=48 um^-1x26 um^3=1248 um^2

Your numbers seem correct.

The fissures in the prism reduce the cell xoumn but greatly increasr th surface area. Look at it this way. Compare a perfect sphere to a piece of paper waddws into an equal sized ball. The volumns are the same, but which has a greater suface area.

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