Examkrackers Bio question

[happily123]

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Can someone please explain this question from EK Bio?

116. Bile allows lipases to work more efficiently by increasing the SA of fat/ If fat globules are assumed to be spherical, then each time bile decreases the diameter of all the fat globules in the small intestine by a factor of 2, the SA of the fat is increased by a factor of:

A. 2
B. 4
C. 8
D. 16



the answer is A

Thanks!

 

[quicheduty]

There might be an easier intuitive way, but here's a basic mathy way:
Say you have an original globule of radius 2. Your volume (4/3πr^3) would be 32/3π and your surface area (4πr^2) would be 16π. I included the constants in the calculation just for absolute clarity, although you really can just ignore them.

Then you bile the globule up which reduces your radius by 1/2, so it's now 1. Check the volume of each globule: it's now 4/3π. Based on the conservation of fat law (jk but you know what I mean) it must mean the bile emulsified the original drop to 8 smaller drops if your total fat volume is to remain 32/3π.

Now check the surface area: one small drop has surface area of 4π. Multiply that by 8 drops and you have 32π total. You now have 2x the total SA as the original drop (16π).

 

[steelersfan1243]

Can you explain where you are getting the 8 drops from? I feel as if I am missing something fundamentally here...

 

[quicheduty]

Can you explain where you are getting the 8 drops from? I feel as if I am missing something fundamentally here...

The volume of the original drop was 32/3π. Bile breaks up fat into smaller bits but you can't gain or lose fat this way. Your total vol must remain 32/3π no matter how much you chop it up. Like slicing up a cake to split with people. Each person gets a small slice of the cake but every slice helps make up the same cake. So if one broken up fat drop has a new vol 4/3π you must have 8 of those to put back together your 32/3π big droplet.

 

[erythrocyte666]

suchbrio explains it well above
when i did this problem, i just thought this: key is that volume is constant before and after emulsification; so halving radius means decreasing volume of one droplet to 1/8 - in other words there will be 8 smaller droplets, each of volume V/8. However, decreasing radius by half reduces surface area of each droplet to 1/4, but since there are 8 droplets you have 8 x (1/4) = 2

 

[quicheduty]

suchbrio explains it well above
when i did this problem, i just thought this: key is that volume is constant before and after emulsification; so halving radius means decreasing volume of one droplet to 1/8 - in other words there will be 8 smaller droplets, each of volume V/8. However, decreasing radius by half reduces surface area of each droplet to 1/4, but since there are 8 droplets you have 8 x (1/4) = 2

aaaaand there's the more intuitive way!!! 🙂