For fusion of a vesicle with it\'s target membrane to occur. The membranes have
ID: 40684 • Letter: F
Question
For fusion of a vesicle with it's target membrane to occur. The membranes have to be brought to within 1.5 nm so the two bilayers can join. Assuming that the relevant portion of the two membranes at the fusion site are circular regions 1.5 nm in diameter, calculate the number of water molecules that would remain between the two membranes ( water is 55.5 m and the volume of a cylinder is ?r^2h). Given that an average phosphlpid occupies a membrane surface of 0.2nm^2, how many phospholipids would be present in each f the opposng monolayers at the fusion site? Are there sufficent water molecules to associate with the hydrophillic head groups of this number of phospholipids (10 phospholipids associate with each headgroup)?
Explanation / Answer
FUSION SITE = 1.5 nm DIAMETER
AVERAGE PHOSPHOLIPID SURFACE AREA = 0.2NM
10 PHOSPHOLIPIDS ASSOCIATE WITH EACH HEAD GROUP
VESICLES ARE SMALL MOLECUES WHICH ARE SURROUNDED BY THE PHOSPHLIPID BILAYER . FORMING HYDROPHILIC HEAD TOWRDS THE WATER , THEY FORM MICELLES
CALUCULATIONS ;
IT is caluculated by the diffusion constant D w ,through eisten equation for diffusion constant .
Dw= pc L/R. Pc is the energy constant L AND R are the radius of the cylinder or water molecule
when they form the micelles the number of water molecules inside the micelles reduces , there is a short reduction the osmostic pressure , so the number of water molecules inside the phospolipids are very less.outside the memberane will be more can be caluculated by
vof interior /total volume of the vesicle= [r-d] 3divided by r3 = where r is the radius of the water molecule , d is the diameter of water , volume i s 0.2 nm average surface area
0.2 / total volume = [ 55 - 1.5]3 divided by 1.53 = 160.5/3.375= 47.9*02= 9.5 total molecules of water are associated with water molecule
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