4) One model of the glomerular membrane is a microporous membrane in which right

Question

4) One model of the glomerular membrane is a microporous membrane in which right cylindrical pores penetrate all the way through the membrane. Assume that the pores have a length of 50 nm and a radius of 3.5 nm. The viscosity of plasma is 0.002 Pa s. The average hydrostatic pressure in the glomerulus is 60 mm Hg, hydrostatic pressure in Bowman's space is 20 mm Hg and the average oncotic pressure of glomerular capillary blood is 28 mm Hg. A. Calculate the flow through a single pore assuming laminar flow (use the Poiseuille flow equation) B. How many pores would there have to be to produce a normal GFR? C. If the total aggregate area of the kidneys for filtration is 1.5 m2, what is the density of the pores (number of pores per unit area)? D. What fraction of the area is present as pores?

Explanation / Answer

In case of glomerular memberane the flow through the single pore is given as

where r, is the radius of the pore=3.5nm

p is the viscosity of the plasma

delta p is the pressure difference of two side of the pore

PGC, average hydrostatic pressure in glomerulus is 60 mm Hg

PBS, hydroststic pressure in bowmwn space is 20 mm Hg

and BBS is average pressure of glomerular capillary blood is 28 mm Hg

so,delta P=PGC-PBS-BBS

=60-20-28

=12 mm Hg or 1599.6 Pa( 1mm Hg=133.3 Pa)

on deriving value in poiseuille flow equation we get

F=5.65*1014 mL min-1

B. Now normal GFR=120 mL per min

on dividing it by flow per pore we get,120/5.65*1014=2.12*1015 pores.

C.Total aggregate area of kidneys for filtration is 1.5 m2

so pore density is 2.12*1015/1.5=1.414*1015 pores /m2

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