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?

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

IntroductionThe glomerular filtration surface will consist of endothelial pores, glomerular basement membrane, endothelial glycocalyx as well as podocyte filtration slit diaphragm. The main function of the glomerulus is that it offers a semipermeable membrane that helps the components of blood to be separated as cellular or macromolecular condensate.A microporous membrane is one that has a very thin structure whose morphology is a precisely interconnected and controlled network consisting of tiny holes called pores. The small impermeable structure helps in allowing water droplets to easily pass through.AnswerA)  Poisseuille's flow equation formula:Q=πr4∆p8η∆rHerer=3.5×10-9m∆r=50×10-9m∆p=60 -20-28 mm Hg=12 mm Hg12 mm Hg=12×133.3 pa=1599.6 paη=0.002 pa.sSubstituting these values in the equation can give usQ=π(3.5×10-9)4×1599.68×0.002×50×10-9Q=9.42×10-22m3.s-1×106cm3.m-3×60s.min-1Q=5.65 ×10-14ml.min-1B) The normal GFR is 120 ml.min-1. The number of pores can be obtained by dividing flow by poresn=120 ml.min-15.65×10-14 ml.min-1=2.12×1015 poresC) Pore density can be defined as the number of pores 'n' in a given unit areap=2.12 ×1015 pores1.5 m2=1.414×1015 pores.m-2D) Fraction of area as pores is given byF=1.414×1015 pores.m-2×πr21m2=1.141 ×1015 pores.m-2.π×(3.5 X 10-9)21m2=0.0544 ...