BI1002 - Urine Formation
- Created by: Gemma
- Created on: 04-02-14 18:47
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Step 1 = ULTRAFILTRATION
- Plasma is forced throught the glomerular capilleries by hydrostatic pressure of blood into Bowman's space
- Small molecules and ions pass though - proteins and blood cells stay in capilleries
- Capillery endothelium pores (fenestrae) = ~100 nM
- Basement membrane = -ve charged (contributes to filtration)
- Tubular pores = ~8nM
- Contain cells called podocytes with finger-like projections called pedicells
- Fluid formed = glomerular flitrate
- To find the composition of tubular fluid = micropunture method
- Use a micropipette to get the contents of the tubule - previously done on turtles
- Autoregulation of renal blood flow (RBF) and glomeruslus filtration rate (GFR)
- Need to match the transport capacity to the filteted load - glomerulotubular balance
- Myogenic control = stretch receptors
- Increase in pressure = stretch in smooth muscle which causes Ca+ entry and the muscle contracts - reducing blood flow
- Humoral factors prostaglandins and nitric oxide so act as vasodilator
- Macula densa also aids in maintains the right rate
- Metabolic control
- Metabolites (H+, adenosine) cause vasodilation. But an increase in pressure = a reduction in metabolities so vasocontricition happens
- Forces to produce glomerulus filterate:
- Starling forces (hydrostatic pressure) acts to force plamsa out the capilleries
- Proteins cannot fit through pores so exert oncotic pressure oppose the starling forces
- Small pressure from capsule that opposes hydrostatic pressure
- Net filtration pressure = 2.5 kPa
- Plasma Clearance = the volume of plasma that is cleared of a substance in a given time
- 125ml filtrate/min = 180l/day BUT only 1.5 L/day
- Use inulin to determine GFR
- freely filtered, not reabsorbed, no secreted, not broken down or synthesised
- rate of excretion = Uc x Vmg.min-1
- rate of excretion = filtered load = rate of filtration = Pc (plasma) x GFR
- Uc x V = Pc x GFR therefore GFR = (Uc x V)/Pc
- GFR measured by the inulin = inulin clearance
- Inulin = filtration only (125 ml/min GFR)
- PAH (para-amiohippuric acid) = filtration and secretion (600 ml/min GFR)
- Glucose = filtration and reabsorption (0 ml/min GFR)
Step 2 = REABSORBTION (PROXIMAL TUBULE)
- Tubular reabsorbtion = the movement of solutes back into the blood of a peritubular capillery
- Proximal tubule reabsorbs =
- 2/3rds of filtered H2O, Na+, K+, Cl- and HCO3-
- All filtered glucose and amino acids
- Reabsorbtion of Na+ is a driving force for reabsorbtion of most solutes
- Reabsorbtion mechainism for HCO3-
- Bicarbonate is freely filtered so must be reabsorbed
- Enzyme carbonic anhydrase:
- Exists in 2 forms = brush border and intracellular
- Brush border = Na+-H+ antiporter exchanges Na+ (tubular fluid) for H+ (intracellular)
- This results in secretion of H+ into the membrane = shift in the equilibrium towards carbonic acid
- Carbonic acid is converted into CO2 and water by enzyme carbonic anydrase (in the brush border)
- H2O…
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