Processes Involved In Urine Formation

Processes involved in urine formation 1) Formation of ultrafiltrate – filtration creating movement from the capillary to Bowman’s space 2) Modification of ultrafiltrate – reabsorption creating movement from the tubule to the capillary AND secretion creating movement from the capillary to the tubule 3) Urinary excretion – substances appearing in urine Glomerular filtration    Glomerular filtration is the starting point for urine formation. Because of the high hydrostatic blood pressure in the glomerular capillaries, filtrate of plasma is pushed out of the capillaries into the urinary space of Bowman’s capsule. This process of filtration is often referred to as ultrafiltration, and filtrate is referred to as an ultrafiltrate of plasma because the glomerular capillary wall, the glomerular filtration barrier, behaves like an ultrafine filter and allows for free passage of small molecules but restricts the passage of large molecules. The glomerular filtration barrier consists of three layers o Fenestrated capillary endothelium o Basement membrane that consists of a meshwork of fine fibrils embedded in a gel-like matrix o Podocytes cell layer (visceral epithelium), which have numerous secondary processes resting on the basement membrane. The narrow space between adjacent foot processes is called the slit pore. A thin membrane stretches across it (the filtration slit diaphragm) and has a fine porous structure. The barrier has a high permeability to water and small molecules (MW < 7000) but restricts macromolecules. The endothelia pores, basement membrane and surface coat of the podocytes contain negatively charged glycoproteins which impede the filtration of negatively charged plasma proteins such as serum albumin. Forces governing glomerular filtration  Thus, net filtration pressure = forces favouring filtration – forces opposing filtration = (PGC + ∏BS) – (PBC + ∏GC)  Note that PGC > ∏GC > PBC while ∏BS is ~ 0  Since ∏GC is the only forces that changes as you do down the capillary as shown by the graph below, it makes sense filtration over a greater length of glomerular capillary would increase GFR. Thus we, can increase surface area and capillary wall conductivity (together these two factors make up Kf Glomerular filtration rate (GFR) Filtration fraction  Filtration fraction (FF) is the volume of filtrate formed from a given volume of plasma entering the glomeruli. The fraction of the renal plasma flow that is filtered (filtration fraction) averages about 0.2, meaning that about 20 per cent of the plasma flowing through the kidney is filtered through the glomerular capillaries. Renal clearance   The renal clearance of a substance is the volume of plasma that is being completely cleared of the substance by the kidneys per unit time – or, renal clearance measures the volume of plasma (calculated /theoretical) from which all of a substance has been removed and excreted in the urine in one minute. Renal clearance provides a useful way of quantifying the excretory function of the kidneys, e.g. to quantify the rate at which blood flows through the kidneys as well as the basic functions of the kidneys: glomerular filtration, tubular reabsorption, and tubular secretion. Measuring GFR – via the clearance rate of inulin    If a substance is freely filtered and is not reabsorbed or secreted by the renal tubules, then the rate at which that substance is excreted in the urine (U x × V) is equal to the filtration rate of substance by the kidneys (GFR × Px). Rearranging this equation will give the equation below. Inulin is a substance that fits this criteria The clearance of creatinine (by-product of muscle metabolism) can also be used to assess GFR. The measurement of creatinine clearance does not require intravenous infusion into the patient, thus this method is much more widely used than inulin for estimating GFR clinically. However, there can be a slight error in measuring plasma creatinine, which leads to an overestimation Measuring renal plasma flow (RPF) – via the clearance rate of PAH   Theoretically, if a substance is completely cleared from the plasma, the clearance rate of that substance is equal to the total renal plasma flow i.e. the amount of the substance delivered to the kidneys in the blood (RPF × Px) would be equal to the amount excreted in the urine (Ux × V). Rearranging this equation will give the equation below. There is no known substance that is completely cleared by the kidneys – PAH (paraamino hippuric acid) is about 90% cleared from the plasma. Therefore, the clearance of PAH can be used as an approximation of renal plasma flow. Measuring renal clearance of glucose   Usually there is no glucose in urine so this would make U x = 0 and thus clearance rate = 0. However, there is a threshold value (T m) at which glucose will be cleared from plasma and excreted in the urine, thus the process is rate dependent