RENAL SYSTEM
Renal Lecture 6 Dr. Janan AlrefaeeRenal clearance: renal clearance of a substance is the volume of plasma that is completely cleared of the substance by the kidneys per unit time. *Renal clearance is useful in measuring (1) the excretory function of the kidneys (2) renal blood flows rate (3) the basic functions of the kidneys: glomerular filtration, tubular reabsorption, and tubular secretion.
Cs= (US*V)/Ps Where Cs is the clearance rate of a substance s, Us is the urine concentration of that substance, and V is the urine flow rate. Ps is the plasma concentration of the substance.
*If a substance is freely filtered and is not reabsorbed or secreted by the renal tubules, then GFR can be calculated as the clearance of this substance as follows: GFR = (US*V)/Ps= Cs
Substances that fits these criteria and used to measure GFR: Inulin, is a polysaccharide which not produced in the body, is found in the roots of certain plants and must be administered intravenously to a patient. Creatinine is used clinically, it is a by-product of muscle metabolism and is cleared from the body fluids almost entirely by glomerular filtration (more suitable because no need for intravenous infusion as in inulin).
Renal Plasma Flow Para-aminohippuric acid (PAH) clearance can be used to estimate renal plasma flow. Theoretically, if a substance is completely cleared from the plasma, the clearance rate of that substance is equal to the total renal plasma flow.
Filtration Fraction (FF) FF is the fraction of plasma that filters through the glomerular membrane. The filtration fraction is calculated as follows: Filtration fraction = GFR (inulin clearance) /Renal plasma flow (PAH clearance). As the normal plasma flow through both kidneys is 650 ml/min and the normal GFR is 125 ml/min or 180 L/day in the average adult human, the filtration fraction 0.19 = 0.2 (20%). FF = 125 ml/min/650 ml/min = 0.19
Calculation of tubular reabsorption or secretion from renal clearances If the filtered load of substance (GFR x plasma concentration of substances) and renal excretion of a substance (urine concentration of substance x urine flow rate) are known, one can calculate whether there is a net reabsorption or a net secretion of that substance by the renal tubules.
If the rate of excretion of the substance is less than the filtered load of the substance, then some of the substance must have been reabsorbed from the renal tubules. Conversely, if the excretion rate of the substance is greater than its filtered load, then there is tubular secretion of the substance, and so can be calculated both of them.
Renal mechanisms for controlling urine concentration Antidiuretic hormone (ADH) (vasopressin): when there is a deficit of water and osmolarity of the body fluids increases above normal, the posterior pituitary gland secretes more ADH, which increases the permeability of the distal tubules and collecting ducts to water and decreases urine volume without obviously alter the rate of renal excretion of the solutes.
When ingestion excess water or when there is excess water in the body and extracellular fluid osmolarity is reduced, the secretion of ADH by the posterior pituitary decreases, thereby reducing the permeability of the distal tubule and collecting ducts to water, which causes excreting large amounts of dilute urine.
The kidneys conserve water by excreting concentrated urine The kidney excretion a small volume of concentrated urine minimizes the intake of fluid required to maintain homeostasis, especially in short water supply. This happen by continuing to excrete solutes while increasing water reabsorption and decreasing the volume of urine formed. The human kidney can produce a maximal urine concentration of 1200 to 1400 mOsm/L.
Obligatory urine volume It is the minimal volume of urine that must be excreted; a normal 70-kilogram human must excrete about 600 milliosmoles of solute each day. If maximal urine concentrating ability is 1200 mOsm/L, the obligatory urine volume, can be calculated as: 600 (mOsm /day) / 1200 (mOsm/ L) = 0.5 L / day
The limited ability of the human kidney to concentrate the urine to a maximal concentration of 1200 mOsm/L explains why severe dehydration occurs if one attempts to drink seawater. Because for every liter of seawater drunk (1200 mOsm/L), there is a net fluid loss of 1 liter (in addition to that the kidney must also excrete other solutes, especially urea, which contribute about 600 mOsm/L, so urine volume should be more) explaining the rapid dehydration that occurs in shipwreck victims who drink seawater.
Requirements for excreting concentrated urine (1) A high level of ADH, (2) a high osmolarity of the renal medullary interstitial fluid (which provides the osmotic gradient necessary for water reabsorption to occur in the presence of high levels of ADH).
Note: The countercurrent mechanism: depends on the special anatomical arrangement of the loops of Henle (countercurrent multiplier) and the vasa recta (countercurrent exchange), in addition to the collecting ducts (discuses later).
The process by which renal medullary interstitial fluid becomes hyperosmotic: 1- The countercurrent multiplier 2-Recirculation of urea from medullary collecting duct to Loop of Henle
1- The countercurrent multiplier steps are: [1]The operation start at loop of Henle assume that it is filled with fluid with a concentration of 300 mOsm/L. [2] Active transport of Na and CI out of thick ascending limb from the tubular lumen to the interstium & increase interstium osmolarity. This creates a high osmotic gradient between the interstial fluid and the fluid in the thin descending loop of Henle which is permeable to water.
[3] Movement of water by osmosis from the thin descending loop of Henle to the interstial fluid leading to increase the osmolarity of tubular fluid equal to interstium osmolarity. [4] The steps are repeated over and over, so the continuous inflow of isotonic tubular fluid from proximal tubule with outflow hypotonic tubular fluid into the distal tubule occurs.
With sufficient time, this process gradually traps solutes in the medulla more than water and multiplies the concentration gradient along the medulla, eventually raising the interstitial fluid osmolarity to 1200 to 1400 mOsm/L in medulla pelvic tip.
2-Recirculation of urea from medullary collecting duct to Loop of Henle When there is water deficit and blood concentrations of ADH are high, water is reabsorbed rapidly from the cortical collecting tubule and inner medullary collecting ducts, causing a higher concentration of urea in the tubular fluid. This lead to large amount of urea is passively reabsorbed from the inner medullary collecting ducts into interstitial fluid.
This urea diffuses from interstitial fluid into the thin loop of Henle, and then passes through the distal tubules, and finally passes back into the collecting duct and so. The recirculation of urea through these terminal parts of the tubular system several times before it is excreted, trap urea in the renal medulla leading to renal medulla hyperosmolarity. This is essential to save body fluid when water shortage
Urea contributes about 40 to 50 % of the renal medullary interstitium osmolarity (500-600 mOsm/L) when the kidney is forming maximally concentrated urine.
The people, who eat a high-protein diet, give up large amounts of urea as a nitrogenous “waste” product, can concentrate their urine much better than people whose protein intake and urea production are low.Malnutrition is associated with great impairment of urine concentrating ability.
Note: About one half of the filtered urea is excreted. Urea excretion rate is determined by two factors: (1) the concentration of urea in the plasma and (2) the glomerular filtration rate.