Aspects of Renal Physiology (12/11/13 lecture)

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Mechanisms for fine-tuning urine osmolarity and its hormonal control, renal clearance, osmolar clearance, free water clearance, the filtration fraction, and titratable acidity. From the 12/11/13 Human Physiology lecture.
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Aspects of Renal Physiology (12/11/13 lecture) Kidneys are Roles of the kidneys include: excretion of waste products (e.g. urea); excretion of foreign compounds (e.g. xenobiotics); control of erythrocyte number via erythropoietin; production of active form of vitamin D; regulation of plasma osmolarity; long term acid/base balance (short term = breathing) and excretion of non-volatile acids; long term control of blood pressure via regulation of H2O and ions. The nephron consists of vascular and tubular elements. The blood enters the afferent arteriole and its diffuse capillary bed under pressure from each ventricular systole and each peristaltic movement, which is sufficient to cross the basement membrane of Bowman’s capsule. It travels through the glomerulus/capsule and the proximal convoluted tubule in the cortex, then the loop of Henle in the medulla, before returning to the distal convoluted tubule in the cortex, and emptying via collection ducts into collection tubules and eventually the renal pelvis, after much reabsorption and secretion of course. Excreted = filtered + secreted – reabsorbed Urine osmolarity is the sum total of the solutes involved. Modifying urine osmolarity conserves or excretes. Hypotonic urine requires Na+ reabsorption independent of H2O, which is achieved via Na+-pumps in the distal convoluted tubule, which is impermeable to H2O. Hypertonic urine, in which H2O reabsorption independent of Na+ occurs, is more difficult. Once filtered, the H2O is technically outside the body, so osmotic forces are required – a hypertonic area inside the body, to set up an H2O concentration gradient. This area is the medullary interstitium, maintained by the loop of Henle. There is a very low water concentration in the medulla (maintained by solutes pumped out) so the H2O concentration is higher in the tubule, forcing it to move out into the interstitium independently of solutes in the tubule. Urine composition (hypotonic or hypertonic) is controlled hormonally. To generate hypertonic urine, in which H2O is resorbed independent of Na+, anti-diuretic hormone is released from the posterior pituitary (neurohypophysis). Its release is sensitive to osmoreceptors in the hypothalamus (specialised cells sensitive to H2O movement which swell or shrink in response to this and their activity is altered because they are nerve cells) and volume receptors in the right atrium (which are stretched if the heart chamber is fuller if more fluid is consumed). Both act to make tubules more permeable to water reabsorption. Alternatively, when the right atrium is stretched, atrial natriuretic peptide (ANP) is released, which inhibits Na+ reabsorption, thereby causing its excretion, in which water follows. To generate hypotonic urine, in which Na+ is reabsorbed independent of H2O, aldosterone is released from the adrenal cortex, where it acts on the kidneys to stimulate Na+ reabsorption via channels. Its release is sensitive to the plasma osmolarity and to angiotensin. Angiotensin is generated via renin release from the kidney. It stimulates Na+ reabsorption, and its release is sensitive to blood pressure and plasma osmolarity. The renin-angiotensin system is based around the juxtaglomerular apparatus. Renin is released from granular juxtaglomerular cells, smooth muscle cells which no longer stretch/contract but secrete renin, and it does so in response to stretch (physical pressure), sympathetic stimulation, and hormones in the blood. The distal convoluted tubule is positioned next to the glomerulus and in between the afferent and efferent arterioles, which is significant because it is the macula densa cells here which feed back to control the release of renin. Angiotensinogen is present in the blood. Signs of lowered blood pressure (decreased renal perfusion pressure – baroreceptor, decreased Na+ at macula densa cells, and increased sympathetic nerve activity – β1) lead to renin being released. This cleaves the last 10 amino acids from angiotensinogen to form angiotensin I. Next, angiotensin converting enzyme (ACE) on the capillaries in the kidneys cleaves angiotensin I to form an octo-peptide, angiotensin II. This produces physiological effects! Lastly, it is converted to the hepta-peptide angiotensin III by aminopeptidase. The physiological effects of angiotensin II to support blood pressure include: vasoconstriction; direct renal sodium retention; aldosterone secretion (H2O reabsorption following Na+ reabsorption); increased thirst; anti-diuretic hormone release; sympathetic activity facilitation to act on heart/vessels; increased cardiac contractility; and long-term cardiac and vascular hypertrophy. Renal clearance The volume of plasma from which the substance is removed, per unit time (typically ml/min). e.g. Curea = 60ml/min : every minute, 60ml of plasma is cleared of urea. Assuming that all substances in the urine come from plasma and blood flow into the kidney = blood flow out (given tiny amount lost), if a substance appears in the urine then kidneys must have decreased its plasma concentration i.e. concentration in venous plasma In order to calculate clearance: empty bladder (so you start off at known point from which you measure the accumulation of urine), perform a timed urine collection (e.g. 24 hours), measure the volume and concentration of the substance in the urine, take a plasma sample, and measure the concentration of the substance in the plasma. Total amount of substance excreted = urinary concentration x urinary volume Minimum volume of plasma that could have supplied the excreted amount = urinary concentration x urinary volume / plasma concentration Rate at which the clearance happens = (urinary conc. x urinary vol.) / (plasma conc. x time) Creatinine is filtered but not secreted nor reabsorbed so its clearance is equal to the glomerular filtration rate (which should be ~120ml/min, a constant). So if Ccreatinine is 2 volumes per minute, then so is the glomerular filtration rate. If a substance undergoes net reabsorption, its clearance will be less than the glomerular filtration rate. e.g. urea, (Curea = 60 ml/min) which is reabsorbed for hypertonicity of the medulla If a substance undergoes net secretion, its clearance will be more than the glomerular filtration rate. Para-aminohippurate (PAH) clearance is used to measured renal plasma flow and renal blow flow. It is filtered and fully secreted, but not reabsorbed (it is fully cleared). This means its clearance is equal to renal plasma flow. Renal blood flow requires one to know the proportion (PCV – proportion of blood volume taken up by erythrocytes) and volume of erythrocytes: Renal blood flow = renal plasma flow / (100 – packed cell volume) * 100 Osmolar clearance – the osmolarity of plasma reflects the concentration of all solute particles, the overall concentration of solute particles that determines water concentration and water movement by osmosis. It is ~300 mOsm (milli osmoles/L). Glucose and NaCl have different molarities so osmoles are used instead. Osmoles = a mole of particles, no matter what they are. The osmolarity of urine and plasma can be measured via a Freezing Point Depression Osmometer as solute particles lower water’s freezing point in a known way. If we know the volume of urine, we can calculate osmolar clearance using the clearance formula (U x V) / (P x T). Free water clearance If you produce 100ml of isotonic (300 mOsm) urine per hour, then the osmolar clearance is 100 ml/hour (the smallest volume of plasma required to account for all the solutes in that 100 ml of isotonic urine is 100ml). If you produce 100ml of hypotonic urine (e.g. 150 mOsm), then the osmolar clearance is 50 ml/hour (the smallest volume of plasma required to account for all the solutes in that 100ml of hypotonic urine is 50ml). The other 50ml is ‘free’ water – it is positive free water clearance. You can see that the kidneys are ‘dumping’ water, because only 50ml of the urine was required for the solutes, the other 50ml is ‘free’ and so may have been ‘dumped’ if blood pressure was too high, for example. If you produce 100ml of hypertonic urine (e.g. 600 mOsm) per hour, then the osmolar clearance is 200ml/hour. The urine is twice as ‘strong’ as the isotonic plasma, so twice the volume of plasma would be required to supply those solutes. Given that 200ml has been cleared, but only 100ml has been excreted, the remaining 100ml is ‘free’ water – but this is negative free water clearance, as it is retained in the body. Free water clearance = urine production rate – osmolar clearance Mostly we are in positive free water states due to good hydration. The filtration fraction is the proportion of the plasma that is filtered when it flows through the glomerular capillaries. The full amount can’t be filtered, or only the erythrocytes would be left in the vessels and blood would be too viscous to flow. As a result, the filtration fraction is ~20%. 4/5 is left behind to dilute erythrocytes. Lastly, urine is the main route for removing non-volatile acids from the body. Therefore, urine is normally acidic as we produce acids all the time. Titratable acidity is the amount of alkali required to bring urine pH back to 7.4 (the more acid excreted, the more base required to return urine to pH 7.4).

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