Physiology (4) 2/4/2018. Wael abu-anzeh

Transcription

1 Physiology (4) 2/4/2018 Wael abu-anzeh

2 In the previous lectures we have discussed the filtration and the reabsorption processes but in this lecture we will talk about the factor that will regulate or control the reabsorption process ( or how much fluid or amount of substances that will be reabsorbed) -: 1- Glomerulotubular balance : which means the ability of the renal tubule to handle the extra load of fluid filtered by increasing its reabsorption. For example: If GFR increases to 150 (ml/min) the reabsorption at the proximal tubule will goes from 81 ml/min to 97.5 ml/min(which is 65% of GFR, "the same percentage when the GFR was normal"). * so this ability to maintain the reabsorption percentage even though the GFR increase is called the Glomerulotubular balance. - This act together with autoregulation to prevent large change in fluid flow in distal tubule. 2- Peritubular capillary and interstitial fluid physical forces: if we can remember that around the loop of Helen there is a capillary which called the peritubluer capillary and this capillary is important for the reabsorption from the renal tubule but the fluid that will be reabsorbed will not go directly from the tubule to the capillary it should first pass to interstitium then from the interstitium to the capillary.

3 fluid physical forces at this point : " the negative sign here indicates that this force will the opposes movement toward the peritubular capillary" a. Hydrostatic pr. In capillary Pc opposes -13. b. Hydrostatic pr. In interstitium Pif favors 6. c. Colloid osmotic pr. in capillary πc favors 32. d. Colloid osmotic pr. In intrst. πif opposes -15. * the summation of these forces is: 10 mmhg in favors or toward the capillary -by simple mathematical calculation Reabsorption = Kf "the filtration coeffiecnt" x Net absorptive pr."which is 10 mmhg" Reabsorption = 12.4 x 10 = 124 ml/min *we have to put in mind that the reabsorption is a two steps procces : - The first step is from the renal tubule to the interstitium. -the second step is from the interstitium to the peritubular capillary. So IN GENERAL FORCES INCREASE PERITUBULAR CAPILLARY REABSORPTION ALSO INCREASE REABSORPTIN FROM RENAL TUBULES AND VICE VERSA. And as we explained : - A. increase Arterial pr. >> "will increase" hydrostatic >>" which will decreases" reabsorption Change resistance of afferent and efferent( or vaso constriction) decreases hydrostatic >> increase reabsorption. B. increase in Plasma colloid osmotic pr. >> incresess coll. In perit. >> ^ reabs. >> ^Filtration fraction ^[protein] >> ^reabsorption

4 C. (A,B) if increase reabsorption >> ^ fluids in interstitium >> ^ hydrostatic pr. and decreases colloid pr. >> decreases reabsorption of fluid from renal tubules. *simple explanation: the main idea of the previous point (point C)is that: In either way or "in case of increase(pointb) or decrease(pointa) absorption toward the blood capillaries", the fluid should pass to the interstium first " as simple as that ## Another explanation Basically, changes in peritubular capillary physical forces influence tubular reabsorption by changing the physical forces in the renal interstitium surrounding the tubules. For example, a decrease in the reabsorptive force across the peritubular capillary membranes, caused by either increased peritubular capillary hydrostatic pressure or decreased peritubular capillary colloid osmotic pressure, reduces the uptake of fluid and solutes from the interstitium into the peritubular capillaries. This action in turn raises renal interstitial fluid hydrostatic pressure and decreases interstitial fluid colloid osmotic pressure because of dilution of the proteins in the renal interstitium. These changes then decrease the net reabsorption of fluid from the renal tubules into the interstitium, especially in the proximal tubules. *at the second part of the picture there were an increase in the hydrostatic capillary pr.(pc) and decrease in the colloid osmotic pr. Of the capillary (πc) : As a result there were decrease in the fluid flow toward the peritubular capillary Backleak). 3- PRESSURE NATIURESIS AND PRESSURE DIURESIS. in a simple way of explanation : when the blood pressure increase beyond the autoregulation range which is ( mmhg ) there will be an increase in the urine output. For example : - in Kidney diseases >> incr BP and GFR >> incr renal interst fliud hydrostatic Pr >> reduce absorption of Na &water >> incr urine output.

5 *Also in some cases :- Kidney diseases reduce angiotensin secretion decrease aldosterone decr Na absorption & increase urine output. 4- Hormonal control : which include 5 different hormones : A-Aldosteron : Increases sodium permeability on luminal side. It's effect is on the on distal tubule and collecting duct >> increase NaCl, H2O reabs. and K+ secretion. *In Addison's disease : no aldosterone >> loss of Na+, accumulation of K+. *Conn's syndrome : excessive production of aldosterone >> Na+ retention K+ depletion. B-Angiotensin II : It's effect is on proximal tubule NaCl,H2O reabsorption and H+ secretion. Other effects: 1. Stimulates aldosterone secretion 2. Constricts efferent art "not the afferent because at the afferent side there is release of vasodilator factors". 3. Stimulates Na+ reabsorption on Na-K pumps, Na-H+ exchanger & Na-bicarbonate co-transporter.

6 C-ADH ( vasopressin): It's effect is on distal, collecting tubule and coll. duct >> ^H2O reabsorption mainly. *it's mechanism of action : simply by binding with its receptor on the tubuler cells (V2>> form cyclic AMP>> which activates protein Kinase A >> which moves aquaporin-2 (AQP-2) to luminal membrane & form water channels. D.Atrial natriuretic peptide ANP : in the atrium of the heart there are receptor which sense the stretch on the wall of the atria and when it is activated it cause the release of ANP which works on reducing the load or the extracelleur fluid and this is important in case of heart failure. -Dilated atria >>Decrease Na+ and H2O reabsorption in distal and collecting duct Also cause decrease renin secretion >> decrease angiotensin formation. E.Parathyroid hormone: It's effect is on Proximal., thick ascending of L.H, distal tubule >> decreases PO4- reabsorption and ^Ca++ reabsorption 5.Sympathetic N.S. (not part of the hormonal control, it is a separate point or regulation factor) : it's direct effect is >> constricting afferent and efferent artery, but here when the blood to the kidney decrease the macula densa which is part of the juxtaglomerular apparatus will increase renin and angiotensin II formation which eventually will increase Na and H2O reabsorption. * A quick recap the regulators of reabsorption are: 1-tubuler balance//2-pericapliiry control//3-blood pressure//4-hormons//5- symathatic innervation.

7 Renal clearness Renal clearance of substance: Volume of plasma that is completely cleared of the substance by the kidneys per unit time. * Useful way of quantitating the effectiveness with which kidneys excrete various substances (used as a kidney function test). a. Freely filtered, not reabs., no secretion excretion rate = filtration rate Example: Creatinine (normally produced in the body by the breakdown of muscle), inulin (Inulin clearance = GFR. // Amt. Filtered = Amt. Excreted) b. Freely filtered, partialy reabs, most electrolytes(slide 113) c. Freely filtered, not excreted completely reabs., a.a, glucose (slide 112) d. Freely filtered, not reabs., but secreted, H, (slide 114) For substance A >> Urinary excretion rate=gfr (because all the filtrated material is excreted not reabsorbed nor secreted ) Creatinine is the most sensitive kidney test and it is better than inulin since it is normally produced in the body and does not need injection. Para-aminohippurate (PAH) is not like substance A is more like substance D so excretion rate does not equal filtration rate (because there is part of it will be secreted), excretion rate equal total renal plasma flow, in other words : * Amt. delivered to kidneys in blood= amt. excreted in urine. Note : The clearance of the PAH is 90% not 100% so in measuring the clearance we should divide by 0.9 Please go to slide and read