Renal Physiology
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Direct answer
The kidneys receive about 1.2 litres of blood per minute — close to a fifth of cardiac output — and filter roughly 125 mL of plasma each minute, some 180 litres a day, of which more than 99 per cent is reabsorbed to leave 1-1.5 litres of urine. Filtration is driven by net filtration pressure near 10 mmHg (glomerular capillary 60 against plasma oncotic 32 and capsular 18) and is held constant between mean arterial pressures of 80 and 180 mmHg by myogenic autoregulation and tubuloglomerular feedback. Clearance analysis — excretion rate divided by plasma concentration — converts urine and blood numbers into functional conclusions, and anchors almost every exam question this topic generates.
What you must remember
- Clearance equation C = U x V / P; inulin, freely filtered with no further traffic, is the gold standard GFR marker, while creatinine slightly overestimates because of tubular secretion.
- PAH is filtered and aggressively secreted, so its clearance approximates effective renal plasma flow near 625 mL/min; filtration fraction is GFR over RPF, about 20 per cent.
- Segmental reabsorption: proximal tubule two-thirds of salt and water with all glucose and amino acids (isotonic), thick ascending limb 25 per cent via NKCC2 (furosemide's target), distal tubule 5-10 per cent via NCC (thiazides), collecting duct under aldosterone (ENaC) and ADH (aquaporin-2).
- Glucose transport maximum: filtered load at normal glycaemia is about 125 mg/min; glycosuria begins above a threshold of 180 mg/dL (the splay) and transport saturates at a Tm near 375 mg/min.
- Counter-current multiplier: the water-impermeable ascending limb pumps sodium without water, building a medullary gradient up to 1200-1400 mOsm/kg, with urea recycling contributing to the inner medulla and the vasa recta acting as exchangers.
- Macula densa sensing of distal sodium chloride adjusts afferent tone (tubuloglomerular feedback, using adenosine), completing autoregulation with the myogenic response.
- Endocrine add-ons: renin from juxtaglomerular cells, erythropoietin from peritubular interstitium, and 1-alpha-hydroxylation of vitamin D — three hormones, one organ.
Working through a clearance problem
The exam version is always numerical. A patient provides urine flow 2 mL/min, urine inulin 60 mg/mL and plasma inulin 1 mg/mL: inulin clearance is (60 x 2)/1 = 120 mL/min, and that number is the GFR. Now add PAH clearance of 600 mL/min: that is effective renal plasma flow, so the filtration fraction is 120/600 = 20 per cent — and with a haematocrit of 0.45, renal blood flow is 600 divided by 0.55, roughly 1.1 L/min. Each substance then tells its own story: a clearance below GFR means net reabsorption, above GFR net secretion, and zero means everything filtered was reclaimed.
Turn the same arithmetic onto disease. A creatinine that doubles from 1 to 2 mg/dL means GFR has roughly halved, because the relationship is inverse and non-linear. Loop diuretics abolish the medullary gradient by blocking NKCC2, so maximum urine osmolality falls and the concentrating defect appears.
Where students slip
The commonest error is confusing clearance with excretion: clearance is the virtual volume of plasma completely cleared of a substance per minute, not the amount leaving in urine. The second is forgetting that glucose clearance is zero in health — every filtered molecule is reabsorbed below the threshold of 180 mg/dL. The third is treating PAH clearance as GFR; it estimates plasma flow precisely because secretion empties peritubular capillaries, the opposite logic to inulin. And the medulla question is missed for a simple reason: candidates memorise the multiplier but forget that urea contributes nearly half the inner-medullary osmoles under ADH action, which is why protein malnutrition impairs urinary concentration — a viva favourite with real Indian clinical weight.
Frequently asked questions
Why is inulin the gold standard for GFR?
It is freely filtered, neither reabsorbed nor secreted nor metabolised, so its clearance equals GFR by definition; creatinine is the clinical proxy with mild secretion overestimation.
What does PAH clearance measure and why?
Effective renal plasma flow (about 625 mL/min), because filtration plus near-complete secretion clear essentially all PAH from plasma in a single renal pass.
Define transport maximum using glucose.
Tm is the maximum reabsorption rate — about 375 mg/min for glucose; filtered loads above it spill into urine, starting at the 180 mg/dL threshold because nephrons differ (the splay).
How is the medullary gradient generated and maintained?
The thick ascending limb pumps sodium without water (multiplier), urea recycles under ADH, and the vasa recta exchange without washing the gradient out — up to 1200-1400 mOsm/kg at the papilla.
What are the two mechanisms of renal autoregulation?
The myogenic response of afferent arterioles to stretch and tubuloglomerular feedback from the macula densa, holding GFR steady between mean arterial pressures of 80 and 180 mmHg.
Why does a small creatinine rise matter so much?
Creatinine and GFR relate inversely and non-linearly, so a rise from 1 to 2 mg/dL represents roughly a halving of filtration — the basis for early detection of acute kidney injury.