The trap that costs marks in urine formation questions: confusing what the proximal convoluted tubule (PCT) reabsorbs versus what the distal convoluted tubule (DCT) and collecting duct handle. Aspirants frequently claim PCT reabsorbs water but not glucose — the reality is the opposite emphasis. PCT reabsorbs nearly all filtered glucose (~99%), along with ~80% of water, Na⁺, and bicarbonate via obligatory reabsorption.
The three processes of urine formation (NCERT Class 11 Biology Chapter 16, page 208):
Glomerular filtration — Blood pressure forces plasma minus proteins through the glomerular capillaries into Bowman's capsule. The filtrate (≈180 L/day) is called the primary urine. GFR ≈ 125 mL/min.
Tubular reabsorption — PCT: bulk reabsorption (glucose, amino acids, Na⁺, water — all nearly complete). Descending limb of Henle's loop: water-permeable, concentrates filtrate. Ascending limb: impermeable to water, actively pumps out NaCl (countercurrent multiplier). DCT: selective reabsorption under hormonal control.
Tubular secretion — H⁺, K⁺, urea, and certain drugs are actively secreted into the tubular fluid at PCT, DCT, and collecting duct. This maintains blood pH and eliminates waste not filtered at the glomerulus.
Concentration mechanism: The medullary gradient (300 → 1200 mOsmol/L from cortex to inner medulla) is maintained by the countercurrent multiplier (loop of Henle) and countercurrent exchanger (vasa recta). ADH increases water permeability of collecting ducts → concentrated urine. Without ADH → dilute urine (diabetes insipidus).
Watch-out for NEET: Questions often test whether a specific substance is reabsorbed or secreted, and at which segment. PCT handles the bulk; DCT/collecting duct handle the fine-tuning under ADH and aldosterone.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The approximate volume of glomerular filtrate formed per day in a healthy adult is:
Answer: B. GFR is ~125 mL/min. Per day: 125 × 60 × 24 = 180,000 mL = 180 L. This is stated in NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because 1.5 L is the approximate volume of final urine produced per day, not the filtrate volume. Confusing output with input is a common similar-term error.
Why C is wrong: C is wrong because 18 L is a factor-of-10 error — likely from miscalculating 125 mL/min × 60 min × 24 hr without proper unit conversion.
Why D is wrong: D is wrong because 1800 L would require a GFR ten times normal (1250 mL/min), which is physiologically impossible.
Which of the following is NOT reabsorbed in the proximal convoluted tubule under normal conditions?
Answer: A. Creatinine is neither reabsorbed nor significantly secreted at PCT in humans; it is used clinically as a GFR marker precisely because it passes through the nephron largely unchanged. NCERT Class 11 Biology Chapter 16.
Why B is wrong: B is wrong because amino acids ARE reabsorbed at the PCT along with glucose via active transport mechanisms.
Why C is wrong: C is wrong because about 50% of filtered urea IS passively reabsorbed in the PCT (and further in the collecting duct). While not complete, it is reabsorbed — unlike creatinine.
Why D is wrong: D is wrong because glucose IS almost completely reabsorbed (~99%) at the PCT via Na⁺-glucose cotransporters. The mistake mistake: nephron reabsorption specifically targets this confusion.
The countercurrent multiplier system in the kidney is formed by:
Answer: C. The loop of Henle's descending limb (permeable to water) and ascending limb (impermeable to water, actively pumps NaCl) create the medullary osmotic gradient — this is the countercurrent multiplier. NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because PCT and DCT are cortical structures that do not create the medullary gradient. They perform reabsorption/secretion but are not arranged as a countercurrent system.
Why B is wrong: B is wrong because afferent and efferent arterioles supply/drain the glomerulus for filtration — they are not tubular structures creating an osmotic gradient.
Why D is wrong: D is wrong because vasa recta is the countercurrent EXCHANGER (maintains the gradient), not the multiplier. The collecting duct responds to ADH but does not generate the gradient.
A patient with damage to the posterior pituitary shows greatly increased urine volume with very low osmolarity. The most likely explanation is:
Answer: B. ADH is stored and released from the posterior pituitary. Without ADH, collecting ducts remain impermeable to water → water is not reabsorbed → large volume of dilute urine (diabetes insipidus). NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because aldosterone promotes Na⁺ reabsorption (and water follows), so EXCESS aldosterone would cause water retention and reduced urine volume — the opposite of the clinical picture.
Why C is wrong: C is wrong because increased GFR would increase filtrate volume but tubular reabsorption would still concentrate the urine normally if ADH is present. The urine would not be dilute.
Why D is wrong: D is wrong because loop of Henle blockage would impair the medullary gradient but would more likely cause oliguria (reduced output) or electrolyte imbalance, not the large-volume dilute urine characteristic of ADH deficiency.
Which segment of the nephron is responsible for the obligatory reabsorption of nearly all filtered glucose?
Answer: D. PCT performs obligatory (non-hormonally regulated) bulk reabsorption including ~99% of glucose via Na⁺-glucose symporters. This is the specific confusion addressed by mistake mistake: nephron reabsorption. NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because the descending limb is permeable to water (concentrates filtrate) but does not have glucose transporters. Glucose reabsorption is complete before filtrate reaches the loop.
Why B is wrong: B is wrong because the collecting duct's role is final water reabsorption (ADH-dependent) and urea recycling. No glucose transport occurs here.
Why C is wrong: C is wrong because DCT handles facultative reabsorption (Na⁺ under aldosterone, water under ADH) — by the time filtrate reaches DCT, glucose is already fully reabsorbed.
Aldosterone acts on which part of the nephron to promote sodium reabsorption?
Answer: C. Aldosterone (from adrenal cortex) acts on DCT and collecting duct to increase Na⁺ reabsorption and K⁺ secretion, representing facultative (hormonally regulated) reabsorption. NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because PCT reabsorption is obligatory (constitutive, not hormone-dependent). Aldosterone does not regulate PCT function.
Why B is wrong: B is wrong because the ascending limb actively pumps NaCl via Na⁺-K⁺-2Cl⁻ cotransporters, but this is constitutive — not under aldosterone control.
Why D is wrong: D is wrong because Bowman's capsule is the site of filtration, not reabsorption. No hormone regulates filtration by acting on Bowman's capsule directly.
The ascending limb of the loop of Henle is described as the "diluting segment." This is because:
Answer: D. The ascending limb is impermeable to water. It actively pumps NaCl into the medullary interstitium. Since solute leaves but water cannot follow, the remaining filtrate becomes progressively hypotonic (dilute). NCERT Class 11 Biology Chapter 16.
Why A is wrong: A is wrong because no nephron segment secretes water into the lumen. Water moves by osmosis, and the ascending limb specifically prevents water movement.
Why B is wrong: B is wrong because urea reabsorption occurs mainly in the inner medullary collecting duct (not ascending limb), and removing solute would make filtrate dilute only if water remained — the key point is the ascending limb's water impermeability.
Why C is wrong: C is wrong because ADH acts on the collecting duct, not the ascending limb. The ascending limb's water impermeability is constitutive (structural), not hormonally regulated.
If GFR is 125 mL/min and 99% of the filtrate is reabsorbed, the approximate volume of urine formed per day is:
Answer: A. Step 1: Filtrate per day = 125 mL/min × 1440 min/day = 180,000 mL = 180 L. Step 2: If 99% reabsorbed, urine = 1% of 180 L = 1.8 L. This matches the normal daily urine output. NCERT Class 11 Biology Chapter 16.
Why B is wrong: B is wrong because 1.25 L is 1% of 125 L: it takes the GFR figure 125 as a daily volume in litres instead of converting 125 mL/min × 1440 min = 180 L, and then takes 1%.
Why C is wrong: C is wrong because 18 L would be 10% of the filtrate — implying only 90% reabsorption. Normal reabsorption is ~99%, not 90%. This is a decimal-place error.
Why D is wrong: D is wrong because 125 L would mean about 70% of the 180 L of filtrate becomes urine. It takes the GFR figure as a daily volume and skips the 99% reabsorption as well.
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Given
• GFR = 125 mL/min• Tubular reabsorption = 124 mL/min• Duration = 24 hours
Required
Calculate: (a) Total filtrate volume per day. (b) Net urine output per day. (c) Percentage of filtrate reabsorbed.
Concept
Urine formation involves filtration at the glomerulus followed by selective reabsorption. The difference between filtration rate and reabsorption rate gives net urine formation rate.
Formula
• Total filtrate = GFR × time• Urine output = (GFR − reabsorption rate) × time• % reabsorbed = (reabsorption rate / GFR) × 100
Substitution
• Total filtrate = 125 mL/min × 1440 min/day• Urine output = (125 − 124) mL/min × 1440 min/day• % reabsorbed = (124/125) × 100
Calculation
• Total filtrate = 180,000 mL = 180 L/day• Urine output = 1 mL/min × 1440 = 1440 mL = 1.44 L/day• % reabsorbed = 99.2%
Note: The numbers 24, 60, and 1440 are exact time-conversion integers and do not limit significant figures.
Final answer
• Daily filtrate: 180 L• Daily urine: ~1.44 L (clinically normal range 1–2 L)• Reabsorption efficiency: 99.2%
Common trap
Students confuse the reabsorption percentage. A common error is stating PCT reabsorbs water but "not glucose" — in reality, PCT reabsorbs nearly all glucose AND most water. If a question states "99% reabsorption," the urine is 1% of filtrate, not 10%.
Similar NEET-style question
"If the GFR of a person is 120 mL/min and the reabsorption rate in the renal tubules is 118.5 mL/min, calculate the amount of urine formed per hour." (Answer: 1.5 mL/min × 60 = 90 mL/hr.)
---
Three steps: glomerular filtration (GFR ~125 mL/min), tubular reabsorption (PCT — Na+/glucose/water; LH — countercurrent), tubular secretion (DCT/CD — H+, K+, NH3). Henle's loop concentrates urine via countercurrent multiplier.
-- NCERT Class 11 Biology, Chapter 16, p. 209These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
PCT reabsorbs ~99% of glucose, ~80% of water + Na+. Loop of Henle: countercurrent multiplier. DCT/CD: fine adjustment under ADH/aldosterone.
More in Human Physiology: 32 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
1 question from NEET 2026. Answers verified against NTA official keys.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →