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Excretory Products and Their Elimination - Function of the tubules

Grade 11CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Proximal Convoluted Tubule (PCT) is lined by simple cuboidal brush border epithelium, which increases the surface area for reabsorption. Nearly 70−80%70-80\% of electrolytes and water are reabsorbed in this segment.

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The PCT maintains the pHpH and ionic balance of body fluids by selective secretion of hydrogen ions (H+H^+), ammonia (NH3NH_3), and potassium ions (K+K^+) into the filtrate and by absorption of HCO3−HCO_3^- from it.

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Henle's Loop plays a significant role in maintaining high osmolarity of medullary interstitial fluid. The descending limb is permeable to water but almost impermeable to electrolytes, concentrating the filtrate. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively, diluting the filtrate.

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Distal Convoluted Tubule (DCT) allows conditional reabsorption of Na+Na^+ and water. It is also capable of reabsorption of HCO3−HCO_3^- and selective secretion of H+H^+, K+K^+, and NH3NH_3 to maintain pHpH and sodium-potassium balance in the blood.

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The Collecting Duct extends from the cortex of the kidney to the inner parts of the medulla. Large amounts of water can be reabsorbed here to produce concentrated urine. It also allows passage of small amounts of urea into the medullary interstitium to maintain osmolarity.

📐Formulae

NFP=GHP−(BCOP+CHP)NFP = GHP - (BCOP + CHP) where NFPNFP is Net Filtration Pressure, GHPGHP is Glomerular Hydrostatic Pressure, BCOPBCOP is Blood Colloid Osmotic Pressure, and CHPCHP is Capsular Hydrostatic Pressure.

GFR≈125 mL/min=180 L/dayGFR \approx 125 \text{ mL/min} = 180 \text{ L/day}

Concentration of Urine≈4×Concentration of Initial Filtrate\text{Concentration of Urine} \approx 4 \times \text{Concentration of Initial Filtrate}

Osmolarity Range=300 mOsmolL−1 to 1200 mOsmolL−1\text{Osmolarity Range} = 300 \text{ mOsmolL}^{-1} \text{ to } 1200 \text{ mOsmolL}^{-1}

💡Examples

Problem 1:

Calculate the Net Filtration Pressure (NFPNFP) if the Glomerular Hydrostatic Pressure is 60 mmHg60 \text{ mmHg}, the Blood Colloid Osmotic Pressure is 30 mmHg30 \text{ mmHg}, and the Capsular Hydrostatic Pressure is 20 mmHg20 \text{ mmHg}.

Solution:

Using the formula NFP=GHP−(BCOP+CHP)NFP = GHP - (BCOP + CHP), we substitute the values: 60−(30+20)10\begin{array}{r} 60 \\ - (30 + 20) \\ \hline 10 \end{array} NFP=10 mmHgNFP = 10 \text{ mmHg}

Explanation:

The Net Filtration Pressure is the total pressure that promotes filtration. It is calculated by subtracting the sum of pressures that oppose filtration (BCOP and CHP) from the pressure that promotes it (GHP).

Problem 2:

If the osmolarity of the filtrate at the beginning of the Loop of Henle is 300 mOsmolL−1300 \text{ mOsmolL}^{-1}, what is the approximate osmolarity at the bend of the loop in the inner medulla?

Solution:

1200 mOsmolL−11200 \text{ mOsmolL}^{-1}

Explanation:

As the filtrate moves down the descending limb of Henle's loop, water is reabsorbed into the medullary interstitium because the limb is permeable to water but not to electrolytes. This results in a four-fold increase in concentration from the cortex (300 mOsmolL−1300 \text{ mOsmolL}^{-1}) to the inner medulla (1200 mOsmolL−11200 \text{ mOsmolL}^{-1}).

Function of the tubules Class 11 Notes & Examples