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Human Physiology - Excretory Products and Their Elimination (Kidney Function, Dialysis)

Grade 11ICSEBiology

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

🔑Concepts

Excretion is the process of removing nitrogenous waste products like Ammonia (NH3NH_3), Urea (NH2CONH2NH_2CONH_2), and Uric acid (C5H4N4O3C_5H_4N_4O_3) from the body.

The functional unit of the kidney is the Nephron, consisting of the Renal Corpuscle (Bowman's capsule + Glomerulus) and the Renal Tubule (PCT, Loop of Henle, DCT).

Urine formation involves three main steps: Ultrafiltration, Selective Reabsorption, and Tubular Secretion.

Ultrafiltration occurs in the Malpighian corpuscle due to the pressure gradient between the afferent and efferent arterioles, resulting in a Glomerular Filtration Rate (GFRGFR) of approximately 125 mL/min125 \text{ mL/min}.

The Counter-current Mechanism in the Loop of Henle and Vasa Recta helps in concentrating urine by maintaining an osmolarity gradient from 300 mOsmol/L300 \text{ mOsmol/L} in the cortex to 1200 mOsmol/L1200 \text{ mOsmol/L} in the inner medulla.

Regulation of kidney function is mediated by Antidiuretic Hormone (ADHADH), the Renin-Angiotensin-Aldosterone System (RAASRAAS), and Atrial Natriuretic Factor (ANFANF).

Haemodialysis is an artificial process to remove urea and excess salts from the blood when kidneys fail, using a dialyzing fluid that is isotonic to plasma but lacks nitrogenous wastes.

📐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 Litres/dayGFR = 125 \text{ mL/min} = 180 \text{ Litres/day}

Excretion=FiltrationReabsorption+Secretion\text{Excretion} = \text{Filtration} - \text{Reabsorption} + \text{Secretion}

CO(NH2)2 (Urea structure formula)CO(NH_2)_2 \text{ (Urea structure formula)}

💡Examples

Problem 1:

If the Glomerular Hydrostatic Pressure (GHPGHP) is 60 mmHg60 \text{ mmHg}, the Blood Colloid Osmotic Pressure (BCOPBCOP) is 32 mmHg32 \text{ mmHg}, and the Capsular Hydrostatic Pressure (CHPCHP) is 18 mmHg18 \text{ mmHg}, calculate the Net Filtration Pressure (NFPNFP).

Solution:

NFP=60(32+18)=10 mmHgNFP = 60 - (32 + 18) = 10 \text{ mmHg}

Explanation:

The Net Filtration Pressure is the total pressure that promotes filtration. It is calculated by subtracting the opposing pressures (osmotic and capsular) from the hydrostatic pressure of the blood in the glomerulus.

Problem 2:

Explain the role of the dialyzing fluid in a dialysis machine regarding the concentration of Urea.

Solution:

The dialyzing fluid contains the same concentration of glucose and salts as normal plasma but contains 0 mg/dL0 \text{ mg/dL} of Urea.

Explanation:

By keeping the Urea concentration at zero in the dialyzing fluid, a steep concentration gradient is maintained, allowing Urea to diffuse out of the patient's blood through the semi-permeable cellophane membrane into the fluid via passive transport.

Excretory Products and Their Elimination (Kidney Function, Dialysis) Revision - Class 11 Biology…