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Life Processes - EX5.5 EX5.5 EX5.5 EX5.5 EX CRETIONCRETIONCRETION

Grade 10CBSE

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

🔑Concepts

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Excretion is the biological process involved in the removal of harmful metabolic wastes (primarily nitrogenous wastes like urea and uric acid) from the body.

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The human excretory system consists of a pair of kidneys, a pair of ureters, a urinary bladder, and a urethra.

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The Nephron is the structural and functional unit of the kidney. Each kidney contains approximately 11 to 1.21.2 million nephrons.

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A nephron consists of a cup-shaped Bowman's capsule (containing the Glomerulus) and a long renal tubule.

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Urine formation involves three main steps: 1. Glomerular Filtration (ultrafiltration of blood), 2. Selective Reabsorption (glucose, amino acids, salts, and water are taken back into the blood), and 3. Tubular Secretion.

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The amount of water reabsorbed depends on how much excess water there is in the body and how much dissolved waste there is to be excreted.

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Hemodialysis (Artificial Kidney) is a device used to remove nitrogenous waste products from the blood through dialysis in case of kidney failure.

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Excretion in plants: Plants remove excess water via transpiration, oxygen is released as a byproduct of photosynthesis, and other wastes are stored in cellular vacuoles, old xylem (resins/gums), or leaves that fall off.

📐Formulae

Net Filtration Pressure (NFP)=Glomerular Hydrostatic Pressure−(Colloid Osmotic Pressure+Capsular Hydrostatic Pressure)\text{Net Filtration Pressure (NFP)} = \text{Glomerular Hydrostatic Pressure} - (\text{Colloid Osmotic Pressure} + \text{Capsular Hydrostatic Pressure})

Urine Excreted=Glomerular Filtration−Tubular Reabsorption+Tubular Secretion\text{Urine Excreted} = \text{Glomerular Filtration} - \text{Tubular Reabsorption} + \text{Tubular Secretion}

6CO2+12H2O→Chlorophyll/SunlightC6H12O6+6H2O+6O2↑6CO_2 + 12H_2O \xrightarrow{\text{Chlorophyll/Sunlight}} C_6H_{12}O_6 + 6H_2O + 6O_2 \uparrow

💡Examples

Problem 1:

If the initial filtrate produced by both kidneys in a healthy adult is 180 L180\text{ L} per day, but the volume of urine actually excreted is only 1.2 L1.2\text{ L} per day, calculate the volume of water reabsorbed by the renal tubules.

Solution:

180.0 L−1.2 L178.8 L\begin{array}{r} 180.0 \text{ L} \\ -1.2 \text{ L} \\ \hline 178.8 \text{ L} \end{array}

Explanation:

The volume of water reabsorbed is calculated by subtracting the final urine volume from the total initial filtrate. This shows that more than 99%99\% of the filtrate is reabsorbed into the bloodstream.

Problem 2:

A patient's kidney produces a Glomerular Filtration Rate (GFR) of 125 mL/min125\text{ mL/min}. Calculate the total filtrate produced in 11 hour.

Solution:

Total Filtrate=125 mL/min×60 min=7500 mL=7.5 L\text{Total Filtrate} = 125\text{ mL/min} \times 60\text{ min} = 7500\text{ mL} = 7.5\text{ L}

Explanation:

To find the hourly filtrate, we multiply the rate per minute by the number of minutes in an hour (6060). This demonstrates the high efficiency of the nephrons.