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Biology - Transpiration

Grade 10ICSE

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

🔑Concepts

Transpiration is the loss of water in the form of water vapor (H2OH_2O) from the aerial parts (leaves and stems) of a plant.

The Mechanism of Stomatal Transpiration involves the turgidity of guard cells. When guard cells absorb water and become turgid, the stomatal pore opens. This is often regulated by the active transport of Potassium ions (K+K^+) into the guard cells.

Types of Transpiration: 1. Stomatal (approx. 8090%80-90\%), 2. Cuticular (loss through the waxy cuticlecuticle), and 3. Lenticular (loss through small openings in woody stems called lenticelslenticels).

Factors affecting the rate include Light (increases rate), Temperature (increases rate), Humidity (decreases rate as the concentration gradient of H2OH_2O molecules decreases), and Wind Velocity.

The Transpiration Pull is a suction force created by the evaporation of water from leaf cells, which facilitates the upward movement of water and minerals (Ascent of Sap) from the roots.

Adaptations to reduce transpiration include sunken stomata, thick cuticles, and leaves modified into spines (as seen in OpuntiaOpuntia or CactiCacti).

📐Formulae

Rate of Transpiration=Distance moved by air bubble in Potometer (cm)Time taken (min)\text{Rate of Transpiration} = \frac{\text{Distance moved by air bubble in Potometer (cm)}}{\text{Time taken (min)}}

6CO2+12H2OLight/ChlorophyllC6H12O6+6H2O+6O26CO_2 + 12H_2O \xrightarrow{\text{Light/Chlorophyll}} C_6H_{12}O_6 + 6H_2O + 6O_2 \uparrow

Transpiration Efficiency=Net PhotosynthesisTranspiration Rate\text{Transpiration Efficiency} = \frac{\text{Net Photosynthesis}}{\text{Transpiration Rate}}

💡Examples

Problem 1:

In a Ganong's Potometer experiment, the air bubble moves a distance of 12 cm12 \text{ cm} in 30 minutes30 \text{ minutes}. Calculate the rate of transpiration in cm/hr\text{cm/hr}.

Solution:

Rate=12 cm30 min×60 min/hr=24 cm/hr\text{Rate} = \frac{12 \text{ cm}}{30 \text{ min}} \times 60 \text{ min/hr} = 24 \text{ cm/hr}

Explanation:

The rate of transpiration is measured by the speed of the air bubble, which indicates the volume of H2OH_2O absorbed by the plant to replace the water lost through transpiration.

Problem 2:

Why does a piece of dry Cobalt Chloride paper change color from Blue to Pink when placed on the underside of a leaf?

Solution:

CoCl2+6H2OCoCl26H2OCoCl_2 + 6H_2O \rightarrow CoCl_2 \cdot 6H_2O

Explanation:

Anhydrous Cobalt Chloride is blue. When it comes into contact with H2OH_2O vapor transpired through the stomata, it becomes hydrated (CoCl26H2OCoCl_2 \cdot 6H_2O), turning pink. This confirms that water vapor is being released from the leaf surface.

Transpiration - Revision Notes & Key Formulas | ICSE Class 10 Science