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Transport in Plants - Water uptake and transpiration

Grade 11IGCSEBiology

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

🔑Concepts

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Water uptake occurs primarily through root hair cells, which provide a large surface area to increase the rate of osmosis of H2OH_2O and active transport of mineral ions.

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Water moves from the soil into the root hair cells down a water potential gradient; soil has a higher Ψ\Psi (water potential) compared to the cell cytoplasm.

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The pathway of water follows: Root hair cells →\rightarrow Root cortex →\rightarrow Xylem →\rightarrow Mesophyll cells →\rightarrow Air spaces →\rightarrow Stomata.

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Xylem vessels are composed of dead cells joined end-to-end with no end walls, forming a continuous tube. Their walls are strengthened with lignin to withstand the negative pressure of the transpiration pull.

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Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata.

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The 'Transpiration Stream' is maintained by cohesion (attraction between H2OH_2O molecules) and adhesion (attraction between H2OH_2O and xylem walls), creating a continuous column of water.

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Environmental factors affecting the rate of transpiration include: Temperature (increases kinetic energy of H2OH_2O molecules), Humidity (decreases the concentration gradient), Wind speed (removes water vapour from the leaf surface), and Light intensity (causes stomata to open for CO2CO_2 diffusion).

📐Formulae

Rate of Transpiration=Distance moved by air bubble (mm)Time (min)\text{Rate of Transpiration} = \frac{\text{Distance moved by air bubble (mm)}}{\text{Time (min)}}

Rate=Volume of water lost (cm3)Time (min)\text{Rate} = \frac{\text{Volume of water lost (cm}^3\text{)}}{\text{Time (min)}}

Ψsoil>Ψroots>Ψleaves>Ψatmosphere\Psi_{soil} > \Psi_{roots} > \Psi_{leaves} > \Psi_{atmosphere}

💡Examples

Problem 1:

In a potometer experiment, the air bubble moves a distance of 45 mm45 \text{ mm} in 15 minutes15 \text{ minutes}. Calculate the rate of transpiration and explain how an increase in humidity would change this value.

Solution:

Rate=45 mm15 min=3 mm/min\text{Rate} = \frac{45 \text{ mm}}{15 \text{ min}} = 3 \text{ mm/min}.

Explanation:

If humidity increases, the concentration of water vapour molecules in the air outside the leaf increases. This reduces the concentration gradient between the air spaces inside the leaf and the external atmosphere, leading to a slower rate of diffusion of H2OH_2O through the stomata.

Problem 2:

Explain the role of H2OH_2O cohesion in the xylem during a hot day.

Solution:

Water molecules are polar and form hydrogen bonds with each other, a property known as cohesion. This allows a continuous, unbroken column of H2OH_2O to be pulled up the xylem from the roots to the leaves.

Explanation:

As H2OH_2O evaporates from the mesophyll cells, it creates a tension (negative pressure) that pulls the entire column upwards. Without cohesion, the column would break under the high tension caused by rapid transpiration on a hot day.

Water uptake and transpiration - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology