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Transport in Plants - Transpiration

Grade 12IGCSEBiology

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

🔑Concepts

Transpiration is the loss of water vapor from the leaves of a plant, primarily through the stomata. It is a consequence of gas exchange as stomata must open to allow CO2CO_2 to enter for photosynthesis.

The Cohesion-Tension Theory explains how water moves upwards through the xylem. Water molecules are polar and form hydrogen bonds, leading to cohesion (water sticking to water). Adhesion occurs when H2OH_2O molecules stick to the lignin or cellulose in the xylem walls.

Water potential (Psi\\Psi) is a measure of the tendency of water to move from one area to another. Water always moves from a region of higher water potential (less negative) to a region of lower water potential (more negative).

The transpiration stream is maintained by the evaporation of H2OH_2O from the mesophyll cell surfaces into the air spaces, followed by diffusion out of the stomata, which lowers the hydrostatic pressure at the top of the xylem.

Environmental factors affecting the rate of transpiration include Light Intensity (increases stomatal opening), Temperature (increases kinetic energy of H2OH_2O molecules), Humidity (decreases the concentration gradient), and Wind Speed (removes the boundary layer of saturated air).

📐Formulae

Ψ=Ψs+Ψp\Psi = \Psi_s + \Psi_p

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

Relative Humidity=Actual vapor pressureSaturation vapor pressure×100%\text{Relative Humidity} = \frac{\text{Actual vapor pressure}}{\text{Saturation vapor pressure}} \times 100\%

💡Examples

Problem 1:

In a potometer experiment, the air bubble moved a distance of 4545 mm in 1515 minutes. Calculate the rate of transpiration in mm/min.

Solution:

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

Explanation:

The potometer measures the rate of water uptake. Under most conditions, water uptake is almost equal to the rate of transpiration, so the movement of the bubble over time gives an estimate of the transpiration rate.

Problem 2:

Compare the water potential of the soil (Psisoil=0.3\\Psi_{soil} = -0.3 MPa) and the root hair cell (Psiroot=0.7\\Psi_{root} = -0.7 MPa). Determine the direction of water movement.

Solution:

Water moves from the soil into the root hair cell.

Explanation:

Water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Since 0.3 MPa>0.7 MPa-0.3 \text{ MPa} > -0.7 \text{ MPa}, the H2OH_2O moves down the gradient into the plant.

Transpiration - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology