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Biology - Transport in Plants

Grade 10IGCSE

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

🔑Concepts

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Xylem: Specialized tissue for the transport of water and mineral ions from the roots to the rest of the plant. It consists of dead, hollow cells strengthened by lignin to withstand low pressures. Movement is unidirectional (upwards).

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Phloem: Living tissue that transports sucrose and amino acids from sources (e.g., leaves) to sinks (e.g., roots, fruits). This process is known as translocation and is multidirectional.

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Root Hair Cells: Adapted for the uptake of water via osmosis and mineral ions via active transport. They possess a large surface area to increase the rate of absorption of H2OH_{2}O.

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Transpiration: The loss of water vapor from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapor through the stomata.

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Water Potential (Psi\\Psi): Water moves from a region of higher water potential (less negative) to a region of lower water potential (more negative) down a water potential gradient.

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Factors affecting Transpiration: Rate increases with higher temperature (increased kinetic energy of H2OH_{2}O molecules), higher light intensity (stomata open), and higher wind speed (removes H2OH_{2}O vapor from leaf surface). It decreases with higher humidity (reduced concentration gradient).

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Source and Sink: A 'source' is an area of the plant where substances are produced (e.g., C6H12O6C_{6}H_{12}O_{6} in leaves during photosynthesis) or released from storage. A 'sink' is an area where these substances are stored or used (e.g., roots, flowers, or developing tubers).

📐Formulae

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

Ψ=Ψs+Ψp\Psi = \Psi_{s} + \Psi_{p}

Surface Area to Volume Ratio=Surface AreaVolume\text{Surface Area to Volume Ratio} = \frac{\text{Surface Area}}{\text{Volume}}

💡Examples

Problem 1:

A student uses a potometer to measure the rate of transpiration. If the air bubble moves 60 mm60\text{ mm} in 5 minutes5\text{ minutes}, calculate the rate of transpiration in mm/min\text{mm/min}.

Solution:

Rate=60 mm5 min=12 mm/min\text{Rate} = \frac{60\text{ mm}}{5\text{ min}} = 12\text{ mm/min}

Explanation:

The rate is determined by dividing the distance the bubble moves (representing H2OH_{2}O uptake) by the time interval recorded.

Problem 2:

Explain the movement of water from the soil into the root hair cell using the concept of water potential (Ψ)(\Psi).

Solution:

Soil (High Ψ)→(\text{High } \Psi) \rightarrow Root Hair Cell (Low Ψ)(\text{Low } \Psi).

Explanation:

Root hair cells contain a concentrated cell sap (solutes like sugars and ions), which lowers their internal Ψ\Psi relative to the soil water. H2OH_{2}O enters via osmosis through the partially permeable membrane.

Problem 3:

During the early spring, a potato tuber begins to grow new shoots. Identify the source and the sink for sucrose in this scenario.

Solution:

Source: Potato tuber; Sink: Developing shoots.

Explanation:

In early spring, stored starch in the tuber is converted back into sucrose and transported to the growing shoots. Thus, the tuber acts as the source and the shoots as the sink.

Transport in Plants - Revision Notes & Key Formulas | IGCSE Grade 10 Science