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Plant Nutrition - Leaf structure

Grade 11IGCSEBiology

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

🔑Concepts

The leaf is the primary organ for photosynthesis, adapted to maximize the absorption of light and the intake of CO2CO_2.

The Waxy Cuticle is a waterproof layer that prevents excessive water loss via evaporation while remaining transparent to allow light through.

The Upper Epidermis is a thin, transparent layer of cells that protects the inner tissues and allows light to reach the photosynthetic layers.

The Palisade Mesophyll consists of tall, columnar cells packed with chloroplasts. They are located near the top of the leaf to maximize light absorption for the reaction 6CO2+6H2OC6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2.

The Spongy Mesophyll contains loosely packed cells with large air spaces, increasing the surface area for the diffusion of CO2CO_2 into cells and O2O_2 out of cells.

Stomata are small pores, mostly on the lower epidermis, that allow gas exchange (CO2CO_2 in, O2O_2 and H2OH_2O vapor out).

Guard Cells control the opening and closing of stomata to balance gas exchange with the conservation of H2OH_2O.

The Vascular Bundle contains the Xylem, which transports H2OH_2O and dissolved mineral ions, and the Phloem, which transports manufactured food like sucrose.

📐Formulae

6CO2+6H2OchlorophylllightC6H12O6+6O26CO_2 + 6H_2O \xrightarrow[chlorophyll]{light} C_6H_{12}O_6 + 6O_2

Carbon Dioxide+WaterGlucose+Oxygen\text{Carbon Dioxide} + \text{Water} \rightarrow \text{Glucose} + \text{Oxygen}

Rate of Photosynthesis1distance2\text{Rate of Photosynthesis} \propto \frac{1}{\text{distance}^2}

💡Examples

Problem 1:

Explain why the palisade mesophyll cells are arranged vertically and packed tightly together.

Solution:

Vertical arrangement and tight packing ensure that a large number of chloroplasts are exposed to direct sunlight. This maximizes the light energy captured to convert H2OH_2O and CO2CO_2 into C6H12O6C_6H_{12}O_6.

Explanation:

The columnar shape allows light to pass through the length of the cell, hitting multiple chloroplasts, which increases the efficiency of the light-dependent stage of photosynthesis.

Problem 2:

A leaf is tested for starch after being kept in the dark for 24 hours. What is the result and why?

Solution:

The iodine test will remain yellow-brown (negative), indicating no starch is present. This is because the plant was 'destarched'.

Explanation:

Without light, the plant cannot perform photosynthesis to produce C6H12O6C_6H_{12}O_6. It converts existing starch reserves back into glucose for respiration to maintain cellular activities, depleting the starch stored in the leaf.

Problem 3:

How does the structure of the spongy mesophyll assist in the movement of CO2CO_2?

Solution:

The large internal surface area and air spaces allow CO2CO_2 to diffuse rapidly from the stomata to the palisade cells.

Explanation:

Diffusion is more efficient in gas than through liquid or cell layers. The air spaces ensure that CO2CO_2 reaches the cell membranes of the mesophyll cells quickly for the photosynthetic reaction.

Leaf structure - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology