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Plant Nutrition - Leaf structure

Grade 12IGCSEBiology

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

🔑Concepts

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Photosynthesis: The process by which plants synthesize carbohydrates from raw materials using energy from light. The balanced equation is 6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2.

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Waxy Cuticle: A transparent, non-cellular layer on the upper surface that reduces water loss via evaporation and acts as a barrier to pathogens.

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Upper Epidermis: A single layer of transparent cells that allows light to pass through to the mesophyll layers while protecting internal tissues.

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Palisade Mesophyll: Tightly packed, columnar cells situated just below the upper epidermis; they contain the highest concentration of chloroplasts to maximize light absorption for photosynthesis.

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Spongy Mesophyll: Loosely arranged cells with large internal air spaces that facilitate the rapid diffusion of CO2CO_2 to palisade cells and the removal of O2O_2.

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Vascular Bundle: Contains Xylem for the transport of H2OH_2O and mineral ions, and Phloem for the translocation of sucrose (C12H22O11C_{12}H_{22}O_{11}) and amino acids.

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Stomata and Guard Cells: Found mainly on the lower epidermis; guard cells control the opening and closing of stomata to regulate gas exchange (CO2CO_2 in, O2O_2 out) and transpiration of H2OH_2O vapor.

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Limiting Factors: Factors such as light intensity, CO2CO_2 concentration, and temperature that can restrict the rate of photosynthesis if they are in short supply.

📐Formulae

6CO2+6H2O→light, chlorophyllC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{light, chlorophyll}} C_6H_{12}O_6 + 6O_2

Rate∝1time taken\text{Rate} \propto \frac{1}{\text{time taken}}

I=P4πr2 (Light Intensity relationship)I = \frac{P}{4\pi r^2} \text{ (Light Intensity relationship)}

💡Examples

Problem 1:

Explain how the internal structure of a leaf is adapted for efficient gas exchange.

Solution:

The spongy mesophyll layer has large air spaces which increase the surface area to volume ratio for the diffusion of gases. CO2CO_2 enters through the stomata and diffuses through these spaces to reach the palisade cells. Conversely, O2O_2 produced during photosynthesis diffuses out of the cells and leaves the leaf via the same route.

Explanation:

Efficient gas exchange relies on short diffusion distances and high surface areas, both provided by the spongy mesophyll and the thinness of the leaf blade.

Problem 2:

A leaf is tested for starch after being partially covered with foil. What result is expected in the covered area, and what is the chemical reason?

Solution:

The covered area will remain brown/orange when tested with iodine solution, indicating no starch is present. The uncovered areas will turn blue-black.

Explanation:

Without light, the covered area cannot perform the reaction 6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2. Since no glucose (C6H12O6C_6H_{12}O_6) is produced, it cannot be converted into the storage carbohydrate starch.

Problem 3:

Calculate the volume of O2O_2 produced if a plant consumes 132132g of CO2CO_2 during photosynthesis at RTP (molar volume =24 dm3= 24 \text{ dm}^3).

Solution:

  1. Molar mass of CO2=12+(16×2)=44 g/molCO_2 = 12 + (16 \times 2) = 44 \text{ g/mol}.
  2. Moles of CO2=13244=3 molCO_2 = \frac{132}{44} = 3 \text{ mol}.
  3. From the equation 6CO2→6O26CO_2 \rightarrow 6O_2, the ratio is 1:11:1.
  4. Moles of O2=3 molO_2 = 3 \text{ mol}.
  5. Volume =3×24=72 dm3= 3 \times 24 = 72 \text{ dm}^3.

Explanation:

Stoichiometry shows that for every molecule of CO2CO_2 fixed, one molecule of O2O_2 is released.

Leaf structure - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology