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Plant Nutrition - Photosynthesis

Grade 12IGCSEBiology

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

🔑Concepts

Photosynthesis is the fundamental process by which plants manufacture carbohydrates from raw materials (CO2CO_2 and H2OH_2O) using energy from light absorbed by chlorophyll.

Chlorophyll is a green pigment found in chloroplasts that absorbs light energy and converts it into chemical energy for the synthesis of C6H12O6C_6H_{12}O_6 (glucose).

The internal structure of the leaf is highly adapted: the palisade mesophyll contains a high density of chloroplasts for maximum light absorption, while the spongy mesophyll allows for gas exchange via air spaces.

Stomata are small pores, primarily on the underside of the leaf, regulated by guard cells to allow CO2CO_2 to diffuse in and O2O_2 to diffuse out.

The fate of glucose: It can be used for respiration, converted into starch for storage, turned into cellulose for cell walls, or combined with nitrates to form amino acids for proteins.

Limiting factors are environmental conditions that restrict the rate of photosynthesis. The three primary limiting factors are light intensity, CO2CO_2 concentration, and temperature.

Temperature affects the rate because photosynthesis is an enzyme-controlled reaction; however, temperatures above 40C40^\circ C may denature the enzymes, causing the rate to drop.

Magnesium ions (Mg2+Mg^{2+}) are required for making chlorophyll, and Nitrate ions (NO3NO_3^-) are required for making amino acids.

📐Formulae

6CO2+6H2Olight and chlorophyllC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{light and chlorophyll}} C_6H_{12}O_6 + 6O_2

Rate of Photosynthesis1time taken\text{Rate of Photosynthesis} \propto \frac{1}{\text{time taken}}

I=1d2I = \frac{1}{d^2}

💡Examples

Problem 1:

In an experiment investigating the rate of photosynthesis using an aquatic plant (e.g., Elodea), a student moves the light source from a distance of 10 cm10\text{ cm} to 20 cm20\text{ cm} from the plant. Calculate the change in light intensity II.

Solution:

The light intensity decreases by a factor of 44.

Explanation:

According to the Inverse Square Law, I1d2I \propto \frac{1}{d^2}. When the distance dd is doubled (from 10 cm10\text{ cm} to 20 cm20\text{ cm}), the light intensity becomes 122=14\frac{1}{2^2} = \frac{1}{4} of the original intensity.

Problem 2:

A plant is kept in the dark for 48 hours before an experiment. What is this process called, and why is it necessary for a starch test?

Solution:

This process is called 'de-starching'.

Explanation:

De-starching ensures that any starch identified in the leaves at the end of the experiment was produced during the experiment, rather than being stored previously. The plant uses up its stored starch (C6H10O5)nC_6H_{10}O_5)_n for respiration while in the dark.

Problem 3:

Explain the shape of a graph where the rate of photosynthesis is plotted against CO2CO_2 concentration, noting that the graph levels off at high concentrations.

Solution:

The graph shows a linear increase initially, followed by a plateau.

Explanation:

Initially, CO2CO_2 is the limiting factor; as its concentration increases, the rate of reaction increases. When the graph levels off, CO2CO_2 is no longer the limiting factor; instead, light intensity or temperature has become the limiting factor restricting the rate.

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