| Lesson Plan |
| Grade: A-Level |
Date: 25/02/2026 |
| Subject: Biology |
| Lesson Topic: illustrate the principle that surface area to volume ratios decrease with increasing size by calculating surface areas and volumes of simple 3-D shapes (as shown in the Mathematical requirements) |
Learning Objective/s:
- Describe how the surface‑area‑to‑volume (SA:V) ratio changes with size for spheres, cubes and cylinders.
- Calculate surface area and volume for given dimensions and derive the SA:V ratio.
- Explain the biological implications of a decreasing SA:V ratio for cells.
- Compare SA:V values across shapes and sizes using numerical data.
- Apply the concept to predict cellular adaptations that maintain efficient exchange.
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Materials Needed:
- Projector and screen
- Whiteboard and markers
- Calculator or spreadsheet software
- Handouts with shape tables and formulas
- Rulers / compasses for drawing 3‑D shapes
- Exit‑ticket cards
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Introduction:
Begin with a quick visual of three spheres of different sizes and ask students what they notice about the relative amount of surface compared to volume. Recall that cells rely on their surface area for exchange, so understanding SA:V is key to cell biology. Today we will calculate SA and V for simple shapes, compare ratios, and link the maths to real cellular adaptations.
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Lesson Structure:
- Do‑now (5 minutes) – Students sketch a small and a large sphere and estimate which has more surface relative to volume.
- Mini‑lecture (10 minutes) – Derive SA and V formulas for sphere, cube, and cylinder; highlight the inverse relationship with size.
- Guided practice (15 minutes) – In pairs, work through the provided table to calculate SA, V and SA:V for each shape at three sizes.
- Concept check (5 minutes) – Quick clicker quiz: identify the shape with the highest SA:V at a given dimension.
- Biological relevance discussion (10 minutes) – Connect the ratios to cell size limits and examples such as microvilli, flattening, and multinucleation.
- Exit ticket (5 minutes) – Write one way a cell could increase its effective surface area to compensate for a low SA:V ratio.
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Conclusion:
Summarise that SA:V decreases as linear dimensions increase, limiting diffusion in larger cells. Highlight how organisms overcome this by flattening cells or adding structures like microvilli and alveoli. For homework, students will locate a real‑world example where SA:V is critical and prepare a short explanation.
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