IGCSE Biology Formula Practice
Worked Examples for Every Calculation | Cambridge IGCSE Biology (0610)
Master calculations with step-by-step solutions
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Worked examples for all biology calculations + printable PDF
Calculating Magnification
Magnification = Image Size ÷ Actual Size
How to Use This Guide
- Study the formula — understand what each variable represents
- Follow the worked example — see how to apply the formula step-by-step
- Try the practice problem — click to reveal the answer
- Check units — always convert to the same units before calculating
1. Magnification Calculations
The most commonly tested Biology calculation. You must be able to find magnification, image size, or actual size - and convert units correctly!
The Magnification Formula
Magnification = Image Size ÷ Actual Size
M = I ÷ A
Remember: Magnification has no units — it's just a number (e.g., ×400)
Finding Magnification
A cell measures 50 μm in real life. In a diagram, it appears as 10 cm. What is the magnification?
Finding Actual Size
A microscope image shows a chloroplast measuring 15 mm. The magnification is ×3000. What is the actual size?
Practice Problem
A red blood cell has an actual diameter of 7 μm. In a microscope image, it measures 28 mm. Calculate the magnification.
Click to reveal answer
Step 1: Convert 28 mm to μm: 28 × 1000 = 28,000 μm
Step 2: M = I ÷ A = 28,000 ÷ 7 = ×4000
2. Percentage Change (Osmosis)
Essential for osmosis experiments. Positive = gain (water entered), negative = loss (water left). Always divide by the INITIAL value.
The Percentage Change Formula
Percentage Change = (Final - Initial) ÷ Initial × 100
Positive result = increase (water gained by osmosis)
Negative result = decrease (water lost by osmosis)
Potato Chip in Water
A potato chip has an initial mass of 2.5 g. After being placed in distilled water for 30 minutes, its final mass is 3.0 g. Calculate the percentage change in mass.
Potato Chip in Concentrated Solution
A potato chip has an initial mass of 3.2 g. After 30 minutes in concentrated sugar solution, its final mass is 2.4 g. Calculate the percentage change.
Practice Problem
A potato cylinder has an initial length of 40 mm. After being in salt solution, its length is 34 mm. Calculate the percentage change.
Click to reveal answer
Step 1: Change = 34 - 40 = -6 mm
Step 2: -6 ÷ 40 × 100 = -15%
The potato lost water by osmosis (solution was hypertonic)
3. Rate Calculations
Used for photosynthesis (bubble counting), respiration, enzyme activity, and transpiration experiments. Rate = amount ÷ time.
The Rate Formula
Rate = Change ÷ Time
Used for: rate of photosynthesis, rate of respiration, rate of enzyme activity
Units: bubbles/min, cm³/min, mm/min, etc.
Rate of Photosynthesis
An aquatic plant produces 45 bubbles of oxygen in 3 minutes. Calculate the rate of photosynthesis.
Rate from Graph (Gradient)
A graph shows gas collected vs time. At 2 minutes, 10 cm³ was collected. At 6 minutes, 50 cm³ was collected. Calculate the rate.
Practice Problem
In an enzyme experiment, 8 cm³ of gas was produced in 40 seconds. Calculate the rate in cm³ per minute.
Click to reveal answer
Step 1: Rate = 8 ÷ 40 = 0.2 cm³/second
Step 2: Convert to per minute: 0.2 × 60 = 12 cm³/min
4. Genetic Ratios & Punnett Squares
Predict offspring ratios using Punnett squares. Key ratios: 3:1 (Bb × Bb) and 1:1 (Bb × bb test cross).
Key Genetic Ratios
Monohybrid Cross
In pea plants, tall (T) is dominant over short (t). Two heterozygous tall plants are crossed. What is the probability of a short offspring?
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
Predicting Number of Offspring
From a cross Tt × Tt, if 80 offspring are produced, how many are expected to be short?
Practice Problem
Brown eyes (B) is dominant over blue eyes (b). Two brown-eyed parents (both Bb) have children. Out of 12 children, how many would you expect to have blue eyes?
Click to reveal answer
Step 1: Bb × Bb gives 1 BB : 2 Bb : 1 bb
Step 2: Probability of blue (bb) = 1/4
Step 3: 12 × 1/4 = 3 children expected with blue eyes
5. Population Sampling
Estimate populations using capture-recapture (mobile animals) or quadrats (plants). Extended content but commonly examined.
Capture-Recapture Method (Lincoln Index)
Population = (n₁ × n₂) ÷ m
n₁ = number caught and marked in first sample
n₂ = total number caught in second sample
m = number of marked individuals in second sample (recaptured)
Estimating Woodlice Population
A scientist catches 40 woodlice, marks them, and releases them. The next day, she catches 50 woodlice, of which 8 are marked. Estimate the population.
Quadrat Sampling
Population = Mean per quadrat × Total area ÷ Quadrat area
Five 0.5 m² quadrats contain 3, 5, 4, 6, and 7 daisies. Estimate the total number of daisies in a 200 m² field.
Practice Problem
30 snails were caught, marked and released. Later, 45 snails were caught, of which 9 were marked. Estimate the population.
Click to reveal answer
Formula: Population = (n₁ × n₂) ÷ m
Calculation: (30 × 45) ÷ 9 = 1350 ÷ 9 = 150 snails
6. Surface Area to Volume Ratio
Explains why cells are small, why organisms need lungs/gills, and why small animals lose heat faster. As size increases, SA:V decreases.
SA:V Ratio for Cubes
Surface Area = 6 × side²
Volume = side³
SA:V = Surface Area ÷ Volume
Key concept: As organisms get larger, SA:V ratio decreases — this affects heat loss, gas exchange, and nutrient uptake
Comparing Cell Sizes
Calculate the SA:V ratio for a cube-shaped cell with sides of 2 μm.
Practice Problem
Calculate the SA:V ratio for a cube with 4 cm sides. Compare this to a 2 cm cube.
Click to reveal answer
4 cm cube: SA = 6 × 16 = 96 cm², V = 64 cm³, Ratio = 96 ÷ 64 = 1.5:1
2 cm cube: SA = 6 × 4 = 24 cm², V = 8 cm³, Ratio = 24 ÷ 8 = 3:1
The smaller cube has a higher SA:V ratio — this is why small organisms lose heat faster and can exchange gases more efficiently across their body surface.
7. Gas Exchange Values
A commonly tested table - know the composition of inhaled vs exhaled air. Examiners expect you to recall these values and explain the differences.
Composition of Inhaled vs Exhaled Air
| Gas | Inhaled Air | Exhaled Air | Change |
|---|---|---|---|
| Oxygen (O₂) | 21% | 16% | ↓ 5% |
| Carbon dioxide (CO₂) | 0.04% | 4% | ↑ ~4% |
| Nitrogen (N₂) | 78% | 78% | No change |
| Water vapour | Variable | Saturated | ↑ Increases |
Why these changes?
- • O₂ decreases: Used in aerobic respiration in cells
- • CO₂ increases: Produced by aerobic respiration in cells
- • Water vapour increases: Evaporates from moist alveoli surfaces
Calculating Percentage Difference
Calculate the percentage decrease in oxygen between inhaled and exhaled air.
Practice Problem
By how many times does the CO₂ concentration increase between inhaled and exhaled air?
Click to reveal answer
Calculation: 4% ÷ 0.04% = 100 times
CO₂ concentration increases 100-fold from inhaled to exhaled air
8. Heart Rate Calculations
Heart rate is measured as beats per minute (bpm). Often measured over 15 or 30 seconds and scaled up.
Heart Rate from Pulse
Heart Rate (bpm) = Number of beats ÷ Time (min)
A student counts 18 pulse beats in 15 seconds. Calculate the heart rate in beats per minute.
Practice Problem
After exercise, a student counts 36 pulse beats in 20 seconds. What is their heart rate?
Click to reveal answer
Step 1: 20 seconds = 20/60 = 1/3 minute
Step 2: 36 ÷ (1/3) = 36 × 3 = 108 bpm
Or: 36 × (60/20) = 36 × 3 = 108 bpm
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