📚 IGCSE CIE Biology: Calculation Practice | IGCSE CIE 生物：计算题专项训练
Mastering calculations is essential for scoring top marks in CIE IGCSE Biology. This training guide walks you through every major calculation topic, from magnification to energy content, with step‑by‑step examples and tips to help you avoid common errors.
掌握计算题是在 CIE IGCSE 生物考试中取得高分的关键。本专项训练带你逐一攻克从放大倍数到能量含量等所有主要计算题型，通过逐步示例和要点提示，助你避开常见陷阱。
1. Magnification and Specimen Size | 放大倍数与标本尺寸
Magnification = Image size ÷ Actual size
放大倍数 = 图像尺寸 ÷ 实际尺寸
st be in the same unit before you calculate. Common conversions: 1 cm = 10 mm, 1 mm = 1000 µm, 1 µm = 1000 nm. The triangle I = M × A (Image = Magnification × Actual) can be used to rearrange the formula.
计算前所有测量值必须使用相同单位。常见换算：1 厘米 = 10 毫米，1 毫米 = 1000 微米 (µm)，1 微米 = 1000 纳米 (nm)。可使用三角形 I = M × A（图像尺寸 = 放大倍数 × 实际尺寸）进行公式变换。
Example: th an image diameter of 24 mm at a magnification of ×600. Calculate the actual diameter in µm.
示例：一张显微照片显示细胞图像直径为 24 毫米，放大倍数为 ×600。计算实际直径，单位为微米。
Solution: Actual size = 24 mm ÷ 600 = 0.04 mm. Convert to µm: 0.04 × 1000 = 40 µm.
解答：实际尺寸 = 24 毫米 ÷ 600 = 0.04 毫米。转换为微米：0.04 × 1000 = 40 µm。
2. Percentage Change | 百分比变化
Percentage change = [(Final value – Initial value) ÷ Initial value] × 100%
百分比变化 = [(终值 – 初值) ÷ 初值] × 100%
A positive result means an increase, while a negative result indicates a decrease. This calculation appears frequently in osmosis experiments where you measure changes in mass or length of potato cylinders.
正值表示增加，负值表示减少。该计算常出现在渗透实验中，用于测量马铃薯条质量或长度的变化。
Example: A potato chip had an initial mass of 5.2 g and a final mass of 4.8 g after soaking in a sugar solution. Find the percentage change in mass.
示例：一条马铃薯片初始质量为 5.2 克，在糖溶液中浸泡后最终质量为 4.8 克。求质量百分比变化。
Solution: Change = (4.8 – 5.2) = –0.4 g. Percentage = (–0.4 ÷ 5.2) × 100% = –7.69% (decrease).
解答：变化 = (4.8 – 5.2) = –0.4 克。百分比 = (–0.4 ÷ 5.2) × 100% = –7.69%（减小）。
3. Rate of Reaction | 反应速率
Rate = Amount of product formed ÷ Time taken
反应速率 = 生成产物的量 ÷ 所用时间
Rate can also be calculated using the amount of substrate used up. Units depend on what is measured, such as cm³ of gas per second (cm³/s) or g of mass lost per minute (g/min).
也可用消耗底物的量来计算速率。单位取决于测量对象，例如每秒产生气体的立方厘米数 (cm³/s) 或每分钟损失的质量克数 (g/min)。
Example: In an enzyme experiment, 18 cm³ of oxygen was produced in 90 seconds. Calculate the rate of reaction.
示例：在一个酶实验中，90 秒内产生了 18 cm³ 氧气。计算反应速率。
Solution: Rate = 18 cm³ ÷ 90 s = 0.2 cm³/s.
解答：速率 = 18 cm³ ÷ 90 s = 0.2 cm³/s。
4. Surface Area to Volume Ratio | 表面积与体积比
Surface area of a cube = 6 × side² | Volume = side³
立方体表面积 = 6 × 边长² | 体积 = 边长³
The ratio (SA:V) decreases as an organism or cell gets larger. This concept explains why large organisms need specialized transport systems and why cells divide to remain small.
随着生物体或细胞变大，表面积与体积比 (SA:V) 会减小。这一概念解释了为何大的生物需要专门的运输系统，以及细胞为何通过分裂保持较小尺寸。
Example: e of side 2 cm. Surface area = 6 × (2²) = 24 cm²; Volume = 2³ = 8 cm³. Ratio = 24:8 → 3:1.
示例：计算边长为 2 厘米的立方体的 SA:V 比。表面积 = 6 × (2²) = 24 cm²；体积 = 2³ = 8 cm³。比值 = 24:8 → 3:1。
5. Genetic Probability | 遗传概率
obability of a genotype = Number of specific offspring ÷ Total offspring in Punnett square
基因型的概率 = 特定后代数量 ÷ 庞纳特方格中的后代总数
Use a Punnett square to determine possible combinations of alleles. Probabilities can be expressed as fractions, percentages or ratios. Always read the question carefully to see which phenotype or genotype is asked for.
使用庞纳特方格确定等位基因的可能组合。概率可以用分数、百分数或比值表示。务必仔细读题，看清题目要求的是哪种表型或基因型。
Example: Two heterozygous tall pea plants (Tt) are crossed. What is the probability of producing a dwarf offspring (tt)?
示例：两株杂合高茎豌豆 (Tt) 杂交。产生矮茎后代 (tt) 的概率是多少？
Solution: Punnett square gives TT, Tt, Tt, tt. Probability of tt = 1/4 = 25%.
解答：庞纳特方格得到 TT、Tt、Tt、tt。tt 的概率 = 1/4 = 25%。
6. Mean, Median, Mode and Range | 平均值、中位数、众数和极差
Mean = Sum of all values ÷ Number of values
平均值 = 所有数值之和 ÷ 数值个数
are ordered from smallest to largest. The mode is the most frequent value. Range = maximum – minimum. These statistics help in analysing experimental data and identifying outliers.
中位数是按从小到大排序后中间的值。众数是出现频率最高的值。极差 = 最大值 – 最小值。这些统计量有助于分析实验数据并识别异常值。
Example: 85, 80. Mean = (72+78+72+85+80) ÷ 5 = 77.4 bpm. Median = 78 bpm. Mode = 72 bpm. Range = 85 – 72 = 13 bpm.
示例：测得的心率（次/分）：72、78、72、85、80。平均值 = (72+78+72+85+80) ÷ 5 = 77.4 次/分。中位数 = 78 次/分。众数 = 72 次/分。极差 = 85 – 72 = 13 次/分。
7. Energy Content of Food | 食物能量含量
Energy transferred (J) = mass of water (g) × 4.2 (J/g°C) × temperature rise (°C)
传递的能量（焦耳）= 水的质量（克） × 4.2 (J/g°C) × 温度升高值 (°C)
d. The specific heat capacity of water is taken as 4.2 J/g°C. Remember that 1 kJ = 1000 J.
要计算每克食物的能量，用总能量除以燃烧的食物质量。水的比热容取 4.2 J/g°C。注意 1 千焦 = 1000 焦。
Example: A 0.6 g piece of bread is burned, heating 20 g of water from 23 °C to 38 °C. Calculate the energy per gram of bread.
示例：燃烧 0.6 克面包，使 20 克水温度从 23 °C 升至 38 °C。计算面包每克的能量。
Solution: Energy = 20 × 4.2 × (38 – 23) = 20 × 4.2 × 15 = 1260 J. Energy per gram = 1260 ÷ 0.6 = 2100 J/g or 2.1 kJ/g.
解答：能量 = 20 × 4.2 × (38 – 23) = 20 × 4.2 × 15 = 1260 J。每克能量 = 1260 ÷ 0.6 = 2100 J/g 或 2.1 kJ/g。
8. Population Estimation – Capture-Recapture | 种群估算 – 标志重捕法
Population size N = (M × C) ÷ R
种群数量 N = (M × C) ÷ R
individuals recaptured in the second sample. This method assumes no migration, no births/deaths and that marks do not affect survival.
M = 首次捕捉并标记的个体数。C = 第二次捕捉的总数。R = 第二次捕捉中已标记的个体数。该方法假设无迁入迁出、无出生死亡，且标记不影响存活。
Example: In a first sample, 30 snails are marked and released. Later, 40 snails are captured, of which 12 are marked. Estimate the snail population.
示例：首次样本中标记并释放 30 只蜗牛。随后捕获 40 只，其中 12 只带标记。估算蜗牛种群数量。
Solution: N = (30 × 40) ÷ 12 = 1200 ÷ 12 = 100 snails.
解答：N = (30 × 40) ÷ 12 = 1200 ÷ 12 = 100 只蜗牛。
9. Graph Skills: Gradient and Interpolation | 图表技能：梯度与插值
Gradient = Change in y ÷ Change in x
梯度 = y 的变化量 ÷ x 的变化量
Select two widely spaced points on the straight section of the graph. Draw a large triangle and read off the units carefully. Interpolation means estimating a value within the range of the plotted data, while extrapolation extends beyond.
在图上直线段选取两个间距较远的点。画一个大的三角形并仔细读取单位。插值是指在已绘制数据范围内估算数值，而外推则是延伸到数据范围之外。
Example: against time. Between 2 min (30 °C) and 6 min (70 °C) the line is straight. Gradient = (70 – 30) ÷ (6 – 2) = 40 ÷ 4 = 10 °C/min.
示例：一幅图显示温度随时间变化。在 2 分钟 (30 °C) 至 6 分钟 (70 °C) 之间直线上升。梯度 = (70 – 30) ÷ (6 – 2) = 40 ÷ 4 = 10 °C/分钟。
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