📚 GCSE Biology Calculation Questions Intensive Training | GCSE 生物：计算题专项训练
Calculation questions in GCSE Biology may seem small in number, but they can be the difference between a grade 6 and a grade 9. From microscope magnifications to energy transfer efficiency, mastering the numeric side of biology will boost your confidence and accuracy across all exam boards — AQA, OCR, Edexcel, and more.
GCSE 生物中的计算题虽然占比不大，却常常是区分 6 分和 9 分的关键。从显微镜放大倍数到能量传递效率，熟练掌握生物中的数字部分，能让你在所有考试局（AQA、OCR、Edexcel 等）中提升信心和准确度。
1. Magnification Calculations | 放大率计算
Magnification tells you how much larger an image appears compared with the real object. The triangle formula is the most reliable tool: Magnification = Image size ÷ Actual size. All three values must share the same unit, so unit conversion is the first step in many exam questions.
放大率告诉你图像比实际物体放大了多少倍。牢记三角形公式：放大率 = 图像尺寸 ÷ 实际尺寸。三个量必须使用相同的单位，因此很多题目第一步就是统一单位。
Magnification = Image size ÷ Actual size Image size = Magnification × Actual size Actual size = Image size ÷ Magnification
放大率 = 图像尺寸 ÷ 实际尺寸 图像尺寸 = 放大率 × 实际尺寸 实际尺寸 = 图像尺寸 ÷ 放大率
Example: A micrograph of a red blood cell is measured as 21 mm across. The actual cell diameter is 7 μm. Calculate the magnification.
例题：一张红细胞显微照片的直径为 21 mm，细胞实际直径为 7 μm。计算放大率。
Step 1: Convert both to the same unit. 21 mm = 21,000 μm. Step 2: Magnification = 21,000 μm ÷ 7 μm = 3000×. Answer: ×3000.
步骤：统一单位，21 mm = 21,000 μm。放大率 = 21,000 μm ÷ 7 μm = 3000 倍。答案：×3000。
2. Microscopy Unit Conversions | 显微镜单位换算
Units often trip up students. You must be fluent in converting between metres (m), millimetres (mm), micrometres (μm), and nanometres (nm). Remember: 1 mm = 1000 μm, 1 μm = 1000 nm, so 1 mm = 1,000,000 nm. Always convert before calculating magnification or actual size.
单位换算是很多学生的失分点。必须熟练掌握米 (m)、毫米 (mm)、微米 (μm) 和纳米 (nm) 之间的转换。记住：1 mm = 1000 μm，1 μm = 1000 nm，因此 1 mm = 1,000,000 nm。计算前务必先统一单位。
| Unit | Equivalent |
| 1 m | = 1000 mm |
| 1 mm | = 1000 μm |
| 1 μm | = 1000 nm |
When changing from a larger unit to a smaller unit, multiply; from smaller to larger, divide. For example, 2.3 mm → μm: 2.3 × 1000 = 2300 μm. 5400 nm → μm: 5400 ÷ 1000 = 5.4 μm.
大单位化小单位要乘，小单位化大单位要除。例如 2.3 mm → μm：2.3 × 1000 = 2300 μm。5400 nm → μm：5400 ÷ 1000 = 5.4 μm。
3. Rate of Enzyme-Controlled Reactions | 酶促反应速率
Rate calculations appear in enzyme practicals, photosynthesis, and respiration. The basic formula is: Rate = Change in quantity ÷ Time. The quantity can be product formed or substrate used. Units could be cm³/min, g/s, or arbitrary units/time.
速率计算出现在酶促实验、光合作用和呼吸作用中。基本公式：速率 = 变化量 ÷ 时间。变化量可以是生成产物的量或消耗底物的量。单位可能是 cm³/min、g/s 或任意单位/时间。
Example: An amylase experiment produced 12 cm³ of reducing sugar in 4 minutes. Calculate the rate of reaction.
例题：淀粉酶实验在 4 分钟内产生了 12 cm³ 的还原糖。计算反应速率。
Rate = 12 cm³ ÷ 4 min = 3 cm³/min. Always include units.
速率 = 12 cm³ ÷ 4 min = 3 cm³/min。一定要写单位。
4. Rate of Photosynthesis from Oxygen or Carbon Dioxide | 光合作用或呼吸速率（气体变化）
In pondweed experiments, you often measure the volume of oxygen bubbles produced per minute. Rate = Total volume ÷ Time. If you count bubbles, rate = number of bubbles ÷ time. Sometimes you are given gas volume measurements over 10 minutes; divide by 10 to get the rate per minute.
在藻类实验中，经常测量每分钟氧气气泡的体积。速率 = 总体积 ÷ 时间。如果计数气泡，速率 = 气泡数 ÷ 时间。有时给出的是 10 分钟内的气体体积，要除以 10 得到每分钟速率。
Example: A pondweed produced 4.5 cm³ of oxygen in 15 minutes. What is the rate per minute?
例题：一棵水草在 15 分钟内产生了 4.5 cm³ 氧气。每分钟速率是多少？
Rate = 4.5 cm³ ÷ 15 min = 0.3 cm³/min. For comparisons, you may need to convert cm³ to mm³ (1 cm³ = 1000 mm³) to make numbers easier.
速率 = 4.5 cm³ ÷ 15 min = 0.3 cm³/min。进行比较时，可能需要将 cm³ 转换为 mm³ (1 cm³ = 1000 mm³) 以便处理。
5. Estimating Population Size (Capture-Recapture) | 估算种群大小（标记重捕法）
The Lincoln Index is used to estimate animal populations. The formula is: Population estimate = (Number caught and marked in first sample × Total number caught in second sample) ÷ Number of marked individuals recaptured in second sample. It assumes no migration, deaths, births, and that marks are not lost.
估算动物种群数量使用林肯指数。公式为：种群估计值 = (第一次标记并释放的个体数 × 第二次捕获的总数) ÷ 第二次捕获中带标记的个体数。假设无迁入迁出、无出生死亡、标记不脱落。
N = (M × C) ÷ R
N 种群估计值 M 第一次标记数 C 第二次捕获总数 R 第二次捕获中含有标记的数量
Example: A student caught 40 woodlice, marked them, and released them. The next day 50 were captured, and 10 had marks. Estimate the population size.
例题：一名学生捕获 40 只鼠妇，标记后释放。第二天捕获 50 只，其中 10 只带有标记。估算种群大小。
N = (40 × 50) ÷ 10 = 2000 ÷ 10 = 200 woodlice. The assumptions need to be stated for full marks.
N = (40 × 50) ÷ 10 = 2000 ÷ 10 = 200 只鼠妇。想要拿满分还需说明假设条件。
6. Surface Area to Volume Ratio | 表面积与体积比
This ratio explains why small organisms can rely on diffusion while larger ones need transport systems. To calculate it, find the surface area and volume (often for a cube) and then divide surface area by volume. The result is often simplified to a ratio in the form X:1.
表面积与体积之比解释了为什么小生物可以靠扩散生存，而大生物需要运输系统。计算时先求表面积和体积（常见的是立方体），再用表面积除以体积。结果通常简化为 X : 1 的形式。
Example: A cube-shaped cell has a side length of 2 cm. Surface area = 6 × (2)² = 24 cm². Volume = 2³ = 8 cm³. SA:V = 24 ÷ 8 = 3:1.
例题：一个立方体形细胞，边长 2 cm。表面积 = 6 × (2)² = 24 cm²。体积 = 2³ = 8 cm³。表面积体积比 = 24 ÷ 8 = 3 : 1。
7. Energy Transfer and Efficiency | 能量传递与效率
In a food chain, only about 10% of the energy in one trophic level is transferred to the next. Efficiency is calculated as: Efficiency = (Energy available to the next level ÷ Energy available to the previous level) × 100%. You may also be given biomass or dry mass figures in kJ or kg.
食物链中，大约只有 10% 的能量能传递到下一个营养级。效率计算公式：效率 = (下一营养级可利用的能量 ÷ 上一营养级可利用的能量) × 100%。题目中可能给出能量或生物量（干重），单位是 kJ 或 kg。
Efficiency (%) = (Energy transferred ÷ Energy input) × 100
传递效率 (%) = (传递的能量 ÷ 输入的能量) × 100
Example: Grass contains 25,000 kJ of energy. A cow that eats the grass absorbs 2,500 kJ. What is the efficiency of energy transfer?
例题：草含有 25,000 kJ 能量，牛吃草后吸收了 2,500 kJ。能量传递效率是多少？
Efficiency = (2500 ÷ 25,000) × 100 = 10%. Low efficiency due to respiration, undigested material, and heat loss.
效率 = (2500 ÷ 25,000) × 100 = 10%。效率低是因为呼吸作用、未消化的物质和散热。
8. BMI Calculation | 体重指数 (BMI) 计算
Body Mass Index assesses whether a person is underweight, healthy, overweight, or obese. Formula: BMI = Body mass (kg) ÷ Height² (m²). It is a population-level indicator and does not account for muscle mass.
体重指数用于评估一个人是否偏瘦、正常、超重或肥胖。公式：BMI = 体重 (kg) ÷ 身高² (m²)。这是群体指标，不区分肌肉与脂肪。
Example: A 1.75 m tall student weighs 70 kg. Calculate her BMI.
例题：一名学生身高 1.75 m，体重 70 kg。计算她的 BMI。
Height² = 1.75 × 1.75 = 3.0625 m². BMI = 70 ÷ 3.0625 = 22.9 kg/m². This falls in the healthy range (18.5–24.9).
身高² = 1.75 × 1.75 = 3.0625 m²。BMI = 70 ÷ 3.0625 = 22.9 kg/m²，属于正常范围 (18.5–24.9)。
9. Percentage Change Calculations | 百分比变化计算
This appears in osmosis experiments (mass change of potato cylinders) and enzyme activity. Formula: Percentage change = (Final value – Initial value) ÷ Initial value × 100%. A negative result means a decrease.
该计算出现在渗透作用实验（土豆条质量变化）和酶活性实验中。公式：百分比变化 = (最终值 – 初始值) ÷ 初始值 × 100%。负值表示减少。
Example: A potato cylinder had an initial mass of 5.2 g and a final mass of 5.7 g after being placed in a sugar solution. Calculate the percentage change in mass.
例题：一个土豆条初始质量为 5.2 g，放入糖溶液后最终质量为 5.7 g。计算质量的百分比变化。
Change = 5.7 – 5.2 = 0.5 g. Percentage change = (0.5 ÷ 5.2) × 100 = 9.6% (increase).
变化量 = 5.7 – 5.2 = 0.5 g。百分比变化 = (0.5 ÷ 5.2) × 100 = 9.6%（增加）。
10. Genetic Probability and Ratios | 遗传概率与比例
Monohybrid crosses produce genotypic and phenotypic ratios. You may be asked to state the probability of an offspring having a particular trait as a fraction, percentage, or decimal. For a heterozygote cross (Aa × Aa), the probability of a recessive phenotype is 1/4 or 25%.
单性状杂交会得出一系列基因型比和表现型比。题目可能会要求用分数、百分比或小数表示某子代表现特定性状的概率。杂合子杂交 (Aa × Aa)，隐性表现型的概率是 1/4 或 25%。
Example: In peas, tall (T) is dominant to short (t). Two heterozygous tall plants are crossed. What is the probability that an offspring will be short?
例题：在豌豆中，高茎 (T) 对矮茎 (t) 为显性。两株杂合高茎豌豆杂交，子代为矮茎的概率是多少？
Punnett square: gametes T and t from each parent. Offspring genotypes: TT, Tt, Tt, tt. Short (tt) appears once out of four. Probability = 1/4 or 0.25 or 25%.
旁氏表：亲本配子 T 和 t。子代基因型：TT、Tt、Tt、tt。矮茎 (tt) 占 1/4。概率 = 1/4 或 0.25 或 25%。
11. Mean, Median, and Mode in Biological Data | 生物数据中的平均数、中位数和众数
Describing the central tendency of results is vital for practical write-ups. Mean is the arithmetic average, median is the middle value when data are ordered, and mode is the most frequent value. Biologists often remove anomalies before calculating the mean to improve accuracy.
描述数据的集中趋势对于实验报告至关重要。均值是算术平均数，中位数是数据排序后中间的值，众数是出现频率最高的值。生物学家通常会在计算均值前剔除异常值以提高准确性。
Example: Red blood cell counts (million per mm³) from 5 samples: 4.2, 4.7, 4.5, 12.1, 4.6. Identify the anomaly and calculate the mean of the valid data.
例题：5 个样本的红细胞计数（百万 / mm³）：4.2, 4.7, 4.5, 12.1, 4.6。找出异常值并计算有效数据的均值。
Anomaly is 12.1. Valid data: 4.2, 4.7, 4.5, 4.6. Sum = 18.0. Mean = 18.0 ÷ 4 = 4.5 million per mm³.
异常值为 12.1。有效数据：4.2, 4.7, 4.5, 4.6。总和 = 18.0。均值 = 18.0 ÷ 4 = 4.5 百万 / mm³。
12. Interpreting Graphs and Calculating Gradients | 图表解读与斜率计算
Exam questions often require you to draw a tangent on a curve to find the rate at a specific time. Gradient = Change in y ÷ Change in x. On a rate-of-reaction graph, a steeper gradient indicates a faster rate. Do not forget to read axes scales and quote units.
考试题常要求你在曲线上画切线，以求出某一时刻的速率。斜率 = y 变化量 ÷ x 变化量。在反应速率曲线图上，斜率越大表示速率越快。别忘了查看坐标轴刻度并标明单位。
Example: A graph shows volume of gas (cm³) against time (s). At 20 s, a tangent touches the curve. The tangent rises from 8 cm³ to 20 cm³ over a horizontal run of 10 s. Calculate the rate at 20 s.
例题：一张图表示气体体积 (cm³) 随时间 (s) 变化。在 20 s 处画切线，切线纵坐标从 8 cm³ 升至 20 cm³，横坐标跨度为 10 s。计算 20 s 时的速率。
Gradient = (20 – 8) cm³ ÷ 10 s = 12 ÷ 10 = 1.2 cm³/s. Always use a ruler and draw a large triangle to reduce error.
斜率 = (20 – 8) cm³ ÷ 10 s = 12 ÷ 10 = 1.2 cm³/s。务必用直尺并画出大三角形以减少误差。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导