📚 A-Level AQA Biology: Calculation Practice Workbook | A-Level AQA 生物：计算题专项训练
Calculations form a core part of the AQA A-level Biology assessments, appearing in both Paper 1 and Paper 2, as well as in the practical endorsement. This workbook systematically covers every type of calculation you could face, from magnification and dilution series to chi-squared tests and water potential. Each section provides a clear method, a worked example, and a short set of practice questions with answers, helping you build confidence and accuracy under timed conditions.
计算题是 AQA A-Level 生物考试的核心组成部分，出现在试卷一、试卷二以及实验考核中。这本专项训练系统地涵盖你可能遇到的所有计算类型，从显微放大、稀释系列到卡方检验和水势计算。每一部分都给出了清晰的方法、范例以及一组带答案的练习题，帮助你在限时条件下建立信心并提升准确度。
1. Magnification, Actual Size and Image Size | 放大倍数、实际尺寸与图像尺寸
The relationship between magnification, image size and actual specimen size is fundamental to microscopy work. The formula triangle is the safest way to rearrange: Image size = Actual size × Magnification. Always convert all lengths to the same unit before calculating – usually micrometres (μm) or millimetres (mm). Remember that 1 mm = 1000 μm, and when using a scale bar, first measure the bar on the image with a ruler.
放大倍数、图像尺寸和实际标本尺寸之间的关系是显微镜工作的基础。使用公式三角形是最安全的转换方式：图像尺寸 = 实际尺寸 × 放大倍数。计算前务必将所有长度单位统一——通常使用微米 (μm) 或毫米 (mm)。记住 1 mm = 1000 μm，使用比例尺时，应先用直尺测量图像上的比例尺长度。
A student measures an image length of 45 mm on a photograph taken at ×2400 magnification. Calculate the actual size of the specimen in μm.
一名学生在一张放大 2400 倍的显微照片上测得图像长度为 45 mm。计算标本的实际尺寸，以 μm 表示。
Actual size = Image size / Magnification = 45 mm / 2400 = 0.01875 mm. Convert to μm: 0.01875 × 1000 = 18.75 μm.
实际尺寸 = 图像尺寸 / 放大倍数 = 45 mm / 2400 = 0.01875 mm。转换为 μm：0.01875 × 1000 = 18.75 μm。
Practice quick-fire: An image of a mitochondrion measures 36 mm when magnified ×8000. Give the actual length in μm. (Answer: 4.5 μm)
快速练习：一个线粒体图像在放大 8000 倍时长 36 mm。以 μm 给出实际长度。（答案：4.5 μm）
2. Percentage Change and Ratios | 百分比变化与比率
Percentage change is used to compare before-and-after measurements, e.g. in osmosis experiments or enzyme rate comparisons. The formula is: (Final value – Start value) / Start value × 100. A negative result indicates a decrease. Ratios are often used to compare surface area to volume or to simplify genetic outcomes. Express ratios in their simplest whole‑number form by dividing both sides by the same factor.
百分比变化用于比较前后测量值，例如在渗透实验或酶速率比较中。公式为：(最终值 – 初始值) / 初始值 × 100。负值表示减少。比率常用于比较表面积与体积，或简化遗传结果。将两边除以相同的因子，用最简整数比表示。
In a beetroot practical, the percentage transmission of light through a solution changed from 82% to 43% after heating. Calculate the percentage change in transmission.
在甜菜根实验中，加热后溶液透光率从 82% 变为 43%。计算透光率的百分比变化。
Percentage change = (43 – 82) / 82 × 100 = −47.6%. The negative sign shows a decrease in transmission, corresponding to more pigment released.
百分比变化 = (43 – 82) / 82 × 100 = −47.6%。负号表示透光率下降，对应释放了更多色素。
A cube with side 3 mm has surface area 6 × (3 × 3) = 54 mm² and volume 27 mm³. Its surface area : volume ratio is 54 : 27, which simplifies to 2 : 1.
一个边长为 3 mm 的立方体，表面积 6 × (3 × 3) = 54 mm²，体积 27 mm³。其表面积与体积比为 54 : 27，简化为 2 : 1。
3. Mean, Median, Mode and Standard Deviation | 平均值、中位数、众数和标准差
The mean is the sum of all values divided by the number of readings. It is the most common measure of central tendency, but can be distorted by outliers. The median is the middle value when data are ordered and is less affected by anomalous results. Standard deviation (s) quantifies the spread of data around the mean. A higher standard deviation indicates greater variability. For AQA, you may be given the formula or asked to interpret calculated values.
平均值是所有数值之和除以读数个数。它是最常用的集中量数，但可能受异常值影响。中位数是排序后位于中间的值，受异常结果影响较小。标准差 (s) 量化数据围绕平均值的分散程度。标准差越大，变异程度越高。AQA 考试中可能会给出公式，或要求解释计算值。
Five replicates for the time taken for catalase to break down hydrogen peroxide are (in seconds): 34, 29, 31, 35, 30. Calculate the mean time.
过氧化氢酶分解过氧化氢所需时间的五次重复数据为（秒）：34、29、31、35、30。计算平均时间。
Mean = (34 + 29 + 31 + 35 + 30) ÷ 5 = 159 ÷ 5 = 31.8 s.
平均值 = (34 + 29 + 31 + 35 + 30) ÷ 5 = 159 ÷ 5 = 31.8 秒。
If standard deviation is small relative to the mean, the data points cluster tightly, indicating high precision. In statistical tests, comparing means often involves standard deviation to see if differences are significant.
如果标准差相对于平均值较小，则数据点紧密聚集，表明精密度高。在统计检验中，比较平均值常结合标准差来判断差异是否显著。
4. Chi‑squared Test (χ²) | 卡方检验
The chi‑squared test is used with categorical data to determine whether the difference between observed and expected frequencies is due to chance or is statistically significant. The formula is: χ² = Σ[(O − E)² / E], where O = observed value, E = expected value. You must then compare your calculated χ² value to a critical value at 5% probability (p = 0.05) for the appropriate degrees of freedom (number of categories − 1, often). If χ² is greater than the critical value, the null hypothesis is rejected, meaning the difference is significant.
卡方检验用于分类数据，以判断观察频数与期望频数的差异是随机产生的还是具有统计显著性。公式为：χ² = Σ[(O − E)² / E]，其中 O = 观察值，E = 期望值。然后需将计算出的 χ² 值与在 5% 概率水平 (p = 0.05) 下、对应自由度（通常为类别数 − 1）的临界值比较。若 χ² 大于临界值，则拒绝零假设，说明差异显著。
A genetics experiment gave the following phenotype counts: round yellow 310, wrinkled yellow 90, round green 95, wrinkled green 28. The expected ratio is 9:3:3:1. Total = 523. Expected values: 9/16 × 523 = 294.2; 3/16 × 523 = 98.1; 3/16 × 523 = 98.1; 1/16 × 523 = 32.7. Calculate χ².
一项遗传学实验得到以下表型计数：圆黄 310、皱黄 90、圆绿 95、皱绿 28。期望比例为 9:3:3:1。总和 = 523。期望值：9/16 × 523 = 294.2；3/16 × 523 = 98.1；3/16 × 523 = 98.1；1/16 × 523 = 32.7。计算 χ²。
χ² = (310−294.2)²/294.2 + (90−98.1)²/98.1 + (95−98.1)²/98.1 + (28−32.7)²/32.7
= 0.85 + 0.67 + 0.10 + 0.68 = 2.30. With 3 degrees of freedom, the critical value at p=0.05 is 7.81. Since 2.30 < 7.81, the null hypothesis is accepted; the differences are due to chance.
χ² = (310−294.2)²/294.2 + (90−98.1)²/98.1 + (95−98.1)²/98.1 + (28−32.7)²/32.7
= 0.85 + 0.67 + 0.10 + 0.68 = 2.30。自由度为 3 时，p=0.05 临界值为 7.81。由于 2.30 < 7.81，接受零假设；差异由偶然导致。
5. Bacterial Growth and Exponential Calculations | 细菌生长与指数计算
Bacteria reproduce by binary fission, leading to exponential growth under ideal conditions. The number of bacteria after a given time can be found using N = N₀ × 2ⁿ, where N₀ is the initial number, and n is the number of generations. The number of generations is total time divided by the generation time (both in the same unit). This calculation is common in aseptic technique and growth curve questions.
细菌通过二分裂繁殖，在理想条件下呈指数增长。指定时间后的细菌数量可用公式 N = N₀ × 2ⁿ 计算，其中 N₀ 为初始数量，n 为世代数。世代数 = 总时间 ÷ 世代时间（单位一致）。该计算在无菌技术及生长曲线题目中常出现。
A single bacterium (N₀ = 1) has a generation time of 20 minutes. How many bacteria will there be after 3 hours? (Log₂ values may be provided if necessary.)
一个单一细菌（N₀ = 1）的世代时间为 20 分钟。3 小时后有多少细菌？（必要时会提供 log₂ 值。）
Total time = 3 × 60 = 180 minutes. Number of generations n = 180 / 20 = 9. N = 1 × 2⁹ = 512 bacteria.
总时间 = 3 × 60 = 180 分钟。世代数 n = 180 / 20 = 9。N = 1 × 2⁹ = 512 个细菌。
If starting with 5000 bacteria and the population size after 2 hours is 160,000, how many generations occurred? 5000 × 2ⁿ = 160,000 → 2ⁿ = 32 → n = 5 (since 2⁵ = 32). Generation time = 120 min / 5 = 24 minutes.
若起始为 5000 个细菌，2 小时后达到 160,000 个，发生了几代？5000 × 2ⁿ = 160,000 → 2ⁿ = 32 → n = 5（因为 2⁵ = 32）。世代时间 = 120 分钟 / 5 = 24 分钟。
6. Dilution Series and Concentration Calculations | 稀释系列与浓度计算
Serial dilutions are used to reduce a concentrated stock solution stepwise, often by mixing 1 part stock with 9 parts diluent to give a 10⁻¹ dilution, then repeating. The dilution factor is the ratio of the final volume to the stock volume transferred. To find the original concentration, multiply the diluted concentration by the total dilution factor. For colourimeter standard curves, the concentration of an unknown is read from the graph using its absorbance.
连续稀释用于逐步降低浓储液的浓度，常按 1 份储液加 9 份稀释液的方式得到 10⁻¹ 稀释液，再重复。稀释因子是终体积与转移储液体积之比。要得到原始浓度，将稀释后的浓度乘以总稀释因子。在比色计标准曲线中，利用未知样品的吸光度从图上读取其浓度。
1 cm³ of bacterial culture is added to 9 cm³ of sterile water (10⁻¹), then 1 cm³ of that is added to another 9 cm³ to create 10⁻². How much is the original culture diluted after five such steps? What if 0.1 cm³ is plated from the final tube and 25 colonies grow – calculate the original CFU per cm³.
将 1 cm³ 细菌培养液加入 9 cm³ 无菌水中（10⁻¹），然后取 1 cm³ 该稀释液加入另一支 9 cm³ 水中制成 10⁻²。经过五次这样的步骤，原培养液被稀释了多少？若从最终试管取 0.1 cm³ 涂板后长出 25 个菌落，计算原始 CFU/cm³。
After five steps: dilution = 10⁻⁵. Plated volume = 0.1 cm³, so the number of colonies from 1 cm³ of that dilution would be 25 × 10 = 250. Original concentration = 250 × 10⁵ = 2.5 × 10⁷ CFU/cm³.
五个步骤后：稀释度 = 10⁻⁵。涂板体积为 0.1 cm³，故该稀释度下每 1 cm³ 的菌落数应为 25 × 10 = 250。原始浓度 = 250 × 10⁵ = 2.5 × 10⁷ CFU/cm³。
7. Birth Rates, Death Rates and Population Growth | 出生率、死亡率与种群增长
Population growth in ecosystems can be calculated using the equation: Population growth rate = (Births + Immigration) − (Deaths + Emigration) over a specific period. Birth rate and death rate are often expressed per 1000 individuals per year. You may need to calculate the percentage growth rate or the change in population size. Remember that a population may show exponential growth when resources are unlimited, but logistic growth when carrying capacity is reached.
生态系统中的种群增长可用公式计算：种群增长率 = (出生数 + 迁入数) − (死亡数 + 迁出数)，应用于特定时间段。出生率和死亡率常以每年每千人的数字表示。你可能需要计算百分比增长率或种群大小的变化量。记住：资源无限时种群可呈指数增长，但达到环境容纳量后则呈逻辑斯谛增长。
In a population of 5000 ducks, 200 chicks are hatched and 50 ducks die in one year. There is no migration. What is the percentage growth rate?
在一个 5000 只鸭子的种群中，一年内孵化出 200 只雏鸭，死亡 50 只。无迁徙。求出百分比增长率。
Change in population = 200 − 50 = 150. Percentage growth = (150 / 5000) × 100 = 3.0%.
种群变化量 = 200 − 50 = 150。增长百分比 = (150 / 5000) × 100 = 3.0%。
8. Surface Area to Volume Ratio | 表面积与体积比
As an organism or cell increases in size, its surface area to volume ratio (SA:V) decreases. This concept is central to understanding heat exchange, nutrient uptake, and waste removal. Calculations require you to find the surface area and volume of simple shapes – cubes, spheres, cylinders – using given formulae (the formula sheet provides these for spheres and cylinders). The ratio is expressed as a number to 1, e.g., 3:1.
随着生物体或细胞体积增大，其表面积与体积比 (SA:V) 下降。这一概念对理解热交换、营养吸收和废物排出至关重要。计算时需根据给定的公式求出简单几何形状（立方体、球体、圆柱体）的表面积和体积（公式表会提供球体和圆柱体的公式）。比值表示为与 1 的比，如 3:1。
A small mammal is modelled as a sphere of radius 2 cm. Surface area of sphere = 4πr², volume = (4/3)πr³. Calculate SA:V ratio.
将一只小型哺乳动物视为半径 2 cm 的球体。球表面积 = 4πr²，体积 = (4/3)πr³。计算 SA:V 比。
Surface area = 4 × π × 4 = 16π ≈ 50.3 cm². Volume = (4/3) × π × 8 = (32/3)π ≈ 33.5 cm³. SA:V = 50.3 / 33.5 ≈ 1.5, so roughly 1.5:1.
表面积 = 4 × π × 4 = 16π ≈ 50.3 cm²。体积 = (4/3) × π × 8 = (32/3)π ≈ 33.5 cm³。SA:V = 50.3 / 33.5 ≈ 1.5，即约为 1.5:1。
9. Rate of Reaction Calculations | 反应速率计算
For enzyme kinetics or photosynthesis experiments, rate is usually calculated as 1 / time taken to reach a defined endpoint (e.g., time for colour to disappear in the starch–amylase practical), or as change in product/substrate concentration per unit time. When a colorimeter is used, initial rate is often determined from the initial linear portion of the absorbance–time graph. Remember to state units, e.g., s⁻¹, or g s⁻¹.
在酶动力学或光合作用实验中，速率通常计算为 1/到达特定终点所需的时间（例如淀粉-淀粉酶实验中颜色消失所需的时间），或单位时间内产物/底物浓度的变化。使用比色计时，初始速率常由吸光度-时间图的初始线性部分确定。记住要写明单位，如 s⁻¹ 或 g s⁻¹。
In an amylase experiment, the time for iodine to stop turning blue-black was 45 seconds at 30°C. Express the rate of reaction in arbitrary units.
在淀粉酶实验中，30°C 时碘液不再变为蓝黑色的时间为 45 秒。以任意单位表示反应速率。
Rate = 1 / time = 1/45 = 0.022 s⁻¹ (often just given as 0.022 arbitrary units). Shorter time equals faster rate.
速率 = 1 / 时间 = 1/45 = 0.022 s⁻¹（通常直接记为 0.022 任意单位）。时间越短，速率越快。
If a photosynthesis experiment produces 2.4 cm³ of oxygen over 3 minutes, the rate is 2.4 / 3 = 0.8 cm³ min⁻¹.
若光合作用实验在 3 分钟内产生 2.4 cm³ 氧气，则速率为 2.4 / 3 = 0.8 cm³ min⁻¹。
10. Water Potential (Ψ = Ψₛ + Ψₚ) | 水势计算
Water potential (Ψ) determines the direction of water movement; water moves from a region of higher (less negative) Ψ to lower (more negative) Ψ. The equation Ψ = Ψₛ + Ψₚ combines solute potential (Ψₛ, always zero or negative) and pressure potential (Ψₚ, usually positive inside plant cells). In a fully turgid cell, Ψ = 0. For a solution in an open container, Ψₚ = 0, so Ψ = Ψₛ. The solute potential of a solution can be calculated using Ψₛ = −iCRT, though at A-level you will usually be given tabulated Ψₛ values for sucrose solutions.
水势 (Ψ) 决定水分移动的方向；水从水势较高（负值较小）的区域移向水势较低（负值较大）的区域。公式 Ψ = Ψₛ + Ψₚ 将溶质势 (Ψₛ，总是零或负值) 与压力势 (Ψₚ，植物细胞内通常为正值) 结合起来。在完全硬胀的细胞中，Ψ = 0。对于开放容器中的溶液，Ψₚ = 0，因此 Ψ = Ψₛ。溶液的溶质势可用 Ψₛ = −iCRT 计算，但在 A-level 考试中，通常会给出蔗糖溶液的溶质势表格。
A plant cell with Ψₛ = −1.8 MPa and Ψₚ = +0.5 MPa is placed in a solution of Ψ = −1.1 MPa. Describe the net movement of water.
一个 Ψₛ = −1.8 MPa、Ψₚ = +0.5 MPa 的植物细胞被放入 Ψ = −1.1 MPa 的溶液中。描述水的净移动方向。
Cell water potential Ψ = Ψₛ + Ψₚ = −1.8 + 0.5 = −1.3 MPa. The external solution has Ψ = −1.1 MPa, which is higher (less negative). Water will move into the cell from the solution.
细胞水势 Ψ = Ψₛ + Ψₚ = −1.8 + 0.5 = −1.3 MPa。外部溶液 Ψ = −1.1 MPa，其水势较高（负值较小）。水将从溶液进入细胞。
11. Genetic Ratios and Probability | 遗传比与概率
Monohybrid and dihybrid crosses often require you to predict phenotypic ratios. The expected ratios (3:1, 1:1, 9:3:3:1) are based on probability rules. You may be asked to calculate the probability of an offspring inheriting a particular genotype and phenotype, or to combine probabilities using the AND (multiply) and OR (add) rules. When pedigree charts are given, use the probability that a specific parent is a carrier as part of the calculation.
单基因和双基因杂交常要求预测表型比。期望比（3:1、1:1、9:3:3:1）基于概率法则。你可能需要计算后代继承特定基因型和表型的概率，或使用“和”（乘）与“或”（加）规则组合概率。当给出系谱图时，需将某一亲本是携带者的概率纳入计算。
Two carriers of the recessive cystic fibrosis allele (Ff) have a child. What is the probability the child will have the disease?
两名隐性囊性纤维化等位基因携带者 (Ff) 生一个孩子。孩子患病的概率是多少？
Punnett square gives genotypes: 1 FF : 2 Ff : 1 ff. Only ff shows the disease, so probability = 1/4 = 0.25 or 25%.
庞纳特方格得出基因型：1 FF : 2 Ff : 1 ff。只有 ff 会患病，因此概率 = 1/4 = 0.25 或 25%。
If a woman whose brother has haemophilia (X‑linked) marries a normal man, the probability their first son will have haemophilia depends on her carrier probability. She has a 1/2 chance of being a carrier (from her mother). If she is a carrier, there is a 1/2 chance of passing the affected X to a son. Combined probability: 1/2 × 1/2 = 1/4.
若一名女性的兄弟患有血友病（X 连锁），她与正常男性结婚，他们第一个儿子患血友病的概率取决于她作为携带者的概率。她有 1/2 的概率是携带者（来自母亲）。如果是携带者，她将致病 X 传给儿子的概率为 1/2。组合概率：1/2 × 1/2 = 1/4。
12. Uncertainty, Error and Percentage Error | 不确定度、误差与百分比误差
Every measurement has an uncertainty. For a single reading on a scale, the absolute uncertainty is usually half the smallest scale division. For a digital instrument, it is ± the last significant digit. Percentage uncertainty helps compare the quality of measurements across different scales. It is calculated as (absolute uncertainty / measured value) × 100. When combining measurements (e.g., calculating a difference or a rate), uncertainties must be added.
每一个测量值都有不确定度。对于标尺上的单次读数，绝对不确定度通常为最小刻度值的一半。对于数字仪器，则为最后一位有效数字的 ±1。百分比不确定度有助于比较不同尺度测量的质量。计算公式为 (绝对不确定度 / 测量值) × 100。当合并测量值（如计算差值或速率）时，不确定度需相加。
A thermometer with 0.5°C divisions reads a temperature of 24.0°C. What is the absolute and percentage uncertainty?
一支分度为 0.5°C 的温度计读数为 24.0°C。其绝对和百分比不确定度各为多少？
Absolute uncertainty = ± half the scale division = 0.25°C. Percentage uncertainty = (0.25 / 24.0) × 100 = 1.04%. This percentage is often rounded appropriately.
绝对不确定度 = ± 最小分度的一半 = 0.25°C。百分比不确定度 = (0.25 / 24.0) × 100 = 1.04%。该百分比常作适当修约。
In a practical, the change in mass is found by subtracting initial mass (50.2 g ±0.05 g) from final mass (53.8 g ±0.05 g). The absolute uncertainty in the mass change is 0.05 + 0.05 = 0.10 g. If the mass change is 3.6 g, then percentage uncertainty = (0.10 / 3.6) × 100 ≈ 2.8%.
在一次实验中，质量变化由最终质量 (53.8 g ±0.05 g) 减去初始质量 (50.2 g ±0.05 g) 得到。质量变化的绝对不确定度为 0.05 + 0.05 = 0.10 g。若质量变化为 3.6 g，则百分比不确定度 = (0.10 / 3.6) × 100 ≈ 2.8%。
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