📚 GCSE Edexcel Biology: Calculation Practice | GCSE Edexcel 生物：计算题专项训练
Calculation questions appear across almost every topic in GCSE Edexcel Biology – from microscopy and enzymes to ecology and respiration. A solid understanding of the key equations and the ability to convert units, handle standard form, and interpret data can make a real difference to your final grade. This article covers the essential calculations, with step-by-step examples, so you can approach any numerical problem with confidence.
在 GCSE Edexcel 生物考试中，计算题几乎遍布每一个主题——从显微镜、酶学到生态学和呼吸作用。扎实掌握关键公式，并能够熟练进行单位换算、处理科学记数法和解读数据，可以切实提升你的最终成绩。本文涵盖必会的计算题型，提供分步示例，让你自信应对任何数值问题。
1. Magnification Calculations | 放大倍数计算
e. Both measurements must be in the same unit. In exams, you often need to convert millimetres (mm) to micrometres (µm) – 1 mm = 1000 µm. Always state the magnification with a '×' sign.
放大倍数表示图像比实物大了多少倍。公式很简单：放大倍数 = 图像尺寸 ÷ 实际尺寸。两个测量值必须使用相同单位。考试中经常需要将毫米 (mm) 换算为微米 (µm) – 1 mm = 1000 µm。计算结果务必带上 '×' 符号。
Example: An image of a cell measures 50 mm across. The actual cell diameter is 5 µm. Convert 50 mm to µm: 50 × 1000 = 50,000 µ on is given and you need actual size, rearrange: Actual size = Image size ÷ Magnification.
示例：一张细胞图像的宽度为 50 mm。实际细胞直径为 5 µm。将 50 mm 转换为 µm: 50 × 1000 = 50,000 µm。放大倍数 = 50,000 ÷ 5 = ×10,000。若已知放大倍数求实际尺寸，变形公式：实际尺寸 = 图像尺寸 ÷ 放大倍数。
Remember: 1 cm = 10 mm, so a measurement of 2.4 cm becomes 24 mm, then 24,000 µm. Edexcel questions frequently test this conversion.
记住：1 cm = 10 mm，所以 2.4 cm 等于 24 mm，进而等于 24,000 µm。Edexcel 试题经常考查这一换算。
2. Unit Conversions and Standard Form | 单位换算与科学记数法
Biology calculations demand comfort with units: millimetre (mm), micrometre (µ ), gram (g), kilogram (kg), litre (L) and millilitre (mL). Common conversions: 1 mm = 1000 µm, 1 µ kg = 1000 g, 1 L = 1000 mL. Remember that 1 cm³ = 1 mL, often used in volume measurements.
生物计算要求你熟练掌握单位：毫米 (mm)、微米 (µm)、纳米 (nm)、克 (g)、千克 (kg)、升 (L) 和毫升 (mL)。常见换算关系：1 mm = 1000 µm, 1 µm = 1000 nm, 1 kg = 1000 g, 1 L = 1000 mL。记住 1 cm³ = 1 mL，体积测量中常用。
s 3.6 × 10⁻³, and 450,000 as 4.5 × 10⁵. In a magnification problem, expressing 0.005 mm in µm gives 5 µm, but sometimes you must show working in standard form, e.g., 5.0 × 10⁻³ mm.
科学记数法用于表示极大或极小的数字。例如 0.0036 可写成 3.6 × 10⁻³，450,000 写成 4.5 × 10⁵。在放大倍数问题中，将 0.005 mm 表示为 µm 等于 5 µm，但有时你需要用科学记数法展示步骤，例如 5.0 × 10⁻³ mm。
Always double-check your conversions. A classic error is leaving image size in mm and actual size in µm, giving a magnification that is a thousand times too small.
始终仔细核对单位换算。经典错误是图像尺寸以 mm 为单位而实际尺寸以 µm 为单位，导致放大倍数缩小一千倍。
3. Surface Area to Volume Ratio | 表面积与体积比
rface area by the volume. For a cube of side length L, surface area = 6L² and volume = L³, so SA:V = 6/L. As the organism or cell gets larger, the ratio decreases.
该比例对理解扩散、渗透和热量交换至关重要。表面积与体积比 (SA:V) 由表面积除以体积得出。对于边长为 L 的立方体，表面积 = 6L²，体积 = L³，因此 SA:V = 6/L。随着生物体或细胞变大，该比例减小。
s shows Cube A has a larger surface area relative to its volume, allowing faster exchange.
示例：立方体 A，边长 1 cm：SA = 6 × 1² = 6 cm², V = 1 cm³, SA:V = 6:1。立方体 B，边长 2 cm：SA = 6 × 4 = 24 cm², V = 8 cm³，比例 = 24:8 = 3:1。这说明立方体 A 相对体积的表面积更大，有利于快速交换。
You may be asked to explain why small organisms do not need a circulatory system, while larger ones do: the low SA:V in big organisms cannot supply all cells by diffusion alone.
你可能会被要求解释为什么小型生物无需循环系统而大型生物需要：大型生物的低 SA:V 意味着单靠扩散无法满足所有细胞的需求。
4. Rate of Enzyme-Controlled Reactions | 酶控反应速率
uantity of product (or substrate change) ÷ Time. On a graph, the initial rate is the steepest part, usually the first 30–60 seconds.
反应速率可以用单位时间内产物的生成量或底物的消耗量来计算。公式为：速率 = 产物量（或底物变化量）÷ 时间。在图表上，初始速率是最陡峭的部分，通常为最初的 30–60 秒。
f the unit asked for is cm³/min, multiply by 60: 0.5 × 60 = 30 cm³/min. Always show units.
例如，一个测量过氧化氢酶产生氧气量的实验：40 秒内收集到 20 cm³ 氧气。速率 = 20 ÷ 40 = 0.5 cm³/s。如果题目要求单位是 cm³/min，则乘以 60: 0.5 × 60 = 30 cm³/min。务必标出单位。
ate gradient = change in y ÷ change in x. The initial rate is found by a tangent at t = 0.
要从产物–时间曲线求速率，需在指定时间点绘制切线，然后计算梯度 = y 变化量 ÷ x 变化量。初始速率通过在 t = 0 时绘制的切线求得。
5. Percentage Change in Mass (Osmosis) | 质量百分比变化（渗透作用）
ctical with potato cylinders requires you to calculate percentage change in mass. Formula: % change = ((Final mass – r entered the tissue (hypotonic solution), a negative value means water left (hypertonic).
经典的渗透实验中使用马铃薯条，要求计算质量百分比变化。公式：% 变化 = ((最终质量 – 初始质量) ÷ 初始质量) × 100。正值表示水分进入组织（低渗溶液），负值表示水分流失（高渗溶液）。
Example: Initial mass = 5.2 g, final mass = 5.8 g. Change = +0.6 g. % change = (0.6 ÷ 5.2) × 100 ≈ +11.5%. If final mass were 4.7 g, change = –0.5 g, % change = (–0.5 ÷ 5.2) × 100 ≈ –9.6%. Plotting % hange against concentration helps identify the isotonic point where the line crosses zero.
示例：初始质量 = 5.2 g，最终质量 = 5.8 g。变化 = +0.6 g。% 变化 = (0.6 ÷ 5.2) × 100 ≈ +11.5%。若最终质量为 4.7 g，变化 = –0.5 g，% 变化 = (–0.5 ÷ 5.2) × 100 ≈ –9.6%。绘制 % 变化对浓度的曲线有助于找出等渗点，即曲线穿越零线的点。
Remember to use the same balance for all measurements and blot the potato to remove surface water before weighing. These details reduce percentage error.
记住所有测量应使用同一台天平，并且在称量前用吸水纸吸去马铃薯表面水分。这些细节可减小百分比误差。
6. Cardiac Output and Heart Rate | 心输出量与心率
asured in beats per minute (bpm), stroke volume in millilitres (mL/beat), and cardiac output is often given in mL/min or L/min.
心输出量是心脏每分钟泵出的血液量。计算公式为：心输出量 = 心率 × 每搏输出量。心率以每分钟心跳次数 (bpm) 表示，每搏输出量以毫升 (mL/beat) 表示，心输出量通常以 mL/min 或 L/min 表示。
n, or 5.25 L/min. During exercise, heart rate may rise to 150 bpm and stroke volume to 110 mL, giving output of 16.5 L/min. Edexcel questions may ask you to calculate the change or state the units.
示例：某人安静时心率为 70 bpm，每搏输出量为 75 mL。心输出量 = 70 × 75 = 5250 mL/min，即 5.25 L/min。运动时心率可能升至 150 bpm，每搏输出量升至 110 mL，此时输出量为 16.5 L/min。Edexcel 题目可能会要求计算变化量或写出单位。
The cardiac output formula also links to rate of oxygen delivery. Higher cardiac output means more oxygen transported to muscles, which is vital for aerobic respiration.
心输出量公式还与氧气输送速率相关。更高的心输出量意味着更多氧气被运输到肌肉，这对有氧呼吸至关重要。
7. Population Size Using Quadrats | 样方法估算种群大小
ber of individuals per quadrat, then multiply by the number of quadrats that fit into the total area: Estimated population = (Mean per quadrat) × (Total area ÷ Area of one quadrat).
在调查栖息地时，使用样方对生物分布进行取样。要估算种群总大小，先计算每个样方的平均个体数量，再乘以整个区域能容纳的样方数量：估算种群 = (每个样方平均数量) × (总面积 ÷ 一个样方面积)。
= 200 ÷ 0.5 = 400. Estimated population = 4.6 × 400 = 1840. Clearly state it is an estimate.
示例：一片 200 m² 的田野，样方面积 0.5 m²。10 个随机样方的计数为：3, 5, 4, 6, 3, 4, 5, 7, 4, 5。平均值 = 46 ÷ 10 = 4.6。田野中的样方数 = 200 ÷ 0.5 = 400。估算种群 = 4.6 × 400 = 1840。务必说明这是估算值。
You also need to calculate the median and mode for the sample, and explain how random sampling (using coordinates) avoids bias.
你还需要计算样本的中位数和众数，并解释随机抽样（使用坐标）如何避免偏差。
8. Biomass Transfer Efficiency | 生物量传递效率
efficiency of biomass transfer between trophic levels can be calculated with: Efficiency (% rophic level ÷ Biomass at lower trophic level) × 100. Efficiency is usually low, around 10% of the organism are eaten, and much energy is lost in respiration, movement and as waste.
在食物链中，营养级之间的生物量传递效率计算公式为：效率 (%) = (较高营养级的生物量 ÷ 较低营养级的生物量) × 100。效率通常很低，约 10%，因为生物体并非所有部分都被取食，且大量能量在呼吸、运动和排泄中丧失。
and the fox contains 200 kJ. Efficiency from grass to rabbit = (2000 ÷ 20,000) × 100 = 10%. From rabbit to fox = (200 ÷ 2000) × 100 = 10%. Edexcel often provides data in kJ or dry mass in grams.
示例：在一条食物链中，草含有 20,000 kJ 生物量，食草的兔子含有 2,000 kJ，狐狸含有 200 kJ。从草到兔子的效率 = (2000 ÷ 20,000) × 100 = 10%。从兔子到狐狸的效率 = (200 ÷ 2000) × 100 = 10%。Edexcel 常以 kJ 或干重克数提供数据。
You can explain why biomass pyramids are almost always upright: because energy is lost at each level, so the mass that can be supported decreases.
你可以解释为什么生物量金字塔几乎总是正立的：因为每个营养级都会损失能量，能够支持的生物量逐级减少。
9. Mean, Median, Mode and Range | 平均值、中位数、众数和范围
Biology investigations generate data that must be summarised mathematically. The mean (average) is sum of values ÷ number of values. The median is the middle value when data are ordered; use it if there are outliers. Mode is the most frequent value. The range (max – min) shows spread, and the interquartile range can be used to describe consistency.
生物学探究会产生需要数学总结的数据。平均值（平均数）是数值总和 ÷ 数值个数。中位数是将数据排序后处于中间的值；当存在异常值时使用。众数是出现频率最高的值。范围（最大值 – 最小值）显示数据散布情况，四分位距可用于描述一致性。
Example: Heart rates after exercise: 120, 125, 122, 118, 200 bpm. The outlier 200 inflates the mean; the median (122) is more representative. Always remove anomalous results before calculating the mean, or use the median.
示例：运动后心率：120, 125, 122, 118, 200 bpm。异常值 200 会拉高平均值；中位数 (122) 更具代表性。计算平均值前务必剔除异常结果，或使用中位数。
You must be able to calculate these values quickly for a set of biological data, and choose the most appropriate statistic to describe the results.
你需要能够快速计算一组生物学数据的这些数值，并选择最合适的统计量来描述结果。
10. Interpreting Graphs and Calculating Gradients | 解读图表与计算梯度
presents the rate of change. Gradient = Change in y ÷ Change in x. For a curve, draw a tangent at the required point and calculate its gradient.
许多 Edexcel 生物题包含显示反应速率、种群增长或气体交换的折线图。直线的梯度（斜率）代表变化率。梯度 = y 变化量 ÷ x 变化量。对于曲线，在所需点绘制切线并计算其梯度。
xygen output. At 5 minutes, oxygen = 10 cm³; at 8 minutes, oxygen = 22 cm³. Gradient = (22 – 10) ÷ (8 – /min. When a curve plateaus, the gradient approaches zero, indicating the rate has slowed.
示例：一个光合作用实验测量氧气产量。在 5 分钟时，氧气 = 10 cm³；在 8 分钟时，氧气 = 22 cm³。梯度 = (22 – 10) ÷ (8 – 5) = 12 ÷ 3 = 4 cm³/min。当曲线趋于平缓时，梯度接近零，表明速率已经减慢。
Always read the axis scales carefully and include units. A negative gradient in a line graph (e.g., mass loss during osmosis) indicates a decrease.
务必仔细读取坐标轴刻度并标出单位。折线图中的负梯度（例如渗透过程中质量减少）表示下降趋势。
Practice these calculations using past paper questions, focusing on drawing large tangents and picking widely spaced points to improve accuracy.
利用往年真题练习这些计算，重点练习画大尺寸切线并选取间隔较远的点以提高准确度。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图