📚 Making Calculations from Data | 基于数据计算
In A-Level Biology, the ability to perform accurate calculations from experimental and observational data is essential. Whether measuring microscopic objects, preparing solutions, analysing population growth, or evaluating statistical significance, students must apply mathematical skills to biological contexts. This article covers the most common calculation types encountered in the Cambridge International A-Level Biology syllabus, providing clear explanations, worked examples, and bilingual guidance to strengthen both understanding and exam technique.
在 A-Level 生物学中，从实验和观察数据中进行精确计算的能力至关重要。无论测量显微物体、配制溶液、分析种群增长还是评估统计显著性，学生都需要将数学技能应用于生物学情境。本文涵盖剑桥国际 A-Level 生物学大纲中最常见的计算类型，提供清晰的解释、例题和双语指导，以强化理解和应试技巧。
1. Calculating Magnification and Actual Size | 计算放大率和实际尺寸
Microscopy calculations often require students to determine magnification, image size, or actual size of a specimen. The fundamental relationship is:
显微镜计算常要求学生确定放大率、图像尺寸或样本的实际大小。基本关系为：
Magnification = Image size ÷ Actual size
放大率 = 图像尺寸 ÷ 实际尺寸
a cell measures 50 mm on a micrograph and its actual diameter is 0.05 mm, the magnification is 50 ÷ 0.05 = ×1000. When the actual size is unknown, rearranging the formula to Actual size = Image size ÷ Magnification allows calculation from a scale bar or known magnification. A chloroplast with an image size of 24 mm at ×4000 magnification has an actual size of 24 mm ÷ 4000 = 0.006 mm = 6 μm.
关键在于使用同一单位，通常将测量值转换为微米（μm）或毫米（mm）。例如，如果显微照片中细胞的图像长度为 50 mm，实际直径为 0.05 mm，则放大率为 50 ÷ 0.05 = ×1000。当实际尺寸未知时，重新整理公式为 实际尺寸 = 图像尺寸 ÷ 放大率，即可通过比例尺或已知放大率进行计算。一个叶绿体在 ×4000 下的图像尺寸为 24 mm，其实际尺寸为 24 mm ÷ 4000 = 0.006 mm = 6 μm。
2. Molarity and Solution Preparation | 摩尔浓度和溶液配制
Preparing solutions of a given molarity is a basic laboratory skill. Molarity (mol dm−3) is calculated as moles of solute divided by volume of solution in dm3. The key formula is:
配制特定摩尔浓度的溶液是基础实验技能。摩尔浓度（mol dm−3）的计算为溶质摩尔数除以溶液的体积（dm3）。关键公式为：
n (mol) = c (mol dm−3) × V (dm3)
物质的量（mol）= 浓度（mol dm−3）× 体积（dm3）
solution from a solid, the mass required is mass (g) = moles × molar mass. If 250 cm3 of 0.2 mol dm−3 cose is needed, the moles = 0.2 × 0.25 = 0.05 mol; with glucose molar mass 180 g mol−1, mass = 0.05 × 180 = 9.0 g. Remember: 1 dm3 = 1000 cm3. Students must also be comfortable using the concentration triangle to rearrange terms.
当用固体配制溶液时，所需质量为 质量（g）= 物质的量 × 摩尔质量。若需配制 250 cm3 的 0.2 mol dm−3 葡萄糖溶液，物质的量 = 0.2 × 0.25 = 0.05 mol；葡萄糖摩尔质量为 180 g mol−1，则 质量 = 0.05 × 180 = 9.0 g。请记住：1 dm3 = 1000 cm3。学生还应熟练使用浓度三角形进行公式变形。
3. Serial Dilution and Dilution Factors | 连续稀释与稀释因子
In many investigations, serial dilutions produce a range of concentrations. A dilution factor is the ratio of the final volume to the initial volume of stock solution. For example, a 1 in 10 dilution means 1 part stock + 9 parts diluent. The key relationship is:
许多研究中，连续稀释可以产生一系列浓度。稀释因子是最终体积与初始储备液体积之比。例如，1 : 10 稀释表示 1 份储备液加 9 份稀释液。关键关系为：
C&sub1;V&sub1; = C&sub2;V&sub2;
初始浓度 × 初始体积 = 最终浓度 × 最终体积
If 2 cm3 of 0.5 mol dm−3 solution is diluted to 10 cm3, final concentration C&sub2; = (0.5 × 2) / 10 = 0.1 mol dm−3. Serial dilutions multiply factors: three successive 10-fold dilutions give an overall dilution of 10−3. When designing dilution series, label each tube clearly and calculate the exact concentration after each step.
若将 2 cm3 0.5 mol dm−3 溶液稀释至 10 cm3，最终浓度 C&sub2; = (0.5 × 2) / 10 = 0.1 mol dm−3。连续稀释将因子相乘：三次 10 倍稀释，总稀释因子为 10−3。在设计稀释系列时，清晰标注各管并计算每一步后的准确浓度。
4. pH and Hydrogen Ion Concentration | pH 与氢离子浓度
s a logarithmic measure of hydrogen ion concentration: pH = −log&sub1;&sub0;[H+
Published by TutorHao | A-Level 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