📚 Cross-Curricular Exam Practice for WJEC Year 10 Biology | WJEC 十年级生物跨学科综合题型训练
In the WJEC Year 10 Biology examination, you will frequently encounter questions that require knowledge and skills from other subjects, such as mathematics, chemistry, physics and geography. These cross-curricular questions test your ability to apply concepts in unfamiliar contexts, interpret data, perform calculations and evaluate experimental methods. This article provides a series of worked examples and practice scenarios to help you master these essential skills, covering topics from magnification and graph interpretation to genetic ratios and energy flow.
在 WJEC 十年级生物考试中，你经常会遇到需要运用其他学科（如数学、化学、物理和地理）知识与技能的题目。这些跨学科的题型考查你在新情境下应用概念、解读数据、进行计算以及评价实验方法的能力。本文通过一系列范例和训练场景，帮助你掌握这些关键技能，内容涵盖放大倍数与图表解读、遗传比例和能量流动等主题。
1. Mathematical Skills: Magnification and Unit Conversions | 数学技能：放大倍数与单位换算
Microscope images often include a scale bar, and you must calculate magnification or actual size. This requires confident unit conversion between millimetres (mm) and micrometres (µm).
显微镜图像通常带有标尺，你需要计算放大倍数或实际大小。这需要熟练地在毫米 (mm) 和微米 (µm) 之间进行单位换算。
Example: A scale bar on a micrograph measures 25 mm and represents an actual length of 5 µm. Calculate the magnification.
示例：显微照片上的标尺长度为 25 mm，代表的实际长度为 5 µm。计算放大倍数。
Step 1: Convert all measurements to the same unit. 25 mm = 25 × 1000 µm = 25,000 µm.
步骤 1：将所有测量值转换为相同单位。25 mm = 25 × 1000 µm = 25,000 µm。
Step 2: Apply the formula:
Magnification = Image size ÷ Actual size
步骤 2：使用公式：
放大倍数 = 图像大小 ÷ 实际大小
Step 3: Substitute values: Magnification = 25,000 µm ÷ 5 µm = 5,000 ×.
步骤 3：代入数值：放大倍数 = 25,000 µm ÷ 5 µm = 5,000 ×。
If a cell in the same image measures 40 mm, find its actual size. 40 mm = 40,000 µm. Actual size = 40,000 µm ÷ 5,000 = 8 µm.
如果同一图像中一个细胞长 40 mm，求其实际大小。40 mm = 40,000 µm。实际大小 = 40,000 µm ÷ 5,000 = 8 µm。
- Always convert to the same unit first; 1 mm = 1,000 µm. 首先统一单位；1 mm = 1,000 µm。
- You can rearrange the formula: Actual size = Image size ÷ Magnification. 公式可变型：实际大小 = 图像大小 ÷ 放大倍数。
2. Graph Interpretation: Enzyme Activity and Temperature | 图线解读：酶活性与温度
Enzymes are proteins that catalyse reactions, and their activity is strongly affected by temperature. You must be able to interpret tabulated data and describe the shape of the resulting graph.
酶是催化反应的蛋白质，其活性受温度影响很大。你必须能够解读表格数据并描述相应图线的形状。
A student measured the activity of an enzyme at different temperatures. The results are shown below.
一名学生测量了不同温度下某种酶的活性。结果如下表所示。
| Temperature (°C) | Enzyme activity / arbitrary units |
|---|---|
| 0 | 0 |
| 10 | 8 |
| 20 | 22 |
| 30 | 38 |
| 40 | 42 |
| 50 | 15 |
| 60 | 0 |
Describe the trend shown by the data. The activity rises gradually from 0 °C, peaks at around 40 °C (the optimum temperature), and then declines sharply as the temperature increases further, reaching zero at 60 °C.
描述数据呈现的趋势。活性从 0 °C 开始逐渐升高，在 40 °C 左右达到峰值（最适温度），随后随着温度继续升高急剧下降，在 60 °C 时降至零。
Explain why enzyme activity falls after 40 °C. High temperatures break the bonds maintaining the enzyme's tertiary structure, causing the active site to change shape. The enzyme denatures and can no longer bind to its substrate.
解释为何酶活性在 40 °C 后下降。高温破坏了维持酶三级结构的化学键，导致活性位点形状改变。酶发生变性，无法再与底物结合。
When drawing such a graph, label the axes correctly (Temperature on x-axis, Enzyme activity on y-axis) and join the points with a smooth curve, not a straight line.
绘制此类图线时，要正确标注坐标轴（x 轴为温度，y 轴为酶活性），并用平滑曲线连接各点，而不是用直线。
3. Percentage Change and Osmosis | 百分比变化与渗透作用
Investigating osmosis often involves measuring mass changes in potato cylinders placed in different sucrose solutions. Calculating percentage change allows you to compare results fairly and identify an isotonic solution.
探究渗透作用通常需要测量马铃薯条在不同蔗糖溶液中的质量变化。计算百分比变化可以让你公平地比较结果，并找出等渗溶液。
A student recorded the following data for potato cylinders of initial mass 2.0 g:
一名学生记录了初始质量为 2.0 g 的马铃薯条的以下数据：
| Sucrose concentration / mol dm⁻³ | Final mass / g |
|---|---|
| 0.0 | 2.4 |
| 0.2 | 2.2 |
| 0.4 | 2.0 |
| 0.6 | 1.8 |
| 0.8 | 1.6 |
| 1.0 | 1.4 |
Calculate the percentage change in mass for the 0.2 mol dm⁻³ solution. Use the formula:
Percentage change = ((Final mass – Initial mass) ÷ Initial mass) × 100%
计算 0.2 mol dm⁻³ 溶液中的质量变化百分比。使用公式：
百分比变化 = ((最终质量 – 初始质量) ÷ 初始质量) × 100%
Substituting: ((2.2 – 2.0) ÷ 2.0) × 100% = (0.2 ÷ 2.0) × 100% = +10%. The positive sign indicates water gained by osmosis.
代入数值：((2.2 – 2.0) ÷ 2.0) × 100% = (0.2 ÷ 2.0) × 100% = +10%。正号表示通过渗透吸水。
Which sucrose concentration is isotonic to the potato cells? 0.4 mol dm⁻³, because there is no change in mass; water molecules enter and leave at equal rates.
哪种蔗糖溶液与马铃薯细胞等渗？0.4 mol dm⁻³，因为质量没有变化；水分子进出速率相等。
Similarly, for the 1.0 mol dm⁻³ solution: ((1.4 – 2.0) ÷ 2.0) × 100% = –30%. The negative value shows a net loss of water.
同理，1.0 mol dm⁻³ 溶液：((1.4 – 2.0) ÷ 2.0) × 100% = –30%。负值表明水净流失。
4. Genetics and Probability | 遗传学与概率
Monohybrid crosses use Punnett squares to predict the genotypes and phenotypes of offspring. WJEC questions often ask you to state probabilities as fractions, percentages or ratios.
单基因杂交使用庞纳特方格来预测后代的基因型和表现型。WJEC 试题经常要求用分数、百分比或比例表示概率。
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). Two heterozygous tall plants (Tt) are crossed.
在豌豆植株中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。两株杂合高茎植株 (Tt) 杂交。
Construct a Punnett square:
构建庞纳特方格：
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
Genotypic ratio: 1 TT : 2 Tt : 1 tt. Phenotypic ratio: 3 tall : 1 short.
基因型比例：1 TT : 2 Tt : 1 tt。表现型比例：3 高茎 : 1 矮茎。
Probability that an offspring is tall = 3/4 or 75%. Probability that an offspring is homozygous recessive = 1/4 or 25%.
后代为高茎的概率 = 3/4 或 75%。后代为隐性纯合子的概率 = 1/4 或 25%。
Remember to use the same letter for an allele pair, and always write the dominant allele first (e.g., Tt, not tT).
记住一对等位基因要用同一个字母，并且总是把显性等位基因写在前面（如 Tt，而不是 tT）。
5. Sampling and Population Estimation | 采样与种群数量估算
Ecologists use the capture-mark-recapture method to estimate population size of mobile organisms. The calculation relies on a simple proportion, a skill shared with mathematics.
生态学家使用标记重捕法来估算移动生物种群的大小。该计算基于简单的比例，属于与数学共通的技能。
The Lincoln index formula is:
N = (M × C) ÷ R
林肯指数公式为：
N = (M × C) ÷ R
r captured in second sample, R = number of marked individuals recaptured in second sample.
其中：N = 种群总量估计值，M = 首次捕获并标记的数量，C = 第二次捕获的总数，R = 第二次捕获中有标记的个体数。
Example: In a pond, 20 fish are caught, marked and released. Later, 30 fish are caught, of which 5 are marked. Estimate the fish population.
示例：在一个池塘中，捕获了 20 条鱼，标记后放回。后来捕获了 30 条鱼，其中 5 条带有标记。估算鱼的总数。
N = (20 × 30) ÷ 5 = 600 ÷ 5 = 120 fish.
N = (20 × 30) ÷ 5 = 600 ÷ 5 = 120 条鱼。
Assumptions to mention: marked individuals mix randomly with the population, no births, deaths or migration, and marking does not affect survival
Published by TutorHao | Year 10 Biology Revision Series | aleveler.com
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