📚 IB WJEC Biology: Typical Exam Questions Solved | IB WJEC 生物：典型例题详解
Mastering IB and WJEC Biology requires not only memorising content but also developing the skill to apply concepts to exam-style questions. This article walks you through typical worked examples drawn from both syllabi, covering magnification, osmosis, genetics, enzymes, photosynthesis, ecology, population genetics, experimental design, and DNA replication. Each problem is fully solved with step-by-step reasoning, making it an ideal revision companion.
掌握 IB 和 WJEC 生物不仅需要记忆知识点，更需要培养将概念应用到考试题型中的能力。本文带领你一步步解析来自两个考试体系的典型例题，涵盖放大倍数、渗透作用、遗传、酶、光合作用、生态、群体遗传学、实验设计和 DNA 复制。每道题都给出完整解答和逐步推理，是理想的复习伴侣。
1. Magnification Calculations | 放大倍数计算
A common IB question: A student observes an onion epidermal cell under a light microscope. The image of the cell measures 50 mm across. The actual diameter of the cell is 20 μm. Calculate the magnification used. Show your working.
一道常见的 IB 题目：一名学生在光学显微镜下观察洋葱表皮细胞，图像中细胞的直径为 50 mm，细胞的实际直径为 20 μm。请计算所使用的放大倍数，写出计算过程。
Solution steps: First, convert both measurements to the same unit. Recall that 1 mm = 1000 μm. Therefore, the image size 50 mm = 50 × 1000 = 50,000 μm. Then apply the standard formula.
解题步骤：首先，将所有测量值换算为相同单位。由于 1 mm = 1000 μm，图像尺寸 50 mm = 50 × 1000 = 50,000 μm。然后应用标准公式。
Magnification = Image size ÷ Actual size
放大倍数 = 图像尺寸 ÷ 实际尺寸
Magnification = 50,000 μm ÷ 20 μm = 2500. So the magnification is 2500×.
放大倍数 = 50,000 μm ÷ 20 μm = 2500。因此放大倍数为 2500 倍。
WJEC frequently asks students to calculate actual size given magnification. For instance, an image of a chloroplast measures 30 mm under a magnification of 6000×. Actual length = 30 mm ÷ 6000 = 0.005 mm = 5 μm. Always show unit conversion.
WJEC 常要求根据放大倍数计算实际尺寸。例如，叶绿体图像在 6000× 放大下测量为 30 mm，实际长度 = 30 mm ÷ 6000 = 0.005 mm = 5 μm。务必展示单位换算过程。
2. Osmosis and Potato Strips | 渗透作用与土豆条
A WJEC practical exam often provides data on potato strips placed in different sucrose concentrations. The table below shows the percentage change in mass after 24 hours. Determine the water potential of the potato tissue and explain the results.
WJEC 实验考试常给出土豆条在不同蔗糖浓度中的质量变化数据。下表为24小时后的质量变化百分比。请确定土豆组织的水势并解释结果。
| Sucrose concentration (mol/dm³) | % Change in mass |
|---|---|
| 0.0 | +12.3 |
| 0.2 | +6.8 |
| 0.4 | +1.1 |
| 0.6 | -3.9 |
| 0.8 | -9.6 |
| 1.0 | -14.2 |
The point where the percentage change is zero occurs between 0.4 and 0.6 mol/dm³ with no exact zero crossing. Interpolating, the isotonic concentration is approximately 0.45 mol/dm³. At this concentration, the water potential inside the potato cells equals that of the external solution, so there is no net movement of water.
质量变化为零的点出现在 0.4 至 0.6 mol/dm³ 之间，没有精确的零交叉。通过内插法，等渗浓度约为 0.45 mol/dm³。在该浓度下，土豆细胞内的水势与外界溶液相等，因此没有水的净移动。
When the solution is hypotonic (lower sucrose concentration), water enters the cells by osmosis, causing an increase in mass. Conversely, in hypertonic solutions, water leaves the cells, and the mass decreases. The WJEC rubric expects reference to water potential gradients and use of the terms ‘hypotonic’ and ‘hypertonic’.
当溶液为低渗（蔗糖浓度较低）时，水通过渗透进入细胞，质量增加。反之，在高渗溶液中，水离开细胞，质量减少。WJEC 评分标准要求提及水势梯度，并使用“低渗”和“高渗”术语。
3. Monohybrid Cross and Phenotypic Ratios | 单基因杂交与表型比例
In both IB and WJEC, a classic genetics question: In fruit flies, normal wings (A) are dominant over vestigial wings (a). A heterozygous fly is crossed with another heterozygous fly. Determine the expected genotypic and phenotypic ratios of the offspring.
在 IB 和 WJEC 中，一道经典的遗传学题目：在果蝇中，正常翅（A）对残翅（a）为显性。杂合子果蝇与另一杂合子果蝇杂交，请确定后代的预期基因型比例和表型比例。
Set up a Punnett square. Parental genotypes: Aa × Aa. Gametes: A or a from each parent.
绘制庞纳特方格。亲本基因型：Aa × Aa。配子：每个亲本产生 A 或 a。
| A | a | |
| A | AA | Aa |
| a | Aa | aa |
The genotypic ratio is 1 AA : 2 Aa : 1 aa. The phenotypic ratio is 3 normal wings : 1 vestigial wing, because both AA and Aa express the dominant trait.
基因型比例为 1 AA : 2 Aa : 1 aa。表型比例为 3 正常翅 : 1 残翅，因为 AA 和 Aa 都表达显性性状。
WJEC may extend this by introducing co-dominance or sex-linked inheritance, but the Punnett square method remains fundamental. Always define the symbols clearly and state the ratios in the simplest form.
WJEC 可能会引入共显性或伴性遗传，但庞纳特方格法始终是基础。务必明确定义符号，并用最简形式表述比例。
4. Enzyme Activity and Temperature | 酶活性与温度
A typical IB data-analysis question provides the rate of an enzyme-catalysed reaction at different temperatures. Catalase was used to decompose hydrogen peroxide, and the volume of oxygen produced per minute was recorded. Interpret the data and explain the shape of the graph.
一道典型的 IB 数据分析题会给出不同温度下酶促反应的速率。使用过氧化氢酶分解过氧化氢，记录每分钟产生的氧气体积。请解读数据并解释曲线形状。
| Temperature (°C) | Rate of O₂ production (cm³/min) |
|---|---|
| 10 | 2.5 |
| 20 | 5.1 |
| 30 | 6.8 |
| 40 | 7.0 |
| 50 | 3.2 |
| 60 | 0.5 |
The optimum temperature is around 40 °C, where the rate peaks at 7.0 cm³/min. Below the optimum, increasing temperature raises the kinetic energy of enzyme and substrate molecules, leading to more frequent successful collisions and faster product formation.
最适温度约为 40 °C，此时速率峰值为 7.0 cm³/min。低于最适温度时，温度升高增加酶和底物分子的动能，导致更频繁的有效碰撞和更快的产物生成。
Above 40 °C, the rate declines sharply because the enzyme denatures. The hydrogen and ionic bonds maintaining the tertiary structure break, altering the active site so that the substrate can no longer bind. Denaturation is irreversible, explaining why the rate at 60 °C is almost zero.
超过 40 °C 后，速率急剧下降，因为酶发生变性。维持三级结构的氢键和离子键断裂，改变了活性位点，使底物无法结合。变性不可逆，因此 60 °C 时速率几乎为零。
5. Limiting Factors in Photosynthesis | 光合作用中的限制因素
IB questions often ask students to identify the limiting factor from a graph of photosynthetic rate against light intensity. A typical scenario: an aquatic plant experiment measured oxygen production at various light intensities under constant CO₂ and temperature. The graph shows an initial linear increase, a gradual curve, and a plateau. Explain the pattern.
IB 试题常要求学生根据光合作用速率与光照强度的关系图识别限制因素。一个典型场景：在水生植物实验中，在恒定 CO₂ 和温度条件下，测量不同光照强度下的氧气产量。图形显示初始线性上升、逐渐弯曲并达到平台期。请解释该模式。
At low light intensity, light is the limiting factor because the rate is directly proportional to light intensity. As light intensity continues to rise, the rate still increases but less steeply, showing that another factor – such as CO₂ concentration or temperature – is beginning to limit the process. At the plateau, light is no longer limiting; the rate is now limited by CO₂ availability or by the capacity of the enzymes involved in the Calvin cycle.
在低光照强度下，光照是限制因素，因为速率与光照强度成正比。随着光照强度持续升高，速率仍然增加但斜率减小，表明另一个因素——如 CO₂ 浓度或温度——开始限制过程。在平台期，光照不再限制；速率现在受限于 CO₂ 供应或卡尔文循环中相关酶的能力。
To score full marks in WJEC, mention the concept of ‘limiting factor’ and explain how increasing the suspected limiting factor (e.g., adding more CO₂) would allow further increase in rate. Use specific terminology such as ‘light-dependent reactions’ and ‘Rubisco activity’.
要在 WJEC 中获得满分，需提到“限制因素”的概念，并解释如何通过提高疑似限制因素（如增加 CO₂）使速率进一步提高。应使用特定术语，例如“光反应”和“Rubisco 活性”。
6. Interpreting Ecological Data: Simpson’s Index | 生态数据解读：辛普森指数
Both IB and WJEC require students to calculate and interpret biodiversity indices. In a survey, a student counted the number of individuals of each species in two woodland sites. Site A: species counts: 25, 25, 25, 25. Site B: counts: 70, 10, 10, 10. Calculate Simpson’s Diversity Index (D) for each site and discuss which has higher diversity.
IB 和 WJEC 都要求学生计算和解读生物多样性指数。在一次调查中，一名学生统计了两个林地中每种物种的个体数量。样地 A：物种数量：25, 25, 25, 25。样地 B：数量：70, 10, 10, 10。分别计算两个样地的辛普森多样性指数（D），并讨论哪个具有更高的多样性。
D = 1 – Σ (n/N)²
Where n is the number of individuals of a species and N is the total number of organisms. For Site A: N = 100. Σ (n/N)² = (25/100)² + (25/100)² + (25/100)² + (25/100)² = 4 × (0.25)² = 4 × 0.0625 = 0.25. So D = 1 – 0.25 = 0.75.
其中 n 为某物种个体数，N 为总生物体数。样地 A：N = 100。Σ (n/N)² = (25/100)² + (25/100)² + (25/100)² + (25/100)² = 4 × (0.25)² = 4 × 0.0625 = 0.25。因此 D = 1 – 0.25 = 0.75。
For Site B: N = 100. Σ (n/N)² = (70/100)² + (10/100)² + (10/100)² + (10/100)² = 0.49 + 0.01 + 0.01 + 0.01 = 0.52. D = 1 – 0.52 = 0.48. Site A has a higher index, indicating greater evenness and higher biodiversity even though species richness is the same. This illustrates how dominance by one species lowers diversity.
样地 B：N = 100。Σ (n/N)² = (70/100)² + (10/100)² + (10/100)² + (10/100)² = 0.49 + 0.01 + 0.01 + 0.01 = 0.52。D = 1 – 0.52 = 0.48。样地 A 的指数更高，表明均匀度更高，因此尽管物种丰富度相同，生物多样性更高。这说明了单一物种的优势如何降低多样性。
7. Hardy-Weinberg Equilibrium | 哈代-温伯格平衡
A WJEC or IB question may test population genetics: In a population of 500 plants, 80 individuals show the recessive phenotype (white flowers). Assuming Hardy-Weinberg equilibrium, calculate the frequency of the dominant allele and the frequency of heterozygous individuals.
WJEC 或 IB 题目可能会考查群体遗传学：在500株植物的群体中，80株个体表现为隐性表型（白花）。假设哈代-温伯格平衡，计算显性等位基因的频率和杂合子个体的频率。
Let the recessive allele frequency be q and dominant allele be p, where p + q = 1. The frequency of homozygous recessive (q²) = 80/500 = 0.16. Thus, q = √0.16 = 0.4. Then p = 1 – 0.4 = 0.6. The frequency of heterozygous individuals (2pq) = 2 × 0.6 × 0.4 = 0.48. Therefore, 48% of the population are carriers.
设隐性等位基因频率为 q，显性等位基因为 p，p + q = 1。隐性纯合子的频率（q²）= 80/500 = 0.16。因此 q = √0.16 = 0.4，p = 1 – 0.4 = 0.6。杂合子频率（2pq）= 2 × 0.6 × 0.4 = 0.48。因此，群体中 48% 为携带者。
Always state the assumptions of the equilibrium: large population, random mating, no mutation, no migration, no natural selection. Failing to mention these can lose marks in extended response questions.
务必阐明哈代-温伯格平衡的假设：大群体、随机交配、无突变、无迁移、无自然选择。未提及这些假设可能会在扩展回答题目中失分。
8. Experimental Design: The Effect of pH on Enzyme Activity | 实验设计：pH 对酶活性的影响
IB Internal Assessment and WJEC practical exams frequently ask students to design an investigation. Design an experiment to determine the effect of pH on the activity of amylase, using starch as the substrate. Identify the independent, dependent and controlled variables.
IB 内部评估和 WJEC 实验考试常要求学生设计研究方案。设计一个实验以确定 pH 对淀粉酶活性的影响，使用淀粉作为底物。明确自变量、因变量和控制变量。
Independent variable: pH of the buffer solution (e.g., pH 4, 5, 6, 7, 8, 9). Dependent variable: the time taken for starch to be fully broken down, indicated by the loss of blue-black colour with iodine solution. Controlled variables: temperature (water bath at 37 °C), enzyme concentration, substrate concentration, volume of solutions.
自变量：缓冲溶液的 pH（如 pH 4, 5, 6, 7, 8, 9）。因变量：淀粉完全分解所需的时间，通过碘液检测蓝黑色消失来判断。控制变量：温度（37 °C 水浴）、酶浓度、底物浓度、溶液体积。
Procedure: Place test tubes with starch solution and buffer in a water bath. Add amylase and start timing. At regular intervals, withdraw a sample and add to iodine on a spotting tile. Record the time when the iodine remains yellow-brown. Repeat at each pH three times for reliability. Calculate mean rates (1/time). Plot a graph of rate against pH. The optimum pH is where the rate is highest. Include a risk assessment and comment on validity.
步骤：将含淀粉溶液和缓冲液的试管置于水浴中。加入淀粉酶并开始计时。每隔一定时间取出样品，滴加到白瓷板上的碘液中。记录碘液保持黄褐色的时间。各 pH 值重复三次以保证可靠性。计算平均速率（1/时间）。绘制速率-pH 图。最适 pH 为速率最高处。务必编写风险评估并评价实验有效性。
9. Semi-Conservative DNA Replication Calculations | 半保留 DNA 复制计算
A possible IB or WJEC question: A DNA molecule contains 2000 bases, of which 30% are adenine. It replicates three times. How many guanine bases must be supplied from the free nucleotide pool during the entire replication process?
一道可能的 IB 或 WJEC 题目：一个 DNA 分子含有 2000 个碱基，其中 30% 是腺嘌呤。它复制三次，整个复制过程中需要从游离核苷酸库中供应多少个鸟嘌呤碱基？
In double-stranded DNA, A pairs with T, and C pairs with G. Therefore, percentage of A = percentage of T = 30%. The remaining 40% is split equally between C and G, so G = 20% of total bases. Number of guanines in one parental DNA = 20% × 2000 = 400.
在双链 DNA 中，A 与 T 配对，C 与 G 配对。因此，A 的百分比 = T 的百分比 = 30%。剩余的 40% 平分给 C 和 G，所以 G 占总碱基的 20%。一个亲代 DNA 分子中的鸟嘌呤数量 = 20% × 2000 = 400。
After three rounds of semi-conservative replication, the total number of DNA molecules = 2³ = 8. The total number of DNA strands = 16. However, the parental strands are retained, leaving 14 newly synthesised strands. But a simpler method: total guanines in all 8 molecules = 8 × 400 = 3200. The parental molecule already contributed 400 guanines. Therefore, free guanines required = 3200 – 400 = 2800.
经过三轮半保留复制，DNA 分子总数 = 2³ = 8。总链数为 16。然而，亲代链被保留，剩下 14 条新合成链。更简单的方法是：8 个分子中的鸟嘌呤总数 = 8 × 400 = 3200。亲代分子已提供 400 个鸟嘌呤。因此，所需游离鸟嘌呤 = 3200 – 400 = 2800。
Always clarify that replication is semi-conservative, meaning each new DNA consists of one original and one new strand. Common mistakes include forgetting to subtract the parental nucleotides or miscalculating the number of generations.
务必阐明复制是半保留的，即每个新 DNA 分子包含一条原链和一条新链。常见错误包括忘记减去亲代核苷酸，或错误计算复制次数。
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