📚 Typical OCR A-Level Biology Example Questions Explained | A-Level OCR 生物：典型例题详解
This article walks through typical exam-style questions for OCR A-Level Biology, covering key topics from microscopy to ecology. Each worked example models the step-by-step reasoning expected in high-mark answers, helping you master command words such as ‘calculate’, ‘describe’, ‘explain’ and ‘suggest’.
本文精选 OCR A-Level 生物考试中常见的典型例题，覆盖从显微镜到生态学的重要主题。每道例题都逐步展示高分答案所需的推理过程，帮助你掌握“计算”“描述”“解释”“建议”等指令词的要求。
1. Microscopy & Magnification Calculations | 显微镜与放大倍数计算
A student examines a red blood cell under a light microscope. The image diameter is 4 mm, but the actual cell diameter is 7 µm. Calculate the magnification. Show your working.
一名学生用光学显微镜观察一个红细胞。图像直径为 4 mm，但细胞实际直径为 7 µm。请计算放大倍数，并写出计算过程。
First, convert all measurements to the same unit. 4 mm = 4000 µm.
首先，把所有测量值转换为相同单位。4 mm = 4000 µm。
Apply the formula: Magnification = Image size ÷ Actual size.
使用公式：放大倍数 = 图像大小 ÷ 实际大小。
Substitute the values: Magnification = 4000 µm ÷ 7 µm ≈ 571.4. The image is magnified about 571 times.
代入数值：放大倍数 = 4000 µm ÷ 7 µm ≈ 571.4。图像被放大了约 571 倍。
When units differ, always convert to µm or mm consistently. Marks are awarded for clear working and the correct unit-less answer.
当单位不同时，务必统一换算为 µm 或 mm。清晰的解题步骤和无单位答案都能得分。
2. Biochemical Tests & Biological Molecules | 生物化学检测与生物分子
An unknown solution is tested. Iodine solution stays yellow, biuret reagent turns purple, and Benedict’s solution forms a brick-red precipitate after heating. Identify the biomolecules present.
一种未知溶液被检测。碘液保持黄色，双缩脲试剂变为紫色，本尼迪特试剂加热后产生砖红色沉淀。请判断其中含有的生物分子。
Iodine staying yellow indicates starch is absent. Starch would give a blue-black colour.
碘液保持黄色说明没有淀粉。若有淀粉会变为蓝黑色。
Biuret turning purple confirms the presence of protein. Peptide bonds cause the colour change.
双缩脲试剂变为紫色证明蛋白质存在。肽键引发颜色变化。
Benedict’s test producing a brick-red precipitate indicates a reducing sugar, such as glucose.
本尼迪特试剂产生砖红色沉淀说明存在还原糖，例如葡萄糖。
Thus, the solution contains protein and reducing sugar, but no starch.
因此，溶液中含有蛋白质和还原糖，但没有淀粉。
3. Enzyme Activity & Graph Analysis | 酶活性与图表分析
The graph shows the rate of an enzyme-controlled reaction at different temperatures. The rate rises to a peak at 40 °C and then drops sharply. Explain the shape of the graph.
图表显示了某种酶促反应在不同温度下的反应速率。速率在 40 °C 达到峰值，然后急剧下降。请解释该图线的形状。
At low temperatures, the substrate and enzyme molecules have low kinetic energy. Fewer successful collisions occur, so the rate is slow.
在低温下，底物和酶分子的动能较低。成功碰撞较少，因此反应速率较慢。
As temperature increases, kinetic energy rises. More enzyme-substrate complexes form, increasing the rate up to the optimum temperature (40 °C).
随着温度升高，动能增加。更多的酶-底物复合物形成，使反应速率上升，直至最适温度（40 °C）。
Beyond the optimum, the high temperature breaks hydrogen and ionic bonds in the enzyme’s tertiary structure. The active site changes shape and denatures, so the substrate no longer fits. The rate falls sharply.
超过最适温度后，高温破坏了酶三级结构中的氢键和离子键。活性位点形状改变并变性，底物不再匹配。因此反应速率急剧下降。
4. Membrane Transport: Osmosis Data Analysis | 膜运输：渗透数据分析
Potato chips are placed in sucrose solutions of different concentrations. Their percentage change in mass is recorded. At 0.35 mol dm⁻³, the mass change is zero. Deduce the water potential of the potato tissue.
马铃薯条被放入不同浓度的蔗糖溶液中，记录其质量变化百分比。在 0.35 mol dm⁻³ 时，质量变化为零。请推断马铃薯组织的水势。
A zero change in mass means there is no net movement of water by osmosis. The water potential of the potato tissue equals the water potential of the external solution.
质量无变化意味着没有渗透造成的净水分移动。马铃薯组织的水势等于外部溶液的水势。
Therefore, the water potential of the potato tissue is the same as that of 0.35 mol dm⁻³ sucrose solution at the same temperature.
因此，在相同温度下，马铃薯组织的水势与 0.35 mol dm⁻³ 蔗糖溶液的水势相同。
If given a calibration curve or table, you would read the water potential value in kPa from the known sucrose concentration.
若给出校准曲线或表格，可根据已知蔗糖浓度，读出以 kPa 为单位的水势值。
5. DNA Replication & Meselson-Stahl Experiment | DNA复制与Meselson-Stahl实验
Meselson and Stahl grew bacteria in ¹⁵N medium, then transferred them to ¹⁴N medium. After one round of replication, the DNA formed a single hybrid band in a centrifuge tube. Explain why this result supports semi-conservative replication.
Meselson 和 Stahl 在含 ¹⁵N 的培养基中培养细菌，然后将其转移到含 ¹⁴N 的培养基中。经过一轮复制后，DNA 在离心管中形成单条杂合带。请解释该结果为何支持半保留复制。
In semi-conservative replication, each new DNA molecule contains one original (‘heavy’) strand and one newly synthesised (‘light’) strand.
在半保留复制中，每个新 DNA 分子包含一条原有的“重链”和一条新合成的“轻链”。
If replication were conservative, after one round the original heavy DNA would stay together and a separate light DNA molecule would appear. Two bands would be seen: one heavy, one light. This was not observed.
若复制是全保留的，一轮复制后原有的重 DNA 会保持在一起，同时出现一个单独的轻 DNA 分子。会观察到两条带：一条重、一条轻。但实际并非如此。
The single hybrid band of intermediate density confirms that each daughter molecule has one ¹⁵N strand and one ¹⁴N strand, supporting the semi-conservative model.
单条中等密度的杂合带证实每个子代分子都有一条 ¹⁵N 链和一条 ¹⁴N 链，这支持了半保留复制模型。
6. Protein Synthesis: Codons & Mutations | 蛋白质合成：密码子与突变
A short section of DNA template strand reads: TAC-GCA-TTA-GGT-ATC. Using the mRNA codon table, identify the amino acid sequence. Then, a mutation substitutes the third base T with C in the second triplet. Describe the effect.
一段 DNA 模板链为：TAC-GCA-TTA-GGT-ATC。使用 mRNA 密码子表，确定氨基酸序列。随后，第二个三联体中第三个碱基 T 突变为 C。请描述该突变的影响。
Transcription produces a complementary mRNA. DNA template TAC transcribes to AUG. GCA transcribes to CGU, TTA to AAU, GGT to CCA, ATC to UAG.
转录产生互补的 mRNA。DNA 模板 TAC 转录为 AUG。GCA 转录为 CGU，TTA 为 AAU，GGT 为 CCA，ATC 为 UAG。
The mRNA codons are: AUG (Met), CGU (Arg), AAU (Asn), CCA (Pro), UAG (stop). The polypeptide is Met-Arg-Asn-Pro.
mRNA 密码子为：AUG（甲硫氨酸）、CGU（精氨酸）、AAU（天冬酰胺）、CCA（脯氨酸）、UAG（终止）。多肽链为 Met-Arg-Asn-Pro。
The mutation changes DNA triplet GCA to GCC. The mRNA codon becomes CGG instead of CGU. CGG also codes for arginine due to the degenerate nature of the genetic code. This is a silent mutation; the amino acid sequence remains unchanged.
突变使 DNA 三联体 GCA 变为 GCC。mRNA 密码子由 CGU 变为 CGG。由于遗传密码的简并性，CGG 同样编码精氨酸。这是一种沉默突变，氨基酸序列不变。
7. Monohybrid & Dihybrid Inheritance | 单基因与双基因遗传
In pea plants, tall (T) is dominant over dwarf (t), and round seed (R) is dominant over wrinkled (r). A plant heterozygous for both traits is crossed with a plant homozygous recessive for both. Predict the phenotypic ratio of the offspring.
在豌豆中，高茎（T）对矮茎（t）显性，圆粒（R）对皱粒（r）显性。一株双因子杂合的植株与一株双隐性纯合的植株杂交。请预测后代的表型比例。
Parent genotypes: TtRr × ttrr. The gametes from the heterozygous parent are TR, Tr, tR, tr. The homozygous recessive parent produces only tr gametes.
亲本基因型：TtRr × ttrr。杂合亲本产生的配子为 TR、Tr、tR、tr。双隐性亲本只产生 tr 配子。
Offspring genotypes result from combining gametes: TtRr, Ttrr, ttRr, ttrr. All offspring receive recessive alleles from the second parent.
配子结合产生的后代基因型为：TtRr、Ttrr、ttRr、ttrr。所有后代都从第二个亲本获得隐性等位基因。
Phenotypes: TtRr – tall, round; Ttrr – tall, wrinkled; ttRr – dwarf, round; ttrr – dwarf, wrinkled. Each genotype occurs with equal probability, giving a 1:1:1:1 phenotypic ratio.
表型：TtRr – 高茎圆粒；Ttrr – 高茎皱粒；ttRr – 矮茎圆粒；ttrr – 矮茎皱粒。每种基因型概率相等，表型比例为 1:1:1:1。
8. Natural Selection & Antibiotic Resistance | 自然选择与抗生素抗性
Explain how the widespread use of antibiotics has led to the evolution of resistant bacterial strains, using the principles of natural selection.
请运用自然选择原理，解释广泛使用抗生素如何导致耐药菌株的进化。
Within a bacterial population, random mutation creates genetic variation. Some bacteria possess alleles that confer resistance to a specific antibiotic.
在细菌种群中，随机突变产生遗传变异。一些细菌携带能赋予特定抗生素抗性的等位基因。
When antibiotics are applied, susceptible bacteria are killed. Resistant bacteria survive and reproduce without competition, passing on the resistance alleles to offspring (vertical gene transmission).
使用抗生素时，敏感菌被杀死。耐药菌存活并在没有竞争的情况下繁殖，将抗性等位基因传递给后代（垂直基因传递）。
Over many generations, the frequency of the resistance allele increases in the population. This is directional natural selection favouring resistant phenotypes.
经过多代后，群体中抗性等位基因的频率增加。这是有利于抗性表型的定向自然选择。
Horizontal gene transfer, such as conjugation, can also spread resistance genes rapidly between different bacterial species.
水平基因转移，例如接合作用，也可在不同细菌物种间快速传播抗性基因。
9. Energy Transfer & Ecological Pyramids | 能量传递与生态金字塔
In a field, grass absorbs 500 000 kJ m⁻² yr⁻¹ of solar energy. Grasshoppers consume grass containing 25 000 kJ m⁻² yr⁻¹. Of this, 4 000 kJ m⁻² yr⁻¹ is stored in new grasshopper biomass. Calculate the percentage efficiency of energy transfer from grass to grasshoppers.
在一块田地中，草每年吸收 500 000 kJ m⁻² 的太阳能。蝗虫食用含有 25 000 kJ m⁻² yr⁻¹ 能量的草，其中 4 000 kJ m⁻² yr⁻¹ 储存在蝗虫的新生物量中。计算从草到蝗虫的能量传递效率百分比。
Energy transfer efficiency = (energy stored in biomass of the next trophic level ÷ energy consumed from the previous trophic level) × 100.
能量传递效率 = （下一营养级生物量中储存的能量 ÷ 从上一营养级摄取的能量） × 100。
Substitute: (4 000 ÷ 25 000) × 100 = 16%. The efficiency is 16%.
代入：（4 000 ÷ 25 000）× 100 = 16%。效率为 16%。
The remaining 84% is lost mainly through respiration, excretion and uneaten parts. This inefficiency explains why food chains are rarely longer than five trophic levels.
其余 84% 的能量主要通过呼吸作用、排泄和未被取食的部分而流失。这种低效率解释了为何食物链很少超过五个营养级。
10. Immune Response: Phagocytosis & Antibodies | 免疫反应：吞噬作用与抗体
Describe the role of phagocytes in the non-specific immune response and explain how B lymphocytes contribute to specific immunity during a primary infection.
描述吞噬细胞在非特异性免疫应答中的作用，并解释在初次感染期间 B 淋巴细胞如何参与特异性免疫。
Phagocytes, such as neutrophils and macrophages, are attracted to pathogens by chemotaxis. They engulf the pathogen by phagocytosis, enclosing it in a phagosome. Lysosomes fuse with the phagosome, releasing lysozyme and other enzymes to digest the pathogen. This response is non-specific.
吞噬细胞（如中性粒细胞和巨噬细胞）通过趋化作用被吸引到病原体周围。它们通过吞噬作用包裹病原体，将其封闭在吞噬体中。溶酶体与吞噬体融合，释放溶菌酶等消化酶，将病原体分解。这种应答是非特异性的。
In specific immunity, B lymphocytes have specific receptors on their surface. During a primary infection, only the B cell with a complementary receptor to the pathogen’s antigen becomes activated (clonal selection).
在特异性免疫中，B 淋巴细胞表面有特异性受体。初次感染期间，只有带有与病原体抗原互补受体的 B 细胞会被激活（克隆选择）。
The activated B cell divides by mitosis to form clones (clonal expansion). Most differentiate into plasma cells that secrete large amounts of specific antibodies. Some become memory B cells, providing immunological memory for a faster secondary response.
活化的 B 细胞通过有丝分裂形成克隆（克隆扩增）。大部分分化为浆细胞，分泌大量特异性抗体。一部分成为记忆 B 细胞，提供免疫记忆，以便在二次应答中更快反应。
Antibodies are proteins that bind to antigens, agglutinating pathogens and marking them for destruction by phagocytes, or neutralising toxins.
抗体是一种能与抗原结合的蛋白质，可使病原体凝集并标记其被吞噬细胞破坏，或中和毒素。
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