Typical IB & CIE Biology Questions: Detailed Solutions | IB & CIE 生物典型例题详解

📚 Typical IB & CIE Biology Questions: Detailed Solutions | IB & CIE 生物典型例题详解

Mastering IB and CIE Biology exams requires not only solid knowledge but also the ability to apply concepts to unfamiliar scenarios. This article walks you through eight typical exam‑style questions covering microscopy, enzyme kinetics, genetics, photosynthesis, ecology, DNA replication, neurobiology and immunology. Each question is followed by a step‑by‑step solution with clear explanations in both English and Chinese, helping you build exam technique and confidence.

掌握 IB 与 CIE 生物考试，不仅需要扎实的知识储备，更需将概念运用于陌生情境的能力。本文精选八道典型考题，涵盖显微镜、酶动力学、遗传学、光合作用、生态学、DNA 复制、神经生物学和免疫学，每道题均配有逐步详解（中英双语），帮助你建立答题技巧与信心。

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1. Microscope Calculation: Determining Actual Size | 显微镜计算：确定实际大小

Question: A student observes a cell under a light microscope using a 40× objective lens and a 10× eyepiece lens. She captures an image and measures the length of the cell on the print as 25 mm. Calculate the actual length of the cell in micrometres (µm).

问题： 某学生使用 40 倍物镜与 10 倍目镜的显微镜观察细胞，拍摄图像后在打印照片上测得细胞长度为 25 毫米。请计算细胞的实际长度，以微米（µm）为单位。

Step 1: Determine the total magnification. Total magnification = eyepiece magnification × objective magnification = 10 × 40 = 400×.

步骤 1：确定总放大倍数。 总放大倍数 = 目镜倍数 × 物镜倍数 = 10 × 40 = 400×。

Step 2: Convert all measurements to the same unit. Since actual cell dimensions are usually given in micrometres, convert the image size from mm to µm: 25 mm = 25 × 1000 = 25 000 µm (1 mm = 1000 µm).

步骤 2：统一单位。 细胞实际尺寸通常以微米表示，因此将图像大小由毫米转换为微米：25 mm = 25 × 1000 = 25 000 µm（1 mm = 1000 µm）。

Step 3: Calculate actual size. Actual size = Image size ÷ Magnification = 25 000 µm ÷ 400 = 62.5 µm.

步骤 3：计算实际大小。 实际大小 = 图像大小 ÷ 放大倍数 = 25 000 µm ÷ 400 = 62.5 µm。

Answer: The actual cell length is 62.5 µm.

答案： 细胞实际长度为 62.5 微米。

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2. Enzyme Inhibition: Interpreting Lineweaver‑Burk Plots | 酶抑制：解读 Lineweaver‑Burk 图

Question: A researcher measures enzyme activity with and without an inhibitor and plots the data as a Lineweaver‑Burk plot (1/V against 1/[S]). In the absence of inhibitor, the y‑intercept is 0.02 s µM⁻¹ and the x‑intercept is −0.05 µM⁻¹. In the presence of inhibitor, the y‑intercept remains 0.02 s µM⁻¹ but the x‑intercept shifts to −0.025 µM⁻¹. Identify the type of inhibitor and calculate Vmax and Km in both conditions.

问题： 研究人员在无抑制剂和有抑制剂条件下测定酶活性，绘制 Lineweaver‑Burk 图（1/V 对 1/[S]）。无抑制剂时，y 截距为 0.02 s µM⁻¹，x 截距为 −0.05 µM⁻¹；有抑制剂时，y 截距保持 0.02 s µM⁻¹，但 x 截距变为 −0.025 µM⁻¹。请判断抑制剂类型，并计算两种条件下的 Vmax 和 Km。

Step 1: Recall the Lineweaver‑Burk equation. The plot gives a straight line: 1/V = (Km/Vmax)(1/[S]) + 1/Vmax. The y‑intercept equals 1/Vmax, and the x‑intercept equals −1/Km.

步骤 1：回顾 Lineweaver‑Burk 方程。 该图为直线：1/V = (Km/Vmax)(1/[S]) + 1/Vmax。y 截距 = 1/Vmax，x 截距 = −1/Km。

Step 2: Analyse the intercepts. The y‑intercept is unchanged (0.02 s µM⁻¹), so Vmax is unaffected. The x‑intercept changes from −0.05 to −0.025 µM⁻¹, meaning the apparent Km has increased. This pattern is characteristic of competitive inhibition.

步骤 2：分析截距变化。 y 截距不变（0.02 s µM⁻¹），说明 Vmax 无变化。x 截距从 −0.05 变为 −0.025 µM⁻¹，表明表观 Km 增大，符合竞争性抑制的特征。

Step 3: Calculate Vmax. Vmax = 1 / (y‑intercept) = 1 / 0.02 = 50 µM s⁻¹.

步骤 3：计算 Vmax。 Vmax = 1 / (y 截距) = 1 / 0.02 = 50 µM s⁻¹。

Step 4: Calculate Km without inhibitor. −1/Km = −0.05 ⇒ Km = 1/0.05 = 20 µM.

步骤 4：计算无抑制剂时的 Km。 −1/Km = −0.05 ⇒ Km = 1/0.05 = 20 µM。

Step 5: Calculate apparent Km with inhibitor. −1/Km’ = −0.025 ⇒ Km’ = 1/0.025 = 40 µM. The inhibitor doubles the apparent Km while Vmax stays at 50 µM s⁻¹, confirming competitive inhibition.

步骤 5：计算有抑制剂时的表观 Km。 −1/Km’ = −0.025 ⇒ Km’ = 1/0.025 = 40 µM。抑制剂使表观 Km 加倍，而 Vmax 仍为 50 µM s⁻¹，证实为竞争性抑制。

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3. Genetics: ABO Blood Type Inheritance | 遗传学：ABO 血型遗传

Question: The ABO blood group is controlled by three alleles: I^A, I^B and i. A man with blood type A (heterozygous) and a woman with blood type B (heterozygous) are expecting a child. What is the probability that their child will have blood type O? Show your working using a Punnett square.

问题： ABO 血型由三个等位基因控制：I^A、I^B 和 i。一对夫妇，丈夫血型 A 型（杂合），妻子血型 B 型（杂合），即将生育。他们的孩子为 O 型血的概率是多少？请用庞纳特方格展示推导过程。

Step 1: Write the parental genotypes. The man is A‑heterozygous: I^Ai. The woman is B‑heterozygous: I^Bi.

步骤 1：写出亲代基因型。 丈夫为杂合 A 型：I^Ai。妻子为杂合 B 型：I^Bi。

Step 2: Determine the gametes. The man produces gametes carrying either I^A or i. The woman produces gametes carrying either I^B or i.

步骤 2：确定配子类型。 丈夫产生含 I^A 或 i 的配子。妻子产生含 I^B 或 i 的配子。

Step 3: Construct the Punnett square.

步骤 3：绘制庞纳特方格。

	IB	i
IA	IAIB	IAi
i	IBi	ii

Step 4: Identify the O‑type offspring. Only the genotype ii gives blood type O. It appears in 1 out of 4 boxes.

步骤 4：确定 O 型血后代。 只有基因型 ii 表现为 O 型血，出现在 4 个格子中的 1 个。

Answer: The probability of having a blood type O child is 1/4 or 25%.

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