Case Study Analysis in Pre-U OCR Biology: Practical Workout | Pre-U OCR 生物案例分析实战演练

📚 Case Study Analysis in Pre-U OCR Biology: Practical Workout | Pre-U OCR 生物案例分析实战演练

The Pre-U OCR Biology course demands not only knowledge recall but also the skilful application of concepts to novel scenarios. Case study analysis is a powerful revision tool that sharpens these abilities. In this workout, we explore eight diverse case studies spanning biochemistry, genetics, ecology and physiology, with guided questions and model answers designed to build your exam confidence and deepen your understanding.

Pre-U OCR 生物课程不仅要求记忆知识，更强调将概念灵活应用于新情境。案例分析是一种能够磨砺这些能力的强大复习工具。在本次实战演练中，我们将探索涵盖生物化学、遗传学、生态学和生理学的八个不同案例，并配有引导性问题和参考答案，旨在建立你的考试信心并加深理解。

1. Enzyme Inhibition and Drug Design | 酶抑制与药物设计

A pharmaceutical company screens two novel compounds, X and Y, for their effect on an enzyme involved in blood pressure regulation. Initial rates of reaction were measured at a fixed enzyme concentration and varying substrate concentrations. The data are shown below. Determine the type of inhibition produced by each compound and calculate the approximate Kₘ and Vₘₐₓ for the uninhibited enzyme.

一家制药公司筛选两种新型化合物 X 和 Y，研究它们对一种参与血压调节的酶的影响。在固定酶浓度和不同底物浓度下测定了反应的初始速率，数据如下。请判断每种化合物产生的抑制类型，并计算无抑制剂时酶的近似 Kₘ 和 Vₘₐₓ。

To distinguish between inhibition types, convert the data to reciprocal values for a Lineweaver–Burk plot (1/rate vs 1/[S]). The y-intercept equals 1/Vₘₐₓ and the x-intercept equals –1/Kₘ. For the uninhibited enzyme, the intercepts give 1/Vₘₐₓ ≈ 2.0, so Vₘₐₓ ≈ 0.5 units s⁻¹; the x-intercept is about –2.0, hence Kₘ ≈ 0.5 mmol dm⁻³.

为了区分抑制类型，将数据转化为倒数用于绘制双倒数图（1/速率 对 1/[S]）。y 轴截距等于 1/Vₘₐₓ，x 轴截距等于 –1/Kₘ。对于无抑制酶的酶，截距给出 1/Vₘₐₓ ≈ 2.0，因此 Vₘₐₓ ≈ 0.5 单位 s⁻¹；x 截距约为 –2.0，因此 Kₘ ≈ 0.5 mmol dm⁻³。

Compound X produces lines that share the same y-intercept as the uninhibited enzyme but with a steeper slope, which means the x-intercept moves closer to zero. This is characteristic of competitive inhibition: Vₘₐₓ is unchanged while the apparent Kₘ increases. Compound Y causes the y-intercept to increase (1/Vₘₐₓ becomes larger), indicating a decrease in Vₘₐₓ; the x-intercept also changes but in a pattern where the lines intersect on the x-axis, typical of non-competitive inhibition, where Kₘ remains constant but Vₘₐₓ is reduced.

化合物 X 产生的直线与无抑制剂酶有相同的 y 截距，但斜率更陡，意味着 x 截距向零移动。这是竞争性抑制的特征：Vₘₐₓ 不变，而表观 Kₘ 增大。化合物 Y 导致 y 截距增大（1/Vₘₐₓ 变大），表明 Vₘₐₓ 降低；x 截距也发生变化，但所有直线相交于 x 轴上，这是非竞争性抑制的典型特征，即 Kₘ 保持不变而 Vₘₐₓ 减小。

Such kinetic analysis is vital in rational drug design. Competitive inhibitors, like compound X, typically resemble the substrate and occupy the active site, whereas non-competitive inhibitors like Y bind at an allosteric site, causing a conformational change that lowers catalytic efficiency. Both strategies can reduce blood pressure by partially inhibiting the enzyme pathway.

这种动力学分析在理性药物设计中至关重要。像化合物 X 这样的竞争性抑制剂通常与底物相似并占据活性位点，而像 Y 这样的非竞争性抑制剂则结合于别构位点，引起构象改变从而降低催化效率。这两种策略都可以通过部分抑制酶促途径来降低血压。

2. Predator–Prey Dynamics and the Lotka–Volterra Model | 捕食者–猎物动态与 Lotka–Volterra 模型

Ecologists in a boreal forest have tracked the population densities of snowshoe hares (prey) and lynx (predator) over a 12-year period. The data exhibit regular cycles. Using the Lotka–Volterra model, explain the coupled oscillations and the observed phase lag between prey and predator peaks.

北方森林中的生态学家追踪了雪兔（猎物）和猞猁（捕食者）在 12 年间的种群密度，数据呈现规律性周期。请运用 Lotka–Volterra 模型解释这种耦合振荡以及猎物与捕食者峰值之间的相位滞后。

The Lotka–Volterra equations describe the rates of change: dN/dt = rN – aNP for prey, and dP/dt = bNP – mP for predator. Here N is prey density, P is predator density, r is prey intrinsic growth rate, a is predation rate coefficient, b is conversion efficiency of prey into predator offspring, and m is predator mortality. Isoclines (dN/dt = 0 and dP/dt = 0) intersect at the equilibrium point, but the system oscillates around it.

Lotka–Volterra 方程描述了变化速率：对于猎物有 dN/dt = rN – aNP，对于捕食者有 dP/dt = bNP – mP。式中 N 为猎物种群密度，P 为捕食者密度，r 为猎物的内禀增长率，a 为捕食率系数，b 为将猎物转化为捕食者后代的效率，m 为捕食者死亡率。零增长等斜线（dN/dt = 0 和 dP/dt = 0）在平衡点相交，但系统围绕该点振荡。

The phase lag occurs because predator numbers respond to prey density with a delay determined by reproduction time and energy transfer. When hares are abundant, lynx have more food, so their birth rate rises and death rate falls – but this takes several months. As lynx numbers climb, they over-exploit hares, causing the hare decline; lynx then crash due to starvation, allowing hares to recover. The predicted lag is typically a quarter-cycle, and the data show the lynx peak approximately 1–2 years after the hare peak.

相位滞后是由于捕食者数量对猎物密度的响应存在由繁殖时间和能量传递决定的延迟。当野兔数量丰富时，猞猁获得更多食物，因此其出生率上升、死亡率下降，但这需要数月时间。随着猞猁数量攀升，它们过度捕食野兔，导致野兔数量下降；随后猞猁因饥饿而崩溃，使野兔得以恢复。预测的滞后通常为四分之一周期，数据中猞猁峰值大约在野兔峰值之后 1–2 年出现。

Although the classic model assumes no density dependence and an unlimited prey herbivory, the principal dynamics remain valid. Field data often show damped oscillations due to additional factors like territoriality and alternative food sources, enriching the analysis for Pre-U discussions.

尽管经典模型假设无密度制约和猎食物无限，但其主要动力学仍然有效。野外数据常因领域性和替代食物来源等额外因素而呈现阻尼振荡，这为 Pre-U 讨论丰富了分析内容。

3. Pedigree Analysis for Cystic Fibrosis | 囊性纤维化的家系分析

Cystic fibrosis (CF) is caused by a recessive allele of the CFTR gene. A healthy couple, John and Mary, are expecting a child. John’s younger sister has CF, and Mary’s brother also has CF. Neither John nor Mary shows any symptoms, and there is no consanguinity. Calculate the probability that their child will be born with cystic fibrosis.

囊性纤维化 (CF) 由 CFTR 基因的隐性等位基因引起。一对健康夫妇 John 和 Mary 正期待一个孩子。John 的妹妹患有 CF，Mary 的哥哥也患有 CF。John 和 Mary 均无症状，且无近亲结婚。请计算他们的孩子出生时患有囊性纤维化的概率。

Since CF is autosomal recessive, an affected individual must be homozygous for the mutant allele. John’s sister is affected, so both of John’s parents must be heterozygous carriers. John is phenotypically normal, so he has a 2/3 chance of being a carrier (heterozygote) and a 1/3 chance of being homozygous wild-type. The same logic applies to Mary, because her brother is affected: their parents are both carriers, giving Mary a 2/3 carrier probability.

由于 CF 为常染色体隐性遗传，患者必定为突变等位基因的纯合子。John 的妹妹患病，因此 John 的父母必定均为杂合子携带者。John 表型正常，因此他成为携带者（杂合子）的概率为 2/3，成为野生型纯合子的概率为 1/3。同理适用于 Mary，因为她的哥哥患病，其父母均为携带者，Mary 成为携带者的概率也是 2/3。

If both John and Mary are carriers, each pregnancy carries a 1/4 risk of producing an affected child (homozygous recessive). Therefore, the overall probability is the product of the three independent probabilities: (2/3) × (2/3) × (1/4) = 4/36 = 1/9. Thus, the unborn child has roughly an 11.1% chance of inheriting CF.

若二人均为携带者，则每次妊娠有 1/4 的几率生出患病孩子（隐性

Published by TutorHao | Pre-U Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）