📚 Calculation Question Intensive Practice for A-Level OCR Biology | A-Level OCR 生物：计算题专项训练
This revision guide focuses on the quantitative skills required for A-Level OCR Biology. Each section provides essential formulas, worked examples, and step-by-step solutions. Mastering these calculations will build confidence for both the written exam and practical endorsement.
本复习指南专注于A-Level OCR生物所需的定量技能。每个小节提供关键公式、范例和逐步解答。掌握这些计算将为笔试和实践认可建立信心。
1. Magnification & Actual Size Calculations | 放大倍数与实际尺寸计算
ze is fundamental in microscopy. The formula is: Magnification = Image size ÷ Actual size.
放大倍数、图像尺寸和实际尺寸之间的关系是显微镜学的基础。公式为：放大倍数 = 图像尺寸 ÷ 实际尺寸。
ert both measurements to the same unit before calculating. Common conversions: 1 mm = 1000 µm, 1 µm = 1000 nm.
计算前务必将两个测量值转换为相同单位。常见换算：1 mm = 1000 µm，1 µm = 1000 nm。
Example: A student measures the diameter of a cell in a micrograph as 24 mm. The actual diameter of the cell is 60 µm. Calculate the magnification.
例题：一名学生在显微照片中测得某细胞的直径为24 mm。该细胞的实际直径为60 µm。计算放大倍数。
Step 1: Convert image size to µm: 24 mm × 1000 = 24 000 µm.
步骤1：将图像尺寸转换为微米：24 mm × 1000 = 24 000 µm。
Step 2: Apply the formula: Magnification = 24 000 µm ÷ 60 µm = ×400.
步骤2：应用公式：放大倍数 = 24 000 µm ÷ 60 µm = ×400。
Image size ÷ Magnification. If an image measures 15 mm at ×5000, the actual size is 15 000 µm ÷ 5000 = 3 µm.
要计算实际尺寸，重新排列：实际尺寸 = 图像尺寸 ÷ 放大倍数。若某图像在×5000下测得15 mm，则实际尺寸为15 000 µm ÷ 5000 = 3 µm。
2. Cell Counting & the Haemocytometer | 细胞计数与血球计数板
A haemocytometer allows direct counting of cells in a known volume. The improved Neubauer chamber has a depth of 0.1 mm and each corner square has an area of 1 mm².
血球计数板可对已知体积中的细胞直接计数。改良的Neubauer计数室深度为0.1 mm，每个角方格的面积为1 mm²。
Volume of one corner square = 1 mm × 1 mm × 0.1 mm = 0.1 mm³ = 1 × 10⁻⁴ cm³ = 1 × 10⁻⁴ mL.
一个角方格的体积 = 1 mm × 1 mm × 0.1 mm = 0.1 mm³ = 1 × 10⁻⁴ cm³ = 1 × 10⁻⁴ mL。
The general formula: Cells per mL = average count per square × dilution factor × 10⁴.
通用公式：每毫升细胞数 = 每个方格的平均计数 × 稀释因子 × 10⁴。
Example: A yeast culture is diluted 10 times. The counts from five corner squares are 28, 32, 29, 31, 30. Calculate the cell concentration in the original culture.
例题：一份酵母培养液稀释10倍。五个角方格的计数为28、32、29、31、30。计算原始培养液中的细胞浓度。
Step 1: Find the mean count: (28+32+29+31+30) ÷ 5 = 30 cells.
步骤1：求均值： (28+32+29+31+30) ÷ 5 = 30 个细胞。
Step 2: Apply the formula: 30 × 10 × 10⁴ = 3.0 × 10⁶ cells mL⁻¹.
步骤2：应用公式：30 × 10 × 10⁴ = 3.0 × 10⁶ 个细胞/毫升。
When staining is used, only viable cells exclude the dye; count only those to estimate live cells.
使用染色时，只有活细胞排斥染料；仅计数这些以估算活细胞。
3. Bacterial Growth & Exponential Calculations | 细菌生长与指数计算
mber of bacteria after a given time is Nₜ = N₀ × 2ⁿ, where n is the number of generations.
细菌以二分裂繁殖，导致指数增长。经过给定时间后的细菌数量为 Nₜ = N₀ × 2ⁿ，其中 n 为代数。
= total time ÷ generation time (doubling time). This equation assumes no limiting factors.
代数 n = 总时间 ÷ 代时（倍增时间）。该方程假设无限制因素。
Example: A single bacterium has a generation time of 20 minutes. Calculate how many bacteria will be present after 3 hours.
例题：一个细菌的代时为20分钟。计算3小时后有多少细菌存在。
Step 1: Total time = 3 × 60 = 180 minutes; n = 180 ÷ 20 = 9 generations.
步骤1：总时间 = 3 × 60 = 180 分钟；n = 180 ÷ 20 = 9 代。
Step 2: N = 1 × 2⁹ = 512 cells.
步骤2：N = 1 × 2⁹ = 512 个细胞。
ivisions per hour = 1 generation / generation time (in hours). Here, 3 divisions per hour.
你还可能被要求计算平均分裂速率：每小时分裂次数 = 1代 / 代时（小时）。此处为每小时3次。
4. Enzyme Activity & Rate Calculations | 酶活性与反应速率计算
ss curve. Rate = change in product amount ÷ time (or change in substrate consumed ÷ time).
酶控反应的速率通常从进程曲线的初始线性部分计算。速率 = 产物变化量 ÷ 时间（或底物消耗变化量 ÷ 时间）。
Common units are cm³ min⁻¹, µmol min⁻¹ or arbitrary units such as absorbance change per minute (ΔA min⁻¹).
常用单位为 cm³ min⁻¹、µmol min⁻¹ 或任意单位如每分钟吸光度变化 (ΔA min⁻¹)。
Example: A catalase experiment recorded the following data: at 0 s, O₂ volume 0.0 cm³; at 30 s, 4.5 cm³; at 60 s, 7.5 cm³. Calculate the initial rate of reaction over the first 30 seconds.
例题：一个过氧化氢酶实验记录如下数据：0 s时O₂体积0.0 cm³；30 s时4.5 cm³；60 s时7.5 cm³。计算前30秒的初始反应速率。
Initial rate = (4.5 – ) cm³ ÷ 30 s = 0.15 cm³ s⁻¹. Alternatively, convert to minutes: 0.15 × 60 = 9.0 cm³ min⁻¹.
初始速率 = (4.5 – 0.0) cm³ ÷ 30 s = 0.15 cm³ s⁻¹。或转换为分钟：0.15 × 60 = 9.0 cm³ min⁻¹。
To compare enzyme activity under different conditions, always measure the initial rate before substrate depletion or product inhibition occurs.
要比较不同条件下的酶活性，始终在底物耗尽或产物抑制发生前测量初始速率。
5. Heart Rate, Stroke Volume & Cardiac Output | 心率、每搏输出量与心输出量
entricle per minute. CO (mL min⁻¹) = heart rate (beats min⁻¹) × stroke volume (mL beat⁻¹).
心输出量 (CO) 是一个心室每分钟泵出的血液体积。CO (mL min⁻¹) = 心率 (次每分) × 每搏输出量 (mL每搏)。
Resting values for a typical human are: heart rate 70 bpm, stroke volume 70 mL, giving a CO of 4900 mL min⁻¹ (4.9 L min⁻¹).
典型人体的静息值为：心率70 bpm，每搏输出量70 mL，得出CO为4900 mL min⁻¹ (4.9 L min⁻¹)。
Example: An athlete has a resting heart rate of 45 bpm and a cardiac output of 5.4 L min⁻¹. Calculate the stroke volume.
例题：一名运动员静息心率45 bpm，心输出量5.4 L min⁻¹。计算每搏输出量。
First, convert CO to mL: 5.4 L = 5400 mL. Rearrange: SV = CO ÷ HR = 5400 ÷ 45 = 120 mL.
首先，转换CO为mL：5.4 L = 5400 mL。重新排列：SV = CO ÷ HR = 5400 ÷ 45 = 120 mL。
R rises to 180 bpm and SV remains 120 mL, CO = 180 × 120 = 21 600 mL min⁻¹ (21.6 L min⁻¹).
运动时，CO可增加五倍。若HR升至180 bpm而SV保持120 mL，则CO = 180 × 120 = 21 600 mL min⁻¹ (21.6 L min⁻¹)。
6. Respiratory Quotient (RQ) | 呼吸商
o oxygen consumed during respiration. RQ = volume of CO₂ produced ÷ volume of O₂ consumed.
呼吸商 (RQ) 是呼吸时产生的二氧化碳与消耗的氧气的比值。RQ = 产生的CO₂体积 ÷ 消耗的O₂体积。
RQ values indicate the respiratory substrate: carbohydrate = 1.0, fat ~0.7, protein ~0.9. Mixed diets give intermediate values.
RQ值指示呼吸底物：碳水化合物 = 1.0，脂肪约0.7，蛋白质约0.9。混合饮食给出中间值。
Example: A respirometer experiment shows that a maggot consumed 8.0 cm³ of O₂ and released 6.4 cm³ of CO₂ over 10 minutes. Calculate RQ.
例题：一个呼吸计实验显示，一只蛆在10分钟内消耗了8.0 cm³的O₂，释放了6.4 cm³的CO₂。计算RQ。
RQ = 6.4 ÷ 8.0 = 0.80.
RQ = 6.4 ÷ 8.0 = 0.80。
An RQ of 0.80 suggests a mixture of substrates, possibly carbohydrate and protein. Complete lipid oxidation would yield an RQ close to 0.7.
RQ为0.80表明是多种底物的混合，可能是碳水化合物和蛋白质。完全脂肪氧化会产生接近0.7的RQ。
7. Genetic Probabilities & Chi-Squared Tests | 遗传概率与卡方检验
Genetic crosses produce predictable ratios. The actual offspring often deviate from expected ratios due to chance. The chi-squared (χ²) test determines if the deviation is significant.
遗传杂交产生可预测的比例。实际后代常因偶然性偏离预期比例。卡方 (χ²) 检验判断偏差是否显著。
χ² = Σ ((O – E)² ÷ E), where O = observed frequency, E = expected frequency.
χ² = Σ ((O – E)² ÷ E)，其中O = 观察频数，E = 期望频数。
Example: In a monohybrid cross, a 3:1 phenotypic ratio is expected. Out of 160 offspring, 115 show the dominant trait and 45 the recessive trait. Test the fit to a 3:1 ratio.
例题：在一个单基因杂交中，预期表型比为3:1。在160个后代中，115个表现显性性状，45个表现隐性性状。检验是否符合3:1比例。
numbers. Total 160, expected dominant = ¾ × 160 = 120; expected recessive = ¼ × 160 = 40.
步骤1：计算期望数。总计160，期望显性 = ¾ × 160 = 120；期望隐性 = ¼ × 160 = 40。
Step 2: χ² = ((115–120)² ÷ 120) + ((45–40)² ÷ 40) = (25 ÷ 120) + (25 ÷ 40) = 0.208 + 0.625 = 0.833.
步骤2：χ² = ((115–120)² ÷ 120) + ((45–40)² ÷ 40) = (25 ÷ 120) + (25 ÷ 40) = 0.208 + 0.625 = 0.833。
Degrees of freedom = number of categories – 1 = 2 – 3.84. Since 0.833 < 3.84, the difference is not significant; the observed ratio fits 3:1.
自由度 = 类别数 – 1 = 2 – 1 = 1。p=0.05时的临界值为3.84。由于0.833 < 3.84，差异不显著；观察比例符合3:1。
8. Ecological Sampling & Simpson’s Index of Diversity | 生态取样与辛普森多样性指数
versity, accounting for both species richness and evenness. For OCR, the formula is: D = 1 – re n = number of individuals of a species, N = total number of individuals of all species.
辛普森多样性指数 (D) 衡量生物多样性，兼顾物种丰富度和均匀度。OCR使用的公式为：D = 1 – Σ (n ÷ N)²，其中 n = 某一物种的个体数，N = 所有物种的个体总数。
D ranges from 0 (low diversity) to nearly 1 (very high diversity). A high D indicates a stable, resilient ecosystem.
D值介于0（低多样性）至近1（极高多样性）之间。高D值表明稳定、有弹性的生态系统。
Example: A woodland survey gives: oak (15), birch (8), holly (4), hazel (3). Calculate Simpson’s Index.
例题：某林地调查得到：橡树 (15)、桦树 (8)、冬青 (4)、榛树 (3)。计算辛普森指数。
birch (8/30)² ≈ 0.0711; holly (4/30)² ≈ 0.0178; hazel (3/30)² = 0.01. Sum = 0.3489. D = 1 – 0.3489 = 0.6511.
N = 15+8+4+3 = 30。计算每种的 (n/N)²：橡树 (15/30)² = 0.25；桦树 (8/30)² ≈ 0.0711；冬青 (4/30)² ≈ 0.0178；榛树 (3/30)² = 0.01。总和 = 0.3489。D = 1 – 0.3489 = 0.6511。
This value represents moderate diversity. In exam questions, you may be provided with a partially completed table to fill in.
该值代表中等多样性。考题中，你可能会得到一个部分完成的表格并要求填写。
9. Water Potential & Pressure Potential | 水势与压力势
Ψₚ (pressure potential). All values are usually in kilopascals (kPa) or megapascals (MPa).
水势 (Ψ) 决定了水分移动的方向。在植物细胞中，Ψ = Ψₛ (溶质势) + Ψₚ (压力势)。所有数值通常以千帕 (kPa) 或兆帕 (MPa) 表示。
. The addition of solutes lowers Ψₛ (becomes negative). Turgor pressure makes Ψₚ positive.
标准大气压下的纯水Ψ = 0 kPa。加入溶质降低Ψₛ（变为负值）。膨压使Ψₚ为正值。
Example: A plant cell has a solute potential of –700 kPa and a pressure potential of 400 kPa. Calculate its water potential.
例题：一个植物细胞的溶质势为–700 kPa，压力势为400 kPa。计算其水势。
Ψ = –700 + 400 = – kPa. Water will move from a region of higher Ψ (less negative) to lower Ψ (more negative).
Ψ = –700 + 400 = –300 kPa。水分将从水势较高（负值较小）的区域移向水势较低（负值较大）的区域。
When a cell is fully turgid, Ψₚ increases until Ψ equals zero, preventing further water entry.
当细胞完全膨胀时，Ψₚ增加直至Ψ等于零，阻止进一步吸水。
10. Surface Area to Volume Ratio & Diffusion | 表面积体积比与扩散
The surface area to volume ratio (SA:V) affects the rate of exchange of substances. As an organism or cell increases in size, its SA:V decreases, making diffusion less efficient.
表面积与体积之比 (SA:V) 影响物质交换的速率。随着生物体或细胞增大，其SA:V减小，使扩散效率降低。
cube, SA = 6 × (side length)²; Volume = (side length)³. The ratio is SA divided by volume.
对于立方体，SA = 6 × (边长)²；体积 = (边长)³。比值为SA除以体积。
Example: Compare the SA:V of a cube with side 1 cm and a cube with side 3 cm.
例题：比较边长为1 cm和3 cm的立方体的SA:V。
Small cube: SA = 6 × 1² = 6 cm²; V = 1 cm³; ratio = 6:1.
小立方体：SA = 6 × 1² = 6 cm²；V = 1 cm³；比值 = 6:1。
Large cube: SA = 6 × 3² = 54 cm²; V = 27 cm³; ratio = 54:27 = 2:1.
大立方体：SA = 6 × 3² = 54 cm²；V = 27 cm³；比值 = 54:27 = 2:1。
The smaller cube has a much larger SA:V. This is why single-celled organisms can rely on diffusion, while larger multicellular organisms require transport systems.
较小的立方体具有大得多的SA:V。这就是为什么单细胞生物可以依赖扩散，而较大的多细胞生物需要运输系统。
11. Calibration Curves & Quantitative Food Tests | 标准曲线与定量食物测试
Colorimetry can quantify substances such as reducing sugars using a Benedict’s test variant or biuret. A calibration curve is constructed by measuring absorbance of known concentrations, then used to interpolate unknown concentrations.
比色法可利用本尼迪克特测试变体或双缩脲反应定量检测如还原糖等物质。通过测量已知浓度的吸光度构建标准曲线，然后用于内插未知浓度。
Create a table of concentration against absorbance. Plot absorbance on the y-axis, concentration on the x-axis, and draw a line of best fit.
制作浓度对应吸光度的表格。将吸光度标于y轴，浓度标于x轴，并绘制最佳拟合线。
Example: A glucose standard gave these results:
例题：葡萄糖标准品得出以下结果：
| Concentration (mmol L⁻¹) | Absorbance (540 nm) |
|---|---|
| 0.0 | 0.00 |
| 0.2 | 0.15 |
| 0.4 | 0.31 |
| 0.6 | 0.47 |
| 0.8 | 0.63 |
| 1.0 | 0.79 |
An unknown solution gives an absorbance of 0.45. Interpolate its concentration.
某未知溶液的吸光度为0.45。内插出其浓度。
Reading from the line of best fit, an absorbance of 0.45 corresponds to a glucose concentration of approximately 0.57 mmol L⁻¹. Always state the range of error based on the graph resolution.
从最佳拟合线上读取，吸光度0.45对应的葡萄糖浓度约为0.57 mmol L⁻¹。总是根据图表分辨率说明误差范围。
12. Statistical Analysis: Mean, Standard Deviation & t-test | 统计分析：均值、标准差与t检验
Published by TutorHao | A-Level Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图