📚 IB & WJEC Biology: Calculation Practice | IB WJEC 生物：计算题专项训练
Calculations form an integral part of both IB and WJEC Biology examinations. From microscopy to genetics and ecology, being able to perform accurate calculations and interpret data is essential for top marks. This guide provides a focused drill covering the most common calculation types appearing in these syllabi, with step-by-step examples to build your confidence.
计算题是 IB 和 WJEC 生物考试的重要组成部分。从显微镜测量到遗传学和生态学，能够准确进行计算并解读数据是取得高分的关键。本指南针对这些大纲中最常见的计算类型进行专项训练，通过分步示例帮助你建立信心。
1. Magnification and Size Calculations | 放大倍数与尺寸计算
Magnification (M) is defined by the equation M = image size ÷ actual size. Always convert both measurements to the same unit before dividing, typically millimetres (mm) or micrometres (µm). Recall that 1 mm = 1000 µm.
放大倍数 (M) 由公式 M = 图像尺寸 ÷ 实际尺寸 定义。计算前务必将两个数值转换为相同单位，通常用毫米 (mm) 或微米 (µm)。记住 1 mm = 1000 µm。
Example: A student draws an animal cell with a length of 30 mm on paper. The actual cell length is 60 µm. What is the magnification of the drawing?
示例：一名学生画了一个动物细胞，在纸上的长度为 30 mm。实际细胞长度为 60 µm。该绘图放大倍数为多少？
Solution: Convert image size to µm: 30 mm = 30 × 1000 = 30 000 µm. Then M = 30 000 µm ÷ 60 µm = 500 ×. The drawing is 500 times larger than the real cell.
解答：将图像尺寸转换为 µm：30 mm = 30 × 1000 = 30 000 µm。然后 M = 30 000 µm ÷ 60 µm = 500 倍。绘图比真实细胞大 500 倍。
If you need to find the actual size, rearrange the formula: actual size = image size ÷ magnification. For example, a micrograph has a magnification of ×2000 and shows a mitochondrion 4 cm long. The true length is 4 cm ÷ 2000 = 0.002 cm = 20 µm.
如果需要求实际尺寸，可变形公式：实际尺寸 = 图像尺寸 ÷ 放大倍数。例如，一张显微照片放大 2000 倍，显示线粒体长 4 cm。其真实长度为 4 cm ÷ 2000 = 0.002 cm = 20 µm。
2. Haemocytometer Cell Counting | 血球计数板细胞计数
A haemocytometer is a specialised slide with a grid of known dimensions, used to count cells in a suspension. A typical chamber depth is 0.1 mm. The central counting area often has 25 large squares, each with 16 smaller squares. You count the cells in a set number of squares and then calculate the concentration of cells per unit volume.
血球计数板是一种带有已知尺寸网格的专用载玻片，用于计数悬浮液中的细胞。典型计数室深度为 0.1 mm。中央计数区通常有 25 个大格，每个大格包含 16 个小格。你计数一定数量的方格中的细胞，然后计算单位体积中的细胞浓度。
Formula: cell concentration (cells per cm³) = (total cells counted ÷ number of squares counted) × (1 ÷ volume of one square in cm³). The volume of one large square (1 mm × 1 mm × 0.1 mm) = 0.1 mm³ = 1 × 10⁻⁴ cm³.
公式：细胞浓度（每 cm³ 细胞数）= (计数的细胞总数 ÷ 计数的方格数) × (1 ÷ 一个方格的体积 cm³)。一个大方格（1 mm × 1 mm × 0.1 mm）的体积 = 0.1 mm³ = 1 × 10⁻⁴ cm³。
Example: A student counts 480 yeast cells in 5 large squares. Calculate the concentration in cells per cm³.
示例：学生在 5 个大方格中数出 480 个酵母细胞。计算每 cm³ 细胞浓度。
Solution: Average per large square = 480 ÷ 5 = 96 cells. Volume of one large square = 0.1 mm³ = 1×10⁻⁴ cm³. So cells per cm³ = 96 ÷ (1×10⁻⁴) = 960 000 cells cm⁻³.
解答：每个大方格平均数 = 480 ÷ 5 = 96 个细胞。一个大方格体积 = 0.1 mm³ = 1×10⁻⁴ cm³。因此每 cm³ 细胞数 = 96 ÷ (1×10⁻⁴) = 960 000 cells cm⁻³。
3. Serial Dilutions and Concentration | 系列稀释与浓度计算
Serial dilutions are used to produce a range of decreasing concentrations from a stock solution. A common dilution factor might be 1:10, meaning one part stock plus nine parts diluent. The concentration after n dilutions is C₀ × (dilution factor)⁻ⁿ, where C₀ is the original concentration.
系列稀释用于从储备液中制备一系列浓度递减的溶液。常见的稀释倍数如 1:10，即 1 份原液加 9 份稀释液。经过 n 次稀释后的浓度为 C₀ × (稀释倍数)⁻ⁿ，其中 C₀ 为初始浓度。
Example: A glucose solution has an initial concentration of 0.8 mol dm⁻³. 1 cm³ of this solution is added to 9 cm³ of water, and the process is repeated twice. What is the final concentration?
示例：某葡萄糖溶液初始浓度为 0.8 mol dm⁻³。取 1 cm³ 溶液加入 9 cm³ 水中，重复该步骤两次。最终浓度是多少？
Each 1:10 dilution reduces concentration by a factor of 10. After three such dilutions: final concentration = 0.8 × (1/10)³ = 0.8 × 0.001 = 0.0008 mol dm⁻³.
每次 1:10 稀释使浓度缩小 10 倍。三次后：最终浓度 = 0.8 × (1/10)³ = 0.8 × 0.001 = 0.0008 mol dm⁻³。
4. Genetics: Monohybrid and Dihybrid Crosses | 遗传学：单基因与双基因杂交概率
A monohybrid cross considers one gene with two alleles. Use a Punnett square to determine genotypic and phenotypic ratios. Probabilities can be multiplied for independent events. For a dihybrid cross involving two unlinked genes, the expected phenotypic ratio is normally 9:3:3:1 when both parents are heterozygous at both loci.
单基因杂交考虑一个基因的两个等位基因。利用棋盘格法确定基因型和表现型比例。独立事件的概率可以相乘。对于涉及两个非连锁基因的双基因杂交，当双亲在两对基因上都是杂合时，预期表现型比例通常为 9:3:3:1。
Example: In guinea pigs, black coat (B) is dominant to white (b), and rough coat (R) is dominant to smooth (r). Two heterozygous black, rough guinea pigs (BbRr) are crossed. Calculate the probability of an offspring being white and smooth.
示例：在豚鼠中，黑毛 (B) 对白毛 (b) 为显性，粗毛 (R) 对光毛 (r) 为显性。两只杂合黑毛粗毛豚鼠 (BbRr) 交配。计算后代为白毛光毛的概率。
Both parents produce gametes BR, Br, bR, br with equal probability (1/4 each). The only way to be white and smooth is genotype bbrr, which requires a br sperm and a br egg. Probability = (1/4) × (1/4) = 1/16.
双亲均产生配子 BR、Br、bR、br，各占 1/4。白毛光毛的唯一基因型是 bbrr，需要 br 精子和 br 卵细胞。概率 = (1/4) × (1/4) = 1/16。
5. Hardy-Weinberg Equilibrium | 哈代-温伯格平衡
The Hardy-Weinberg principle predicts allele and genotype frequencies in a non-evolving population. For a gene with two alleles A and a, let p = frequency of A and q = frequency of a, where p + q = 1. Then the genotype frequencies are p² (AA), 2pq (Aa), and q² (aa). This only holds under conditions such as no mutation, large population, random mating, no selection and no gene flow.
哈代-温伯格定律用于预测非进化种群中等位基因和基因型频率。假设一个基因有 A 和 a 两个等位基因，设 p = A 的频率，q = a 的频率，且 p + q = 1。则基因型频率为 p² (AA)、2pq (Aa) 和 q² (aa)。该定律仅在无突变、大种群、随机交配、无选择和无基因流等条件下成立。
Example: In a population, 1 in 2500 individuals has a recessive genetic disorder (aa). Assuming Hardy-Weinberg equilibrium, calculate the frequency of carriers (Aa).
示例：在一个种群中，每 2500 个个体中有 1 个患有隐性遗传病 (aa)。假设哈代-温伯格平衡，计算携带者 (Aa) 的频率。
q² = 1/2500, so q = √(1/2500) = 1/50 = 0.02. Then p = 1 – 0.02 = 0.98. Frequency of carriers = 2pq = 2 × 0.98 × 0.02 = 0.0392, or about 3.92%.
q² = 1/2500，所以 q = √(1/2500) = 1/50 = 0.02。则 p = 1 – 0.02 = 0.98。携带者频率 = 2pq = 2 × 0.98 × 0.02 = 0.0392，约 3.92%。
6. Chi-squared Test | 卡方检验
The chi-squared (χ²) test is used to determine whether there is a significant difference between observed and expected frequencies. The formula is χ² = Σ (O – E)² ÷ E, where O = observed value and E = expected value. After calculating χ², compare it to a critical value from a table at a chosen probability level (usually p = 0.05) with degrees of freedom (df = number of categories – 1).
卡方 (χ²) 检验用于判断观察频率与预期频率之间是否存在显著差异。公式为 χ² = Σ (O – E)² ÷ E，其中 O = 观察值，E = 期望值。计算 χ² 后，将它与选定概率水平（通常 p = 0.05）和自由度（df = 类别数 – 1）下的临界值比较。
Example: In a genetics experiment, a cross of heterozygous plants gives phenotypes round yellow (144), wrinkled yellow (48), round green (50), wrinkled green (18). Total = 260. The expected ratio is 9:3:3:1. Calculate expected numbers, χ², and interpret the result.
示例：在遗传学实验中，杂合植株杂交产生表现型：圆黄 144，皱黄 48，圆绿 50，皱绿 18。总数 260。预期比例为 9:3:3:1。计算期望数值、χ² 并解读结果。
Expected: 9/16 × 260 = 146.25, 3/16 × 260 = 48.75, 3/16 × 260 = 48.75, 1/16 × 260 = 16.25. χ² = (144-146.25)²/146.25 + (48-48.75)²/48.75 + (50-48.75)²/48.75 + (18-16.25)²/16.25 ≈ 0.0346 + 0.0115 + 0.0321 + 0.1885 = 0.2667. With df = 3, critical value at p=0.05 is 7.815. Since 0.267 < 7.815, there is no significant difference; the data fit the expected ratio.
期望值：9/16 × 260 = 146.25，3/16 × 260 = 48.75，3/16 × 260 = 48.75，1/16 × 260 = 16.25。χ² = (144-146.25)²/146.25 + (48-48.75)²/48.75 + (50-48.75)²/48.75 + (18-16.25)²/16.25 ≈ 0.0346 + 0.0115 + 0.0321 + 0.1885 = 0.2667。自由度 df=3，p=0.05 时临界值为 7.815。由于 0.267 < 7.815，无显著差异；数据符合预期比例。
7. Enzyme Kinetics: Rate of Reaction | 酶动力学：反应速率
The rate of an enzyme-catalysed reaction can be calculated from the change in product concentration over time or the disappearance of substrate. Rate = change in quantity ÷ time taken. You may need to interpret graphs of product formed vs time and calculate initial rate using the tangent at time zero.
酶催化反应的速率可通过产物浓度随时间的变化或底物消耗计算。速率 = 变化量 ÷ 所用时间。你可能需要解读产物生成量-时间图，并用零点处切线计算初始速率。
Example: In an experiment measuring hydrogen peroxide breakdown by catalase, 12 cm³ of oxygen was produced in the first 30 seconds. Calculate the initial rate in cm³ s⁻¹.
示例：在过氧化氢酶催化过氧化氢分解的实验中，前 30 秒产生 12 cm³ 氧气。以 cm³ s⁻¹ 为单位计算初始速率。
Initial rate = 12 cm³ ÷ 30 s = 0.4 cm³ s⁻¹. If the substrate concentration was 0.2 mol dm⁻³ and the rate is proportional, this can be used to compare enzyme activity under different conditions.
初始速率 = 12 cm³ ÷ 30 s = 0.4 cm³ s⁻¹。若底物浓度为 0.2 mol dm⁻³ 且速率与之成正比，可用于比较不同条件下的酶活性。
8. Osmolarity and Water Potential | 渗透压与水势
Water potential (Ψ) determines the direction of water movement. It is the sum of solute potential (Ψₛ) and pressure potential (Ψₚ): Ψ = Ψₛ + Ψₚ. Ψₛ is always negative or zero, and adding solutes makes it more negative. Cells in pure water have high water potential; water moves from a region of higher Ψ to lower Ψ.
水势 (Ψ) 决定水分移动方向。它是溶质势 (Ψₛ) 和压力势 (Ψₚ) 之和：Ψ = Ψₛ + Ψₚ。Ψₛ 总是为负值或零，溶质越多其值越负。细胞处在纯水中具有较高水势；水从 Ψ 较高的区域移向较低的区域。
Calculation example: A plant cell with Ψₛ = -0.7 MPa and Ψₚ = +0.3 MPa. Its water potential Ψ = -0.7 + 0.3 = -0.4 MPa. If surrounding solution has Ψ = -0.2 MPa, water will enter the cell because -0.2 > -0.4.
计算示例：某植物细胞 Ψₛ = -0.7 MPa，Ψₚ = +0.3 MPa。其水势 Ψ = -0.7 + 0.3 = -0.4 MPa。若周围溶液 Ψ = -0.2 MPa，水将进入细胞，因为 -0.2 > -0.4。
9. Population Ecology: Growth Rate and Capture-Recapture | 种群生态：增长率和标记重捕法
Population growth rate can be calculated from change in numbers over time: growth rate = (births + immigration) – (deaths + emigration). For exponential growth, the intrinsic rate of increase (r) is used. To estimate population size of motile organisms, the Lincoln index (capture-recapture) is applied: N = (M × C) ÷ R, where M = number initially captured and marked, C = total captured in second sample, R = number of marked individuals recaptured.
种群增长率可由数量随时间的变化计算：增长率 = (出生 + 迁入) – (死亡 + 迁出)。指数增长时使用内禀增长率 (r)。对于可移动生物，用林肯指数（标记重捕法）估计种群大小：N = (M × C) ÷ R，其中 M = 首次捕获并标记数，C = 第二次捕获总数，R = 重捕的标记个体数。
Example: Researchers catch 80 woodlice, mark them and release. A week later they catch 120 woodlice, of which 15 are marked. Estimate population size.
示例：研究人员捕获 80 只鼠妇，标记后放回。一周后捕获 120 只，其中 15 只带有标记。估算种群大小。
N = (80 × 120) ÷ 15 = 9600 ÷ 15 = 640. Assumptions include no births, deaths, immigration or emigration between samples, and marks are not lost.
N = (80 × 120) ÷ 15 = 9600 ÷ 15 = 640。假设条件包括两次取样间无出生、死亡、迁入或迁出，且标记不会脱落。
10. Energy Flow and Ecological Efficiency | 能量流动与生态效率
In ecosystems, energy is transferred between trophic levels. The efficiency of energy transfer can be calculated using: efficiency (%) = (energy available to the next trophic level ÷ energy available to the current trophic level) × 100. Typical ecological efficiencies range from 5 to 20%.
在生态系统中，能量在营养级之间传递。能量传递效率可通过下式计算：效率 (%) = (传递到下一营养级的能量 ÷ 当前营养级可利用的能量) × 100。典型的生态效率在 5% 至 20% 之间。
Example: In a food chain, producers fix 25 000 kJ m⁻² yr⁻¹. Primary consumers receive 3 000 kJ m⁻² yr⁻¹. Calculate the efficiency of energy transfer from producers to primary consumers.
示例：在某食物链中，生产者固定 25 000 kJ m⁻² yr⁻¹。初级消费者获得 3 000 kJ m⁻² yr⁻¹。计算生产者到初级消费者的能量传递效率。
Efficiency = (3 000 ÷ 25 000) × 100 = 12%. This relatively low efficiency can be explained by respiration, egestion, and uneaten parts.
效率 = (3 000 ÷ 25 000) × 100 = 12%。这一相对较低的效率可用呼吸作用、排遗以及未被取食部分的能量损失来解释。
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