📚 IB Biology: Calculation Practice | IB 生物：计算题专项训练
Quantitative skills are an integral part of the IB Biology course. You are expected to perform calculations related to microscopy, water potential, enzyme rates, cell counting, dilutions, genetics, statistics, energy flow, population ecology, and molecular biology. This article provides a focused revision on the most common calculation question types, with worked examples in both English and Chinese to help you master the techniques.
定量技能是IB生物课程的重要组成部分。你需要完成显微镜、水势、酶反应速率、细胞计数、稀释、遗传学、统计学、能量流动、种群生态学和分子生物学等相关计算。本文针对最常见的计算题型进行专项训练，并提供中英双语例题与解答，帮助你掌握解题方法。
1. Magnification and Actual Size | 放大倍数与实际大小
Microscope calculations are fundamental in IB Biology. The core formula is:
Actual size = Measured size / Magnification
显微镜计算是IB生物的基础。核心公式为：实际大小 = 测量大小 / 放大倍数。
You must ensure all lengths are in the same unit, usually micrometres (µm). 1 mm = 1000 µm.
你必须保证所有长度单位统一，通常为微米 (µm)。1 mm = 1000 µm。
Example: A student measures a cell image as 20 mm when printed at a magnification of ×400. Calculate the actual size.
示例：一位学生在放大×400的条件下打印细胞图像，测量长度为20 mm。请计算实际大小。
Solution: Actual size = 20 mm / 400 = 0.05 mm = 50 µm.
解答：实际大小 = 20 mm / 400 = 0.05 mm = 50 µm。
2. Scale Bars and Cell Size | 比例尺与细胞大小
Many IB exam images include a scale bar. If the scale bar measures a known length on the image, use proportions.
许多IB考试图像中包含比例尺。如果比例尺在图像上的测量长度已知，可以用比例计算。
For instance, a scale bar of 15 mm on the diagram represents 5 µm. A mitochondrion measures 6 mm on the same diagram. Its actual length = 6 mm × (5 µm / 15 mm) = 2 µm.
例如，图像上15 mm长的比例尺代表5 µm。同一图像中线粒体测量长度为6 mm，其实际长度 = 6 mm × (5 µm / 15 mm) = 2 µm。
Always work with the ratio: actual size / measured size = scale actual / scale measured.
始终使用比例关系：实际大小 / 测量大小 = 比例尺实际值 / 比例尺测量值。
3. Water Potential (Ψ) | 水势
Water potential determines the direction of water movement. It is given by:
Ψ = Ψₛ + Ψₚ
水势决定水分移动的方向。计算公式为：Ψ = Ψₛ + Ψₚ。
Ψₛ is solute potential (osmotic potential); pure water has Ψₛ = 0. Ψₚ is pressure potential (turgor pressure). Units are usually kPa or MPa.
Ψₛ 是溶质势（渗透势），纯水的 Ψₛ = 0；Ψₚ 是压力势（膨压）。单位通常为 kPa 或 MPa。
Example: A plant cell has a solute potential of -750 kPa and a pressure potential of 500 kPa. Calculate the cell’s water potential.
示例：一个植物细胞的溶质势为 -750 kPa，压力势为 500 kPa。计算该细胞的水势。
Solution: Ψ = -750 + 500 = -250 kPa. Water will tend to enter the cell from a solution of higher water potential (e.g. 0 kPa).
解答：Ψ = -750 + 500 = -250 kPa。水分会从水势更高的溶液（如 0 kPa）进入细胞。
4. Enzyme Activity and Initial Rate | 酶活性与初始反应速率
Initial reaction rate is the rate at the very beginning of the reaction, before substrate becomes limiting. It is obtained by drawing a tangent to the progress curve at time zero.
初始反应速率是反应刚刚开始、底物还未成为限制因素时的速率。通过在产品生成量-时间曲线上零时刻处作切线求得。
Calculate the slope: Initial rate = Change in product concentration / Change in time.
计算斜率：初始速率 = 产物浓度变化 / 时间变化。
Example: The tangent shows that absorbance increases by 0.30 units in 120 s. If the calibration indicates 1 absorbance unit = 0.5 µmol, the initial rate = (0.30 × 0.5) µmol / 120 s = 0.00125 µmol s⁻¹.
示例：切线显示吸光度在120 s内增加了0.30单位。若校准曲线表明1吸光度单位 = 0.5 µmol，则初始速率 = (0.30 × 0.5) µmol / 120 s = 0.00125 µmol s⁻¹。
5. Haemocytometer Cell Counting | 血球计数板细胞计数
A haemocytometer is used to count cells under the microscope. The formula for cell concentration is:
Cells per mL = (Total cells counted / Number of squares) × Dilution factor × 10⁴
血球计数板用于在显微镜下计数细胞。细胞浓度的计算公式为：每毫升细胞数 = (计数细胞总数 / 方格数) × 稀释倍数 × 10⁴。
The factor 10⁴ arises because each square has a volume of 0.1 mm³, and 1 mL = 1000 mm³. Typical IB problems use the central grid (5 × 5 squares).
因子 10⁴ 是因为每个大格的体积为 0.1 mm³，而 1 mL = 1000 mm³。典型的IB题目使用中央网格（5 × 5 大格）。
Example: A student counts 120 cells in 5 large squares with a dilution factor of 1. Cell concentration = (120 / 5) × 1 × 10⁴ = 2.4 × 10⁵ cells mL⁻¹.
示例：某学生在5个大格中共计120个细胞，未稀释。细胞浓度 = (120 / 5) × 1 × 10⁴ = 2.4 × 10⁵ 个/mL。
6. Dilution Calculations | 稀释计算
Serial dilutions and simple dilutions are common in enzyme and microbiology practicals. The equation is:
C₁V₁ = C₂V₂
系列稀释和简单稀释在酶与微生物实验中很常见。使用公式：C₁V₁ = C₂V₂。
C is concentration, V is volume. When performing a 1 in 10 dilution, the dilution factor is 10⁻¹. Repeating this three times gives 10⁻³.
C 为浓度，V 为体积。进行1:10稀释时，稀释因子为 10⁻¹；连续三次即 10⁻³。
Example: You need 100 mL of 0.5 mol dm⁻³ sucrose from 2 mol dm⁻³ stock. Volume of stock = (0.5 × 100) / 2 = 25 mL. Add 75 mL water.
示例：需从2 mol dm⁻³ 的储备液配制100 mL 0.5 mol dm⁻³ 蔗糖溶液。所需储备液体积 = (0.5 × 100) / 2 = 25 mL，加水75 mL。
7. Hardy-Weinberg Equation | 哈代-温伯格方程
The Hardy-Weinberg principle predicts allele frequencies when a population is not evolving. The two key equations are:
p + q = 1
p² + 2pq + q² = 1
哈代-温伯格原理用于预测不发生进化时种群的等位基因频率。两个关键方程为：p + q = 1 和 p² + 2pq + q² = 1。
p = frequency of dominant allele, q = frequency of recessive allele. p² = frequency of homozygous dominant, 2pq = heterozygous, q² = homozygous recessive.
p = 显性等位基因频率，q = 隐性等位基因频率。p² = 显性纯合子频率，2pq = 杂合子频率，q² = 隐性纯合子频率。
Example: In a population, 1 in 2500 individuals has a recessive condition. Find the carrier frequency. q² = 1/2500, so q = 1/50 = 0.02. p = 0.98. Carrier frequency 2pq = 2 × 0.98 × 0.02 = 0.0392 (about 3.92%).
示例：某人群中，每2500人有1人患隐性遗传病。求携带者频率。q² = 1/2500，q = 1/50 = 0.02，p = 0.98。携带者频率 2pq = 2 × 0.98 × 0.02 = 0.0392（约 3.92%）。
8. Chi-squared Test | 卡方检验
The chi-squared test (χ²) is used to determine if there is a significant difference between observed and expected results. The test statistic is calculated by:
χ² = Σ (O – E)² / E
卡方检验 (χ²) 用于判断观测值与预期值之间是否存在显著差异。统计量计算公式为：χ² = Σ (O – E)² / E。
O = observed frequency, E = expected frequency. Compare the calculated χ² with a critical value from the table (at p=0.05) for the appropriate degrees of freedom (df).
O = 观测频率，E = 预期频率。将计算得到的 χ² 值与卡方分布表（p=0.05）中的临界值比较，自由度 (df) 对应合适值。
Example: You test a hypothesis of a 1:1 ratio in the offspring of a genetic cross. Observed: 35 purple, 25 white. Total 60. Expected: 30 each. χ² = (35-30)²/30 + (25-30)²/30 = 25/30 + 25/30 ≈ 1.67. df=1, critical value = 3.84. Since 1.67 < 3.84, the difference is not significant — accept the hypothesis.
示例：检验遗传杂交中1:1的比例假设。观测值：35紫花、25白花。总数60。预期值各30。χ² = (35-30)²/30 + (25-30)²/30 = 25/30 + 25/30 ≈ 1.67。自由度=1，临界值=3.84。因为1.67 < 3.84，差异不显著，接受原假设。
9. Energy Transfer and Trophic Efficiency | 能量传递与营养级效率
Ecological efficiencies are usually calculated as a percentage:
Efficiency = (Energy in higher trophic level / Energy in lower trophic level) × 100%
生态效率通常以百分比计算：效率 = (较高营养级的能量 / 较低营养级的能量) × 100%。
This can refer to gross production or biomass. Most IB questions give data in kJ m⁻² yr⁻¹.
此处的能量可以指总产量或生物量。大多数IB题目给出的数据单位为 kJ m⁻² yr⁻¹。
Example: Plants absorb 8000 kJ m⁻² yr⁻¹ of light energy. Primary consumers receive 500 kJ m⁻² yr⁻¹. Efficiency = (500 / 8000) × 100% = 6.25%.
示例：植物吸收8000 kJ m⁻² yr⁻¹ 的光能，初级消费者获得500 kJ m⁻² yr⁻¹。效率 = (500 / 8000) × 100% = 6.25%。
10. Population Growth | 种群增长
Population growth rate can be expressed as:
Growth rate = (Births – Deaths) / Initial population
种群增长率可表示为：增长率 = (出生数 – 死亡数) / 初始种群数量。
For exponential growth, the per capita growth rate r is constant: dN/dt = rN. In logistic growth, the rate slows as carrying capacity K is approached.
指数增长时，人均增长率r恒定：dN/dt = rN。逻辑斯谛增长中，当种群接近环境容纳量K时，增速放缓。
Example: A population of 2000 mice has 400 births and 300 deaths in one year. r = (400 – 300) / 2000 = 0.05 (5% per year). After one year, population would be roughly 2000 × 1.05 = 2100.
示例：一个2000只小鼠的种群，一年内出生400只，死亡300只。r = (400 – 300) / 2000 = 0.05（每年5%）。一年后种群数量约为 2000 × 1.05 = 2100。
11. Standard Deviation and Standard Error | 标准差与标准误
Standard deviation (s) measures the spread of data around the mean. Standard error (SEM) estimates the precision of the sample mean.
标准差 (s) 衡量数据围绕平均值的分散程度。标准误 (SEM) 估计样本平均值的精确度。
Formulae:
s = √[ Σ(x – x̄)² / (n – 1) ]
SEM = s / √n
公式：s = √[ Σ(x – x̄)² / (n – 1) ]，标准误 SEM = s / √n。
In IB exams you are often given s and asked to calculate SEM, then represent it as error bars. When SEM error bars do not overlap, the difference may be significant.
IB考试常给出标准差s，要求计算标准误，并画出误差棒。当标准误的误差棒不重叠时，差异可能显著。
Example: For a sample of n=16, s=8. SEM = 8 / √16 = 2. A 95% confidence interval is approximately mean ± 2 × SEM.
示例：样本量 n=16，s=8。标准误 = 8 / √16 = 2。95% 置信区间约为平均值 ± 2 × SEM。
12. DNA Length from Base Pairs | 从碱基对数计算DNA长度
In B-form DNA, each base pair is separated by 0.34 nm along the helix axis. To find the total length of a DNA molecule:
Length (nm) = Number of base pairs × 0.34 nm
在B型DNA中，每个碱基对沿双螺旋轴相距0.34 nm。计算DNA分子的总长度：长度 (nm) = 碱基对数 × 0.34 nm。
One complete turn of the helix contains 10 base pairs and spans 3.4 nm. The diameter of the double helix is 2 nm.
双螺旋的一整圈包含10个碱基对，跨距3.4 nm。双螺旋直径为2 nm。
Example: The human haploid genome contains about 3 × 10⁹ base pairs. Total DNA length per cell ≈ 3 × 10⁹ × 0.34 nm = 1.02 × 10⁹ nm = 1.02 m. That is longer than many cells, which is why DNA is tightly packed.
示例：人类单倍体基因组约有 3 × 10⁹ 碱基对。每个细胞DNA总长度 ≈ 3 × 10⁹ × 0.34 nm = 1.02 × 10⁹ nm = 1.02 m。这比许多细胞的直径长得多，因此DNA必须高度折叠。
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