📚 Common Mistakes in Biology Exam Questions: IB & OCR | IB OCR 生物易错题精讲
Biology exams demand precision in language and clarity in conceptual understanding. Whether you’re studying for the IB Diploma or OCR A Level, certain misconceptions repeatedly catch students off guard. This article dissects the most common pitfalls in genetics, cell biology, metabolism, and ecology, blending illustrative examples with exam-style reasoning to help you convert confusion into marks.
生物考试对语言精准度和概念理解的清晰度要求极高。不论你正在备考IB文凭课程还是OCR A Level，总有一些误解会反复让学生掉进陷阱。本文详细剖析遗传学、细胞生物学、新陈代谢和生态学中最常见的易错点，结合示范性例题和考试风推理，帮你把曾经的困惑变成实实在在的分数。
1. Diffusion and Osmosis: Directional Confusion | 扩散与渗透：方向迷惑
A frequent error is stating that water moves from a region of low solute concentration to high solute concentration but failing to clarify that it is the water potential gradient, not the solute gradient, that drives osmosis. In exam answers, you must refer to ‘water potential’ (with more negative values meaning lower potential) and specify that water moves from a higher water potential to a lower water potential across a partially permeable membrane.
一个常见错误是说水从溶质浓度低的区域向溶质浓度高的区域移动，却没有阐明驱动渗透的是水势梯度而不是溶质梯度。在考试答案中，你必须提及“水势”（数值越负表示水势越低），并明确指出水是顺着水势梯度从水势较高的地方穿过部分透性膜向水势较低的地方移动。
Consider a typical question: ‘Explain why a plant cell placed in a concentrated salt solution becomes plasmolysed.’ Many students answer, ‘Because water leaves the cell by osmosis,’ without linking to water potential. A full mark response would state: ‘The salt solution has a lower (more negative) water potential than the cell sap. Therefore, water moves out of the cell by osmosis from a region of higher water potential inside the vacuole to a region of lower water potential outside, causing the protoplast to shrink away from the cell wall.’
试想一道典型题目：“解释为什么将植物细胞放在浓盐溶液中会发生质壁分离。”许多学生回答“因为水通过渗透作用离开细胞”，却没有联系到水势。能拿满分的答案应表述为：“盐溶液的水势比细胞液的水势更低（更负）。因此，水通过渗透作用从液泡内水势较高的区域移动到细胞外水势较低的区域，导致原生质体收缩并与细胞壁分离。”
2. Enzyme Kinetics: Initial Rate vs. Equilibrium | 酶动力学：初速率与平衡的混淆
When describing the effect of substrate concentration on the rate of an enzyme-catalysed reaction, students often draw a curve that keeps rising, forgetting that the reaction reaches a maximum velocity (Vmax) when all active sites are saturated. Moreover, they may confuse the ‘initial rate’ with the rate at some other time point. Marks are awarded for specifying that the initial rate is measured at the very start of the reaction, when substrate concentration is effectively constant and product accumulation has not yet caused significant inhibition.
在描述底物浓度对酶促反应速率的影响时，学生们画出的曲线常常持续上升，忘了当所有活性位点都被饱和时，反应会达到最大速率（Vmax）。此外，他们可能把“初速率”与其他时间点的速率搞混。要拿分必须明确指出初速率是在反应刚刚开始时测量的，此时底物浓度基本恒定，产物积累尚未引起明显的抑制作用。
A common graph-labeling mistake: plotting rate on the y-axis and substrate concentration on the x-axis, students continue the curve linearly beyond Vmax. The correct shape is a rectangular hyperbola for Michaelis-Menten kinetics, leveling off at Vmax. In IB and OCR exams, you may be asked to calculate the Michaelis constant (Km) from such a graph. Km is the substrate concentration at which the reaction rate is half of Vmax, and it indicates the affinity of the enzyme for its substrate — a lower Km means higher affinity. Confusing Km with Vmax is a classic error.
一个常见的作图错误是：Y轴为速率，X轴为底物浓度，学生却让曲线越过Vmax后继续线性上升。正确的曲线形状应该是米氏动力学所描述的矩形双曲线，最终在Vmax处趋于平缓。在IB和OCR考试中，你可能会被要求根据这类图计算米氏常数（Km）。Km是反应速率达到Vmax一半时的底物浓度，它反映了酶对底物的亲和力——Km越低表示亲和力越高。混淆Km和Vmax是一个经典错误。
3. Mitosis vs. Meiosis: Chromosome Number and Genetic Variation | 有丝分裂与减数分裂：染色体数目与遗传变异
A persistent misunderstanding is that mitosis produces daughter cells with the same number of chromosomes as the parent cell, but students then mistakenly apply this to meiosis, stating that meiosis also produces identical diploid cells. Meiosis involves two divisions and reduces the chromosome number by half, producing haploid gametes that are genetically distinct due to independent assortment and crossing over. In OCR questions, you might be asked to compare these processes in a table, while IB exams expect you to draw and label chiasmata at prophase I.
一个顽疾误解是，有丝分裂产生的子细胞染色体数与亲代细胞相同，但学生随后错误地将此套用到减数分裂上，声称减数分裂也产生相同的二倍体细胞。实际上减数分裂包含两次分裂，将染色体数目减半，产生单倍体配子，并且因为自由组合和交叉互换，配子在遗传上各不相同。在OCR考题中，你可能被要求用表格比较这两个过程；而IB考试则希望你画出并标注前期的交叉（chiasmata）。
When explaining how meiosis generates variation, avoid simply saying ‘crossing over mixes genes.’ Be precise: during prophase I, non-sister chromatids of homologous chromosomes form bivalents and exchange alleles at chiasmata, creating new combinations of alleles on the same chromosome. Independent assortment refers to the random orientation of homologous pairs at metaphase I, leading to millions of possible combinations in gametes. Another typical pitfall is confusing ‘sister chromatids’ with ‘homologous chromosomes’ — sister chromatids are identical copies of a single chromosome joined by a centromere, while homologous chromosomes are a pair of chromosomes (one maternal, one paternal) that have the same genes but potentially different alleles.
在解释减数分裂如何产生变异时，不要简单说“交叉互换混合了基因”。要精准表述：在前期I，同源染色体的非姐妹染色单体形成二价体，在交叉点交换等位基因，从而在同一条染色体上产生新的等位基因组合。自由组合则指中期I同源染色体对在赤道板上的随机排列，这导致配子中可能出现数百万种组合。另一个常见误区是混淆“姐妹染色单体”和“同源染色体”——姐妹染色单体是由着丝粒连接的一条染色体的两个完全相同的拷贝，而同源染色体是一对染色体（一条来自母方，一条来自父方），它们拥有相同的基因但可能携带不同的等位基因。
4. Genetic Crosses: Linked Genes and Epistasis | 遗传杂交：连锁基因与上位效应
Dihybrid crosses become trickier when genes are linked on the same chromosome. The expected 9:3:3:1 phenotypic ratio only applies if the two genes assort independently — that is, they are on different chromosomes or far apart on the same chromosome. If they are tightly linked, the offspring will largely show parental phenotypes, with a small percentage of recombinants due to crossing over. In an exam, if you obtain a ratio like 60:9:11:62, recognise it as linkage with crossing over, and be ready to calculate the recombination frequency.
当基因位于同一条染色体上连锁时，双因子杂交会变得更复杂。预期的9:3:3:1表型比例仅在两个基因独立分配时才适用——即它们位于不同染色体上，或虽在同一染色体上但相距甚远。如果它们紧密连锁，后代将大部分表现出亲本表型，只有因交叉互换产生的小部分重组子代。考试中如果你得到诸如60:9:11:62的比例，要能识别出这是有交换的连锁，并准备好计算重组频率。
Epistasis is another common source of error. Students often mislabel a gene as ‘dominant’ when it is actually epistatic. In recessive epistasis, the homozygous recessive allele of one gene masks the expression of another gene, yielding a 9:3:4 ratio (e.g., coat colour in Labrador retrievers). In dominant epistasis, a dominant allele at one locus masks the other gene, producing a 12:3:1 ratio. Do not rely on memorising ratios alone; be able to explain the molecular mechanism: perhaps Gene A codes for an enzyme that produces a pigment precursor, and Gene B codes for an enzyme that modifies the pigment; if Gene A is non-functional, no pigment is produced regardless of Gene B’s alleles.
上位效应是另一个常见错误源。学生常把一个基因错误地标记为“显性”，实际上它却是上位的。在隐性上位效应中，一个基因的纯合隐性等位基因会掩盖另一个基因的表达，产生9:3:4的比例（例如拉布拉多犬的毛色）。在显性上位效应中，一个基因座上的显性等位基因会掩盖另一个基因，产生12:3:1的比例。不要只靠记比例；要能解释分子机制：可能基因A编码一种产生色素前体的酶，而基因B编码一种修饰色素的酶；如果基因A功能丧失，那么不论基因B携带什么等位基因，都不会产生色素。
5. DNA Replication and Transcription: Direction Misconceptions | DNA复制与转录：方向性误解
A common exam blunder is writing that DNA polymerase synthesises new strands in a 3′ to 5′ direction. This is incorrect. DNA polymerase can only add nucleotides to the 3′ end of the growing strand, so synthesis always proceeds in a 5′ to 3′ direction. The template strand is read in the 3′ to 5′ direction. Because the two template strands are antiparallel, one new strand (leading) is synthesised continuously, while the other (lagging) is made in short Okazaki fragments that are later joined by DNA ligase.
一个常见的考试失误是写DNA聚合酶以3’至5’方向合成新链。这是错的。DNA聚合酶只能在生长链的3’端添加核苷酸，因此合成方向始终是5’至3’。模板链则是以3’至5’的方向被读取。由于两条模板链是反向平行的，其中一条新链（前导链）连续合成，而另一条链（后随链）则以短的冈崎片段合成，之后再被DNA连接酶连接起来。
Similarly, during transcription, RNA polymerase reads the template DNA strand in the 3′ to 5′ direction and synthesises the mRNA transcript in the 5′ to 3′ direction. Students often draw the mRNA as complementary to the coding (sense) strand, but it is actually complementary to the template (antisense) strand and has the same sequence as the coding strand, with U replacing T. When an exam asks for the sequence of the transcribed mRNA, always check which strand is provided.
类似地，在转录过程中，RNA聚合酶以3’至5’的方向阅读模板DNA链，并沿着5’至3’的方向合成mRNA转录本。学生们经常把mRNA画成与编码链（有义链）互补，但实际上它与模板链（反义链）互补，并且序列与编码链相同，只是用U替代了T。当考题要求写出转录出的mRNA序列时，一定要检查提供的是哪条链。
6. Respiration: Substrate-level Phosphorylation and the Four Stages | 细胞呼吸：底物水平磷酸化与四大阶段
Many students confuse substrate-level phosphorylation with oxidative phosphorylation. Substrate-level phosphorylation occurs when a phosphate group is transferred directly from a phosphorylated intermediate to ADP, forming ATP; this happens in glycolysis (conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate, and phosphoenolpyruvate to pyruvate) and in the Krebs cycle (succinyl-CoA to succinate). Oxidative phosphorylation, by contrast, uses energy released from the electron transport chain to create a proton gradient that drives ATP synthase — it occurs only in the presence of oxygen on the inner mitochondrial membrane.
许多学生混淆了底物水平磷酸化和氧化磷酸化。底物水平磷酸化是指磷酸基团直接从磷酸化的中间产物转移到ADP上形成ATP；这发生在糖酵解中（1,3-二磷酸甘油酸转化为3-磷酸甘油酸，以及磷酸烯醇式丙酮酸转化为丙酮酸）以及克雷布斯循环中（琥珀酰CoA转化为琥珀酸）。与之相反，氧化磷酸化利用电子传递链释放的能量建立质子梯度，驱动ATP合酶工作——它只发生在有氧条件下的线粒体内膜上。
Another tricky area is the link reaction. Students often forget that pyruvate is transported into the mitochondrial matrix, where it undergoes decarboxylation (CO₂ removed) and oxidation (NAD⁺ reduced to NADH) to form acetyl-CoA, which then combines with oxaloacetate in the Krebs cycle. If asked to state the number of carbon atoms in citrate, remember it is a 6-carbon compound, regenerating the 4-carbon oxaloacetate. Also, be able to distinguish between the roles of NADH and FADH₂: both carry electrons to the electron transport chain, but FADH₂ enters at a lower energy level, yielding fewer ATP molecules.
另一个棘手之处是连接反应。学生常常忘记丙酮酸先被运进线粒体基质，在那里进行脱羧（释放CO₂）和氧化（NAD⁺还原为NADH），形成乙酰CoA，然后乙酰CoA在克雷布斯循环中与草酰乙酸结合。如果被问到柠檬酸中的碳原子数，要记住它是6碳化合物，循环最终再生了4碳的草酰乙酸。另外，要能区分NADH和FADH₂的作用：两者都将电子带入电子传递链，但FADH₂的能级较低，最终产生的ATP分子数也更少。
7. Photosynthesis: Calvin Cycle and Photorespiration | 光合作用：卡尔文循环与光呼吸
Students often incorrectly locate the Calvin cycle in the thylakoid membrane, but it takes place in the stroma of the chloroplast. The light-dependent reactions occur on the thylakoid membrane, where photosystems II and I absorb light energy to split water, generate ATP, and reduce NADP⁺ to NADPH. The Calvin cycle uses ATP and NADPH to fix CO₂ into glycerate 3-phosphate (GP) and then reduce GP to triose phosphate (TP), which can be used to regenerate RuBP or produce glucose. A missing detail: the enzyme rubisco catalyses the fixation of CO₂ onto RuBP, but it can also fix O₂, leading to photorespiration — a wasteful process especially in C3 plants under hot, dry conditions.
学生们常错误地把卡尔文循环定位在类囊体膜上，但它实际发生在叶绿体基质中。光依赖反应在类囊体膜上进行，光系统II和I吸收光能，分解水，生成ATP，并将NADP⁺还原为NADPH。卡尔文循环利用ATP和NADPH将CO₂固定到甘油酸-3-磷酸（GP）中，接着将GP还原为磷酸丙糖（TP），TP可用于再生RuBP或生成葡萄糖。一个易遗漏的细节：Rubisco酶催化CO₂固定到RuBP上，但它也能固定O₂，导致光呼吸——一种浪费的过程，在炎热干燥条件下的C3植物中尤为突出。
In OCR and IB essays, questions may ask you to explain why plants need both ATP and NADPH for the Calvin cycle. ATP provides the energy for the reduction of GP to TP and for the regeneration of RuBP, while NADPH provides the reducing power (hydrogen atoms). A common mistake is to say that the Calvin cycle is ‘light-independent’; it is more accurate to call it the ‘light-independent stage’ because it relies on the products of the light-dependent stage and several Calvin cycle enzymes are indirectly activated by light. Moreover, the term ‘dark reaction’ is outdated and discouraged.
在OCR和IB的论述题中，可能会问你为什么卡尔文循环既需要ATP也需要NADPH。ATP为GP还原成TP以及RuBP的再生提供能量，而NADPH则提供还原力（氢原子）。一个常见错误是说卡尔文循环“不需要光”；更准确的叫法是“光非依赖阶段”，因为它依赖于光依赖阶段的产物，并且几个卡尔文循环中的酶是间接由光激活的。再者，“暗反应”这个术语已经过时，考试中不应使用。
8. Immunology: B Cells, T Cells and the Specificity of Antibodies | 免疫学：B细胞、T细胞与抗体的特异性
Students frequently confuse the roles of B lymphocytes and T lymphocytes. B cells are responsible for humoral immunity, producing antibodies that bind to specific antigens on pathogens in the blood and lymph. Helper T cells stimulate B cells to divide and differentiate into plasma cells and memory cells, while cytotoxic T cells destroy infected body cells by releasing perforin and granzymes. A significant error is to say that antibodies kill pathogens directly — they do not. Antibodies neutralise pathogens by agglutination, precipitation, or by marking them for destruction by phagocytes (opsonisation).
学生们经常弄混B淋巴细胞和T淋巴细胞的功能。B细胞负责体液免疫，产生抗体与血液和淋巴中病原体的特定抗原结合。辅助性T细胞刺激B细胞分裂并分化为浆细胞和记忆细胞，而细胞毒性T细胞则通过释放穿孔素和颗粒酶来摧毁被感染的体细胞。一个严重错误是说抗体直接杀死病原体——它们并不直接杀死。抗体通过凝集、沉淀或标记病原体以供吞噬细胞消灭（调理作用）的方式来中和病原体。
Monoclonal antibodies are another area littered with misconceptions. Exam answers should explain that a mouse is injected with an antigen, its spleen cells (B cells) are fused with myeloma cells to produce hybridomas, which are then screened for the desired antibody. The resulting hybridoma cells can divide indefinitely and secrete a single type of antibody specific to that antigen. Common slips: forgetting that hybridomas are made by fusion, not just by culturing B cells; or thinking that monoclonal antibodies are identical because they come from a single clone of plasma cells — true, but must be linked to the fact they all recognise the same epitope.
单克隆抗体是另一个充满误解的领域。考试答案应解释为：先给小鼠注射某种抗原，取出其脾细胞（B细胞）与骨髓瘤细胞融合，生成杂交瘤细胞，再从中筛选出能产生所需抗体的细胞。由此得到的杂交瘤细胞可以无限分裂，分泌出单一且对该抗原特异的抗体。常见失误：忘记杂交瘤细胞是通过融合而非单纯培养B细胞得来的；或者认为单克隆抗体之所以相同是因为它们来自一个浆细胞克隆——这点没错，但必须联系到它们都识别同一种抗原表位。
9. Ecological Succession and Sampling: Interpreting Data | 生态演替与取样：解读数据
In questions about succession, students often describe pioneer species like lichens and mosses breaking down rock, but they fail to mention that these species change the abiotic environment (e.g., soil formation, increase in organic matter), making conditions more favourable for subsequent species. The climax community is not always a forest — in some biomes it can be grassland or bog, depending on climate and other factors. Also, do not confuse primary succession (starting on bare rock) with secondary succession (after a disturbance on existing soil).
在关于演替的题目中，学生们常描述地衣和苔藓这类先锋物种分解岩石，却没有提到这些物种会改变非生物环境（比如形成土壤、增加有机物），为后来的物种创造更有利的条件。顶极群落并不总是森林——在某些生物群落中，它可能是草原或沼泽，取决于气候和其他因素。另外，切勿混淆原生演替（从裸岩上开始）和次生演替（在已有土壤上经扰动后发生）。
Sampling techniques using quadrats and transects often cause confusion when calculating biodiversity indices. Simpson’s Diversity Index (D = 1 – (∑(n/N)²)) measures species diversity, taking both richness and evenness into account. A common error in interpreting results is to compare only the number of species (richness) and conclude one habitat is more diverse, while the index may show the other habitat is more diverse due to greater evenness. When using a belt transect to study zonation, be explicit that the transect is laid perpendicular to the environmental gradient, and quadrats are placed at regular intervals.
使用样方和样线进行取样时，计算生物多样性指数常常容易混乱。辛普森多样性指数（D = 1 – (∑(n/N)²)）测量物种多样性，同时考虑了丰富度和均匀度。在解读结果时，一个常见错误是仅仅比较物种数量（丰富度）就下结论说某个生境更多样，但实际上由于更高的均匀度，辛普森指数可能显示另一个生境更为多样。当使用带状样线研究成带现象时，要明确样线应垂直于环境梯度布设，并沿着样线等距放置样方。
10. Graphing and Statistical Analysis: Common Pitfalls in Data Presentation | 作图与统计分析：数据展示中的常见陷阱
In IB Internal Assessment and OCR practical components, graph drawing carries substantial marks but is frequently mishandled. Axes must be labelled with both variable name and unit (e.g., ‘Rate of reaction / s⁻¹’), linear scales must be consistent and cover more than half of the graph paper, and points must be plotted with small, precise crosses. A line of best fit should be smooth, ignoring anomalous points. Students often connect point-to-point with straight lines, which is only acceptable for certain graph types explicitly requested.
在IB内部评估和OCR实验部分中，绘图占有相当分数，但常被错误处理。坐标轴必须标记变量名称和单位（例如“反应速率 / s⁻¹”），线性刻度必须均匀一致且占据超过半张坐标纸的范围，数据点要用细小的精准十字叉绘制。最佳拟合线应该是平滑的，忽略异常点。学生们经常用直线把点逐点连接起来，这只有在题目明确要求某类图表时才可接受。
Statistical tests like the t-test and chi-squared test appear in both specifications. For chi-squared, students often fail to state the null hypothesis correctly: ‘There is no significant difference between observed and expected frequencies.’ They also may forget to calculate degrees of freedom (number of categories minus 1) and to compare the calculated value with the critical value at p=0.05. If the calculated value exceeds the critical value, you reject the null hypothesis; the difference is statistically significant. A common mistake is saying ‘accept the null hypothesis’ when the result is not significant; you should say ‘fail to reject the null hypothesis.’
t检验和卡方检验等统计方法在两大课程中都有出现。对于卡方检验，学生们经常未能正确写出零假设：“观察到的频数与其期望频数之间不存在显著差异。”他们还可能忘记计算自由度（分类数减一），并忘了将计算值与p=0.05时的临界值进行比较。如果计算值大于临界值，就拒绝零假设，意味着差异在统计上显著。一个常见错误是在结果不显著时说“接受零假设”；应该说“未能拒绝零假设”。
11. Gene Technology: PCR, Gel Electrophoresis, and GMOs | 基因技术：PCR、凝胶电泳与转基因生物
When describing the polymerase chain reaction (PCR), students often miss the three temperature-dependent steps: denaturation (95°C separates DNA strands), annealing (55-65°C allows primers to bind), and extension (72°C for Taq polymerase to synthesise new strands). A classic error is stating that DNA polymerase from E. coli is used; in fact, Taq polymerase from Thermus aquaticus is used because it is thermostable and does not denature at the high temperatures used. Also, primers are short single-stranded DNA sequences complementary to the target region, not random.
在描述聚合酶链式反应（PCR）时，学生们常常遗漏三个温度依赖步骤：变性（95°C分开DNA双链）、退火（55-65°C使引物结合）和延伸（72°C让Taq聚合酶合成新链）。一个经典错误是说使用了大肠杆菌的DNA聚合酶；事实上使用的是嗜热水生菌的Taq聚合酶，因为它耐热，不会在高温下变性。同时，引物是与靶区域互补的短单链DNA序列，并非随机序列。
Gel electrophoresis questions commonly ask for interpretation of DNA banding patterns. DNA is negatively charged and moves toward the positive anode; smaller fragments migrate faster and further through the gel. A common confusion is to think the thickest band always represents the longest fragment; thickness indicates the quantity of DNA, while position indicates size. When explaining how GMOs are created, specify the roles of restriction enzymes cutting sticky ends, DNA ligase joining fragments, and vectors (like plasmids) carrying the gene of interest into the host cell. Do not forget to mention the use of antibiotic resistance marker genes for selection.
凝胶电泳的题目常要求解读DNA带型图谱。DNA带负电，会向正极移动；较小的片段在凝胶中迁移得更快更远。一个常见混淆是以为最粗的带总是代表最长的片段；粗细反映的是DNA的数量，而位置才反映大小。在解释如何创制转基因生物时，要具体说明限制性内切酶切割出黏性末端、DNA连接酶连接片段、载体（如质粒）将目标基因带入宿主细胞这些作用。别忘了提到用抗生素抗性标记基因来进行筛选。
12. Transport in Plants: Xylem, Phloem, and the Cohesion-Tension Theory | 植物运输：木质部、韧皮部与内聚力-张力学说
The cohesion-tension theory of water transport in xylem vessels is frequently misapplied. Students remember that transpiration at the leaf creates tension, but they sometimes state that the tension pulls water up continuously from the roots by active transport. In fact, the tension generated by evaporation pulls the water column under negative pressure; cohesion between water molecules (due to hydrogen bonding) and adhesion to the xylem walls (capillarity) prevent the column from breaking. There is no active pumping in xylem. Root pressure can push water up at night, but it is a minor force.
木质部导管中水分运输的内聚力-张力学说常被误用。学生们记得叶片蒸腾作用会产生张力，但他们有时说这种张力是通过主动运输把水从根部连续拉上来的。实际上，蒸发产生的张力以负压形式牵拉水柱；水分子之间的内聚力（源于氢键）以及水与木质部壁的附着力（毛细现象）防止水柱断裂。木质部中并没有主动泵送作用。根压在夜间可能推动水分上升，但那只是次要力量。
Phloem translocation is explained by the mass flow hypothesis. Source cells (e.g., leaves) actively load sucrose into sieve tubes, lowering the water potential. Water enters by osmosis from adjacent xylem, creating a high hydrostatic pressure. At the sink (e.g., roots), sucrose is unloaded and the water potential rises, so water leaves the phloem, reducing pressure. The pressure gradient drives the bulk flow of phloem sap from source to sink. A typical exam mistake is to describe this as a bidirectional movement along the same individual sieve tube; actually, flow is always from source to sink, and individual sieve tubes might carry sap in different directions at different times.
韧皮部的输导作用由压力流动假说来解释。源端细胞（如叶片）主动装载蔗糖进入筛管，降低了水势。水分从相邻木质部以渗透方式进入，形成高静水压。在库端（如根部），蔗糖被卸载，水势上升，于是水分离开韧皮部，压力降低。压力梯度驱动韧皮部汁液从源端向库端整体流动。一个典型的考试错误是将此描述为同一筛管中可同时进行双向运输；实际上，流动总是从源到库，而且同一筛管在不同时间可能沿不同方向运输汁液，但不可以在同一时刻双向流动。
Published by TutorHao | Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导