📚 Year 10 CAIE Biology: Unit Test Mock Paper Analysis | Year 10 CAIE 生物：单元测试模拟卷解析
Mock unit tests are one of the most effective ways to prepare for your CAIE Biology assessments. By working through typical exam-style questions and analysing the reasoning behind each answer, you strengthen your understanding of key concepts and avoid common pitfalls. This article takes you through ten high-frequency topic areas, presenting a sample question from each, followed by a detailed breakdown of the correct approach and underlying theory. Whether you are checking your revision progress or learning how to structure a 4‑mark answer, this analysis will sharpen your exam technique.
单元模拟测试是备考 CAIE 生物考试最有效的方式之一。通过演练典型的考试题目并解析每道答案背后的逻辑，你可以加深对核心概念的理解，避开常见失分点。本文覆盖十个高频考查主题，每个主题都配有一道样题，随后详细拆解正确的解题思路和理论基础。无论你是想检测复习进度，还是学习如何组织一道 4 分简答题，这份解析都能帮助你优化应试技巧。
1. Cells and Magnification Calculations | 细胞与放大倍数计算
A common Year 10 question states: “A student observes a plant cell under a light microscope at ×400 magnification. The image length on the drawing is 60 mm. Calculate the actual length of the cell in micrometres (µm). Show your working.” The formula sheet reminds you that actual size = image size ÷ magnification.
一道常见的十年级考题：“一名学生用光学显微镜在 400 倍下观察植物细胞。绘图上的图像长度为 60 mm。计算细胞的真实长度，以微米（µm）表示，并写出计算过程。”公式表提示：实际尺寸 = 图像尺寸 ÷ 放大倍数。
First, express the image size in the same unit you need for the actual size. Since the final answer must be in µm and 1 mm = 1000 µm, the image length is 60 × 1000 = 60 000 µm. Then apply the formula: actual size = 60 000 µm ÷ 400 = 150 µm. Always write the steps: conversion, substitution, calculation and unit. A common mistake is forgetting to convert mm to µm, which gives a tiny and incorrect value.
首先，将图像尺寸换算成与实际尺寸一致的单位。因为最终答案要以 µm 表示，而 1 mm = 1000 µm，图像长度为 60 × 1000 = 60 000 µm。再代入公式：实际尺寸 = 60 000 µm ÷ 400 = 150 µm。务必写出步骤：换算、代入、计算和单位。常见错误是忘记将 mm 换算成 µm，结果会得出非常小且不正确的数值。
If the question involves a scale bar, measure the bar length on the diagram, convert to suitable units, and use the same formula. For example, a 10 µm scale bar drawn as 20 mm represents a magnification of ×2000, but the actual size of the bar remains 10 µm. Use the relationship: actual size of object = (size of object on diagram ÷ size of scale bar on diagram) × scale bar value.
如果题目涉及比例尺，先测量图中比例尺的长度，换算为合适单位，再使用相同的公式。例如，一个 10 µm 比例尺画成 20 mm，代表放大倍数为 2000 倍，但比例尺的真实尺寸仍是 10 µm。使用关系式：物体实际尺寸 =（图上物体尺寸 ÷ 图上比例尺尺寸）× 比例尺数值。
2. Enzyme Activity and Temperature | 酶活性与温度
A typical enzyme question provides a graph of rate of reaction against temperature for amylase. It asks: “Explain why the rate of reaction increases up to the optimum temperature and then falls sharply after 40 °C.” The answer must link enzyme structure, kinetic energy and denaturation.
典型的酶考题会给出淀粉酶反应速率随温度变化的曲线，问：“解释为什么反应速率在达到最适温度前上升，而在 40 °C 后急剧下降。”答案必须联系酶的结构、动能和变性作用。
As temperature rises from 0 °C to the optimum, the enzyme and substrate molecules gain more kinetic energy. They move faster, collide more often, and are more likely to form enzyme‑substrate complexes, so the rate increases. Once the optimum temperature is exceeded, the weak bonds (hydrogen, ionic and hydrophobic interactions) holding the tertiary structure of the enzyme break. The active site changes shape irreversibly, so the substrate can no longer fit. This is denaturation, and the rate drops sharply.
当温度从 0 °C 升至最适温度时，酶和底物分子获得更多动能。它们运动速度加快，碰撞更频繁，更容易形成酶‑底物复合物，因此反应速率上升。一旦超过最适温度，维持酶三级结构的弱键（氢键、离子键和疏水作用）发生断裂。活性位点不可逆地改变形状，底物无法再契合。这就是变性，反应速率急剧下降。
For a 4‑mark answer, state: (1) more kinetic energy, (2) more successful collisions, (3) bonds break above optimum, (4) active site shape changes and enzyme is denatured. Do not use the term ‘killed’ because enzymes are not alive; always say ‘denatured’. Also, note that at very low temperatures enzymes are inactivated but not denatured; they resume activity when temperature rises back to optimum.
在 4 分答案中，要写出：(1) 更多动能，(2) 更多有效碰撞，(3) 超过最适温度后键断裂，(4) 活性位点形状改变，酶变性。不要使用“杀死”一词，因为酶不是活的；始终说“变性”。另外要注意，极低温度下酶只是失活而未变性，温度回升到最适范围后酶活性可以恢复。
3. Food Tests: Reagents, Methods and Colour Changes | 食物测试：试剂、方法与颜色变化
Question: “Describe how you would test a sample of milk for the presence of reducing sugar, starch and protein. Include the name of each reagent and the expected positive result.” This tests your knowledge of standard food tests: Benedict’s, iodine and biuret.
题目：“描述你如何检测一份牛奶样品中是否含有还原糖、淀粉和蛋白质。写出每种试剂的名称和预期的阳性结果。”这考查你对标准食物测试的掌握：本尼迪克特试剂、碘液和双缩脲试剂。
For reducing sugar, add an equal volume of Benedict’s solution to the food sample in a test tube. Heat the mixture in a water bath at about 80 °C for 5 minutes. A positive result shows a colour change from blue to green, yellow, orange or brick‑red, depending on the concentration of reducing sugar. Milk contains lactose, a reducing sugar, so a brick‑red precipitate will form.
检测还原糖：在试管中向食品样品加入等体积的本尼迪克特溶液。将混合液放入约 80 °C 的水浴中加热 5 分钟。阳性结果会从蓝色变为绿色、黄色、橙色或砖红色，具体取决于还原糖浓度。牛奶含有乳糖，是一种还原糖，因此会形成砖红色沉淀。
To test for starch, add a few drops of iodine solution (iodine in potassium iodide) directly to a room‑temperature sample. If starch is present, the iodine changes from yellow‑brown to blue‑black. Milk usually does not contain starch, so no colour change is expected. For protein, add a few drops of biuret reagent (sodium hydroxide followed by copper sulfate) to the sample; a positive test turns from pale blue to purple or mauve. Milk contains casein protein, giving a purple result.
检测淀粉：向室温样品中直接加入几滴碘液（碘溶于碘化钾）。如果存在淀粉，碘液会从黄棕色变为蓝黑色。牛奶通常不含淀粉，因此不会变色。检测蛋白质：向样品中加入几滴双缩脲试剂（氢氧化钠溶液和硫酸铜溶液），阳性测试会从淡蓝色变为紫色或淡紫色。牛奶含有酪蛋白，因此会出现紫色结果。
Common exam pitfalls include using Benedict’s reagent without heating or expecting a biuret colour change without strong alkaline conditions. Remember that Benedict’s and biuret reagent must be distinguished: Benedict’s tests reducing sugars with heating, while biuret tests peptide bonds in proteins at room temperature.
常见考试失分点包括：使用本尼迪克特试剂却不加热，或期望在非强碱性条件下出现双缩脲颜色变化。务必分清：本尼迪克特试剂在加热条件下检测还原糖，而双缩脲试剂在室温下检测蛋白质中的肽键。
4. Human Digestive System and Enzyme Action | 人体消化系统与酶的作用
Question: “Explain how the digestion of starch in bread is completed as it passes through the human alimentary canal. Name the enzymes involved and the end products formed.” This requires a sequential account from the mouth to the small intestine.
题目：“解释面包中的淀粉在人体消化道中如何被完全消化。说出参与消化的酶和形成的终产物。”这要求从口腔到小肠依次叙述。
Digestion of starch begins in the mouth where salivary amylase (ptyalin) catalyses the breakdown of starch into maltose, a disaccharide. The food is then swallowed and enters the stomach, where the acidic pH denatures salivary amylase, stopping further starch digestion temporarily. In the small intestine, pancreatic juice containing pancreatic amylase is secreted from the pancreas; this continues hydrolysing starch into maltose under alkaline conditions maintained by bile.
淀粉的消化始于口腔，唾液淀粉酶催化淀粉分解为二糖麦芽糖。食物随后被吞咽进入胃，酸性 pH 使唾液淀粉酶变性，暂时停止对淀粉的消化。在小肠中，胰脏分泌含有胰淀粉酶的胰液；在胆汁维持的碱性条件下，胰淀粉酶继续将淀粉水解为麦芽糖。
The final step occurs on the membranes of the epithelial cells lining the small intestine. Here, the enzyme maltase (a disaccharidase) breaks maltose into glucose, a monosaccharide. Glucose is small enough to be absorbed into the bloodstream by active transport. A full answer must also mention that the end product for starch digestion is glucose, and that it is soluble and can be transported away for respiration.
最后一步发生在小肠上皮细胞的细胞膜上。这里的麦芽糖酶（一种二糖酶）将麦芽糖分解为单糖葡萄糖。葡萄糖足够小，可以通过主动运输吸收进血液。完整的答案还必须提到淀粉消化的终产物是葡萄糖，并且葡萄糖可溶于水，可以被运输到身体各处用于呼吸作用。
Marks are often lost by omitting the conditions (pH) or failing to separate the roles of amylase and maltase. Remember to specify that endopeptidases and proteases digest proteins, but for starch only carbohydrates are involved. Using a table to summarise the enzymes, substrates, products and locations can be an excellent revision tool.
常见失分原因是遗漏条件（pH）或未能区分淀粉酶和麦芽糖酶的作用。记住，内肽酶和蛋白酶消化蛋白质，而消化淀粉只涉及碳水化合物酶。可以用表格汇总酶、底物、产物和作用部位，作为很好的复习工具。
| Enzyme | Substrate | Product | Site |
| Salivary amylase | Starch | Maltose | Mouth |
| Pancreatic amylase | Starch | Maltose | Small intestine |
| Maltase | Maltose | Glucose | Small intestine (epithelium) |
5. Photosynthesis: Limiting Factors and Experiments | 光合作用：限制因素与实验
An exam question may ask: “A student places pondweed in a beaker of water and counts the number of oxygen bubbles produced per minute at different distances from a lamp. Identify the variable being changed, explain the shape of the graph, and state how you could confirm that carbon dioxide is the limiting factor at high light intensities.”
考试中可能会问：“一名学生将水蕴草放在装水的烧杯中，在距离光源不同的距离处记录每分钟产生的氧气气泡数。指出改变的自变量，解释曲线形状，并说明如何证明在较高光强度下二氧化碳是限制因素。”
The independent variable is light intensity, controlled by altering the distance from the lamp. As light intensity increases, the rate of photosynthesis rises proportionally at first because more light energy is available for the light‑dependent reactions, increasing ATP and reduced NADP production. The graph then levels off as another factor, such as carbon dioxide concentration or temperature, becomes limiting.
自变量是光强度，通过改变与灯的距离来控制。光强度增加时，光合作用速率起初成正比上升，因为有更多光能可用于依赖光的反应，从而增加 ATP 和还原型辅酶 II 的合成。然后曲线趋于平缓，因为另一个因素（如二氧化碳浓度或温度）成为限制因素。
To confirm that CO₂ is limiting at the plateau, increase the bicarbonate concentration in the water (source of CO₂). If the rate of bubbling rises again, CO₂ was the limiting factor. To make this a valid investigation, a control run with no added hydrogencarbonate is needed, and all other variables – temperature, wavelength of light, type of pondweed – must be kept constant. A common error is failing to use the term ‘limiting factor’ precisely or not linking the bubbling rate to the dependent variable correctly.
要确认平台期 CO₂ 是限制因素，可以提高水中的碳酸氢盐浓度（CO₂ 的来源）。如果气泡速率再次上升，就说明 CO₂ 是限制因素。为使探究有效，需要设置一组不加碳酸氢盐的对照实验，同时保持所有其他变量——温度、光的波长、水蕴草种类——恒定。常见错误是未能准确使用“限制因素”一词，或没有将气泡速率正确关联到因变量上。
Rate of photosynthesis ∝ O₂ bubbles min⁻¹
6. Aerobic and Anaerobic Respiration in Humans | 人体有氧呼吸与无氧呼吸
Question: “Comparing aerobic and anaerobic respiration in muscle cells, describe the differences in oxygen requirement, products and ATP yield. Explain why anaerobic respiration cannot be sustained for long.”
题目：“比较肌细胞有氧呼吸与无氧呼吸在需氧情况、产物和 ATP 产量上的差异。解释无氧呼吸为什么无法长时间维持。”
Aerobic respiration uses oxygen to break down glucose completely into carbon dioxide and water, releasing approximately 36–38 ATP molecules per glucose. Anaerobic respiration does not require oxygen. In humans, it converts glucose into lactic acid (lactate) and yields only 2 ATP molecules per glucose. The word equations you must know: for aerobic, glucose + oxygen → carbon dioxide + water (+ energy); for anaerobic, glucose → lactic acid (+ little energy).
有氧呼吸利用氧气将葡萄糖彻底分解为二氧化碳和水，每分子葡萄糖释放大约 36–38 个 ATP。无氧呼吸不需要氧气。在人体中，它将葡萄糖转化为乳酸（乳酸盐），每分子葡萄糖只产生 2 个 ATP。必须掌握的词方程：有氧呼吸，葡萄糖 + 氧气 → 二氧化碳 + 水（+ 能量）；无氧呼吸，葡萄糖 → 乳酸（+ 少量能量）。
Anaerobic respiration cannot continue for long because lactic acid accumulates in the muscles, lowering the pH. This inhibits enzyme activity and causes muscle fatigue and cramps. Furthermore, the low ATP yield is insufficient to sustain vigorous activity. To recover, the body repays an ‘oxygen debt’ by continuing to breathe deeply after exercise, which oxidises lactic acid back to pyruvate and then to carbon dioxide and water in the liver.
无氧呼吸无法长时间持续，因为乳酸在肌肉中积累，pH 下降。这会抑制酶活性，导致肌肉疲劳和抽筋。此外，低 ATP 产量不足以维持剧烈运动。为了恢复，身体需要通过运动后持续深呼吸来偿还“氧债”，此时肝脏中将乳酸氧化为丙酮酸，再分解为二氧化碳和水。
In yeast, anaerobic respiration produces ethanol and carbon dioxide instead of lactic acid. This is fermentation. Exams often ask you to link the products to practical uses: ethanol for biofuels, CO₂ for bread rising. When writing comparisons, using a table can help secure full marks.
酵母的无氧呼吸则产生乙醇和二氧化碳，而不是乳酸，这叫做发酵。考试常要求将产物与实际用途联系起来：乙醇用于生物燃料，二氧化碳用于面包发泡。在写比较时，使用表格有助于获得满分。
7. The Heart and Double Circulation | 心脏与双循环
Question: “Describe the pathway of a red blood cell from the right atrium to the aorta, naming the chambers, valves and vessels it passes through. Explain why the left ventricle has a thicker muscular wall than the right ventricle.”
题目：“描述一个红细胞从右心房到主动脉的路径，说出所经过的心腔、瓣膜和血管名称。解释为什么左心室壁肌肉比右心室壁厚。”
The red blood cell, deoxygenated, is already in the right atrium. It passes through the tricuspid (right atrioventricular) valve into the right ventricle. When the ventricle contracts, the tricuspid valve closes and the semi‑lunar valve opens; blood is forced into the pulmonary artery. It travels to the lungs where it becomes oxygenated, returns via the pulmonary veins to the left atrium, passes through the bicuspid (mitral) valve into the left ventricle, and finally is pumped through the aortic semi‑lunar valve into the aorta.
这个红细胞是缺氧的，已在右心房中。它经过三尖瓣（右房室瓣）进入右心室。当心室收缩时，三尖瓣关闭，半月瓣打开；血液被压入肺动脉。它到达肺部进行氧合，通过肺静脉返回左心房，经过二尖瓣进入左心室，最终被泵入主动脉半月瓣进入主动脉。
The left ventricle has a much thicker muscular wall because it must pump blood at high pressure throughout the entire body (systemic circuit). The right ventricle only pumps blood to the nearby lungs (pulmonary circuit) at a lower pressure, so less force is required. The thickness of the myocardium reflects the workload. A common misconception is that both ventricles pump the same volume of blood per beat – they do, but the left ventricle generates higher pressure.
左心室壁肌肉厚得多，因为它必须以高压将血液泵送到全身（体循环）。右心室只需将血液低压泵送到邻近的肺部（肺循环），所需力量较小。心肌厚度反映了工作负荷。一个常见误解是认为两个心室每次搏动泵出血量不同——其实相同，但左心室产生的压力更高。
Always use the correct names: tricuspid on the right, bicuspid on the left. Semilunar valves are found at the bases of the pulmonary artery and aorta. When labelling diagrams, show the septum separating the two sides; any hole in the septum (septal defect) would cause mixing of oxygenated and deoxygenated blood, a topic that frequently appears in data‑analysis questions.
始终使用正确名称：右侧是三尖瓣，左侧是二尖瓣。半月瓣位于肺动脉和主动脉的根部。在示意图中标注出分隔两侧的隔膜；如果隔膜出现缺损（隔膜缺损），会导致富氧血和缺氧血混合，这也是数据分析题中常出现的内容。
8. Transport in Plants: Transpiration and Translocation | 植物运输：蒸腾作用与输导作用
Question: “Explain how water moves from the soil into the root hair cell and is then transported to the mesophyll cells of a leaf. Describe the role of transpiration in this process.”
题目：“解释水如何从土壤进入根毛细胞，然后被运输到叶片叶肉细胞。描述蒸腾作用在这一过程中的角色。”
Water enters root hair cells by osmosis because the mineral ions actively transported into the root hair lower the water potential inside the cell compared with the soil solution. From the root hair, water moves via osmosis and through three pathways – apoplast, symplast and vacuolar – across the cortex to the xylem in the stele. Once inside the dead, hollow xylem vessels, water is pulled upwards in a continuous column, driven by transpiration pull.
水通过渗透作用进入根毛细胞，因为主动运输进根毛的矿物离子降低了细胞内的水势，使之低于土壤溶液。从根毛开始，水通过渗透经由三条途径——质外体、共质体和液泡途径——穿过皮层到达中柱的木质部。进入死细胞构成的中空木质部导管后，水在蒸腾拉力的作用下形成连续水柱向上运输。
Transpiration is the evaporation of water from the surfaces of mesophyll cells into the air spaces of the leaf, followed by diffusion out through stomata. This water loss creates a tension (suction) that pulls more water molecules up the xylem because of cohesion (water molecules stick to each other by hydrogen bonds) and adhesion (water molecules stick to xylem walls). The result is the transpiration stream. A potometer can measure the rate of water uptake as an estimate of transpiration rate.
蒸腾作用是水从叶肉细胞表面蒸发进入叶片气室，再通过气孔扩散出去的过程。这种水分流失产生一种张力（吸力），由于内聚力（水分子间通过氢键相互吸引）和粘附力（水分子附着于木质部壁），会把更多水分子拉上木质部。这样就形成了蒸腾流。可以使用蒸腾计测量吸水速率，以估计蒸腾速率。
Translocation describes the movement of sucrose and amino acids in the phloem from sources (leaves) to sinks (roots, fruits). This requires energy, unlike the passive transpiration stream. When comparing xylem and phloem, remember: xylem vessels are dead, lignified and carry water and minerals upwards; phloem sieve tubes are living, have companion cells and transport sugars bidirectionally.
输导作用描述的是蔗糖和氨基酸在韧皮部从源（叶片）向库（根、果实）的运输，这需要能量，与被动运输的蒸腾流不同。比较木质部和韧皮部时记住：木质部导管是死细胞、已木质化，向上运输水分和矿物质；韧皮部筛管是活细胞，伴有伴胞，双向运输糖类。
9. DNA, Genes and Chromosomes | DNA、基因与染色体
A typical genetics question may ask: “Explain the relationship between DNA, a gene and a chromosome. Describe how DNA’s structure enables it to store genetic information.”
典型的遗传学考题可能会问：“解释 DNA、基因和染色体之间的关系。描述 DNA 的结构如何使其能够储存遗传信息。”
DNA (deoxyribonucleic acid) is a long molecule composed of two strands twisted into a double helix. A gene is a specific sequence of nucleotide bases along the DNA that codes for a particular protein. A chromosome is a tightly coiled package of DNA wrapped around histone proteins, visible during cell division. The hierarchy is: DNA makes up genes, and many genes are linked together to form a chromosome.
DNA（脱氧核糖核酸）是由两条链缠绕成双螺旋结构的长分子。基因是 DNA 上编码特定蛋白质的特定核苷酸碱基序列。染色体是 DNA 缠绕在组蛋白上紧密盘绕形成的包装结构，在细胞分裂时可见。层次关系是：DNA 构成基因，许多基因连在一起形成染色体。
DNA stores information in the sequence of its four bases: adenine (A), thymine (T), cytosine (C) and guanine (G). The bases pair complementarily – A with T, C with G – held together by weak hydrogen bonds. A sequence of three bases (a triplet) codes for one amino acid. The order of triplets determines the order of amino acids in a polypeptide, thus determining the protein’s shape and function. The double‑stranded structure and complementary base pairing allow DNA to replicate precisely during mitosis and meiosis.
DNA 通过四种碱基的序列来储存信息：腺嘌呤 (A)、胸腺嘧啶 (T)、胞嘧啶 (C) 和鸟嘌呤 (G)。碱基互补配对——A 与 T，C 与 G——由弱氢键连接。三个碱基序列（三元组）编码一个氨基酸。三元组的顺序决定了多肽链中氨基酸的顺序，从而决定蛋白质的形状和功能。双链结构和互补碱基配对使 DNA 能在有丝分裂和减数分裂时精确复制。
Diploid human cells contain 23 pairs of homologous chromosomes (46 in total). Gametes are haploid, containing only one chromosome from each pair, ensuring that fertilisation restores the diploid number. Misunderstanding often occurs when students confuse a gene with an allele; an allele is a variant form of a gene. Use the example of eye colour: the gene codes for colour, while different alleles produce brown, blue or green iris colour.
人体二倍体细胞含有 23 对同源染色体（共 46 条）。配子是单倍体，每对染色体只含一条，确保受精后恢复二倍体数目。学生常混淆基因与等位基因：等位基因是基因的不同形式。以眼睛颜色为例：基因编码色彩，而不同等位基因则产生棕色、蓝色或绿色虹膜。
10. Ecology: Sampling, Food Chains and Energy Flow | 生态学：取样、食物链与能量流动
Question: “Describe how a student could use a quadrat and a transect line to investigate the distribution of daisy plants across a trampled football field. Explain why energy transfer between trophic levels is inefficient.”
题目：“描述学生如何使用样方和样线调查雏菊在被踩踏的足球场上的分布。解释为什么营养级之间的能量传递效率低下。”
To sample daisy distribution, a student would lay a transect line (a measuring tape) across the field from a heavily trampled area to an undisturbed edge. At regular intervals along the line, a quadrat (usually 0.5 m × 0.5 m or 1 m²) would be placed. The number of daisy plants in each quadrat would be counted, or percentage cover estimated. Repeating the transect at three parallel lines improves reliability. This provides data on how daisy abundance changes with trampling pressure.
要调查雏菊的分布，学生可以将一条样线（卷尺）从踩踏严重的区域铺向未受干扰的边缘。沿样线每隔固定间隔放置一个样方（通常是 0.5 m × 0.5 m 或 1 m²）。计数每个样方内雏菊的数量，或估算覆盖百分比。在三条平行样线上重复操作可提高可靠性。这样就能获得雏菊丰度随踩踏压力变化的数据。
Energy transfer between trophic levels is inefficient because organisms use most of the energy they consume for respiration, movement, growth and maintaining body temperature. A large proportion is lost as heat to the environment, and not all parts of the organism are eaten or digested (bones, cellulose). Typically, only about 10% of energy is passed from one trophic level to the next. This loss limits the length of food chains, usually to four or five levels.
营养级之间的能量传递效率低下，因为生物体将摄入的大部分能量用于呼吸、运动、生长和维持体温。很大一部分以热量形式散失到环境中，而且并非生物体的所有部分都被取食或消化（如骨头、纤维素
Published by TutorHao | Year 10 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