WJEC GCSE生物 细胞 酶 渗透 光合作用 遗传
WJEC GCSE Biology 的结构决定了复习节奏该往哪边压：Unit 1 覆盖 Cells, Organ Systems and Ecosystems，Unit 2 覆盖 Variation, Homeostasis and Micro-organisms，两份各占 45%、各 80 分，Unit 3 是占 10% 的实践评估、30 分、不分层，通常安排在最后一年的一月到二月完成，这些数字都写在 WJEC 官方 specification 3400 里。换句话说，笔试决定最终等级，而两张大题里既有考记忆的短答，也有要给数据找解释的长答。很多学生以为生物就是背书，翻完教科书直接进考场，结果在 describe 与 explain 的评分差别上丢分。下面按考试实际用到的顺序过一遍：细胞结构与放大倍数、酶、渗透与运输、光合作用限制因子、呼吸作用、单基因遗传、自然选择、生态取样，每一块都给出算法和常见失分点。
WJEC GCSE Biology splits its assessment in a way that should shape your revision. Unit 1 covers Cells, Organ Systems and Ecosystems and Unit 2 covers Variation, Homeostasis and Micro-organisms, and together they carry 90 per cent of the marks at 80 marks each, while Unit 3 is a 30-mark practical assessment worth 10 per cent, taken in the final year between January and February, as the WJEC specification 3400 sets out. Written papers therefore decide your grade, and they mix short recall items with longer questions that ask you to interpret data. Memorising the textbook on its own will not do, since command words such as describe and explain are marked by different criteria. What follows works through the topics that actually appear, with the calculations and the places where marks go missing.
显微镜放大倍数与细胞器功能
放大倍数计算是几乎每卷都有的小题。目镜 10 倍、物镜 40 倍时总放大倍数是两者相乘的 400 倍，不是相加。图像长度换算回实际长度用同一个倍数：公式是实际长度 = 图像长度 / 放大倍数，所以在 400 倍下量出 20 mm 的图像，实际长度是 0.05 mm，也就是 50 微米。细胞器按功能配对记最省力：细胞核控制细胞活动并存放遗传物质，线粒体通过有氧呼吸释放能量、在肌肉细胞里数量明显更多，核糖体负责合成蛋白质，植物细胞另外有细胞壁、叶绿体和永久液泡，动物细胞没有这三样。考比较题时要把共同点和差异都写出来，只写植物细胞含叶绿体会丢掉对比分。很多学生以为放大倍数越大看到的细胞越多，其实视野范围反而缩小。
Magnification is arithmetic rather than description; with a 10x eyepiece and a 40x objective the total is 400x, obtained by multiplying the two values instead of adding them, and the same factor converts a measured image length back to real size, so a 20 mm image at 400x represents 0.05 mm, or 50 micrometres. Organelles are marked by function, so examiners expect each structure to be linked to its job, which means the nucleus controlling activity and holding genetic material, mitochondria releasing energy through aerobic respiration and appearing in large numbers in muscle cells, and ribosomes building proteins. In comparison questions, shared features and differences both have to be named, and mentioning chloroplasts without stating that animal cells lack them scores partially at best. Higher magnification narrows the field of view, so a bigger image does not mean more cells on the slide.
酶的温度pH曲线与淀粉酶实验
酶是蛋白质，作用方式常用锁与钥匙解释：底物形状与活性位点互补，所以一种酶通常只催化一种反应。温度升高加快分子运动，反应速率随之上升，超过最适温度后酶变性，活性位点形状改变，速率陡降而且不可逆，这在图上表现为断崖式下降而不是缓慢回落。人体多数酶的最适温度约 37 摄氏度，胃蛋白酶最适 pH 约为 2，小肠里的酶接近 pH 8，偏离最适值同样会引起变性。解释速率曲线时要同时写出"分子运动加快"和"超过最适温度后变性"两层意思，只写一层通常只得一半分。淀粉与淀粉酶的实验里，每分钟取一滴加到碘液上，颜色由蓝黑转为黄褐色说明淀粉已被分解。实验设计题必须写清控制变量，温度、pH、底物体积都要固定，变量交代不清会被扣分。
Enzymes are proteins whose action is usually explained with the lock and key model, because the substrate fits the active site only when the shapes are complementary; raising the temperature speeds up molecular movement and the rate rises with it, but past the optimum the enzyme denatures, the active site changes shape, and the rate collapses sharply and permanently. Most human enzymes work best near 37 degrees Celsius; pepsin is adapted to about pH 2 in the stomach, while enzymes in the small intestine work near pH 8. Explaining a rate curve needs both ideas, faster collisions and denaturation beyond the optimum, and answers that give only one of them usually score half the marks. The standard practical uses starch with amylase and takes an iodine sample every minute, turning from blue-black to yellow-brown once the starch has gone.
扩散渗透与马铃薯条质量变化
扩散是分子从高浓度区域向低浓度区域净移动，速率取决于浓度差、温度、表面积和扩散距离。渗透是水通过部分透性膜从水势高的一侧移向水势低的一侧。植物组织放进纯水里吸水膨胀，膨压升高让组织变硬；放进浓糖水里水从细胞流出，细胞膜与细胞壁分离，出现质壁分离。马铃薯条实验的算法是把质量变化除以初始质量再乘 100，得到百分比变化，然后把浓度对百分比变化作图，直线与横轴相交的那个浓度就是细胞内部溶液的近似浓度。表面积与体积之比也是必考计算：边长 1 cm 的立方体比值是 6，边长 2 cm 的立方体比值是 3，比值越大扩散效率越高，这正是小肠绒毛和肺泡长成那种形态的原因。主动运输逆浓度梯度进行，需要载体蛋白和 ATP，缺氧或加入抑制剂后就会停止。
Diffusion is the net movement of particles from a region of higher concentration to one of lower concentration; osmosis is the special case in which water crosses a partially permeable membrane from the side of higher water potential. Plant tissue in pure water takes water in until the cells are turgid, whereas tissue in a strong sugar solution loses water and the membrane pulls away from the cell wall, which is plasmolysis. In the potato cylinder practical, percentage change is mass change divided by starting mass multiplied by 100, and the concentration at which the plotted line crosses the axis matches the cell contents. Surface area to volume ratio belongs in the same topic: a 1 cm cube has a ratio of 6 against 3 for a 2 cm cube, which is why villi and alveoli are shaped the way they are; active transport works against the gradient and needs carrier proteins plus energy from respiration.
光合作用限制因子与气泡计数
光合作用的文字方程式是二氧化碳 + 水 -> 葡萄糖 + 氧气，能量由光提供。叶绿素吸收红光和蓝紫光、反射绿光，所以叶片看上去是绿的。限制因子指当前供应最不足的那一个条件：光照弱时提高二氧化碳浓度看不出任何变化，把灯移近或调亮立刻有反应；光照充足后二氧化碳成了限制因子，此时继续加光曲线就变平。答题要写成"该因素成为限制因子"，而不是罗列一堆条件不够。沉水植物数气泡的实验中，计时起点要统一，最初 30 秒常作为适应期不计入结果。实验题要求写变量控制和重复次数，三次重复取平均值是标准写法。叶片上气孔由保卫细胞控制开闭，干旱时关闭以减少水分散失，代价是进入叶片的二氧化碳随之减少。
The word equation for photosynthesis is carbon dioxide + water -> glucose + oxygen, with light supplying the energy; chlorophyll absorbs red and blue-violet light and reflects green, which is why a leaf looks green. A limiting factor is whichever condition is in shortest supply: in dim light, raising the carbon dioxide concentration changes nothing at all, but moving the lamp closer produces an immediate response, and once light is plentiful carbon dioxide takes over as the limiting factor, so the curve flattens. Answers should name the factor that is limiting rather than listing conditions that are merely adequate. Counting bubbles from pondweed needs an agreed starting point, and the first 30 seconds is normally treated as an adaptation period and excluded. Leaves lose less water when their stomata close in drought, at the cost of reduced carbon dioxide entry.
有氧与无氧呼吸及气体检验
有氧呼吸把葡萄糖和氧气转成二氧化碳和水，释放的能量明显多于无氧呼吸，方程式写作葡萄糖 + 氧气 -> 二氧化碳 + 水。无氧呼吸在人体肌肉里把葡萄糖转成乳酸，在酵母里转成乙醇和二氧化碳，两者释放的能量都少得多。运动时的现象用这条线就能解释：强度上去以后氧气供应跟不上，肌肉转向无氧途径，乳酸累积引起酸痛。检验气要用对试剂和词句，石灰水变浑浊说明有二氧化碳，余烬复燃说明有氧气，写成"有气体产生"拿不到分。种子呼吸实验要先给种子消毒，避免微生物呼吸干扰结果，这一条在实验设计题里反复出现。能量比较题要按质量算，同样质量的不同营养物质释放的能量不同，按体积比较会算错。
Aerobic respiration converts glucose and oxygen into carbon dioxide and water and releases far more energy than the anaerobic route, summarised as glucose + oxygen -> carbon dioxide + water; anaerobic respiration produces lactic acid in human muscle and ethanol plus carbon dioxide in yeast, and both routes release much less energy. This explains what happens during hard exercise, because once oxygen supply cannot keep up the muscles switch pathway and lactic acid accumulates. Gas tests are marked on wording as much as on observation: limewater turning cloudy confirms carbon dioxide, a glowing splint confirms oxygen, and writing that a gas was produced scores nothing. Seed respiration experiments require sterilising the seeds so that microbes do not respire instead, a point that examiners keep testing in practical design questions; energy comparisons should be made per unit mass, since equal volumes of different nutrients do not carry equal energy.
单基因杂交与庞尼特方格概率
遗传计算先定符号：显性等位基因用大写字母，隐性用小写字母，Bb 表示杂合子，表现型是显性性状。庞尼特方格是概率题最稳的工具，两个杂合子 Bb 与 Bb 交配，四个格子分别是 BB、Bb、Bb、bb，表现型比例 3 比 1，基因型比例 1 比 2 比 1，答案写 3/4 或 75% 都算对。读题时要注意问的是某一特定后代的概率，不是整个人群的比例。很多学生以为显性等位基因一定更常见，实际并不成立，显性只说明杂合状态下性状会表现出来。系谱图题按代次逐个推基因型：隐性遗传病患者的基因型一定是纯合隐性，如果父母表现正常，那么两人都是携带者。另外记住性别决定，男性是 XY，儿子的 Y 染色体来自父亲。
Genetic crosses start with symbols, a capital letter for the dominant allele and a lower case letter for the recessive, so Bb marks a heterozygote that shows the dominant characteristic. A Punnett square remains the safest tool for probability: crossing Bb with Bb gives BB, Bb, Bb and bb across the four cells, a 3 to 1 phenotypic ratio and a 1 to 2 to 1 genotypic ratio. Read the question carefully, because it normally asks about one particular offspring rather than a whole population; students often assume a dominant allele must also be the common one, and that assumption is simply wrong. Pedigree questions are worked person by person, since an affected individual with a recessive condition is homozygous recessive, which makes both unaffected parents carriers; males are XY, so a son inherits his Y chromosome from his father.
自然选择与细菌抗药性演化
自然选择写成一条逻辑链就顺了：种群中的变异本来已经存在，某些变异带来生存优势，带有优势的个体更可能繁殖并把等位基因传下去，于是该等位基因的频率逐代上升。细菌抗药性是最常用的例子，抗生素杀死敏感菌株，剩下的抗药菌株继续分裂，种群很快以抗药个体为主。答题关键是把"变异是随机的、抗生素只起筛选作用"讲清楚，写成"细菌为了生存而产生抗药性"会被判错，因为后天获得的性状不会遗传。演化速度与繁殖世代的长度直接相关，条件合适时细菌二十来分钟就能分裂一次，一天之内可以完成几十代筛选，所以抗药性在临床上出现得很快。这类题还常要求区分单体抗药与整个种群抗药，种群层面的比例变化才是自然选择的证据。
Natural selection is best written as a chain of reasoning: variation already exists in the population, some variants carry a survival advantage, the individuals with that advantage are more likely to reproduce and pass on their alleles, and the frequency of those alleles rises from generation to generation; antibiotic resistance is the clearest illustration, because the antibiotic kills the susceptible bacteria and leaves the resistant ones to divide, shifting the population towards resistance quite quickly. The decisive phrase is that variation is random and the antibiotic acts only as a selecting agent; writing that bacteria develop resistance in order to survive loses the marks, since acquired characteristics are not inherited; speed matters here, because bacteria can divide roughly every 20 minutes, so many generations of selection fit inside a single day. Many questions also ask you to separate resistance in one bacterium from resistance in the population.
样方取样估算与命令词答题
样方取样的核心是随机和可重复。0.5 米乘 0.5 米的样方框要先随机定点再放框，不能挑长得最茂盛的地方放，否则估算值偏高。种群总数按下式估算：总数 = 每个样方的平均个体数 x 总面积 / 样方面积。一块 200 平方米的草地，10 个样方平均每框 8 株，估算就是 8 x 200 / 0.25，等于 6400 株，漏掉最后一步除法是常见错误。样线法适合观察随距离变化的分布，从高潮线向陆地每隔 2 米记录一次物种，能看出明显的分带。答题时 describe 要写观察到什么，explain 要写为什么，两者互换直接扣分，单位漏写和有效数字位数不对在生物计算题里同样算错。WJEC 考试局官网的实验题评分说明一直强调变量控制与重复次数，写清这两点比多写一段结论更有用。
Quadrat sampling depends on being random and repeatable; with a 0.5 m by 0.5 m quadrat, points are chosen before the frame is placed, because dropping it on the lushest patch biases the estimate upwards. Population size is estimated as the mean count per quadrat multiplied by the total area and divided by the quadrat area, so ten quadrats averaging eight plants in a 200 square metre field give 8 x 200 / 0.25, or 6,400 plants. Belt transects suit questions about change with distance, such as recording species every 2 metres from the high water mark inland to expose zonation. Command words decide the marking, since describe asks for what you observe and explain asks for why it happens, and missing units or wrong significant figures are penalised in calculations; a frequent slip is stopping after the mean and never scaling it to the whole site.
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