Topic 4: Bioenergetics
AQA · GCSE · Biology الموضوع 4 34:41 سرد باللغة الإنجليزية · ترجمة مدمجة بالإنجليزية + الصينية
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النص الصوتي
Plants and animals respire.
植物和动物都呼吸。
Photosynthesis uses light energy to make glucose; respiration transfers energy for living processes.
光合作用利用光能合成葡萄糖，呼吸作用为生命活动传递能量。
This lesson develops equations, experimental evidence and rate reasoning.
本课练习方程、实验证据及速率推理。
Keep the chemical substances separate from the energy input: light goes above the arrow, not among the reactants.
把化学物质和能量输入分开：光写在箭头上方，而不是写进反应物。
Light transfers energy; it is not a chemical substance used as a reactant.
光传递能量，并非化学反应物。
All three missing formulas are needed for the mark.
三个缺失的化学式都要写对才能得分。
The symbol equation is: 6 C O 2 plus 6 H 2 O gives C 6 H 12 O 6 plus 6 O 2.
化学方程式为 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂。
Carbon dioxide and water may be listed in either order.
CO₂ 和 H₂O 的书写次序可以互换。
Give both parts: where the energy comes from, and where in the cell the reaction happens.
写明两部分：能量从哪里来，以及反应在细胞的哪个部位进行。
Chlorophyll absorbs light in chloroplasts. An organelle name alone does not explain the energy transfer.
叶绿体中的叶绿素吸收光，细胞器名称不能代替能量解释。
All three leaves were on the same plant, and each was tested for starch after 48 hours.
三片叶都在同一株植物上，48小时后分别测试淀粉。
Link each treatment to photosynthesis first, and then to starch.
先把每种处理与光合作用联系起来，再联系到淀粉。
The supplied investigation tests consequences of removing requirements for photosynthesis. Transparent does not mean air can enter.
透明不代表空气可进入，实验考查缺少光合条件的后果。
Two separate marks: no glucose was made, and so none could be converted to starch.
两个得分点分开：没有合成葡萄糖，因此也没有葡萄糖能转化成淀粉。
The specification includes amount of chlorophyll as well as light, CO₂ and temperature.
大纲包括叶绿素含量，不只有光照、二氧化碳和温度。
Teacher models have uncalibrated axes. The temperature optimum depends on the system, not a universal 40°C rule.
教师曲线没有数值标定，最适温度取决于具体系统。
The optimum label belongs at the peak of the curve, not on the falling side.
最适温度标在曲线峰顶，而不是下降段。
Temperature initially speeds enzyme-controlled reactions; excessive heat changes enzyme shape and lowers rate.
升温最初加快酶促反应，过热改变酶形状使速率下降。
Read the axes and their units before answering anything about this graph.
先读坐标轴和单位，再回答关于这幅图的问题。
Lower-part zoom of the original graph, with no changed points or scale.
原图下部放大，数据和刻度不变。
The overview retains the upper curve and complete vertical-axis label.
总图保留上部曲线和纵轴名称。
Use the first straight section and the original thousands-of-lux scale.
读初始直线及千勒克斯横轴。
y equals m x plus c describes a straight line, so look for the interval where the graph is straight, not just rising.
y等于mx加c描述的是直线，所以要找图像呈直线的区间，而不只是上升的区间。
The answer is nought to five thousand lux.
答案是零到五千勒克斯。
That initial section of the graph is straight, with a constant gradient.
图像开头这一段是直线，斜率不变。
Beyond it the graph bends and then plateaus: at the plateau, more light does not increase the rate under these conditions.
之后图像弯曲，然后趋于平稳：在平台期，在这些条件下增加光照不会再提高速率。
Read the four condition labels and find point A before explaining what limits the rate.
先读四条曲线的条件，找到A点，再解释限制因素。
The question already tells you that light is limiting at A. Now say what a limiting factor means.
题目已说明A点光照是限制因素，现在要解释限制因素的含义。
The definition applies under the current conditions. Temperature is not helped indefinitely by increases.
定义只适用于当前条件，不能无限升温。
Compare curves in matched pairs, where only one condition changes.
用只改变一个条件的成对曲线来比较。
One: at fixed carbon dioxide, raising the temperature from fifteen to twenty-five degrees raises the rate.
第一，二氧化碳浓度不变时，温度从十五度升到二十五度，速率提高。
Two: a higher temperature increases particle movement and the rate of enzyme-controlled reactions.
第二，温度升高，粒子运动加快，酶促反应速率提高。
Three: more carbon dioxide supplies more reactant at the same temperature.
第三，在同一温度下，更多的二氧化碳提供更多的反应物。
Four: every curve plateaus when another factor limits the rate, so more light then has no effect.
第四，当另一个因素限制速率时，每条曲线都达到平台，此时增加光照没有作用。
Light is not what limits the plateaus; something else, such as chlorophyll, does.
限制平台的不是光照，而是叶绿素等其他因素。
Calculate the change in light intensity. Do not simply assume the photosynthesis rate changes by the same factor.
计算光强的变化，不能直接认为光合速率按相同比例变化。
For the same lamp, intensity is proportional to the inverse square of distance. Units cancel in the distance ratio.
同一灯源光强与距离平方成反比，距离比的单位相消。
Separate the physical law for light intensity from a biological prediction, which depends on conditions.
区分光强的物理规律与有条件的生物预测。
The simple proportional prediction needs a region where rate is proportional to intensity and other conditions remain suitable.
只有速率与光强成正比且其他条件合适时才能按比例预测。
For each change give three points: what is done, how it increases photosynthesis, and its economic problem.
每种改进写三点：采取什么措施、它如何提高光合作用，以及它的经济问题。
Do not substitute the cheapest input for evidence that it limits production. These are teacher-rubric examples.
不能只挑便宜投入，须判断限制条件，以下为教师评分示例。
Teacher bubble-count setup, not to scale. This model does not claim to collect a measured gas volume.
图示为数气泡法，不声称已收集定量气体。
A full plan includes what you measure, how you calculate the rate, and which variables you control.
完整的实验设计要包含测量什么、如何计算速率，以及控制哪些变量。
Use the same pondweed and comparable solution at each distance. Allow adjustment before recording.
每个距离使用可比黑藻及溶液，计时前先适应。
Two controls should be named and explained, not merely “keep it a fair test”.
应写具体控制量及原因，不能只说公平实验。
Describe an actual change to how the gas is measured, not just the name of a piece of equipment.
要写出对气体测量方法的实际改进，而不只是写一个仪器名称。
Repeated counts reduce random variation but cannot turn unequal bubbles into equal volumes.
重复降低随机波动，不能让气泡体积相同。
The measured quantity here is the time taken for ten bubbles.
这里测量的是产生十个气泡所需的时间。
The water baths were set, then the method waited two minutes.
设定水浴后先等两分钟。
Name the independent variable before calculating the rate.
先说出自变量，再计算速率。
The independent variable is temperature. The measured dependent variable is time; the calculated rate uses its reciprocal relationship.
自变量温度，测量因变量时间，速率需由气泡数除以时间。
Read the actual times first. Do not assume that forty degrees must have destroyed the enzymes.
先读实际时间再解释，不能想当然地认为40度时酶一定失活。
In this experiment 25 seconds is faster than 84 seconds for ten bubbles. Do not infer a universal optimum from the sampled points.
同样十个气泡25秒比84秒快，离散点不能证明普适最适温度。
This investigation changes the colour of the light, not the distance or the temperature.
这个实验改变的是光的颜色，而不是距离或温度。
How you change the colour is a method; naming the variables is a separate answer.
如何改变颜色是方法，说出变量名称是另一个答案。
Use different coloured LEDs or bulbs, or transparent coloured filters.
使用不同颜色的LED灯或灯泡，或透明的彩色滤光片。
The independent variable is the colour, or wavelength, of the light, not 'the colour of the filter'.
自变量是光的颜色或波长，而不是“滤光片的颜色”。
The dependent variable is the time for ten leaf discs to float to the surface of the solution.
因变量是十片叶圆片浮到溶液表面所需的时间。
Explain each step of the method in turn: removing the air, supplying carbon dioxide, and why the discs rise.
依次解释每一步：除去空气、供应二氧化碳，以及叶片圆片为何上浮。
CO₂ is supplied for photosynthesis; oxygen produced during photosynthesis makes the discs rise.
供二氧化碳进行光合，光合产生的氧气使叶片上浮。
Keep the dashed and solid lines apart, and read the percentage axis. Leaf colour alone is not evidence.
分清虚线和实线，读百分数坐标，不能只凭叶片颜色推断。
In this investigation red light was slower than blue, and faster than green.
在本实验中，红光下比蓝光慢，比绿光快。
Use the absorption curves and the wavelength ranges as evidence. The times alone are not enough.
要用吸收曲线和波长范围作证据，只比较时间是不够的。
Approximately 80% absorption near 490 nm is within the original blue range. Avoid claiming all blue wavelengths are absorbed at 80%.
约490纳米吸收80%，不能说所有蓝光都吸收80%。
Read where each chlorophyll has its peaks, then explain why having two types helps.
读出每种叶绿素的吸收峰，再解释为何两种叶绿素有益。
'Absorbs all wavelengths' earns no mark, and neither does a vague claim about absorbing the most light.
写“吸收所有波长”不得分，笼统地说“吸收最多的光”也不得分。
Glucose is a starting material for several processes, not only a stored fuel.
葡萄糖有多种用途，不只是储存燃料。
Link this to mineral deficiency: what else, besides glucose, does a plant need to make proteins?
联系植物缺素：除了葡萄糖，植物合成蛋白质还需要什么？
Nitrate ions are absorbed from soil. They are not a replacement energy source for light.
硝酸根从土壤吸收，不是光能的替代品。
Grey printing marks the green region; read the labels rather than the shading.
灰色印刷表示绿色部分，应读标签，而不是看深浅。
The plant was uncovered, in the light, for 48 hours.
植物在光下未遮盖48小时。
Predict each region's result, then explain the difference.
预测两个区域的结果，再解释差别。
A positive iodine test is blue-black; a negative result remains yellow/orange/brown. The prediction belongs to the supplied conditions.
碘液阳性蓝黑色，阴性保持黄橙棕色，结论适用于指定条件。
Starch is not in the equation, yet testing for it shows that photosynthesis has happened, because glucose is stored as starch.
方程里没有淀粉，但测淀粉可以证明发生了光合作用，因为葡萄糖会以淀粉形式储存。
Link this to photosynthesis: in the dark, does a plant still respire?
联系光合作用：在黑暗中，植物还进行呼吸吗？
Plants respire in light and darkness. Photosynthesis requires light. Do not say respiration creates energy.
植物在明暗条件下均呼吸，呼吸作用释放而不是创造能量。
Write the word equation first, then check the atoms balance in the symbol equation.
先写文字方程，再核对符号方程两边原子是否守恒。
Energy is transferred by the reaction; it is not a chemical substance to balance. The exam asks for mitochondria.
能量不是需要配平的化学物质，主要场所为线粒体。
Compare where each type happens, its products, and the energy released per glucose molecule, in human muscle cells.
比较人体肌肉细胞中两种呼吸的场所、产物，以及每个葡萄糖分子释放的能量。
The question rules out oxygen. Make each difference a matched pair about the same feature.
题目排除氧气比较，每个差异都要围绕同一特征成对作答。
Aerobic respiration produces carbon dioxide and water, while anaerobic respiration in muscle produces lactic acid.
有氧呼吸产生二氧化碳和水，而肌肉中的无氧呼吸产生乳酸。
Aerobic respiration transfers much more energy per glucose molecule.
每分子葡萄糖经有氧呼吸转移的能量多得多。
Another comparison: aerobic respiration happens mainly in the mitochondria, anaerobic in the cytoplasm.
另一组比较：有氧呼吸主要在线粒体中进行，无氧呼吸在细胞质中进行。
Pair each point with its partner, and give two comparisons for two marks.
每一点都要与对应的另一点成对比较，两组比较得两分。
During exercise both types can happen at once. Explain why the anaerobic one releases less energy from each glucose.
运动时两种呼吸可同时进行，解释无氧呼吸为何从每个葡萄糖释放的能量较少。
Keep the comparison per glucose molecule; do not claim energy is released for respiration.
无氧呼吸仍释放能量，比较基准是每分子葡萄糖。
Separate the products in muscle from those in plants and yeast.
区分肌肉与植物、酵母无氧呼吸的产物。
Yeast can also respire aerobically when oxygen is available. These equations describe its anaerobic pathway.
酵母有氧时也能有氧呼吸，此处为无氧途径。
Explain it yourself first, then check the link between the gas from fermentation and the rising dough.
先独立解释，再查看发酵产生的气体与面团膨胀之间的联系。
Warm conditions can support enzyme activity; increasing temperature indefinitely would not keep improving fermentation.
温暖可促进酶活动，但不能无限升温。
Original 2021 Figure 6. Well lit for five days, then completely covered with black paper.
先读装置：水草与蜗牛共存，五天后完全遮光。
Original 2021 Figure 7. Carbon dioxide concentration is in arbitrary units, not a respiration rate.
浓度不是呼吸速率，注意第五天遮光和第十天蜗牛死亡。
A flat line does not mean nothing is happening; explain the balance between carbon dioxide taken in and released.
水平线不代表没有反应，要解释二氧化碳吸收与释放之间的平衡。
The one-mark dark interval requires the linked explanation, not merely the word respiration.
遮光阻止光合作用，但两者仍呼吸并释放二氧化碳。
Separate the dead snail from the living microorganisms that decompose it.
区分死去的蜗牛和分解它的活微生物。
Do not claim the dead snail keeps respiring, or that muscle anaerobic respiration releases carbon dioxide.
不是死蜗牛继续呼吸，而是分解者呼吸放出二氧化碳。
Breathing rate and the volume of each breath are different measurements.
呼吸频率和每次呼吸量是不同的测量值。
Explain fatigue with the course model: too little oxygen, lactic acid building up, then how the blood carries it to the liver.
按本课程模型解释：供氧不足、乳酸积累，再说明血液如何把乳酸运到肝脏。
This does not mean all aerobic respiration switches off. Liver conversion and oxygen debt are distinct statements.
并非有氧呼吸全部停止，肝脏转化与氧债定义应分开。
This exam graph shows two quantities during exercise and recovery.
这幅考题图显示运动和恢复期间的两个量。
The two vertical axes measure different quantities and have no number scale, so compare the changes within each curve, not the heights of the two lines.
两条纵轴代表不同的量，而且没有数值刻度，所以要比较每条曲线自身的变化，而不是比较两条线的高低。
Use the evidence that lactic acid is still high to explain why extra oxygen is needed after exercise.
利用乳酸仍然很高的证据，解释运动后为何需要额外的氧气。
During anaerobic respiration, lactic acid builds up: there is an oxygen debt.
无氧呼吸时乳酸积累，形成氧债。
After exercise, extra oxygen is needed to react with and remove the lactic acid.
运动后需要额外的氧气与积累的乳酸反应，把它除去。
Keep this separate: the blood carries lactic acid to the liver to be converted to glucose, but do not say the extra oxygen is a reactant in that conversion.
要区分开：血液把乳酸运到肝脏转化为葡萄糖，但不要说额外的氧气是这一转化的反应物。
Choose two effects, and give each one its own biological reason.
选两个效应，每个都要配上对应的生物学原因。
These are two valid pairs. Increased post-exercise heart rate with oxygen-debt reasoning is another permitted pair.
两个效应但无理由不足四分，持续心率升高与偿还氧债也是可用组合。
A six-mark method is marked as one connected, valid plan, not as six separate facts.
六分方法题按整体是否连贯有效来评分，而不是六条互不相关的事实。
Same athletes control many individual differences. Randomising exercise order is a teacher improvement to reduce order effects.
同一组受试者控制个体差异，恢复静息心率后再进行下一次。
Teacher-created example data, all heart rates in beats per minute. A and B are different exercise types of equal duration.
教师自拟数据，所有心率单位为次每分钟，先算每人增量。
Calculate the change separately for each trial. Do not compare final heart rates while ignoring different baselines.
分别计算每次试验增量，不能忽略静息基线不同而只比最终心率。
Sequence the method: rest, measure a baseline, exercise for a fixed time, measure promptly, recover, then repeat with the other exercise.
按顺序安排方法：休息，测量基线值，进行固定时长的运动，及时测量，恢复，然后换另一种运动重复。
For analysis, calculate each athlete's increase, then the mean increase for each type of exercise, and compare the means.
分析时，计算每位运动员的增加量，再求每种运动的平均增加量并加以比较。
Using the same athletes, the same duration and timing, and recovery between trials makes the comparison valid.
使用相同的运动员、相同的时长和测量时间，并在两次之间充分恢复，能使比较有效。
The numbers here are an example, not exam data.
这里的数字只是示例，不是考题数据。
Connect what you learned about blood: haemoglobin, oxygen, aerobic respiration and energy.
联系已学的血液运输：血红蛋白、氧气、有氧呼吸和能量。
A source-specific biological explanation, not advice on changing blood composition.
解释原题生物学机制，不把红细胞增多说成促进无氧呼吸。
Use the information given, that muscle mass increases, to explain the better performance.
利用题目给出的“肌肉质量增加”解释表现提升。
These drugs are banned in competitive sport.
这类药物在竞技运动中是禁用的。
Explain only the model the question gives; add no mechanism it does not supply.
只解释题目给定的模型，不添加题目没有给出的机制。
Metabolism is every reaction in the body, not only respiration.
代谢包括体内所有反应，不只是呼吸。
Connect the building reactions with the breakdown reactions.
把合成反应和分解反应联系起来。
Metabolism is the sum of all reactions in a cell or body. These enzyme-controlled pathways are linked by materials and energy.
代谢是细胞或体内所有反应总和，酶控制途径由物质与能量相连。
Making proteins is one process; breaking down excess amino acids into urea is another.
合成蛋白质是一个过程，把多余氨基酸分解成尿素是另一个过程。
Proteins do not turn straight back into urea.
蛋白质不会直接变回尿素。
Recognise sugars, amino acids, fatty acids and glycerol in synthesis and breakdown; keep within GCSE scope.
识别糖、氨基酸、脂肪酸和甘油的作用，保持GCSE深度。
Before you call this topic done — write both photosynthesis equations cold, and say endothermic. Read a limiting-factor graph at a glance, and at higher tier halve the light intensity with the inverse-square law. Describe RP6 with its controls. Give the anaerobic equations for muscle and yeast, explain exercise and the oxygen debt, and list what metabolism includes.
收尾自查：默写光合作用两个方程并说对“吸热”；一眼读懂限制因素图像；高阶要用平方反比算光照减半；带控制变量描述 RP6；写出肌肉和酵母的无氧呼吸方程；解释运动与氧债；列举代谢包含的反应。