Topic 7: Ecology
AQA · GCSE · Biology Tema 7 1:00:14 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
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Transcripción
Topic 7 looks outward — the Sun's energy flows through ecosystems while carbon and water cycle between the living and non-living world. Communities compete and depend on one another, adaptations fit each species to its conditions, and human activity — waste, land use, deforestation and global warming — threatens biodiversity.
第七主题向外看：太阳能流经生态系统，碳和水在生命与非生命世界间循环；群落内既竞争又相互依赖，适应性让物种契合环境；人类活动——废物、土地利用、森林砍伐和全球变暖——正威胁生物多样性。
The biology-only sections connect biomass transfer to food security.
仅生物方向还要求用营养级解释食物链为何短、粮食安全需要什么。
What name describes all the organisms living in the same habitat?
生活在同一栖息地的所有生物叫什么？
Choose between population, community and ecosystem.
在种群、群落和生态系统中选择。
The answer is community.
答案是群落。
A population is one species; an ecosystem adds the interactions with the non-living surroundings.
种群只包括一个物种；生态系统还要加上生物与非生物环境之间的相互作用。
An individual is one organism, such as one rabbit.
个体是一个生物，例如一只兔子。
A population is one species in an area: the rabbits in the field.
种群是一个区域内的同一物种：田野里的兔子。
A community is the populations of different species: the plants, rabbits, birds and microbes.
群落是不同物种的种群：植物、兔子、鸟和微生物。
An ecosystem is the community interacting with its non-living surroundings, such as the soil, water and conditions.
生态系统是群落与非生物环境（例如土壤、水和环境条件）相互作用的整体。
A list of animals alone is not an ecosystem.
只列出一些动物并不构成生态系统。
Define an ecosystem, for two marks.
给生态系统下定义，共两分。
Identify its living part, its non-living part, and give an example of an interaction between them.
说出它的生物部分和非生物部分，并举一个二者相互作用的例子。
Pause and try it.
暂停并试着回答。
The living part is a community: populations of different organisms.
生物部分是群落：不同生物的种群。
That community interacts with the non-living, or abiotic, environment.
群落与非生物环境相互作用。
For example, plants absorb water and mineral ions from the soil, and light affects photosynthesis.
例如，植物从土壤中吸收水和矿质离子，光照影响光合作用。
Naming living and non-living things without connecting them leaves the definition incomplete.
只列出生物和非生物而不把它们联系起来，定义就不完整。
A simple community model.
一个简单的群落模型。
Solid arrows show food passing along: plants feed rabbits and insects, and predators eat those animals.
实线箭头表示食物的传递：植物被兔子和昆虫取食，捕食者又吃掉这些动物。
The lower panel shows two different links between flowers and bees: bees feed on the flowers, and the dashed arrow shows that bees pollinate them.
下方展示了花和蜜蜂之间的两种不同联系：蜜蜂以花为食，虚线箭头表示蜜蜂为花传粉。
Give three things animals compete for.
说出动物竞争的三种东西。
Then, in the model, which resource do the foxes and owls share?
然后看模型：狐狸和猫头鹰共享哪种资源？
And which resources can competing plants share?
相互竞争的植物又会共享哪些资源？
Pause and try it.
暂停并试着回答。
Plants compete for light, space, water and mineral ions, so neighbouring plants can reduce each other's supply.
植物竞争光、空间、水和矿质离子，所以相邻的植物会减少彼此的供应。
Animals compete for food, mates and territory.
动物竞争食物、配偶和领地。
In the model, foxes and owls both eat rabbits, so that shared food can limit both.
在模型中，狐狸和猫头鹰都吃兔子，这一共享的食物会限制二者。
Competition needs no fight: using a resource leaves less for another organism.
竞争不一定要打斗：一个生物使用了资源，另一个就会少一些。
Competition for mates is usually within one species.
对配偶的竞争通常发生在同一物种内。
Explain how removing one species can affect a community, using a named example.
用一个具体的例子，解释除去一个物种会如何影响群落。
Start with the bees and flowers in the model, connect pollination to plant reproduction, then extend it to the animals that eat the plants.
从模型中的蜜蜂和花讲起，把传粉与植物繁殖联系起来，再推广到以这些植物为食的动物。
Pause and try it.
暂停并试着回答。
Bees get food from flowers, and pollinate them.
蜜蜂从花中获取食物，同时为花传粉。
Fewer bees may mean less pollination of the plants that depend on them.
蜜蜂减少，依赖它们的植物得到的传粉可能减少。
Fewer seeds or fruits can reduce the plants' reproduction.
种子或果实减少，会降低植物的繁殖。
And less plant food may reduce the survival or reproduction of the animals that eat them.
植物性食物减少，又可能降低以它们为食的动物的存活或繁殖。
Each step depends on these relationships; other pollinators can change the outcome.
每一步都取决于这些关系；其他传粉者可能改变结果。
A new predator arrives in a woodland food web.
一种新的捕食者进入林地食物网。
Explain two effects.
解释两种影响。
Use the rabbit example, and give a biological cause for each effect.
用兔子的例子，并为每种影响给出生物学原因。
Pause and try it.
暂停并试着回答。
First, the new predator eats more rabbits, so fewer rabbits survive and their population may fall.
第一，新的捕食者吃掉更多兔子，存活的兔子减少，兔子种群可能下降。
Second, the existing rabbit predators now face more competition for prey, so less food may reduce their survival and reproduction.
第二，原有的兔子捕食者面临更激烈的猎物竞争，食物减少可能降低它们的存活和繁殖。
Another linked effect: less grazing could let the rabbits' food plants increase.
另一个相关影响：啃食减少，兔子吃的植物可能增多。
Daisies grow better further from a tree.
雏菊在离树较远的地方长得更好。
Suggest three other abiotic factors that could be responsible; water is already excluded.
提出三个可能起作用的其他非生物因素；水分已被排除。
Choose three named non-living factors that fit plants near a tree.
选出三个适用于树旁植物的具体非生物因素。
Pause and try it.
暂停并试着回答。
Light intensity: the tree's shade can change the light reaching the plants, so say intensity, not just 'sun'.
光照强度：树荫会改变到达植物的光照，所以要写“光照强度”，而不只是“阳光”。
Soil pH, its acidity or alkalinity, affects plant growth.
土壤pH，即酸碱度，会影响植物生长。
Mineral ion content: growth depends on the mineral supply, so name minerals or ions, not vague 'nutrients'.
矿质离子含量：植物生长依赖矿质供应，要写矿质或离子，而不是笼统的“营养”。
Temperature, soil type, wind and soil oxygen are also accepted; carbon dioxide and space are not, in this context.
温度、土壤类型、风和土壤含氧量也可以；在这一情境中，二氧化碳和空间不得分。
Suppose the nitrate in the soil falls.
假设土壤中的硝酸盐减少。
Nitrate supplies the nitrogen for amino acids and proteins.
硝酸盐为合成氨基酸和蛋白质提供氮。
So less protein can be made, which can reduce the plant's growth.
于是能合成的蛋白质减少，植物生长可能变慢。
Magnesium has a different role, in chlorophyll; not every ion makes amino acids.
镁的作用不同，它用于合成叶绿素；并不是每种离子都用来合成氨基酸。
This exam graph shows four bird species through one year, by month.
这幅考题图显示四种鸟一年中各月的数量。
Notice that the vertical axis has no number scale, so you can describe trends but not read exact counts.
注意纵轴没有数字刻度，所以只能描述趋势，不能读出准确的数量。
Describe what happens to the number of birds of species A.
描述A种鸟的数量如何变化。
Find its dotted line, split the year into intervals with different trends, and give the months.
找出它的虚线，把一年分成趋势不同的几段，并写出月份。
Can this graph give an exact percentage?
这幅图能给出准确的百分比吗？
Pause and try it.
暂停并试着回答。
Species A stays constant from January to March, or early April.
A种鸟的数量从一月到三月（或四月初）保持不变。
It increases during April, towards May.
四月间增加，一直到五月。
Then it decreases from May or June towards December.
然后从五月或六月起减少，直到十二月。
Give the months, not just the seasons.
要写出月份，而不只是季节。
With no numbered scale, there are no exact counts or percentages.
纵轴没有数字刻度，所以不能给出准确数量或百分比。
Two more questions.
再看两个问题。
A disease affects the birds in June and July: which species shows the lowest resistance?
六月和七月有一种疾病影响这些鸟：哪种鸟的抵抗力最低？
And one species migrates between countries: which one, and what in the time pattern shows it?
有一种鸟在国家之间迁徙：是哪一种？ 时间上的什么规律说明了这一点？
Pause and try it.
暂停并试着回答。
Species B has the lowest resistance in the June and July outbreak.
在六七月的疾病暴发中，B种鸟的抵抗力最低。
Species D migrates: it is present for only part of the year, from about May to September or October.
D种鸟会迁徙：它一年中只有一段时间在，大约从五月到九月或十月。
Give the reason with the name.
说出名称时要给出理由。
And do not assume every fall is disease, or every absence is migration.
不要认为每次数量下降都是疾病造成的，也不要认为每次不出现都是迁徙。
Less soil moisture is an abiotic change: plants get less water.
土壤水分减少是非生物变化：植物得到的水分减少。
A new pathogen is biotic: disease can reduce survival.
新的病原体是生物因素：疾病会降低存活率。
A new predator is biotic: more prey may die.
新的捕食者是生物因素：更多猎物可能死亡。
And competition for limited food is biotic: some organisms get less food.
对有限食物的竞争也是生物因素：有些生物得到的食物更少。
Classify a factor by what it is, then explain its effect.
先按因素的性质分类，再解释它的影响。
Name the three types of adaptation, with one example of each.
说出三种适应类型，并各举一个例子。
Separate a body feature from an action, and choose an internal process for the functional example.
把身体特征与行为区分开，功能适应的例子要选一个体内过程。
Pause and try it.
暂停并试着回答。
Structural: thick blubber reduces heat transfer in cold conditions.
结构适应：厚厚的脂肪层在寒冷环境中减少热量散失。
Behavioural: migration changes the conditions or resources an animal meets.
行为适应：迁徙改变动物所面对的环境条件或资源。
Functional: venom production is an internal process that can help catch prey.
功能适应：产生毒液是一种体内过程，有助于捕获猎物。
The shape of a fang would be structural.
毒牙的形状则属于结构适应。
Organisms do not choose an inherited feature because they need it.
生物不会因为需要就选择某种遗传特征。
This exam drawing shows a hoverfly and a wasp.
这幅考题图画的是食蚜蝇和黄蜂。
Both have yellow and black markings, but the hoverfly has no sting.
二者都有黄黑相间的斑纹，但食蚜蝇没有螫针。
Both insects have yellow and black markings and live in the same habitat.
两种昆虫都有黄黑斑纹，生活在同一栖息地。
The wasp has a sting; the hoverfly does not.
黄蜂有螫针，食蚜蝇没有。
Explain how the markings help the hoverfly survive.
解释这些斑纹如何帮助食蚜蝇生存。
Give three linked reasons, connecting its appearance to how predators behave.
给出三个相互联系的理由，把它的外观与捕食者的行为联系起来。
Pause and try it.
暂停并试着回答。
The hoverfly looks like the wasp, because their markings are similar.
食蚜蝇看起来像黄蜂，因为它们的斑纹相似。
Predators avoid wasps, so that they do not get stung.
捕食者会避开黄蜂，以免被螫。
So predators are less likely to attack or eat the hoverfly.
所以捕食者攻击或吃掉食蚜蝇的可能性降低。
Saying only that the predator 'thinks it is a wasp' is not enough: explain the avoided sting and the reduced attack.
只说捕食者“以为它是黄蜂”是不够的，要解释避免被螫以及受攻击减少。
An animal keeps a steady body temperature in cold water, and has a thick insulating layer.
一种动物在冷水中保持稳定的体温，并有一层厚厚的隔热层。
Classify this adaptation, explain how insulation affects heat transfer, and explain why a different environment could change the benefit.
判断这是哪种适应，解释隔热层如何影响热量传递，并解释为什么环境不同时，这种好处可能改变。
Pause and try it.
暂停并试着回答。
A thick insulating layer is a structural adaptation.
厚厚的隔热层属于结构适应。
Insulation reduces the rate of heat transfer to the cold surroundings.
隔热层降低向寒冷环境传递热量的速率。
So less energy is needed to keep the body warm in that setting.
所以在这种环境中，维持体温所需的能量较少。
An adaptation's benefit depends on the environment; it is not the best feature everywhere.
适应的好处取决于环境；它并不是在任何地方都是最好的特征。
Bacteria living at deep-sea vents face high temperature, high pressure and no light.
生活在深海热泉口的细菌面临高温、高压和无光的环境。
Explain what an extremophile is, and suggest where these bacteria could get their energy.
解释什么是极端微生物，并推测这些细菌可能从哪里获得能量。
Pause and try it.
暂停并试着回答。
An extremophile survives conditions that are extreme for most organisms.
极端微生物能在对大多数生物来说极端的环境中生存。
High temperature and pressure fit the vent bacteria example.
高温和高压正符合热泉细菌这个例子。
As an extension: some vent bacteria use energy released by chemical reactions to make their food, which is called chemosynthesis; they need no sunlight.
拓展内容：有些热泉细菌利用化学反应释放的能量来制造食物，这叫化能合成作用，不需要阳光。
Chemical reactions release energy; they do not create it.
化学反应释放能量，而不是创造能量。
Name the organism at the start of every food chain, and what it produces.
说出每条食物链开头的生物，以及它产生什么。
Give its trophic role and the substance it makes, not just the name of a plant.
要写出它的营养级角色和它制造的物质，而不只是一种植物的名称。
Pause and try it.
暂停并试着回答。
A green plant or alga is the producer: it makes glucose by photosynthesis and builds biomass.
绿色植物或藻类是生产者：通过光合作用制造葡萄糖，形成生物量。
For example: grass, to rabbit, to fox.
例如：草、兔子、狐狸。
The arrows show food passing to the consumer, so the rabbit is the primary consumer, and the fox is the secondary consumer in this chain.
箭头表示食物传给消费者，所以在这条食物链中，兔子是初级消费者，狐狸是次级消费者。
A model of predator and prey numbers over time.
一个捕食者和猎物数量随时间变化的模型。
Look at where the peaks fall: each prey peak comes before the matching predator peak.
看看峰值的位置：每个猎物数量的峰值都出现在对应的捕食者峰值之前。
Explain, in order, why predator and prey numbers rise and fall out of step.
按顺序解释，为什么捕食者和猎物的数量升降不同步。
Start when prey are plentiful, and continue until the prey can recover.
从猎物很多时讲起，一直讲到猎物数量能够恢复。
Pause and try it.
暂停并试着回答。
One: when prey are plentiful, more food supports predator survival and reproduction, so predator numbers rise.
第一，猎物多时，充足的食物使捕食者更易存活和繁殖，捕食者数量上升。
Two: more prey are eaten, so prey numbers fall.
第二，更多猎物被吃掉，猎物数量下降。
Three: predators now have less food, so their population falls.
第三，捕食者的食物减少，种群数量下降。
Four: fewer predators let the prey recover, and the predators respond after a delay.
第四，捕食者减少，猎物得以恢复，而捕食者的变化要滞后一段时间。
In the model, the first prey peak is at month three, and the matching predator peak is at month five.
在模型中，第一个猎物峰值在第三个月，对应的捕食者峰值在第五个月。
Calculate the delay.
计算二者之间的延迟。
Pause and try it.
暂停并试着计算。
The prey peak is at three months, and the predator peak at five months.
猎物峰值在三个月，捕食者峰值在五个月。
Delay equals predator peak time minus prey peak time.
延迟等于捕食者峰值时间减去猎物峰值时间。
Five months minus three months is two months.
五个月减三个月，得两个月。
The vertical axis has no scale, so this model gives no population counts.
纵轴没有刻度，所以这个模型不提供种群数量。
Two ways to place quadrats.
放置样方的两种方式。
Random positions spread across an area estimate a population.
在整个区域内随机放置，用来估算种群数量。
Regular positions along a transect line show how a species changes along a gradient.
沿样线等距放置，用来显示物种沿某种梯度的变化。
Choose the positions to match the question.
根据问题选择样方位置。
Describe a transect investigation of how light intensity affects the distribution of a plant species.
描述用样线法探究光照强度如何影响某种植物分布的实验。
Say where the samples go, what you measure, and how you use repeats.
说明样方放在哪里、测量什么，以及如何利用重复实验。
Pause and plan it.
暂停并构思。
Lay a tape from shaded ground into open ground, and place a quadrat at regular distances along it.
从阴影处向开阔地拉一条卷尺，沿卷尺每隔一定距离放一个样方。
At each position, count the plants or estimate their percentage cover, and measure the light intensity with a light meter.
在每个位置，数出植物数量或估计覆盖百分比，并用光度计测量光照强度。
Repeat the transect, calculate means at matching positions, and relate abundance to the measured light.
重复做几条样线，计算对应位置的平均值，把植物的多少与测得的光照联系起来。
Keep the quadrat size and counting rules the same.
样方大小和计数规则要保持一致。
How would you choose quadrat positions to estimate a population across a whole field?
要估算整块田地中的种群数量，你会如何选择样方位置？
Describe how the positions are selected, and explain why that choice reduces sampling bias.
描述如何选定位置，并解释这种选择为什么能减少取样偏差。
Pause and try it.
暂停并试着回答。
For a population estimate, use random coordinates over the whole area, then multiply the mean density by the total area.
估算种群数量时，在整个区域内用随机坐标取样，再用平均密度乘以总面积。
For a distribution along a gradient, use regular positions on a transect, and relate local abundance to the factor.
研究沿梯度的分布时，沿样线等距取样，把各处的数量与该因素联系起来。
To improve precision, take more samples or repeat the transects: means reduce the effect of variation.
要提高精确度，就多取样或重复样线：平均值能减小变异的影响。
Five quadrats gave these plant counts, already in ascending order: four, six, six, eight and eleven.
五个样方的植物计数如下，已按从小到大排列：四、六、六、八、十一。
The counts are four, six, six, eight and eleven.
计数为四、六、六、八、十一。
Calculate the mean, the mode and the median.
计算平均数、众数和中位数。
Use all five counts for the mean.
求平均数时要用上全部五个计数。
Pause and try it.
暂停并试着计算。
The mean is the sum of the counts divided by the number of quadrats.
平均数等于计数之和除以样方数。
Four plus six plus six plus eight plus eleven is thirty-five; divided by five, that is seven plants per quadrat.
四加六加六加八加十一等于三十五，除以五，得到每个样方七株植物。
Six occurs twice and every other count once, so the mode is six.
六出现两次，其他计数各出现一次，所以众数是六。
There are five ordered values, so the third is the middle: the median is also six.
共有五个排好序的数值，第三个位于中间，所以中位数也是六。
They match here, but they are different statistics: the mean uses every value, the median is a position, and the mode is the most frequent value.
这里两者相同，但它们是不同的统计量：平均数用到每个数值，中位数是一个位置，众数是出现最多的数值。
Now an extension.
拓展题。
Random quadrats of nought point two five square metres have a mean of seven plants.
面积为零点二五平方米的随机样方，平均每个有七株植物。
The field is one hundred square metres.
田地面积为一百平方米。
Estimate its population: find the plants per square metre first, then scale to the field.
估算种群数量：先求每平方米的株数，再按田地面积放大。
Pause and try it.
暂停并试着计算。
We know the mean is seven plants per quadrat, and each quadrat is nought point two five square metres.
已知平均每个样方七株植物，每个样方面积为零点二五平方米。
Density equals the mean count divided by the quadrat area.
密度等于平均计数除以样方面积。
Seven divided by nought point two five gives twenty-eight plants per square metre.
七除以零点二五，得到每平方米二十八株。
The estimated population equals the density times the total area.
估算的种群数量等于密度乘以总面积。
Twenty-eight plants per square metre, times one hundred square metres, is two thousand eight hundred plants.
每平方米二十八株，乘以一百平方米，等于两千八百株。
This assumes the random samples represent the field; it is an estimate, not a complete count.
这假定随机样本能代表整块田地；它只是估计值，而不是完整的计数。
These exam results come from samples taken at increasing distances from a tree.
这些考题数据来自距离一棵树越来越远的取样点。
For each distance, the table gives the soil masses, the water percentage and the number of daisies.
表格给出每个距离处的土壤质量、含水百分比和雏菊数量。
Soil was heated at one hundred degrees Celsius, cooled and reweighed, and this was repeated until the mass stopped changing.
土壤在一百摄氏度下加热，冷却后再称量，如此重复直到质量不再变化。
Why repeat?
为什么要重复？
Think about whether water might remain after one heating.
想一想加热一次之后是否还可能残留水分。
Pause and try it.
暂停并试着回答。
Repeating to constant mass makes sure all the water is removed, so the soil is completely dry.
重复加热至质量恒定，是为了确保除去全部水分，使土壤完全干燥。
If a container is used, subtract its mass; this table already gives soil masses.
如果使用了容器，要减去容器质量；这张表已经给出土壤本身的质量。
Just writing 'remove water' does not explain why the drying is repeated.
只写“除去水分”不能解释为什么要重复干燥。
At twenty metres, the fresh soil is two hundred and fifty-two grams, the dried soil one hundred and eighty-one grams, and the water lost seventy-one grams.
在二十米处，新鲜土壤为二百五十二克，干土为一百八十一克，失去的水分为七十一克。
Calculate X, the water percentage, to the nearest whole number.
计算含水百分比X，结果取整数。
Use the fresh mass as the denominator.
分母要用新鲜土壤的质量。
Pause and try it.
暂停并试着计算。
The water mass is the fresh mass minus the dry mass.
水的质量等于新鲜质量减去干燥质量。
Two hundred and fifty-two grams minus one hundred and eighty-one grams is seventy-one grams.
二百五十二克减一百八十一克，等于七十一克。
The water percentage is the water mass divided by the fresh mass, times one hundred.
含水百分比等于水的质量除以新鲜质量，再乘以一百。
Seventy-one grams divided by two hundred and fifty-two grams, times one hundred, is twenty-eight point one seven percent.
七十一克除以二百五十二克，再乘以一百，得到百分之二十八点一七。
Round only at the end: X is twenty-eight percent to the nearest whole number.
只在最后一步取舍：X取整数为百分之二十八。
Two more parts.
还有两问。
Describe the water percentage from four to sixteen metres.
描述从四米到十六米的含水百分比。
Then suggest two changes that would improve the conclusion that more water is linked to more daisies.
然后提出两项改进，使“水分越多雏菊越多”这一结论更可靠。
Keep the description separate from any explanation.
描述和解释要分开。
Pause and try it.
暂停并试着回答。
From four to sixteen metres, the water percentage increases with distance, from ten to twenty-six percent.
从四米到十六米，含水百分比随距离增加，由百分之十升到百分之二十六。
An increasing trend is not the same as direct proportion.
上升趋势并不等于正比关系。
To improve the conclusion: repeat at each distance and calculate a mean, and sample more trees or different directions, or add intermediate distances.
改进结论的方法：在每个距离重复取样并计算平均值；对更多树或不同方向取样，或增加中间的距离。
Water and daisies increasing together is an association in these samples, because both vary with distance.
水分和雏菊同时增加，只是这些样本中的相关关系，因为二者都随距离变化。
That water caused the increase is not established by this alone: light and other factors may change too.
仅凭这些数据不能证明是水分导致雏菊增多：光照等其他因素也可能变化。
And one transect cannot estimate the whole field, because a chosen gradient is not a random sample.
一条样线也不能估算整块田地，因为特意选定的梯度不是随机样本。
The carbon cycle.
碳循环。
Each arrow is a transfer of carbon by a process: photosynthesis takes carbon dioxide from the air, feeding passes carbon compounds along, and respiration, decay and combustion return carbon dioxide.
每个箭头表示由某一过程完成的碳转移：光合作用从空气中吸收二氧化碳，取食使含碳化合物沿食物链传递，呼吸作用、分解作用和燃烧把二氧化碳返回大气。
Trace a carbon atom from carbon dioxide in the air into a plant, then into an animal, and back to the atmosphere.
追踪一个碳原子：从空气中的二氧化碳进入植物，再进入动物，最后回到大气。
Name the process at each transfer.
说出每一次转移的过程。
Pause and try it.
暂停并试着回答。
Photosynthesis: the plant uses carbon dioxide to make glucose and other carbon compounds.
光合作用：植物利用二氧化碳合成葡萄糖和其他含碳化合物。
Feeding: the animal eats plant material, so carbon compounds pass to the consumer.
取食：动物吃植物，含碳化合物传给消费者。
Respiration: plants, animals and decomposers all respire, releasing carbon dioxide to the air.
呼吸作用：植物、动物和分解者都进行呼吸，释放二氧化碳到空气中。
Respiration is not only for animals.
呼吸作用不只是动物才有。
Explain how decomposers return the carbon in fallen leaves so that living plants can use it again.
解释分解者如何把落叶中的碳归还，使活的植物能再次利用。
Describe how large molecules are broken down, then the route back to plants.
描述大分子如何被分解，以及碳回到植物的途径。
Pause and try it.
暂停并试着回答。
Microorganisms secrete enzymes onto the dead material, breaking large molecules into small soluble ones that they absorb.
微生物把酶分泌到死亡的物质上，把大分子分解成可溶的小分子，再吸收它们。
The decomposers respire, releasing carbon dioxide into the air.
分解者进行呼吸，把二氧化碳释放到空气中。
Plants then take in that carbon dioxide and use it in photosynthesis to make glucose.
植物再吸收这些二氧化碳，通过光合作用合成葡萄糖。
Keep three routes apart.
要区分三条途径。
Carbon in dead material: decomposition and respiration return carbon dioxide to the air.
死亡物质中的碳：分解和呼吸作用把二氧化碳返回空气。
Mineral nutrients in dead material: decomposers recycle them, returning mineral ions to the soil.
死亡物质中的矿质营养：由分解者回收，矿质离子回到土壤。
Carbon in fuel: combustion returns carbon dioxide to the air.
燃料中的碳：燃烧把二氧化碳返回空气。
A six-mark question.
一道六分题。
Fallen leaves contain carbon and nitrogen compounds.
落叶中含有碳化合物和氮化合物。
Describe how these are recycled and used by living trees.
描述它们如何被回收，并被活的树利用。
Plan both routes: carbon to new plant material, and nitrogen to new plant material.
两条途径都要构思：碳如何变成新的植物物质，氮如何变成新的植物物质。
Pause and plan it.
暂停并构思。
Bacteria and fungi secrete enzymes that digest the large molecules in the dead leaves.
细菌和真菌分泌酶，消化落叶中的大分子。
These microorganisms respire, releasing carbon dioxide into the air.
这些微生物进行呼吸，把二氧化碳释放到空气中。
The carbon dioxide enters the tree's leaves through the stomata and is used in photosynthesis to make glucose, then other plant material.
二氧化碳通过气孔进入树叶，用于光合作用合成葡萄糖，再合成其他植物物质。
Connect each step to the next; a list of process names is not enough.
每一步都要与下一步联系起来；只罗列过程名称是不够的。
The nitrogen in dead leaves is recycled into nitrate ions in the soil.
落叶中的氮被回收，成为土壤中的硝酸根离子。
The roots absorb the nitrate ions, partly by active transport.
根吸收硝酸根离子，部分通过主动运输。
The tree uses the nitrogen to make amino acids and proteins for growth.
树利用这些氮合成氨基酸和蛋白质，用于生长。
This answer does not need the detailed nitrogen cycle.
这道题不需要写出详细的氮循环。
Explain how water from the sea can become fresh water available to plants on land.
解释海水如何变成陆地上植物可以利用的淡水。
Explain how water enters and leaves the air, and how it returns to the sea.
说明水如何进入和离开空气，又如何回到海洋。
Pause and try it.
暂停并试着回答。
Evaporation forms water vapour and leaves the dissolved salts behind; plants also lose water vapour by transpiration.
蒸发形成水蒸气，溶解的盐留在海里；植物也通过蒸腾作用散失水蒸气。
Cooling causes condensation into droplets, and precipitation returns water to the land and sea.
冷却使水蒸气凝结成小水滴，降水使水回到陆地和海洋。
Run-off and drainage carry water back towards the sea, and plants absorb water from the soil.
地表径流和排水把水带回海洋，植物则从土壤中吸收水分。
Four corrections.
四处纠正。
Plants do not only photosynthesise: they also respire.
植物不只进行光合作用，也进行呼吸作用。
Decay does not send carbon straight to animals: decomposers absorb the breakdown products and respire.
分解作用不会把碳直接送到动物体内：分解者吸收分解产物并进行呼吸。
Mineral ions belong to the soil, not to the atmospheric carbon dioxide route.
矿质离子属于土壤，与大气中二氧化碳的途径不同。
And materials recycle, but energy does not: it transfers and dissipates.
物质可以循环，能量却不能：能量只会传递和散失。
Name the gas produced by anaerobic decay, and give one use for it.
说出厌氧分解产生的气体，并说出它的一种用途。
Say how it can supply useful energy.
说明它如何提供有用的能量。
Pause and try it.
暂停并试着回答。
Anaerobic means without oxygen.
厌氧是指没有氧气。
Anaerobic decay produces methane, which a biogas generator collects.
厌氧分解产生甲烷，由沼气发生器收集。
The methane can be burned as a fuel for heating or for generating electricity.
甲烷可以作为燃料燃烧，用于供暖或发电。
The generator produces and collects the gas; burning the fuel releases its energy.
发生器产生并收集气体；燃烧燃料才释放能量。
A gardener wants vegetable waste to decay quickly into compost.
一位园丁想让菜叶等废料快速分解成堆肥。
Explain three conditions to provide, for four marks.
解释应提供的三个条件，共四分。
Consider temperature and moisture, and the value of aerating the heap.
考虑温度和湿度，以及给堆肥通气的作用。
Pause and try it.
暂停并试着回答。
Warmth, near the enzymes' optimum, speeds up enzyme-controlled reactions; too hot can denature the enzymes.
温暖、接近酶的最适温度，能加快酶促反应；温度过高会使酶变性。
Moisture supports microbial activity, but waterlogging can restrict oxygen.
湿润有利于微生物活动，但积水会限制氧气。
Turning the heap supplies oxygen, so aerobic decomposers can respire.
翻动堆肥可以提供氧气，使需氧分解者能进行呼吸。
Required practical ten: the effect of temperature on decay.
必做实验十：温度对分解速率的影响。
Each beaker holds twenty-five cubic centimetres of the same fresh milk, kept at five, fifteen or twenty-five degrees Celsius.
每个烧杯装二十五立方厘米同一种鲜奶，分别保持在五、十五和二十五摄氏度。
The pH is measured each day for four days, at consistent times.
连续四天，每天在固定时间测量pH。
A falling pH shows the milk decaying.
pH下降说明牛奶在分解。
Every sample starts at pH six point eight.
所有样品起始pH都是六点八。
After one day, the twenty-five degree milk has fallen to five point three, while the five degree milk is still at six point five.
一天后，二十五摄氏度的牛奶降到五点三，而五摄氏度的仍为六点五。
By day four, the warmest milk has levelled off at four point eight, the fifteen degree milk has reached four point eight, and the coldest is at five point six.
到第四天，最温暖的牛奶稳定在四点八，十五摄氏度的也降到四点八，最冷的是五点六。
Suggest one improvement to the milk-decay method.
为牛奶分解实验提出一项改进。
What would repeats let you calculate?
重复实验能让你计算什么？
Could more frequent measurements show the curve better?
更频繁的测量能否更好地显示曲线？
Pause and try it.
暂停并试着回答。
Repeat at each temperature and calculate a mean.
在每个温度下重复实验并计算平均值。
Or measure the pH at shorter time intervals, or log it continuously.
或者缩短测量pH的时间间隔，或用数据记录器连续测量。
Testing more temperatures is also accepted.
测试更多温度也可以。
One valid improvement earns the mark.
答出一项合理的改进即可得分。
This graph shows the five and fifteen degree results redrawn, with pH against time in days.
这幅图重画了五摄氏度和十五摄氏度的结果，纵轴为pH，横轴为时间（天）。
The twenty-five degree curve is still missing.
二十五摄氏度的曲线还没有画。
Add the twenty-five degree points from the table, for days nought to four, and draw a curved line of best fit.
把表中二十五摄氏度第零到第四天的数据点画上，并画一条平滑的最佳拟合曲线。
Plot carefully on the labelled scales, and choose a curve that follows the data.
按标注的刻度仔细描点，选择贴合数据的曲线。
Pause and try it.
暂停并试着画。
Check your graph against this one.
把你的图和这幅图对照。
The blue points are the twenty-five degree results, joined by a smooth curve that falls steeply at first and then levels off.
蓝色的点是二十五摄氏度的结果，用一条平滑曲线连起来：先急剧下降，然后趋于平稳。
The twenty-five degree points are six point eight, five point three, four point nine, four point eight and four point eight, for days nought to four.
二十五摄氏度第零到第四天的数据为六点八、五点三、四点九、四点八和四点八。
Draw a smooth curved line of best fit; a single straight line is not suitable.
画一条平滑的最佳拟合曲线；一条直线是不合适的。
The last two values are equal, so do not force the curve to keep falling.
最后两个数值相等，不要强行让曲线继续下降。
Plots for days one to three earn up to two marks, and the curve earns one.
第一到第三天的描点最多得两分，曲线得一分。
At day two, how many times faster is the fifteen degree rate than the five degree rate, which is given as nought point three pH units per day?
在第二天，十五摄氏度时的速率是五摄氏度时的多少倍？ 五摄氏度的速率已给出，为每天零点三个pH单位。
Draw a tangent to the fifteen degree curve at day two, find its gradient from well-separated points, then divide.
在十五摄氏度曲线的第二天处画切线，用相距较远的两点求斜率，再相除。
Pause and try it.
暂停并试着计算。
Here is an example tangent, touching the fifteen degree curve at day two.
这是一条示例切线，在第二天处与十五摄氏度曲线相切。
Its gradient is the rate at that instant, not the average over all four days.
它的斜率是那一时刻的速率，而不是四天的平均速率。
Read two points on the tangent, not on the curve: day nought at pH six point five, and day four at pH four point five.
在切线上读两个点，而不是在曲线上：第零天pH六点五，第四天pH四点五。
The rate of decrease is the fall in pH divided by the time interval.
下降速率等于pH的下降量除以时间间隔。
Six point five minus four point five is two, over four days, giving nought point five pH units per day.
六点五减四点五等于二，除以四天，得每天零点五个pH单位。
The ratio is the rate at fifteen degrees divided by the rate at five degrees.
比值等于十五摄氏度的速率除以五摄氏度的速率。
Nought point five over nought point three is about one point six seven.
零点五除以零点三约为一点六七。
So the fifteen degree rate is about one point six seven times as fast; the ratio has no unit.
所以十五摄氏度的速率约是五摄氏度的一点六七倍；这个比值没有单位。
The gradient is negative because pH falls, so compare the sizes of the decreases.
因为pH在下降，斜率为负，所以要比较下降量的大小。
Milk contains lipids.
牛奶中含有脂质。
Explain why the pH changes more quickly at the higher temperature tested.
解释为什么在所测的较高温度下，pH变化得更快。
Explain what happens to lipid breakdown, and name the product that lowers the pH.
说明脂质分解发生了什么变化，并说出使pH降低的产物。
Pause and try it.
暂停并试着回答。
Within the range tested, the enzymes are more active at the higher temperature.
在所测的温度范围内，温度较高时酶的活性更强。
So the lipids are broken down more quickly.
因此脂质分解得更快。
Fatty acids are produced faster, so the pH falls faster.
脂肪酸产生得更快，pH也就下降得更快。
Amino acids are not products of lipid breakdown.
氨基酸不是脂质分解的产物。
The twenty-five degree pH levels off near four point eight: that shows only that the measured pH changes little after day three.
二十五摄氏度的pH在四点八附近趋于平稳：这只说明第三天之后测得的pH变化很小。
It does not show that all the bacteria have died.
它不能说明所有细菌都已死亡。
Nor does it show that more heating always speeds decay, because enzymes can denature.
也不能说明温度越高分解总是越快，因为酶会变性。
And a different tangent gives a slightly different rate, so use a valid tangent consistently.
换一条切线会得到略有不同的速率，所以要始终使用一条合理的切线。
For higher tier: give a seasonal change, a geographical change and a human-caused change that could alter where a species is found.
高级卷内容：举出一个季节性变化、一个地理性变化和一个人为变化，它们都可能改变某个物种的分布。
Name each change, then link it to survival, growth or movement.
说出每一种变化，并把它与存活、生长或迁移联系起来。
Pause and try it.
暂停并试着回答。
Seasonal: temperature changes through the year, so survival or seasonal movement may change.
季节性：一年中温度变化，存活情况或季节性迁移可能改变。
Geographical: water availability differs between areas, so plant growth may differ.
地理性：不同地区的水分条件不同，植物生长可能不同。
Human-caused: draining a wetland reduces the water, so water-dependent plants may decline locally.
人为：排干湿地使水分减少，依赖水的植物可能在当地减少。
These are possibilities to test with evidence, not proof of how one species responds.
这些都是需要用证据检验的可能性，并不能证明某个物种一定这样反应。
Define biodiversity.
给生物多样性下定义。
Could a field of a million wheat plants still have low variety of species?
一块有一百万株小麦的田地，物种多样性是否仍可能很低？
Pause and try it.
暂停并试着回答。
Biodiversity is the variety of different species on Earth, or within an ecosystem.
生物多样性是指地球上或某一生态系统中不同物种的多样程度。
A million wheat plants are still only one species.
一百万株小麦仍然只是一个物种。
So a species count measures variety, but a count of all the organisms does not.
所以物种数能衡量多样性，而生物个体总数不能。
Use the word species in your definition.
定义中要用到“物种”这个词。
Give two reasons why high biodiversity makes an ecosystem more stable.
给出两个理由，说明为什么高生物多样性使生态系统更稳定。
Think about food and shelter, and about maintaining the physical environment.
想一想食物和栖身之所，以及对物理环境的维持。
Pause and try it.
暂停并试着回答。
A wider variety reduces how much organisms depend on one species for food or shelter.
物种更多样，生物对某一个物种作为食物或栖身之所的依赖就更小。
And different species help to maintain the physical environment.
不同的物种还有助于维持物理环境。
This supports stability, but it is no guarantee: losing a species can still have serious effects.
这有利于稳定，但并不能保证：失去一个物种仍可能造成严重影响。
A five-mark question.
一道五分题。
Explain how human population growth leads to more pollution, and name where the pollution can occur.
解释人口增长如何导致更多污染，并说出污染可能发生在哪里。
Link resource use to waste, then name the water, air and land routes.
把资源消耗与废物联系起来，再说出水、空气和土地这三种途径。
Pause and try it.
暂停并试着回答。
A larger population uses more resources, and rising living standards can raise consumption further.
人口增加会消耗更多资源，生活水平提高还会进一步增加消费。
More waste is produced, and without suitable treatment or disposal it can pollute.
产生的废物更多，如果没有适当的处理或处置，就会造成污染。
Pollution can kill organisms and reduce biodiversity.
污染会杀死生物，降低生物多样性。
In water: sewage, fertiliser and toxic chemicals can harm aquatic organisms.
水体中：污水、化肥和有毒化学物质会危害水生生物。
In air: smoke and acidic gases can harm plants and animals.
空气中：烟尘和酸性气体会危害植物和动物。
On land: landfill and toxic chemicals can remove or contaminate habitat.
土地上：垃圾填埋和有毒化学物质会破坏或污染栖息地。
Another five-mark question.
又一道五分题。
Explain why large-scale deforestation happens, and give three environmental consequences.
解释为什么会发生大规模砍伐森林，并给出三个环境后果。
Name the farming uses of the cleared land, then explain the habitat and carbon consequences.
说出被清理土地的农业用途，再解释对栖息地和碳的影响。
Pause and try it.
暂停并试着回答。
The reason: land is needed to raise cattle, or to grow rice or biofuel crops.
原因：需要土地养牛，或种植水稻和生物燃料作物。
The consequences: habitat loss, so local populations may die or migrate; fewer photosynthesising trees, so less carbon dioxide is removed; and burning or decay of the wood, which releases carbon dioxide and adds to warming.
后果：栖息地丧失，当地种群可能死亡或迁走；进行光合作用的树木减少，吸收的二氧化碳减少；木材燃烧或腐烂释放二氧化碳，加剧全球变暖。
This exam diagram shows a section through peat, with layer A above layer B.
这幅考题图显示泥炭的一个剖面，A层在B层上方。
The conditions in each layer are listed on the next frame.
每一层的条件列在下一页。
Layer A contains a lot of air.
A层含有大量空气。
Layer B is made of the dead remains of plants.
B层由植物的遗体组成。
Its conditions are a low pH and very little oxygen, and it stores carbon dioxide and methane.
它的条件是pH低、氧气极少，其中储存着二氧化碳和甲烷。
Explain why the dead plant remains in layer B do not decay, for three marks.
解释为什么B层中的植物遗体不会分解，共三分。
Use the low oxygen concentration; another route uses the low pH and enzymes.
利用氧气浓度低这一点；另一种思路是利用pH低和酶。
Pause and try it.
暂停并试着回答。
Very little oxygen limits aerobic respiration in the decomposers.
氧气极少，限制了分解者的有氧呼吸。
So less energy is released for the activity of microorganisms such as bacteria and fungi.
因此，细菌和真菌等微生物进行活动所释放的能量减少。
Alternatively, the low pH can reduce enzyme activity, and so slow the microbes' metabolic reactions.
另一种解释是：pH低会降低酶的活性，从而减慢微生物的代谢反应。
The respiration that matters is in the decomposers, not the dead plants.
这里关键的是分解者的呼吸，而不是死去植物的呼吸。
Peat is dug up for fuel and for compost.
人们挖掘泥炭作燃料和堆肥。
Explain how removing it can affect the temperature of the atmosphere, for four marks.
解释移除泥炭如何影响大气温度，共四分。
Separate the release of trapped gas from combustion, and complete the link to warming.
把被困气体的释放与燃烧区分开，并完整说明与变暖的联系。
Pause and try it.
暂停并试着回答。
Removing peat can release the carbon dioxide and methane trapped inside it.
移除泥炭会释放其中被困的二氧化碳和甲烷。
Burning peat also produces carbon dioxide, and microbial decay can release more.
燃烧泥炭也会产生二氧化碳，微生物分解还会释放更多。
These greenhouse gases trap heat, so the temperature of the atmosphere increases.
这些温室气体吸收热量，使大气温度升高。
A six-mark question.
一道六分题。
Explain how destroying peat affects biodiversity, both locally and globally.
解释破坏泥炭如何影响本地和全球的生物多样性。
Plan two linked chains: locally, from habitat to populations; globally, from greenhouse gases to habitat change.
构思两条相互联系的链条：本地方面，从栖息地到种群；全球方面，从温室气体到栖息地变化。
Pause and plan it.
暂停并构思。
Destroying peat removes a specialised habitat.
破坏泥炭会毁掉一种特殊的栖息地。
Food and shelter are lost, so local populations may die or migrate.
食物和栖身之所丧失，当地种群可能死亡或迁走。
Local species variety can fall, though losing a local population is not automatically a global extinction.
当地的物种多样性可能下降，但本地种群消失并不等于全球灭绝。
Removal releases trapped carbon dioxide and methane, and burning peat produces carbon dioxide.
移除泥炭会释放被困的二氧化碳和甲烷，燃烧泥炭会产生二氧化碳。
Decomposers respire, releasing carbon dioxide, and anaerobic decay can produce methane.
分解者呼吸释放二氧化碳，厌氧分解还会产生甲烷。
The warming changes conditions such as rainfall, temperature or sea level.
变暖改变降雨、气温或海平面等条件。
Habitats and food change elsewhere, so migration or extinction of species may follow.
其他地方的栖息地和食物随之改变，物种可能迁徙甚至灭绝。
Name two programmes that help to maintain biodiversity.
说出两项有助于维持生物多样性的计划。
One could target a species at risk; another could target habitat loss or wasted resources.
一项可以针对濒危物种，另一项可以针对栖息地丧失或资源浪费。
Pause and try it.
暂停并试着回答。
Breeding programmes for endangered species support populations at risk.
濒危物种繁育计划帮助处于危险中的种群。
Protecting or regenerating rare habitats keeps or restores suitable conditions.
保护或恢复稀有栖息地，可以保持或恢复适宜的条件。
Field margins and hedgerows add habitat to single-crop farmland.
田边地带和树篱为单一作物农田增加栖息地。
Reducing deforestation and carbon dioxide emissions cuts habitat loss and warming.
减少砍伐森林和二氧化碳排放，可以减少栖息地丧失和变暖。
And recycling rather than landfill reduces resource extraction and waste.
用回收代替填埋，可以减少资源开采和废物。
A simple model of a single-crop field, before and after hedgerows are planted around it.
一个单一作物农田的简单模型，显示在周围种上树篱之前和之后的情况。
It shows added habitat, not measured species counts.
它表示增加的栖息地，而不是实测的物种数量。
Farmers replant hedgerows around single-crop fields.
农民在单一作物农田周围重新种植树篱。
Explain how this affects biodiversity, for four marks.
解释这对生物多样性有什么影响，共四分。
What can different plants provide?
不同的植物能提供什么？
And how might connected hedgerows affect movement?
相连的树篱又会如何影响动物的移动？
Pause and try it.
暂停并试着回答。
Hedgerows can add a variety of plant species.
树篱能增加多种植物。
Different foods, shelter and nesting sites can then support different animal species.
不同的食物、栖身之所和筑巢地点就能养活不同的动物物种。
Connected hedgerows can let animals move between habitats, though the effects depend on how they are managed.
相连的树篱能让动物在栖息地之间移动，但效果取决于如何管理。
This exam graph comes from rainforest areas where the trees were cut down.
这幅考题图来自树木被砍伐后的雨林地区。
Each point compares the tree species in a regenerating area with a nearby undisturbed control.
每个点都把恢复区的树木物种数与附近未受干扰的对照区进行比较。
A closer view of the lower part of the graph, where the recently cleared areas sit.
放大图的下半部分，这里是最近被砍伐的地区。
The percentages and years are unchanged; refer back to the full graph for the axes and key.
百分比和年份都没有改变；坐标轴和图例请参照完整的图。
And the upper part, from eighty to one hundred and sixty percent.
再看上半部分，从百分之八十到百分之一百六十。
The points keep their positions along the time axis.
各点沿时间轴的位置不变。
The horizontal axis shows the years since the trees were cut down.
横轴表示树木被砍伐后的年数。
The vertical axis shows the number of tree species as a percentage of the control.
纵轴表示树木物种数占对照区的百分比。
Each circle is one result for one study area.
每个圆圈代表一个研究区域的一个结果。
One hundred percent means the same number of species as the control, but not necessarily the same species.
百分之一百表示物种数与对照区相同，但不一定是相同的物种。
Two questions.
两个问题。
Why are the results given as percentages?
为什么结果用百分比表示？
And give a conclusion other than 'biodiversity increases'.
除了“生物多样性增加”之外，再给出一个结论。
Think about the different starting numbers of tree species, and how the rate of recovery changes.
想一想各地区原有的树木物种数不同，以及恢复速度如何变化。
Pause and try it.
暂停并试着回答。
Percentages let you compare areas that started with different numbers of tree species.
用百分比可以比较原有树木物种数不同的地区。
Recovery is rapid at first and then slows, and results vary between areas at similar times.
恢复起初很快，随后变慢，而且在相近时间不同地区的结果差别较大。
Matching the control's species count does not prove that the original species have returned.
物种数与对照区相同，并不能证明原来的物种已经恢复。
Say species, not just trees.
要写“物种”，而不只是“树木”。
Explain why tree biodiversity affects animal biodiversity.
解释为什么树木的多样性会影响动物的多样性。
What kinds of food might become available, and what kinds of habitat might be added?
可能会有哪些种类的食物？ 又可能增加哪些种类的栖息地？
Pause and try it.
暂停并试着回答。
Different tree species can provide more types of food. They can also provide more types of shelter or nesting sites.
不同的树种可以提供更多种类的食物，也可以提供更多种类的栖身之所或筑巢地点。
These resources can support a wider variety of animal species.
这些资源能养活更多种类的动物。
Saying only 'more food' does not earn the mark: it is the variety that matters.
只说“更多食物”不得分：关键在于种类多样。
Protecting a habitat from farming keeps its species and their resources, but the land may be needed for food or income.
保护栖息地不被开垦，能保留其中的物种及其资源，但这片土地可能需要用来生产粮食或获得收入。
Restoring hedgerows adds habitat and connections, but costs money and crop area.
恢复树篱能增加栖息地和连通性，但要花钱，还会占用耕地。
To make a recommendation, use the evidence you are given, and state a limitation or trade-off.
提出建议时要依据所给的证据，并说明一项局限或权衡。
No costs are given here, so do not invent them.
这里没有给出成本数据，不要自己编造。
Name trophic levels one to four.
说出第一到第四营养级。
Who makes organic food using light, and who eats the herbivores?
谁利用光制造有机物？ 谁吃草食动物？
Pause and try it.
暂停并试着回答。
Level one: producers, the plants and algae that make food.
第一营养级：生产者，即制造食物的植物和藻类。
Level two: primary consumers, herbivores that eat producers.
第二营养级：初级消费者，吃生产者的草食动物。
Level three: secondary consumers, carnivores that eat herbivores.
第三营养级：次级消费者，吃草食动物的肉食动物。
Level four: tertiary consumers, carnivores that eat other carnivores.
第四营养级：三级消费者，吃其他肉食动物的肉食动物。
An apex predator has no predators of its own.
顶级捕食者没有天敌。
Decomposers secrete enzymes onto dead material.
分解者把酶分泌到死亡的物质上。
Small soluble food molecules then diffuse into the microorganisms.
可溶的小分子食物随后扩散进入微生物体内。
The material can come from any trophic level, not only the final predator, so decomposers are not just a last link on the chain.
这些物质可以来自任何营养级，而不只是来自最后的捕食者，所以分解者并不只是食物链的最后一环。
Four thousand two hundred kilograms of maize seeds feed one thousand chickens, with a mean mass of two point two kilograms each.
四千二百千克玉米种子喂养一千只鸡，每只鸡平均质量二点二千克。
Construct the pyramid of biomass.
画出生物量金字塔。
Calculate the total chicken biomass first, and put the producer at the bottom.
先算出鸡的总生物量，并把生产者放在最下面。
Pause and try it.
暂停并试着画。
The total chicken biomass is one thousand times two point two kilograms, which is two thousand two hundred kilograms.
鸡的总生物量等于一千乘以二点二千克，即两千二百千克。
Choose a scale that is symmetrical around zero, and label the axis biomass in kilograms.
选择以零为中心对称的刻度，坐标轴标注为生物量（千克）。
Draw the maize seeds, four thousand two hundred kilograms, at the bottom, and the chickens, two thousand two hundred kilograms, above.
下面画玉米种子，四千二百千克；上面画鸡，两千二百千克。
The bar width shows biomass; label both bars.
条形的宽度表示生物量；两个条形都要标注。
Compare your pyramid with this scaled answer.
把你的金字塔和这个按比例画出的答案对照。
The chicken bar is just over half the width of the maize bar.
鸡的条形宽度略多于玉米条形的一半。
Calculate the ratio of chicken biomass to maize seed biomass, in its simplest form.
计算鸡的生物量与玉米种子生物量之比，并化为最简形式。
Write the chicken mass first, then divide both terms by a common factor.
先写鸡的质量，再用公因数同时除两项。
Use these masses, not an assumed ten percent.
要用题目给出的质量，而不是假定的百分之十。
Pause and try it.
暂停并试着计算。
The ratio is two thousand two hundred to four thousand two hundred.
比为两千二百比四千二百。
Divide both by two hundred, giving eleven to twenty-one.
两项都除以二百，得十一比二十一。
The first term is smaller, as it should be.
第一项较小，这是合理的。
These masses imply about fifty-two percent, not the typical ten percent, so always use the data supplied.
这些质量对应约百分之五十二，而不是通常的百分之十，所以一定要使用题目给出的数据。
Five hundred kilograms of primary consumers support forty kilograms of secondary consumers.
五百千克初级消费者养活四十千克次级消费者。
Calculate the efficiency of the transfer.
计算这一传递的效率。
Write the equation, then substitute both masses and give a percentage.
先写出公式，再代入两个质量，结果用百分比表示。
Pause and try it.
暂停并试着计算。
Efficiency equals the biomass transferred divided by the biomass available, times one hundred percent.
效率等于传递的生物量除以可利用的生物量，再乘以百分之一百。
Forty kilograms over five hundred kilograms, times one hundred, is eight percent.
四十千克除以五百千克，乘以一百，得百分之八。
The mass units cancel, so the answer is a percentage.
质量单位相互抵消，所以答案是百分比。
There is one mark for the method and one for the value.
方法一分，数值一分。
Moss produces three hundred and forty grams per square metre per year.
苔藓每年每平方米产生三百四十克生物量。
Approximately how much becomes primary consumer biomass?
大约有多少会成为初级消费者的生物量？
Use the approximate ten percent transfer, and keep the rate units.
用约百分之十的传递率，并保留速率单位。
Pause and try it.
暂停并试着计算。
Ten percent is nought point one.
百分之十等于零点一。
Three hundred and forty times nought point one is thirty-four grams per square metre per year.
三百四十乘以零点一，等于每年每平方米三十四克。
The other figure you may know, about one percent of sunlight, is the light energy captured by producers, not this feeding transfer.
你可能知道的另一个数据，即约百分之一的阳光，指的是生产者捕获的光能，而不是这里的取食传递。
Explain why a food chain rarely has more than four or five trophic levels, for four marks.
解释为什么食物链很少超过四到五个营养级，共四分。
Which eaten material is not absorbed?
被吃下的物质中，哪些没有被吸收？
And which absorbed material leaves in respiration or in urine?
被吸收的物质中，哪些通过呼吸作用或尿液离开？
Pause and try it.
暂停并试着回答。
Egestion: unabsorbed material leaves in faeces, so it never becomes consumer biomass.
排遗：未被吸收的物质随粪便排出，不会成为消费者的生物量。
Respiration: glucose is used, and the material leaves as carbon dioxide and water.
呼吸作用：葡萄糖被消耗，物质以二氧化碳和水的形式离开。
Excretion: water and urea leave in urine.
排泄：水和尿素随尿液排出。
So only part becomes new growth, leaving less biomass to support the next level.
所以只有一部分变成新的生长，能养活下一营养级的生物量更少。
'About ten percent' is a summary; explain the losses.
“约百分之十”只是概括，要解释这些损失。
Chickens can use eighty percent of the maize biomass they eat for respiration and growth.
鸡吃下的玉米生物量中，有百分之八十可用于呼吸和生长。
What happens to the remaining twenty percent?
剩下的百分之二十去了哪里？
Name the route for unabsorbed material, and give the specific product, not just 'waste'.
说出未被吸收物质的去向，并写出具体的产物，而不只是“废物”。
Pause and try it.
暂停并试着回答。
The remaining material is egested as faeces, because it is never absorbed into the chicken.
剩下的物质以粪便的形式排遗，因为它从未被鸡吸收。
Urine, heat and respiration do not answer this question: respiration is already inside the eighty percent.
尿液、热量和呼吸作用都不是这道题的答案：呼吸作用已经包含在那百分之八十里。
A five-mark calculation.
一道五分计算题。
One thousand chickens each need forty-six megajoules.
一千只鸡每只需要四十六兆焦。
Maize gives sixteen megajoules per kilogram, but only eighty percent is usable, and one hectare yields sixteen point four tonnes.
玉米每千克提供十六兆焦，但只有百分之八十可以利用，每公顷产量为十六点四吨。
Find the area of maize needed in square metres, to three significant figures.
求所需玉米地的面积，单位为平方米，保留三位有效数字。
One hectare is ten thousand square metres.
一公顷等于一万平方米。
Pause and try it.
暂停并试着计算。
The energy required is one thousand times forty-six megajoules, which is forty-six thousand megajoules.
所需能量为一千乘以四十六兆焦，即四万六千兆焦。
The yield of sixteen point four tonnes per hectare is sixteen thousand four hundred kilograms per hectare.
每公顷十六点四吨的产量，就是每公顷一万六千四百千克。
The usable energy is sixteen thousand four hundred, times sixteen, times nought point eight: two hundred and nine thousand nine hundred and twenty megajoules per hectare.
可利用的能量为一万六千四百乘以十六，再乘以零点八，即每公顷二十万九千九百二十兆焦。
The area equals the energy required divided by the usable energy per hectare.
面积等于所需能量除以每公顷可利用的能量。
Forty-six thousand over two hundred and nine thousand nine hundred and twenty, times ten thousand, is two thousand one hundred and ninety-one point three square metres.
四万六千除以二十万九千九百二十，再乘以一万，得两千一百九十一点三平方米。
To three significant figures, the area is two thousand one hundred and ninety square metres.
保留三位有效数字，面积为两千一百九十平方米。
Keep values unrounded until the end.
中间数值不要提前取舍。
Food security means having enough food to feed the population.
粮食安全是指有足够的食物养活人口。
Choose two pressures on it, and explain how each changes the supply of food or the demand for it.
选择两种影响粮食安全的压力，并解释每一种如何改变食物的供应或需求。
Pause and try it.
暂停并试着回答。
An increasing birth rate means more people need food.
出生率上升意味着需要食物的人更多。
Changing diets can mean scarce food is transported elsewhere.
饮食习惯改变可能导致稀缺的食物被运往别处。
And new pests or pathogens can damage crops or farm animals.
新的害虫或病原体会损害农作物或家畜。
When rains fail or the environment changes, production can fall and famine may follow.
降雨不足或环境变化时，产量可能下降，甚至发生饥荒。
High costs of farm inputs can leave farmers unable to get what they need.
农业投入的成本高，农民可能无法得到所需物资。
And conflict can disrupt the supply of water or food.
冲突会破坏水和食物的供应。
Farmers limit the movement of cattle and control the temperature of their sheds.
农民限制牛的活动，并控制牛棚的温度。
Explain how each increases food production, and give an ethical objection.
解释每一项措施如何提高食物产量，并给出一个伦理上的反对理由。
Link the conditions to the animals' activity and respiration.
把这些条件与动物的活动和呼吸作用联系起来。
Pause and try it.
暂停并试着回答。
Less movement means less muscle activity, so less energy is transferred to the surroundings.
活动减少，肌肉活动就少，传递到周围环境的能量也少。
A suitable temperature means less extra respiration is needed to keep the body warm.
温度适宜时，维持体温所需的额外呼吸就少。
So more of the food can go into growth, although respiration still continues.
于是更多的食物可以用于生长，不过呼吸作用仍在进行。
But confinement and restricted movement raise animal-welfare objections.
但圈养和限制活动会引起动物福利方面的反对。
Intensive farming can be more efficient, but it raises animal welfare concerns and restricts natural behaviour.
集约化养殖效率更高，但会引起动物福利问题，并限制动物的自然行为。
Controlled surroundings cost energy and money.
控制环境需要消耗能源和资金。
High-protein feed supports growth, but judge it on the evidence you are given about costs and results.
高蛋白饲料能促进生长，但要依据所给的成本和效果证据来判断。
A recommendation needs both a mechanism and a justified judgement.
提出建议既要说明机制，也要有合理的判断。
A six-mark question.
一道六分题。
Explain how net size and fishing quotas help maintain fish stocks, then describe one biotechnology solution to the demand for food.
解释渔网网眼大小和捕捞配额如何帮助维持鱼类种群，再描述一种满足粮食需求的生物技术方法。
Why let young fish escape?
为什么要让幼鱼逃脱？
And which cultured organism provides a protein-rich food?
哪种培养的生物能提供富含蛋白质的食物？
Pause and try it.
暂停并试着回答。
Larger mesh openings let young fish escape, so they can grow and breed before being caught.
网眼更大，幼鱼就能逃脱，在被捕捞之前长大并繁殖。
A quota limits the total catch, leaving enough breeding adults.
配额限制捕捞总量，留下足够的繁殖成鱼。
The aim is a stock where breeding continues; recovery depends on the limits and how well they are enforced.
目的是使鱼群能够持续繁殖；能否恢复取决于限制措施以及执行的力度。
Name the fungus used to produce mycoprotein.
说出用于生产真菌蛋白的真菌。
Give its genus.
写出它的属名。
Pause and try it.
暂停并试着回答。
Fusarium is a fungus.
镰刀菌是一种真菌。
It is grown on glucose syrup in aerobic conditions.
它在有氧条件下以葡萄糖浆为养料培养。
The biomass is then harvested and purified to produce mycoprotein, a protein-rich food suitable for vegetarians.
然后收获并纯化生物量，制成真菌蛋白，一种适合素食者的高蛋白食物。
Fusarium culture produces mycoprotein, a protein-rich food.
培养镰刀菌可生产真菌蛋白，一种高蛋白食物。
Genetically modified crops such as golden rice can increase the food supply or improve nutrition.
转基因作物（例如黄金大米）可以增加粮食供应或改善营养。
Genetically modified bacteria make human insulin for diabetes: that meets a health need, not a direct food need, so it does not answer a question about hunger.
转基因细菌生产治疗糖尿病的人胰岛素：这满足的是健康需求，而不是直接的粮食需求，所以不能回答关于饥饿的问题。
Before you call this topic done — list what plants and animals compete for, the abiotic and biotic factors, the three adaptation types and the extremophile example. Draw the carbon cycle with decomposers, run the RP9 statistics, do a ten-percent biomass efficiency calculation, and name the six threats to food security with a solution each.
收尾自查——默写动植物的竞争项目、非生物与生物因素、三类适应和极端微生物例子；画出带分解者的碳循环；会做 RP9 的数据统计；完成 10% 生物量传递效率计算；列举粮食安全的六大威胁并各配一个对策。