APBIO-U08-L01 The pill bug and the damp side
Photo: Franco Folini, Wikimedia Commons, CC BY 2.5 (resized).
Here is a shallow dish. The paper on its left half is damp. The paper on its right half is dry. Ten pill bugs are tipped into the middle.
After two minutes, nine of the ten sit on the damp side. What did the pill bugs notice, and what did they do about it?
Unit 8 · Ecology
1What ecology studies
How does a whole animal answer a change around it? The pill bug detected a change, the dry air, and answered it with an action, the walk to the damp side.
The change the pill bug detected is the stimulus. The action is the response.
Stimulus and response are the same two words your body’s feedback loops used in Unit 4, one scale up.
Some responses are things the whole animal does, such as the walk. Other responses are changes inside its body.
Every answer an organism gives to its surroundings belongs to one branch of biology.
In the oak forest of the introduction live oaks, deer, beetles and fungi, on soil that gets rain and sunlight.
Which of the following is a population?
- A. ✓ The oaks of one kind in the forest
- B. The oaks, deer, beetles and fungi of the forest togetherSeveral kinds of organism living in one place together are a community.
A population is all the organisms of one species living in one place. - C. The oaks, deer, beetles and fungi together with the soil, rain and sunlightThe living things of a place together with their non-living surroundings are an ecosystem.
A population is one species in one place.
Why: A population is all the organisms of one species living in one place.
The oaks of one kind in the forest are one species in one place.
So they are a population.
Video: Watch: What ecology studies
The dish from above in time lapse: ten pill bugs tipped into the middle, most of them ending up on the damp half; then a log lifted in a forest and the pill bugs under it, on the damp soil.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L01a.mp4
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Biologists ask how the oaks, deer, beetles and fungi of that forest affect one another. They also ask how the soil, rain and sunlight affect those living things.
Suppose you lift a log in that forest. Under it you find forty pill bugs, all of them on the damp soil.
Two questions follow. How many pill bugs live under this one log is a question about a population of pill bugs. Why they gather where the soil is damp is a question about how pill bugs answer their surroundings.
The study of how living things interact with one another and with their surroundings is called .
Eco- comes from the Greek word for a home. -logy means a study. So ecology is the study of living things at home in their surroundings.
Ecology works at the three levels the introduction sorted the forest into: populations, communities and ecosystems.
Ecology is a branch of biology: the study of how living things interact with one another and with their surroundings.
What you are expected to know State what ecology is: the study of how living things interact with one another and with their surroundings, at the level of populations, communities and ecosystems.
Ecology is one branch of biology.
Which of the following is a question that ecology asks?
- A. How the muscles in one trout’s tail push it through the waterHow one trout’s muscles work is the biology of one organism.
Ecology asks how living things interact with one another and with their surroundings. - B. ✓ Why the trout in a stream gather where the water is coldest
- C. How a town can cut the plastic it throws awayCutting plastic waste is something people do.
Ecology is a branch of biology: the study of how living things interact.
Why: Ecology studies how living things interact with one another and with their surroundings.
Where the trout gather in the stream is a question about the trout and their surroundings.
So it is a question that ecology asks.
18Quick quiz: ecology mixed practice
A biologist counts the pill bugs under fifty logs in a forest and records how damp the soil is under each log.
Which branch of biology is this biologist working in?
- A. The study of how one cell worksCounting pill bugs under logs and recording the soil studies a population and its surroundings.
How living things interact with their surroundings is ecology, not the biology of one cell. - B. The study of how traits pass from parent to offspringCounting pill bugs under logs and recording the soil does not follow a trait from parent to offspring.
It asks how a population and its surroundings go together: ecology. - C. ✓ Ecology
Why: The biologist is counting a population of pill bugs and recording its surroundings, the damp soil.
How living things interact with their surroundings is ecology.
Biology has several branches.
What is ecology?
- A. The study of how one cell makes and uses the energy it needs to stay aliveHow a cell makes and uses energy is the biology of one cell.
Ecology is about living things interacting with one another and with their surroundings. - B. ✓ The study of how living things interact with one another and with their surroundings
- C. The care of the environment by people who recycle waste and save energyCaring for the environment is something people do.
Ecology is a branch of biology: the study of how living things interact.
Why: Ecology is the study of how living things interact with one another and with their surroundings.
Biology has several branches.
(a) State what ecology is. (1 pt)
- Award 1 point for: the study of how living things interact with one another and with their surroundings (accept with or without the three levels).
22The stimulus and the response
A large glass of juice raises a person’s blood glucose from 90 mg/dL to 140 mg/dL. Cells of the pancreas then release insulin, and the glucose falls back toward 90 mg/dL.
Which of the following is the stimulus?
- A. ✓ The rise in blood glucose to 140 mg/dL
- B. The release of insulin by cells of the pancreasReleasing insulin is what the body did about the rise.
The stimulus is the change that moved the glucose away from its set point: the rise.
Why: A stimulus is a change that moves a regulated quantity away from its set point.
The rise to 140 mg/dL moved the glucose away from 90 mg/dL.
So the rise is the stimulus.
Video: Watch: The stimulus and the response
The dish from above; a pill bug on the dry paper starts to walk and stops on the damp paper; the word stimulus appears over the dry half and the word response along the walk.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L01b.mp4
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Here is the dish again: damp paper on the left, dry paper on the right, and nine of the ten pill bugs on the damp side.
A pill bug tipped onto the dry paper stands in dry air. The pill bug detects the dry air.
The pill bug walks. It walks until it reaches the damp paper, and there it stops.
An organism detects a change (the stimulus) and does something or changes something inside itself (the response); the response is kept by selection when it raises survival or offspring.
For the pill bug, the change it detected was the dry air. So the dry air was the stimulus.
What the pill bug did about the dry air was walk to the damp side. So the walk was the response.
Kept by selection means this. The pill bugs that walked to damp air survived and left more offspring. So the walk stayed common. That is natural selection, the process Unit 7 taught.
Stimulus and response are the same two words your body’s feedback loops used.
There the stimulus was the rise in blood glucose. The response was the release of insulin.
Here the whole pill bug detects the change, and the whole pill bug answers it. The two words are the same, one scale up: a whole organism instead of one quantity inside a body.
Here the response is what the whole organism does, or what changes inside its body, about the change: not the response stage inside one cell.
What you are expected to know Identify the stimulus (the change the organism detects) and the response (what the organism then does) in a described case.
A lamp is switched on above an earthworm lying on the soil. The earthworm pulls back into its burrow.
Which of the following is the stimulus?
- A. ✓ The light from the lamp
- B. The earthworm pulling back into its burrowPulling back into the burrow is what the earthworm did about the light.
The stimulus is the change the earthworm detected: the light.
Why: The stimulus is the change the organism detects.
The earthworm detected the light from the lamp.
So the light is the stimulus.
Suppose a shadow passes over a cricket that is chirping. The cricket falls silent.
Which of the following is the response?
- A. The shadow passing over the cricketThe shadow is the change the cricket detected: the stimulus.
The response is what the cricket did about it: falling silent. - B. ✓ The cricket falling silent
Why: The response is what the organism does about the change it detected.
The cricket detected the shadow and fell silent.
So falling silent is the response.
A finger touches a leaf of a sensitive plant (Mimosa). Within seconds the leaf folds up.
Which of the following is the stimulus?
- A. The leaf folding upFolding up is what the plant did about the touch: the response.
The stimulus is the change the plant detected: the touch. - B. ✓ The touch of the finger
Why: The stimulus is the change the organism detects.
The plant detected the touch of the finger.
So the touch is the stimulus.
A twig snaps behind a deer. The deer sprints into the trees.
Which of the following is the response?
- A. ✓ The deer sprinting into the trees
- B. The sound of the twig snappingThe snap is the change the deer detected: the stimulus.
The response is what the deer did about it: sprinting into the trees.
Why: The response is what the organism does about the change it detected.
The deer detected the snap and sprinted into the trees.
So the sprint is the response.
A person steps out of a warm house into cold air and starts to shiver.
Which of the following is the stimulus?
- A. The shiveringShivering is what the person’s body did about the cold: the response.
The stimulus is the change the person detected: the cold air. - B. ✓ The cold air
Why: The stimulus is the change the organism detects.
The person detected the cold air.
So the cold air is the stimulus.
Suppose the water at the surface of a lake warms through the afternoon. A trout near the surface swims down to deeper, cooler water.
Which of the following is the response?
- A. The surface water warmingThe warming is the change the trout detected: the stimulus.
The response is what the trout did about it: swimming down. - B. ✓ The trout swimming down to deeper water
- C. The trout being cooler afterwardsBeing cooler is the result of the swim.
The response is what the trout did: swimming down to deeper water.
Why: The response is what the organism does about the change it detected.
The trout detected the warmer water and swam down.
So the swim to deeper water is the response.
43Behavioral or physiological
Video: Watch: Behavioral or physiological
Two panels: a lizard walking from a sunny rock into shade, and a person standing in the sun with sweat appearing on their skin; each panel sorted under its word.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L01c.mp4
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Suppose a lizard lies on a rock in full sun, and its body warms. The lizard walks into the shade of a bush.
The walk into the shade is something the whole lizard does. Anyone watching could see the lizard move.
A response that the whole organism does, such as moving or calling, is called a . Its behavior is what the animal does.
Now suppose a person stands in full sun, and their body warms. Sweat comes out onto their skin.
Sweating is a change the person’s body makes inside itself. The person has not gone anywhere.
A response that is a change inside the organism’s body is called a . Physiology is the working of the body’s insides.
For example, a person in full sun moves into the shade. This is a behavioral response, because the whole person does something: they move.
But a person in full sun sweats. This is a physiological response, because the person’s body changes something inside itself: it releases sweat onto the skin.
And a person in cold air shivers. This is a physiological response, because the person’s body changes something inside itself: its muscles tighten and loosen fast.
But a person in cold air walks indoors. This is a behavioral response, because the whole person does something: they walk.
And an owl flies out to hunt when the light fades. This is a behavioral response, because the whole owl does something: it flies out.
One stimulus can bring both kinds of response from one organism. A hot person sweats, and the same hot person moves into the shade.
The table below compares the two kinds of response: where each happens, what you would see, and one example of each.
What you are expected to know Classify an organism’s response as behavioral (something the whole organism does) or physiological (a change inside its body).
Suppose a cat hears a dog bark close by.
Which of the following is a physiological response?
- A. The cat leaps onto the fenceLeaping is something the whole cat does.
A physiological response is a change inside the cat’s body. - B. The cat hisses at the dogHissing is something the whole cat does.
A physiological response is a change inside the cat’s body. - C. ✓ The cat’s heart beats faster
Why: A physiological response is a change inside the organism’s body.
A faster heartbeat is a change inside the cat’s body.
So the faster heartbeat is the physiological response.
As the morning sun comes up, a snake slides out onto a warm rock.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responseSliding onto the rock is something the whole snake does.
A response the whole organism does is a behavioral response.
Why: Sliding onto the rock is something the whole snake does.
So it is a behavioral response.
A bright light shines into a person’s eyes, and their pupils narrow.
Which kind of response is this?
- A. A behavioral responseThe narrowing of the pupils is a change inside the person’s body.
A change inside the body is a physiological response. - B. ✓ A physiological response
Why: The pupils narrowing is a change inside the person’s body.
So it is a physiological response.
A fox appears at the edge of a field, and a rabbit’s heart beats faster.
Which kind of response is this?
- A. A behavioral responseA faster heartbeat is a change inside the rabbit’s body.
A change inside the body is a physiological response. - B. ✓ A physiological response
Why: A faster heartbeat is a change inside the rabbit’s body.
So it is a physiological response.
A hedgehog comes out to feed after dark.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responseComing out to feed is something the whole hedgehog does.
A response the whole organism does is a behavioral response.
Why: Coming out to feed is something the whole hedgehog does.
So it is a behavioral response.
The ground dries out, and a snail pulls back into its shell.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responsePulling back into its shell is something the whole snail does.
A response the whole organism does is a behavioral response.
Why: Pulling back into its shell is something the whole snail does.
So it is a behavioral response.
A person smells food cooking, and their mouth waters.
Which kind of response is this?
- A. A behavioral responseThe mouth watering is a change inside the person’s body.
A change inside the body is a physiological response. - B. ✓ A physiological response
Why: The mouth watering is a change inside the person’s body.
So it is a physiological response.
A student watches a person sweating in the sun and says: “Only movements count as responses, so sweating is something else.”
Is the student correct?
- A. YesSweating is a change the body makes inside itself because of the heat.
A change inside the body made about a stimulus is a physiological response. - B. ✓ No
Why: The heat is the stimulus.
The person’s body releases sweat because of the heat.
A change inside the body made about a stimulus is a response.
So sweating is a response, a physiological one, and the student is not correct.
Suppose a cat is out on a cold night. It curls up tightly, and its heart beats a little faster.
Which of the two responses is the physiological response?
- A. ✓ The cat’s heart beating faster
- B. The cat curling up tightlyCurling up is something the whole cat does: a behavioral response.
The change inside the cat’s body is the faster heartbeat.
Why: A physiological response is a change inside the organism’s body.
Curling up is something the whole cat does.
The faster heartbeat is a change inside the cat’s body.
So the faster heartbeat is the physiological response.
Here is the dish again, damp on the left and dry on the right, nine pill bugs on the damp side.
The dry air was the stimulus.
The walk to the damp side was the response. Because the whole animal walked, it was a behavioral response.
71Quick quiz: behavioral response, physiological response mixed practice
Three organisms each answer a stimulus.
Which of the following is a physiological response?
- A. A dog running to the door when it hears a carRunning to the door is something the whole dog does.
A physiological response is a change inside the body. - B. ✓ A person’s face flushing red in the heat
- C. A bee flying back to its hive as the light fadesFlying back to the hive is something the whole bee does.
A physiological response is a change inside the body.
Why: A physiological response is a change inside the organism’s body.
The face flushing red is a change in the blood flow inside the person’s body.
So the flushing is the physiological response.
In cold air, a person’s skin turns pale.
Which kind of response is this?
- A. A behavioral responseThe skin turning pale is a change inside the person’s body.
A change inside the body is a physiological response. - B. ✓ A physiological response
Why: The skin turning pale is a change inside the person’s body.
So it is a physiological response.
As evening comes, a frog begins to call from the edge of a pond.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responseCalling is something the whole frog does.
A response the whole organism does is a behavioral response.
Why: Calling is something the whole frog does.
So it is a behavioral response.
On a hot afternoon, a cow walks to the water trough and drinks.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responseWalking to the trough and drinking are things the whole cow does.
A response the whole organism does is a behavioral response.
Why: Walking to the trough and drinking are things the whole cow does.
So they are a behavioral response.
An organism answers a stimulus.
What is a behavioral response?
- A. A change the organism’s body makes inside itselfA change inside the body is a physiological response.
A behavioral response is something the whole organism does. - B. ✓ Something the whole organism does, such as moving or calling
Why: A behavioral response is something the whole organism does about a stimulus, such as moving or calling.
An organism answers a stimulus.
(a) State what a behavioral response and a physiological response are. (1 pt)
A physiological response is a change inside the organism’s body.
- Award 1 point for: behavioral = something the whole organism does; physiological = a change inside its body.
78Mixed practice mixed practice
A hand comes near a tortoise, and the tortoise pulls its head into its shell.
Which kind of response is this?
- A. ✓ A behavioral response
- B. A physiological responsePulling its head into its shell is something the whole tortoise does.
A response the whole organism does is a behavioral response.
Why: Pulling its head into its shell is something the whole tortoise does.
So it is a behavioral response.
Cockles live buried in the sand of a seashore.
Which of the following is a question in ecology?
- A. Which bases pair up with which in a cockle’s DNA?Which bases pair is a question about one molecule inside a cell.
Ecology asks how living things interact with their surroundings. - B. How does a cell in a cockle’s gut make ATP?How a cell makes ATP is a question about the inside of one cell.
Ecology asks how living things interact with their surroundings. - C. ✓ Why do cockles crowd where the sand stays wet at low tide?
Why: Ecology is the study of how living things interact with one another and with their surroundings.
Why cockles crowd where the sand stays wet asks how the cockles interact with their surroundings.
So it is a question in ecology.
Suppose a slug is on a garden path at dawn. The sun comes up and the path dries out. The slug crawls under a stone.
Which of the following is the stimulus?
- A. ✓ The path drying out
- B. The slug crawling under the stoneCrawling under the stone is what the slug did about the change: the response.
The stimulus is the change the slug detected: the path drying out.
Why: The stimulus is the change the organism detects.
The slug detected the path drying out.
So the drying of the path is the stimulus.
A student stands up to give a talk to the class, and her heart beats faster.
Which kind of response is this?
- A. A behavioral responseA faster heartbeat is a change inside the student’s body.
A change inside the body is a physiological response. - B. ✓ A physiological response
Why: A faster heartbeat is a change inside the student’s body.
So it is a physiological response.
At the first light of dawn, a robin begins to sing.
Which of the following is the response?
- A. ✓ The robin singing
- B. The first light of dawnThe first light is the change the robin detected: the stimulus.
The response is what the robin did about it: singing.
Why: The response is what the organism does about the change it detected.
The robin detected the first light and began to sing.
So the singing is the response.
A student says: “Ecology means protecting the environment, so it is a job rather than a branch of biology.”
Is the student correct?
- A. YesEcology is the study of how living things interact with one another and with their surroundings.
That study is a branch of biology. - B. ✓ No
Why: Ecology is the study of how living things interact with one another and with their surroundings.
A study of living things is biology.
So ecology is a branch of biology, and the student is not correct.
Suppose a horse stands in a field in full sun on a hot afternoon. Its body starts to warm.
(a) Identify one behavioral response and one physiological response the horse could make to the heat. (2 pt)
A physiological response: the horse’s skin releases sweat.
- Award 1 point for a behavioral response: an action the whole horse does, such as walking into shade or standing in a breeze.
- Award 1 point for a physiological response: a change inside the horse’s body, such as sweating.
(b) Explain how the two responses in (a) show that one stimulus can bring both kinds of response from one organism. (1 pt)
Frame They show this because …
The horse’s walk into the shade is something the whole horse does, a behavioral response to the heat.
The horse’s sweating is a change inside its body, a physiological response to the same heat.
So one stimulus, the heat, brought both kinds of response from one horse.
- Award 1 point for: both responses answer the same stimulus (the heat), one as an action of the whole horse and one as a change inside its body.
Glossary
- ecology
- The study of how living things interact with one another and with their surroundings, at the level of populations, communities and ecosystems.
- behavioral response
- A response that the whole organism does about a stimulus, such as moving or calling: a lizard walking into the shade, an owl flying out to hunt as the light fades.
- physiological response
- A response that is a change inside the organism’s body: a person sweating in the heat, or shivering in the cold.
APBIO-U08-L02 A change inside, a change outside, and why the answer helps
Photo: US National Park Service, Wikimedia Commons, public domain.
Here is a bat. It leaves its roost as the light fades at dusk. Now suppose the same bat has caught nothing for two nights. It leaves the roost earlier and hunts for longer.
The first change was outside the bat and the second was inside it. Why is either response worth having?
Unit 8 · Ecology
1A change inside, or a change outside
A lizard sits on a rock in the sun. The rock gets hot, and the lizard walks into the shade.
Which of the following is the stimulus?
- A. ✓ The rock getting hot
- B. The walk into the shadeThe walk is what the lizard does about the change.
The response is what the organism does; the stimulus is the change it detected.
Why: The stimulus is the change the organism detects.
The lizard detected the rock getting hot.
So the rock getting hot is the stimulus.
Video: Watch: A change inside, or a change outside
The bat in its roost as the light fades, then the hungry bat leaving earlier; the two changes labeled where they happen, outside the bat and inside it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L02a.mp4
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Where does the change an organism answers come from? And why does the answer matter?
The fading light is a change outside the bat, in its surroundings.
The bat’s hunger is a change inside its own body. Nothing outside the bat changed, and the hunger is still a stimulus.
Both responses matter for one reason: the bat that answers them stays fed and alive. An animal that stays alive can breed.
One question follows every behavior: does the response keep the organism alive, or let it reproduce?
Here is the bat again, in its roost at dusk. The light fades in the surroundings around it.
The bat leaves the roost. The fading light happened outside the bat.
A change in an organism’s surroundings is called an .
Now suppose the same bat has caught nothing for two nights. Its stomach is empty.
The bat leaves the roost earlier and hunts for longer. The empty stomach happened inside the bat’s own body.
A change inside an organism’s own body is called an .
Nothing outside the bat changed. The bat still detected its empty stomach, and it still answered it.
So an internal change is a stimulus too. The organism does not have to see a change to detect it.
Now suppose the air around the bat gets colder. The change happened in the bat’s surroundings.
So the colder air is an external change.
Now suppose the bat’s own body gets colder. The change happened inside the bat.
So the colder body is an internal change.
The two cases differ in one thing only: where the change happened. To sort any change, ask where it happened: inside the organism’s body, or in its surroundings.
The table below sorts the four changes, two inside the bat and two in its surroundings.
What you are expected to know Sort a change an organism responds to as an internal change or an external change.
Three changes happen around and inside a frog.
Which of the following is an internal change?
- A. The pond where the frog lives freezes overThe pond froze in the frog’s surroundings, not inside its body.
A change in the surroundings is an external change. - B. ✓ The frog’s blood glucose falls
- C. Rain starts to fall on the frog’s pondThe rain fell in the frog’s surroundings, not inside its body.
A change in the surroundings is an external change.
Why: The frog’s blood is inside its body.
The fall in glucose happened inside the frog.
So the fall in blood glucose is an internal change.
A hare has not eaten all day. The glucose in its blood falls.
Which kind of change is the fall in glucose?
- A. ✓ An internal change
- B. An external changeThe glucose fell inside the hare’s own body, not in its surroundings.
A change inside the body is an internal change.
Why: The glucose is in the hare’s blood, inside its body.
The fall happened inside the hare.
So the fall in glucose is an internal change.
In autumn the nights get longer over the forest where a deer lives.
Which kind of change is the lengthening of the nights?
- A. An internal changeThe nights got longer in the deer’s surroundings, not inside its body.
A change in the surroundings is an external change. - B. ✓ An external change
Why: The length of the night is part of the deer’s surroundings.
The change happened outside the deer.
So the lengthening of the nights is an external change.
A hawk’s shadow passes over a mouse feeding in a field.
Which kind of change is the passing shadow?
- A. An internal changeThe shadow passed over the field, in the mouse’s surroundings.
A change in the surroundings is an external change. - B. ✓ An external change
Why: The shadow fell on the field around the mouse.
The change happened outside the mouse.
So the passing shadow is an external change.
A dog chases a ball for ten minutes. Its body temperature rises.
Which kind of change is the rise in body temperature?
- A. ✓ An internal change
- B. An external changeThe temperature rose inside the dog’s own body, not in the air around it.
A change inside the body is an internal change.
Why: The dog’s body temperature is a state of its own body.
The rise happened inside the dog.
So the rise in body temperature is an internal change.
A fox has drunk nothing since morning. The amount of water in its body falls.
Which kind of change is the fall in body water?
- A. ✓ An internal change
- B. An external changeThe water fell inside the fox’s own body, not in its surroundings.
A change inside the body is an internal change.
Why: The water is inside the fox’s body.
The fall happened inside the fox.
So the fall in body water is an internal change.
Frost forms overnight on the grass where a rabbit feeds.
Which kind of change is the frost?
- A. An internal changeThe frost formed on the grass, in the rabbit’s surroundings.
A change in the surroundings is an external change. - B. ✓ An external change
Why: The grass is part of the rabbit’s surroundings.
The frost formed outside the rabbit.
So the frost is an external change.
31Why the answer helps
A pill bug on the dry side of a dish walks until it reaches the damp side.
Which kind of response is the walk?
- A. ✓ A behavioral response
- B. A physiological responseThe walk is something the whole pill bug does.
A change inside the pill bug’s body would be a physiological response.
Why: The whole pill bug walked.
So the walk is a behavioral response.
Video: Watch: Why the answer helps
A bat that stays in its roost and a bat that flies out among the insects; the pill bug in dry air and in damp air, water leaving its body; the outcome written under each: fed, alive.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L02b.mp4
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Why is the bat’s response to the fading light worth having? Ask what happens to a bat that does not respond.
Suppose a bat stays in its roost after the light fades. The insects the bat eats are flying at dusk, outside the roost.
The bat that stays in catches no insects. It goes hungry.
The bat that leaves its roost as the light fades flies among those insects. It catches them and is fed.
A fed bat stays alive. So the response to the fading light keeps the bat alive.
Now suppose the same bat has caught nothing for two nights. It leaves the roost earlier and hunts for longer.
The hungry bat has more time in the air to catch insects. So the response to its hunger keeps it fed and alive too.
Now consider the pill bug in dry air again. In dry air a pill bug loses water through its body surface.
A pill bug that loses too much water dries out and dies.
The pill bug that walks to the damp side loses water more slowly. So it survives.
A response can help in a second way: it can let the organism reproduce.
Suppose a male frog calls at a pond edge on a spring night. Females move toward the calling male.
The calling male mates and fathers tadpoles. So the calling lets the frog reproduce.
A response is worth having when it passes one test: it keeps the organism alive, or it lets the organism reproduce.
An explanation of why a response helps ends at survival or offspring. It does not need to say how the animal senses the change.
What you are expected to know Explain how a response to a stimulus keeps the organism alive or lets it reproduce.
Suppose a snail sits on a dry stone in the sun for an hour.
What happens to the water in the snail’s body?
- A. ✓ The snail loses water
- B. The snail gains waterWater leaves the snail through its skin into the dry air.
No water comes in from dry air. - C. The amount of water stays the sameIn dry air water keeps leaving through the snail’s skin.
So the amount of water in the snail falls.
Why: In dry air water leaves the snail through its skin.
Nothing replaces that water.
So the snail loses water.
A student puts a pill bug on the dry side of a dish. The paper on the other side of the dish is damp. The pill bug walks until it reaches the damp side.
(a) Explain how the walk to the damp side keeps the pill bug alive. (1 pt)
Frame The walk keeps the pill bug alive because …
In damp air the pill bug loses water through its body surface more slowly.
A pill bug that keeps its water does not dry out.
So the pill bug survives.
- Award 1 point for: in damp air the pill bug loses less water (does not dry out), so it survives.
A student watches the pill bug walk to the damp side and says: “The walk is worth having because the pill bug decides to do it.”
Is the student correct?
- A. YesThe walk helps because it ends in damp air, where the pill bug loses less water.
Whether the pill bug decided to walk does not change that. - B. ✓ No
Why: The walk ends in damp air.
There the pill bug loses less water.
So it survives.
The walk is worth having because it keeps the pill bug alive.
Whether the pill bug decided to walk does not change that.
On a hot afternoon the water at the surface of a lake warms. Less oxygen dissolves in warm water than in cool water. A trout swims down into the cooler, deeper water.
Which of the following does the swim into cooler water do for the trout?
- A. ✓ The swim keeps the trout alive
- B. The swim lets the trout reproduceNo mate, eggs or offspring are part of this case.
The swim changes how much oxygen the trout can take in. - C. The swim neither keeps the trout alive nor lets it reproduceIn warm water the trout takes in too little oxygen and can die.
In cool water it takes in enough oxygen to stay alive.
Why: Less oxygen dissolves in the warm surface water.
A trout that stays there takes in too little oxygen and can die.
More oxygen dissolves in the cooler, deeper water.
So the swim keeps the trout alive.
Here is the bat again, leaving its roost as the light fades. Here too is the hungry bat, leaving earlier and hunting for longer.
The fading light was an external change. The hunger was an internal change.
Both responses keep the bat fed and alive. A bat that stays alive can breed.
57Mixed practice mixed practice
In spring a male stork stands at a nest and claps its bill in a display. A female joins him at the nest. The pair raises chicks.
Which of the following does the display do for the male stork?
- A. The display keeps the stork aliveThe display brings no food and removes no danger.
It brings the female, and the pair raises chicks. - B. ✓ The display lets the stork reproduce
- C. The display neither keeps the stork alive nor lets it reproduceThe pair raises chicks after the display.
Chicks are offspring, so the display let the stork reproduce.
Why: The display brings a female to the nest.
The pair raises chicks, which are the male’s offspring.
So the display lets the stork reproduce.
On a cold night a squirrel’s body temperature falls.
Which kind of change is the fall in body temperature?
- A. ✓ An internal change
- B. An external changeThe temperature fell inside the squirrel’s own body, not in the air around it.
A change inside the body is an internal change.
Why: The squirrel’s body temperature is a state of its own body.
The fall happened inside the squirrel.
So the fall in body temperature is an internal change.
A bat roosts by day and flies out at dusk to hunt. On a day it has not eaten, its hunger makes it leave the roost earlier. A student says: “Only a change outside the bat can be a stimulus, so the bat’s hunger is just a feeling.”
Is the student correct?
- A. YesA stimulus is a change the organism detects, inside the body or outside it.
The bat detects its own hunger and answers it, so the hunger is a stimulus. - B. ✓ No
Why: A stimulus is a change the organism detects.
The change can happen inside the organism’s own body.
The bat detects its hunger and leaves the roost earlier.
So the hunger is a stimulus, an internal one.
A lizard sits on a rock at noon. The rock gets hot. The lizard walks into the shade.
Which of the following does the walk into the shade do for the lizard?
- A. ✓ The walk keeps the lizard alive
- B. The walk lets the lizard reproduceNo mate, eggs or offspring are part of this case.
The walk changes how hot the lizard’s body gets. - C. The walk neither keeps the lizard alive nor lets it reproduceIn the shade the lizard’s body stops heating up.
A lizard whose body gets too hot dies, so the walk does help.
Why: On the hot rock the lizard’s body heats up.
A lizard whose body gets too hot dies.
In the shade its body stops heating up.
So the walk keeps the lizard alive.
The stream a deer drinks from dries up in a drought.
Which kind of change is the drying of the stream?
- A. An internal changeThe stream dried up in the deer’s surroundings, not inside its body.
A change in the surroundings is an external change. - B. ✓ An external change
Why: The stream is part of the deer’s surroundings.
The drying happened outside the deer.
So the drying of the stream is an external change.
A student watches a lizard walk off a hot rock into the shade and says: “The walk helps the lizard because the lizard chooses to walk.”
Is the student correct?
- A. YesThe walk helps because of what it does to the lizard’s body.
In the shade the body stops heating up, so the lizard stays alive. - B. ✓ No
Why: In the shade the lizard’s body stops heating up.
A lizard whose body gets too hot dies.
So the walk keeps the lizard alive.
The walk helps for that reason, whether or not the lizard chose it.
Suppose an earthworm lies on a dry pavement in the morning sun. An earthworm loses water through its skin in dry air. The earthworm detects the dry air and crawls into the damp soil beside the pavement.
(a) State whether the dry air is an internal change or an external change for the earthworm. (1 pt)
- Award 1 point for: external change, because the dry air is in the earthworm’s surroundings (outside its body).
(b) Explain how crawling into the damp soil keeps the earthworm alive. (1 pt)
Frame Crawling into the damp soil keeps the earthworm alive because …
An earthworm that loses too much water dries out and dies.
In the damp soil the earthworm loses water more slowly.
So the earthworm does not dry out and survives.
- Award 1 point for: in the damp soil the earthworm loses less water (does not dry out), so it survives.
Glossary
- external change
- A change in an organism's surroundings, such as the light fading at dusk. The organism detects it and responds to it, so it is a stimulus.
- internal change
- A change inside an organism's own body, such as its stomach being empty. Nothing outside the organism need change; the organism still detects it and responds to it, so it is a stimulus too.
APBIO-U08-L03 Turning toward, speeding up: taxis, kinesis and the plant that leans
Photo: Yug, Wikimedia Commons, CC BY-SA 3.0 (cropped and resized).
Here are two dishes. In the first dish, bacteria swim toward a drop of sugar. In the second dish, one half is warm and the other half is cool, and ten woodlice start in the middle. The woodlice on the warm half scurry about fast, in no fixed direction. A woodlouse that happens to reach the cool half slows down. After a few minutes, eight of the ten woodlice are on the cool half.
The bacteria reached the sugar. The woodlice reached the cool half. Did the bacteria steer? Did the woodlice steer?
Unit 8 · Ecology
1Steering, or just speeding up
How can a response get an animal to a better place without steering?
The bacteria swim toward the drop of sugar. Their movement has a direction.
The woodlice head nowhere in particular. Each woodlouse simply moves fast on the warm half and slowly on the cool half.
A woodlouse that moves slowly on the cool half stays there longer. So the woodlice pile up on the cool half, and not one of them steered toward it.
Plants have two responses of their own. A shoot grows toward the light, and a plant times its flowering by the length of the night.
A woodlouse on the warm half of a dish moves fast. On the cool half the woodlouse moves slowly.
Which of the following is the stimulus?
- A. ✓ The warmth of the dish
- B. The woodlouse’s fast movementThe fast movement is what the woodlouse does.
What the woodlouse does is the response.
Why: The stimulus is the change the woodlouse detects.
The woodlouse detects the warmth.
So the warmth is the stimulus.
Look at the first dish. The sugar is the stimulus, and the bacteria swim toward it.
When an organism moves toward or away from the stimulus, the movement is called a .
A taxis has a direction. The bacteria’s swim has one: toward the sugar.
Now look at the second dish. The warmth is the stimulus, and each woodlouse moves fast on the warm half and slowly on the cool half.
No woodlouse heads for the cool half. The movement has no fixed direction.
When an organism changes its speed, or how often it turns, with no fixed direction, the movement is called a .
Kinesis is the Greek word for movement. The animal moves more or moves less, and that is all.
Suppose an insect flies toward a lamp at night.
Which kind of movement is the insect’s flight?
- A. ✓ A taxis
- B. A kinesisA kinesis has no fixed direction.
The insect’s flight has one: toward the lamp.
Why: The lamp is the stimulus.
The insect moves toward the lamp.
A movement toward or away from the stimulus is a taxis.
Video: Watch: Steering, or just speeding up
The two dishes side by side in time lapse: the bacteria heading for the drop of sugar; the woodlice scurrying fast on the warm half and slowing on the cool half until most are gathered there; the two words appearing under the dishes.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L03a.mp4
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A kinesis still gets the woodlice to a better place. On the cool half a woodlouse moves slowly, so it stays there longer.
On the warm half a woodlouse moves fast, so it soon leaves. So the woodlice gather on the cool half by slowing down there, not by steering toward it.
A pill bug is one kind of woodlouse.
Here is the second dish drawn twice. The left drawing shows the woodlice walking toward the cool half, and the right drawing shows what the woodlice actually do.
Imagine each woodlouse walked from the warm half toward the cool half. This is a taxis, because the movement has a direction: toward the cool half.
But the woodlice actually speed up on the warm half and slow down on the cool half. This is a kinesis, because the movement has no fixed direction.
Here are four more cases, drawn with their verdicts.
For example, the bacteria swim toward the drop of sugar. This is a taxis, because the movement has a direction: toward the sugar.
But now imagine the bacteria swimming fast in plain water and slowly near the sugar, with no fixed direction. This is a kinesis, because the movement has no fixed direction.
And suppose a bacterium swims away from a drop of a harmful chemical. This is a taxis too, because the movement has a direction: away from the chemical.
And suppose an animal turns often in bright light and rarely in the shade, with no fixed direction. This is a kinesis too, because the movement has no fixed direction.
Here is a table comparing a taxis with a kinesis.
What you are expected to know Classify an animal’s movement as a taxis or a kinesis.
Suppose beetles walk fast in dry air and slowly in damp air, in no fixed direction.
Which kind of movement is the beetles’ walking?
- A. A taxisA taxis has a direction, toward or away from the stimulus.
These beetles change speed with no fixed direction. - B. ✓ A kinesis
Why: The dryness is the stimulus.
The beetles change their speed with no fixed direction.
A change in speed with no fixed direction is a kinesis.
Suppose a flatworm turns often in bright light and rarely in the shade, in no fixed direction.
Which kind of movement is the flatworm’s turning?
- A. A taxisA taxis has a direction, toward or away from the stimulus.
The flatworm’s turning has no fixed direction. - B. ✓ A kinesis
Why: The light is the stimulus.
The flatworm changes how often it turns, with no fixed direction.
A change in turning with no fixed direction is a kinesis.
Suppose a snail crawls away from a bright light.
Which kind of movement is the snail’s crawl?
- A. ✓ A taxis
- B. A kinesisA kinesis has no fixed direction.
The snail’s crawl has one: away from the light.
Why: The light is the stimulus.
The snail moves away from the light.
A movement toward or away from the stimulus is a taxis.
Suppose a fish swims toward the smell of food.
Which kind of movement is the fish’s swim?
- A. ✓ A taxis
- B. A kinesisA kinesis has no fixed direction.
The fish’s swim has one: toward the smell.
Why: The smell is the stimulus.
The fish moves toward the smell.
A movement toward or away from the stimulus is a taxis.
Suppose a fish swims fast in warm water and slowly in cool water, in no fixed direction, and ends up in the cool water.
Which kind of movement is the fish’s swim?
- A. A taxisEnding up in the cool water does not give the movement a direction.
No fish headed for the cool water; each fish slowed down there. - B. ✓ A kinesis
Why: The warmth is the stimulus.
The fish changes speed with no fixed direction.
The fish slows in the cool water, so it stays there longer.
A change in speed with no fixed direction is a kinesis, even when it ends somewhere better.
The two drawings below show the path of one small animal in a dish with a shaded patch.
Which drawing shows a taxis?
- A. ✓ Drawing 1
- B. Drawing 2In drawing 2 the path bends again and again, with no fixed direction.
That movement is a kinesis.
Why: In drawing 1 the animal moves in one direction: toward the patch.
A movement toward the stimulus is a taxis.
Suppose fish in a tank swim fast at the warm end and slowly at the cool end, in no fixed direction. After an hour most of the fish are at the cool end. A student says: “They must have swum toward the cool end.”
Is the student correct?
- A. YesThe fish do not head for the cool end.
Each one slows down there and stays; at the warm end each one moves fast and soon leaves. - B. ✓ No
Why: Each fish moves fast at the warm end and slowly at the cool end.
A fish that reaches the cool end slows down.
So it stays there.
A fish at the warm end soon leaves it.
So the fish gather at the cool end without swimming toward it.
37Quick quiz: taxis, kinesis mixed practice
Suppose an insect speeds up in dry air, in no fixed direction.
Is the insect’s movement a taxis?
- A. YesA taxis has a direction, toward or away from the stimulus.
This movement has no fixed direction. - B. ✓ No
Why: The insect changes speed with no fixed direction.
A change in speed with no fixed direction is a kinesis, not a taxis.
Suppose a beetle walks toward the smell of rotting fruit.
Is the beetle’s walk a taxis?
- A. ✓ Yes
- B. NoThe beetle moves toward the smell.
A movement toward the stimulus is a taxis.
Why: The smell is the stimulus.
The beetle moves toward the smell.
So the walk is a taxis.
Suppose a snail turns more often in bright light, in no fixed direction.
Is the snail’s turning a kinesis?
- A. ✓ Yes
- B. NoThe snail changes how often it turns, with no fixed direction.
That change is a kinesis.
Why: The snail changes how often it turns.
The turning has no fixed direction.
So the turning is a kinesis.
Suppose a fish swims away from a patch of cold water.
Is the fish’s swim a kinesis?
- A. YesA kinesis has no fixed direction.
The fish’s swim has one: away from the cold water. - B. ✓ No
Why: The fish moves away from the cold water.
A movement away from the stimulus is a taxis, not a kinesis.
An animal responds to a stimulus by moving.
What is a taxis?
- A. ✓ A movement toward or away from the stimulus
- B. A change in speed or turning, with no fixed directionA change in speed or turning with no fixed direction is a kinesis.
- C. A change inside the animal’s bodyA change inside the animal’s body is a physiological response, not a movement.
Why: A taxis is a movement toward or away from the stimulus.
An animal responds to a stimulus by moving.
What is a kinesis?
- A. A change inside the animal’s bodyA change inside the animal’s body is a physiological response, not a movement.
- B. A movement toward or away from the stimulusA movement toward or away from the stimulus is a taxis.
- C. ✓ A change in speed or turning, with no fixed direction
Why: A kinesis is a change in speed or turning, with no fixed direction.
An animal responds to a stimulus by moving.
(a) State what a taxis is. (1 pt)
- Award 1 point for: a movement toward or away from the stimulus.
(b) State what a kinesis is. (1 pt)
- Award 1 point for: a change in speed or in turning, with no fixed direction.
45The plant that leans, and the plant that keeps a calendar
A shoot on a windowsill grows toward the light from the window.
Which kind of change is the light?
- A. An internal changeAn internal change happens inside the plant’s own body.
The light comes from the plant’s surroundings. - B. ✓ An external change
Why: An internal change happens inside the organism’s own body.
An external change happens in its surroundings.
The light comes from the window, in the plant’s surroundings.
So the light is an external change.
Video: Watch: The plant that leans, and the plant that keeps a calendar
An onion shoot bending toward a lamp in time lapse; then a calendar of autumn nights lengthening beside a chrysanthemum, which flowers once the nights are long.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L03b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L03b.mp4
Plants respond to light too. Here is a photograph of onion shoots growing toward the lamp above them.
The shoots grew toward the light. The response is a direction of growth.
When a shoot grows toward the light, the response is called .
Photo means light and tropism means a turning. So phototropism is a turning toward the light.
Now consider a chrysanthemum in a garden through the autumn. The chart beside the photograph shows the hours of darkness each night growing from 9 hours in July to 14 hours in November.
Once the nights are long enough, the chrysanthemum flowers. The response is a timing, not a direction.
When a plant times its flowering by the length of the day or the night, the response is called .
Photo means light and period means a stretch of time. So photoperiodism is a response to how long the light, and so the dark, lasts each day.
The chrysanthemum does not sense the season itself. It responds to the hours of darkness each night.
Those hours grow longer as the autumn goes on. Once they pass the length the chrysanthemum responds to, the plant flowers.
For example, the onion shoot leans toward the lamp. This is phototropism, because the response is a direction of growth.
But the chrysanthemum flowers as the autumn nights lengthen. This is photoperiodism, because the response is a timing.
And imagine a seedling in a dark cupboard with one small hole, its shoot growing toward the hole. This is phototropism, because the response is a direction of growth.
But now imagine a plant that flowers in spring, once the nights have grown short. This is photoperiodism, because the response is a timing.
In phototropism, the light decides which way the shoot grows. In photoperiodism, the length of the night decides when the plant flowers.
What you are expected to know Identify a plant’s response as phototropism or photoperiodism.
Suppose a potted plant on a shelf bends toward the room’s one window.
Which kind of response is the bend?
- A. ✓ Phototropism
- B. PhotoperiodismPhotoperiodism is a timing: when the plant flowers.
This bend is a direction of growth.
Why: The light from the window is the stimulus.
The plant grows toward the light.
A shoot growing toward the light is phototropism.
Suppose a seedling in a dark box grows toward the box’s one small hole.
Which kind of response is the seedling’s growth?
- A. ✓ Phototropism
- B. PhotoperiodismPhotoperiodism is a timing of flowering.
This seedling is growing in a direction: toward the light from the hole.
Why: Light comes through the hole.
The seedling grows toward the light.
A shoot growing toward the light is phototropism.
Suppose a plant flowers in spring, once the nights have grown short.
Which kind of response is the flowering?
- A. PhototropismPhototropism is a shoot growing toward the light.
This plant is timing its flowering by the length of the night. - B. ✓ Photoperiodism
Why: The length of the night is the stimulus.
The plant times its flowering by that length.
A plant timing its flowering by the length of the night shows photoperiodism.
Suppose a vine’s shoots grow toward the sunnier side of a garden.
Which kind of response is the shoots’ growth?
- A. ✓ Phototropism
- B. PhotoperiodismPhotoperiodism is a timing of flowering.
These shoots grow in a direction: toward the sunnier side.
Why: The sunlight is the stimulus.
The shoots grow toward the sunlight.
A shoot growing toward the light is phototropism.
Suppose a plant flowers in autumn, once the nights have grown long.
Which kind of response is the flowering?
- A. PhototropismPhototropism is a direction of growth, toward the light.
Flowering once the nights grow long is a timing. - B. ✓ Photoperiodism
Why: The length of the night is the stimulus.
The plant flowers once the nights are long.
A plant timing its flowering by the length of the night shows photoperiodism.
A chrysanthemum flowers in autumn. A student says: “The plant flowers because it senses that the weather has turned colder.”
Is the student correct?
- A. YesThe chrysanthemum responds to the hours of darkness each night, not to the temperature.
- B. ✓ No
Why: The chrysanthemum responds to the length of the night.
In autumn the nights grow longer.
The longer nights are what the plant responds to, not the colder weather.
Here are the two dishes again: the bacteria swimming toward the drop of sugar, and the woodlice gathered on the cool half of the warm-and-cool dish.
The bacteria’s swim toward the sugar is a taxis. The woodlice’s fast-and-slow scurry, with no fixed direction, is a kinesis.
And here are the onion shoots again, leaning toward the lamp. That lean is phototropism, the plant’s version of moving toward the stimulus.
The chrysanthemum flowering as the autumn nights lengthen is photoperiodism: a timing, not a direction.
74Quick quiz: phototropism, photoperiodism mixed practice
Suppose a sunflower seedling’s shoot bends toward the morning sun.
Is the bend phototropism?
- A. ✓ Yes
- B. NoThe shoot grows toward the light.
A shoot growing toward the light is phototropism.
Why: The sunlight is the stimulus.
The shoot grows toward the sunlight.
So the bend is phototropism.
Suppose a plant flowers in early summer, once the nights have grown short.
Is the flowering photoperiodism?
- A. ✓ Yes
- B. NoThe plant times its flowering by the length of the night.
That timing is photoperiodism.
Why: The plant flowers when the nights are short.
The length of the night sets the timing.
So the flowering is photoperiodism.
Suppose a plant flowers once the nights are longer than 11 hours.
Is the flowering phototropism?
- A. YesPhototropism is a shoot growing toward the light.
This plant is timing its flowering by the length of the night. - B. ✓ No
Why: The plant times its flowering by the length of the night.
That timing is photoperiodism, not phototropism.
A plant responds to light.
What is phototropism?
- A. ✓ A shoot bending to grow toward the light
- B. A plant timing its flowering by the length of the day or nightA plant timing its flowering by the length of the day or night shows photoperiodism.
- C. A shoot growing faster in brighter lightPhototropism is a direction of growth, not a speed of growth.
Why: Phototropism is a shoot growing toward the light.
A plant responds to light.
What is photoperiodism?
- A. A shoot growing faster in brighter lightPhotoperiodism is a timing of flowering, not a speed of growth.
- B. A shoot bending to grow toward the lightA shoot bending to grow toward the light is phototropism.
- C. ✓ A plant timing its flowering by the length of the day or night
Why: Photoperiodism is a plant timing its flowering by the length of the day or night.
A plant responds to light.
(a) State what phototropism is. (1 pt)
- Award 1 point for: a shoot growing toward the light (a direction of growth set by the light).
(b) State what photoperiodism is. (1 pt)
- Award 1 point for: a plant timing its flowering by the length of the day or the night.
81Mixed practice mixed practice
Suppose a plant flowers as the winter nights shorten into spring.
Which of the following is this response?
- A. TaxisA taxis is an animal moving toward or away from the stimulus.
This plant is timing its flowering. - B. KinesisA kinesis is an animal changing speed or turning with no fixed direction.
This plant is timing its flowering. - C. PhototropismPhototropism is a shoot growing toward the light.
This plant is timing when it flowers. - D. ✓ Photoperiodism
Why: The length of the night is the stimulus.
The plant times its flowering by that length.
That timing is photoperiodism.
Suppose an insect moves fast in dry air and slowly in damp air, in no fixed direction.
Which of the following is this response?
- A. TaxisA taxis has a direction, toward or away from the stimulus.
This insect’s movement has no fixed direction. - B. ✓ Kinesis
- C. PhototropismPhototropism is a shoot growing toward the light.
This is an animal changing its speed. - D. PhotoperiodismPhotoperiodism is a plant timing its flowering.
This is an animal changing its speed.
Why: The dryness is the stimulus.
The insect changes speed with no fixed direction.
That change is a kinesis.
Suppose a shoot in a greenhouse bends toward the brightest pane of the glass roof.
Which of the following is this response?
- A. TaxisA taxis is an animal moving toward or away from the stimulus.
This is a shoot growing. - B. KinesisA kinesis is an animal changing speed or turning.
This is a shoot growing. - C. ✓ Phototropism
- D. PhotoperiodismPhotoperiodism is a timing of flowering.
This shoot is growing in a direction: toward the light.
Why: The bright light is the stimulus.
The shoot grows toward the light.
That growth is phototropism.
Suppose a fish swims away from a patch of cloudy water.
Which of the following is this response?
- A. ✓ Taxis
- B. KinesisA kinesis has no fixed direction.
This fish swims in one direction: away from the cloudy water. - C. PhototropismPhototropism is a shoot growing toward the light.
This is an animal swimming. - D. PhotoperiodismPhotoperiodism is a plant timing its flowering.
This is an animal swimming.
Why: The cloudy water is the stimulus.
The fish moves away from the cloudy water.
That movement is a taxis.
Suppose a millipede moves fast in bright light and slowly in the shade, in no fixed direction, and ends up in the shade.
Which of the following is this response?
- A. TaxisEnding up in the shade does not give the movement a direction.
The millipede did not head for the shade; it slowed down there. - B. ✓ Kinesis
- C. PhototropismPhototropism is a shoot growing toward the light.
This is an animal changing its speed. - D. PhotoperiodismPhotoperiodism is a plant timing its flowering.
This is an animal changing its speed.
Why: The light is the stimulus.
The millipede changes speed with no fixed direction.
That change is a kinesis, even though the millipede ends up somewhere better.
Suppose a houseplant’s stem curves toward a skylight.
Which of the following is this response?
- A. TaxisA taxis is an animal moving toward or away from the stimulus.
This is a plant’s stem growing. - B. KinesisA kinesis is an animal changing speed or turning.
This is a plant’s stem growing. - C. ✓ Phototropism
- D. PhotoperiodismPhotoperiodism is a timing of flowering.
This stem is growing in a direction: toward the light.
Why: The light from the skylight is the stimulus.
The stem grows toward the light.
That growth is phototropism.
A student puts twenty small insects into a dish. One half of the dish is brightly lit and the other half is dim. In the bright half each insect moves fast, in no fixed direction. In the dim half each insect moves slowly. After ten minutes, most of the insects are in the dim half.
(a) Explain how the insects come to gather in the dim half. (2 pt)
Frame The insects gather in the dim half because …
In the bright half an insect moves fast, so it soon wanders out of the bright half.
In the dim half an insect moves slowly, so it stays there longer.
So after ten minutes more insects are in the dim half than in the bright half, though no insect moved toward it.
- Award 1 point for: the insects move fast in the bright half and slowly in the dim half, so an insect leaves the bright half sooner and stays in the dim half longer.
- Award 1 point for: the insects collect in the dim half without moving toward it (the movement has no direction; a kinesis).
Glossary
- taxis
- A movement toward or away from the stimulus: the movement has a direction. Bacteria swimming toward a drop of sugar show a taxis.
- kinesis
- A change in an animal's speed, or in how often it turns, with no fixed direction. Woodlice speeding up on the warm half of a dish and slowing on the cool half show a kinesis; they gather on the cool half by slowing down there, not by steering toward it.
- phototropism
- A shoot growing toward the light: a direction of growth. An onion shoot leaning toward a lamp shows phototropism.
- photoperiodism
- A plant timing its flowering by the length of the day or the night: a timing, not a direction. A chrysanthemum flowering as the autumn nights lengthen shows photoperiodism.
APBIO-U08-L04 Testing the pill bug
Photo: Kiloueka, Wikimedia Commons, CC0 (resized).
Here is the dish again: nine pill bugs on the damp side, one on the dry. A friend says they would have ended up like that anyway.
How would you set up a test that shows they chose?
Unit 8 · Ecology
1The one thing you change
What does a fair test of a behavior need? It needs one variable you change, one variable you measure, and a control.
The variable you change here is dampness. The variable you measure is the count of pill bugs on each side.
The control is a dish with damp paper on both sides. It shows how the pill bugs spread when there is nothing to choose between.
Without the control, a lopsided count in the test dish could not be put down to dampness.
In Unit 3 a student compared catalase from potato and from liver. The source of the catalase was the condition she deliberately changed.
Which name does a fair test give to the condition the experimenter deliberately changes?
- A. ✓ The independent variable
- B. The dependent variableThe dependent variable is the quantity measured to see the effect.
- C. A control variableA control variable is a condition kept the same in every tube.
Why: The independent variable is the condition the experimenter deliberately changes.
She deliberately changed the source of the catalase.
So the source of the catalase was the independent variable.
Video: Watch: The one thing you change
The test dish from above, damp paper on one side and dry on the other; the word dampness written across the two sides as the first word is said, and the name independent variable beneath it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04-p1.mp4
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Suppose you set up the test dish. You lay damp paper on the left side and dry paper on the right side, and tip in 10 pill bugs.
Dampness is the one condition you deliberately change from one side to the other.
In Unit 3 the enzyme tubes had one condition the student deliberately changed: the source of the catalase, potato or liver. That condition was called the independent variable.
The same name applies here. Dampness is the independent variable, because you deliberately set one side damp and the other side dry.
The size of the dish is not the independent variable, because you keep it the same for both sides.
The number of pill bugs tipped in is not the independent variable, because you keep it at 10 in every dish.
So the independent variable is the one condition the experimenter deliberately sets: here, damp on one side and dry on the other.
What you are expected to know Identify the independent variable in a described behavior experiment: the one condition the experimenter deliberately changes.
A student covers the left side of a dish so it is dark and leaves the right side in the light. She tips in 12 woodlice and counts the woodlice on each side after 3 minutes.
Which is the independent variable?
- A. The number of woodlice on each sideThe number of woodlice on each side is the result she measures.
That count is the dependent variable. - B. The size of the dishThe size of the dish is the same for both sides.
- C. ✓ Dark or light on a side
Why: The independent variable is the condition the experimenter deliberately sets to two values.
She makes one side dark and leaves the other side in the light.
So light is the independent variable.
A student lays warm sand on the left side of a tray and cool sand on the right side. He tips in 12 mealworms and counts the mealworms on each side after 5 minutes.
Which is the independent variable?
- A. The number of mealworms on each sideThe number of mealworms on each side is the result he measures.
That count is the dependent variable. - B. ✓ Temperature
- C. The size of the trayThe size of the tray is the same for both sides.
Why: The independent variable is the condition the experimenter deliberately sets to two values.
He makes one side warm and the other side cool.
So temperature is the independent variable.
A student lays warm sand on one side of a tray and cool sand on the other side, and tips 12 mealworms into every tray he sets up.
Is the number of mealworms tipped in the independent variable?
- A. YesHe tips 12 mealworms into every tray.
A condition kept the same is not the condition he changes. - B. ✓ No
Why: The independent variable is the one condition the experimenter deliberately changes.
He keeps the number of mealworms at 12 in every tray.
So the number of mealworms is not the independent variable.
Temperature is.
A student lays damp cloth on one side of a board and dry cloth on the other side, tips in 15 snails and counts the snails on each side after 10 minutes.
Is dampness the independent variable?
- A. ✓ Yes
- B. NoShe deliberately sets one side damp and the other side dry.
Dampness is the one condition she changes.
Why: The independent variable is the one condition the experimenter deliberately changes.
She sets one side damp and the other side dry.
So dampness is the independent variable.
20The one thing you measure
In the Unit 3 enzyme tubes, the student collected the oxygen from each tube for 4 minutes and measured its volume.
Which name does a fair test give to the quantity measured to see the effect?
- A. The independent variableThe independent variable is the condition the experimenter deliberately changes.
- B. ✓ The dependent variable
- C. A control variableA control variable is a condition kept the same in every tube.
Why: The dependent variable is the quantity measured to see the effect.
She measured the volume of oxygen to see the effect of the tissue.
So the volume of oxygen was the dependent variable.
Video: Watch: The one thing you measure
The test dish two minutes on, nine pill bugs on the damp side and one on the dry; the count written under each side and the name dependent variable beneath the counts.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04-p2.mp4
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Here is the test dish again, damp paper on the left side and dry paper on the right side, 10 pill bugs tipped in.
After 2 minutes you count the pill bugs on each side: nine on the damp side, one on the dry side.
The number of pill bugs on each side is the one quantity you measure, to see the effect of dampness.
In the enzyme tubes, the quantity the student measured to see the effect was the volume of oxygen, in mL. That quantity was called the dependent variable.
The number of pill bugs on each side is the dependent variable, because you measure it to see the effect of dampness.
Dampness is not the dependent variable, because you set it rather than measuring it as the result.
So the dependent variable is the result you measure: here, a count of pill bugs on each side after 2 minutes.
What you are expected to know Identify the dependent variable in a described behavior experiment: the quantity measured to see the effect.
A student tips 20 fruit flies into a tube lit at one end and dark at the other end. She counts the flies at each end after 3 minutes.
Which is the dependent variable?
- A. ✓ The number of flies at each end
- B. Dark or light at an endLight is the condition she deliberately changes from one end to the other.
Light is the independent variable. - C. The number of flies tipped inShe sets the number of flies, 20, at the start.
She does not measure it as the result.
Why: The dependent variable is the quantity measured to see the effect.
She counts the flies at each end after 3 minutes.
So the number of flies at each end is the dependent variable.
A student lays rough cloth on one side of a board and smooth cloth on the other side. He tips in 15 snails and counts the snails on each side after 10 minutes.
Which is the dependent variable?
- A. The roughness of the clothThe roughness of the cloth is the condition he deliberately changes.
Roughness is the independent variable. - B. The size of the boardThe size of the board is the same for both sides.
- C. ✓ The number of snails on each side
Why: The dependent variable is the quantity measured to see the effect.
He counts the snails on each side after 10 minutes.
So the number of snails on each side is the dependent variable.
A student covers the left side of a dish so it is dark and leaves the right side in the light, then tips in 12 woodlice.
Is light the dependent variable?
- A. YesShe sets one side dark and the other side light.
Light is a condition she sets, not a result she measures. - B. ✓ No
Why: The dependent variable is the quantity measured to see the effect.
She sets the light on each side.
She does not measure it as the result.
So light is not the dependent variable.
Light is the independent variable.
A student lays warm sand on one side of a tray and cool sand on the other side, tips in 12 mealworms and counts the mealworms on each side after 5 minutes.
Is the number of mealworms on each side the dependent variable?
- A. ✓ Yes
- B. NoHe counts the mealworms on each side to see the effect of temperature.
That count is the result he measures.
Why: The dependent variable is the quantity measured to see the effect.
He counts the mealworms on each side after 5 minutes.
So the number of mealworms on each side is the dependent variable.
35The dish you compare against
Unit 3 used two names that sound alike: the control, and a control variable. They are not the same thing.
In the Unit 3 enzyme tubes, one tube of hydrogen peroxide got the same treatment as the others but no tissue.
What is the control in an experiment?
- A. A condition kept the same in every tubeA condition kept the same in every tube is a control variable.
- B. The condition the experimenter deliberately changesThe condition the experimenter deliberately changes is the independent variable.
- C. ✓ The tube that lacks only the factor under test
Why: The control is a whole tube.
It gets the same treatment as the other tubes but lacks the factor under test.
So it shows the result with that factor absent.
Video: Watch: The dish you compare against
A second dish appears beside the test dish, damp paper on both sides, the same ten pill bugs tipped in; the word control written under it and the table of the four parts of the test filling in row by row.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04-p3.mp4
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Now consider the friend's objection. Perhaps pill bugs bunch together on one side of any dish, damp or not.
To answer it, you need a dish with nothing to choose between.
Suppose you set up a second dish beside the first. You lay damp paper on both sides and tip in 10 pill bugs the same way.
The second dish shows how the pill bugs spread when there is no difference to choose between.
In the enzyme tubes, the tube given the same treatment as the others but lacking the factor under test was called the control.
The both-sides-damp dish is the control, because it gets the same paper, the same 10 pill bugs and the same 2 minutes, but lacks the difference in dampness.
Keeping 10 pill bugs in every dish is not the control, because it is a condition kept the same, not a dish.
A control is a whole dish. A condition kept the same in every dish is a control variable.
A control variable is not the control.
The table below compares the four parts of the test: what you change, what you measure, what you keep the same, and the control.
What you are expected to know Identify the control in a described behavior experiment: the dish given the same treatment but with no difference to choose between.
A student tests whether woodlice gather in the dark. Four dishes are drawn below, numbered 1 to 4, each with paper on both sides. The shaded paper is dark. Dishes 1, 2 and 4 hold 12 woodlice each; dish 3 holds 20.
Which dish is the control?
- A. Dish 1Dish 1 has dark paper on one side and light on the other.
The woodlice have a difference to choose between, so dish 1 is a test dish. - B. Dish 2Dish 2 has dark paper on one side and light on the other, the sides swapped.
The woodlice still have a difference to choose between. - C. Dish 3Dish 3 has dark paper on one side and light on the other, with more woodlice.
More woodlice does not remove the difference between the sides. - D. ✓ Dish 4
Why: The control is the dish given the same treatment but with no difference to choose between.
Dish 4 has dark paper on both sides.
So dish 4 is the control.
A student lays warm sand on one side of a tray and cool sand on the other side, tips in 12 mealworms, and sets up a second tray with warm sand on both sides and 12 mealworms. He says: “The control in this experiment is keeping 12 mealworms in every tray.”
Is the student correct?
- A. YesKeeping 12 mealworms in every tray is a condition kept the same: a control variable.
The control is a whole tray. - B. ✓ No
Why: The control is a whole tray given the same treatment but with no difference to choose between.
Keeping 12 mealworms in every tray is a condition kept the same, a control variable.
So the student is not correct.
The control is the tray with warm sand on both sides.
A student lays damp cloth on one side of a board and dry cloth on the other side, tips in 15 snails and counts the snails on each side after 10 minutes.
Which of the following boards is the control for this experiment?
- A. ✓ A second board with damp cloth on both sides
- B. A second board with twice as many snailsDoubling the snails leaves a damp side and a dry side.
The snails still have a difference to choose between. - C. A second board left for twice as longDoubling the time leaves a damp side and a dry side.
The snails still have a difference to choose between.
Why: The control is the board given the same treatment but with no difference to choose between.
A board with damp cloth on both sides has no difference in dampness.
So the board with damp cloth on both sides is the control.
53Why you need the second dish
In the Unit 3 enzyme tubes, a tube of hydrogen peroxide with no tissue sat beside the potato tubes.
What did the no-tissue tube show?
- A. That the potato contains catalaseThe no-tissue tube on its own does not show what the potato does.
Only the comparison with the potato tubes shows that. - B. ✓ How much oxygen the peroxide releases on its own
- C. How much peroxide each tube held at the startThe no-tissue tube does not measure the starting volume of peroxide.
Why: The tested factor was the tissue.
The no-tissue tube was the control, the tube lacking the tested factor.
So it showed how much oxygen the peroxide releases on its own.
Video: Watch: Why you need the second dish
The two dishes side by side with their counts, nine and one against five and five; the justification written beneath them one line at a time: without the control dish, the lopsided count could not be put down to dampness.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04-p4.mp4
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Here are the two dishes again: the test dish, damp paper on one side and dry on the other, and the control dish, damp paper on both sides.
Suppose the control dish shows 5 pill bugs on each side after 2 minutes.
In the control dish there was nothing to choose between. The pill bugs spread evenly, 5 and 5.
In the test dish the only difference was dampness. Nine pill bugs sat on the damp side.
So the lopsided count in the test dish can be put down to dampness.
Now imagine the control dish had also shown nine pill bugs on one side and one on the other.
Then the pill bugs bunch on one side even with nothing to choose between. The lopsided count in the test dish could not be put down to dampness.
The control rules out one thing: the pill bugs gathering on one side anyway.
To justify a control, say what it rules out. Without the control dish, you could not tell whether the pill bugs would have gathered on one side with no difference in dampness.
What you are expected to know Justify a control in a behavior experiment by stating what it rules out.
A student tips 20 fruit flies into a tube lit at one end and dark at the other end. Beside it she sets up a second tube lit at both ends with 20 fruit flies.
What does the second tube show?
- A. That the flies prefer the darkThe second tube on its own does not show what light does to the flies.
Only the comparison with the first tube shows that. - B. How many flies were tipped inThe student sets the number of flies at the start.
The second tube does not measure it. - C. ✓ How the flies spread when both ends are lit
Why: The second tube has no difference in light between its ends.
So it shows how the flies spread when there is nothing to choose between.
A student lays rough cloth on one side of a board and smooth cloth on the other side, tips 15 snails into the middle and counts the snails on each side after 10 minutes. Beside it she sets up a second board with rough cloth on both sides and 15 snails.
(a) Explain why the student needs the second board before she can put a lopsided count on the first board down to the roughness of the cloth. (1 pt)
Frame Without the second board, the student could not tell whether …
The second board has rough cloth on both sides.
So the snails have nothing to choose between.
It shows how the snails spread when nothing differs.
So a lopsided count on the first board, and an even spread on the second, can be put down to the roughness of the cloth.
- Award 1 point for: the second board shows how the snails spread with no difference between the sides, so a lopsided count on the first board can be put down to the roughness of the cloth (without it, the student could not tell whether the snails gather on one side anyway).
- Accept: ‘it rules out the snails bunching on one side of any board’.
A student tips 20 fruit flies into a tube lit at one end and dark at the other end, and 20 fruit flies into a second tube lit at both ends. After 3 minutes, 17 flies sit at the dark end of the first tube. In the second tube, 17 flies sit at one end.
Can the lopsided count in the first tube be put down to light?
- A. YesIn the second tube the flies bunched at one end with no difference in light.
So bunching happens anyway. - B. ✓ No
Why: The second tube has no difference in light.
The flies still bunched at one end, 17 against 3.
So the flies bunch at one end anyway.
The lopsided count in the first tube cannot be put down to light.
Here is the dish again, nine pill bugs on the damp side and one on the dry, and the friend who says it was chance.
Dampness was the variable you changed. The count on each side was the variable you measured.
The both-sides-damp dish was the control. Without it, the lopsided count could not be put down to dampness.
72Mixed practice mixed practice
A student covers one side of a box so it is dark and leaves the other side lit. She tips in 20 crickets and counts the crickets on each side after 5 minutes.
Which is the dependent variable?
- A. ✓ The number of crickets on each side
- B. Dark or light on a sideLight is the condition she deliberately changes from one side to the other.
Light is the independent variable. - C. The number of crickets tipped inShe sets the number of crickets, 20, at the start.
She does not measure it as the result.
Why: The dependent variable is the quantity measured to see the effect.
She counts the crickets on each side after 5 minutes.
So the number of crickets on each side is the dependent variable.
A student covers one side of a box so it is dark and leaves the other side lit, tips in 20 crickets, and sets up a second box lit on both sides with 20 crickets.
Is keeping 20 crickets in every box the control?
- A. YesKeeping 20 crickets in every box is a condition kept the same: a control variable.
The control is a whole box. - B. ✓ No
Why: The control is a whole box given the same treatment but with no difference to choose between.
Keeping 20 crickets in every box is a condition kept the same, a control variable.
So it is not the control.
The control is the box lit on both sides.
A student covers one side of a box so it is dark and leaves the other side lit. She tips in 20 crickets and counts the crickets on each side after 5 minutes.
Which is the independent variable?
- A. ✓ Dark or light on a side
- B. The number of crickets on each sideThe number of crickets on each side is the result she measures.
That count is the dependent variable. - C. The size of the boxThe size of the box is the same for both sides.
Why: The independent variable is the condition the experimenter deliberately sets to two values.
She makes one side dark and leaves the other side lit.
So light is the independent variable.
A student covers one side of a box so it is dark and leaves the other side lit, tips in 20 crickets, and sets up a second box lit on both sides with 20 crickets.
What does the box lit on both sides show?
- A. That crickets prefer the darkThe box lit on both sides on its own does not show what the dark does to the crickets.
Only the comparison with the first box shows that. - B. How many crickets were tipped inThe student sets the number of crickets at the start.
The second box does not measure it. - C. ✓ How the crickets spread with no difference in light
Why: The box lit on both sides has no difference in light.
So it shows how the crickets spread when there is nothing to choose between.
A student covers one side of a box so it is dark and leaves the other side lit, tips in 20 crickets, and sets up a second box lit on both sides with 20 crickets. A classmate says: “The control is the box lit on both sides, and keeping 20 crickets in every box is one of the experiment’s control variables.”
Is the classmate correct?
- A. ✓ Yes
- B. NoKeeping 20 crickets in every box is a condition kept the same in every box.
A condition kept the same is a control variable; the control is a whole box.
Why: The control is a whole box with no difference to choose between.
A control variable is a condition kept the same in every box.
The student keeps 20 crickets in every box.
So it is a control variable and not the control, and the classmate is correct.
A student covers one side of a box so it is dark and leaves the other side lit, and tips in 20 crickets.
Which of the following boxes is the control for this experiment?
- A. A second box with twice as many cricketsDoubling the crickets leaves a dark side and a lit side.
The crickets still have a difference to choose between. - B. ✓ A second box lit on both sides
- C. A second box left for twice as longDoubling the time leaves a dark side and a lit side.
The crickets still have a difference to choose between.
Why: The control is the box given the same treatment but with no difference to choose between.
A box lit on both sides has no difference in light.
So the box lit on both sides is the control.
A student covers one side of a box so it is dark and leaves the other side lit. She tips in 20 crickets and counts the crickets on each side after 5 minutes. She sets up a second box lit on both sides with 20 crickets, counted the same way.
(a) Identify the control in the student’s experiment. (1 pt)
- Award 1 point for: the box lit on both sides (the box with no difference in light).
(b) Justify the student’s decision to set up a second box. (1 pt)
It shows how the crickets spread when there is nothing to choose between.
Without it, the student could not tell whether the crickets would have gathered on one side anyway.
So a lopsided count in the first box, and an even spread in the second, can be put down to light.
- Award 1 point for: the second box shows how the crickets spread with no difference in light, so a lopsided count in the first box can be put down to light (without it, the student could not tell whether the crickets bunch on one side anyway).
APBIO-U08-L04B Nine to one, or chance?
Here is the test dish again: damp paper on one side, dry paper on the other, and 20 pill bugs tipped in. After two minutes, 16 pill bugs sit on the damp side and 4 on the dry side.
If the pill bugs had no preference, about 10 would sit on each side. Is 16 to 4 far enough from 10 to 10 to say they chose?
Unit 8 · Ecology
1No preference: the null hypothesis and the expected counts
How do you tell a real choice from a lucky spread?
You start from the null hypothesis that the pill bugs have no preference. So the expected count on each side is the total divided by two: 10 and 10.
Chi-square measures how far the observed counts sit from the expected counts. Your formula sheet prints its equation below, the same equation you used on the fruit-fly cross in Unit 5.
Chi-square, exactly as the AP formula sheet writes it.
is chi-square, the one number for the gap
is the observed count on one side
is the expected count for that side
means add the sides
With 16 and 4 against 10 and 10, chi-square is 7.2. The critical value at p = 0.05 with one degree of freedom is 3.84.
7.2 is larger than 3.84. So the null hypothesis is rejected: the pill bugs chose.
The only new step is where the expected counts come from: an even split, never “they prefer damp”.
In Unit 5 a breeder predicted a ratio from a Punnett square and counted the offspring.
Which of the following was the null hypothesis for the cross?
- A. The offspring really differ from the predicted ratio, for a reason the breeder can nameA claim that the counts really differ from the model is the alternative hypothesis.
The null hypothesis claims no real difference. - B. ✓ The offspring occur in the ratio the model predicts, and any difference from it is due to chance
Why: The null hypothesis is the no-difference statement.
For a breeding cross, the no-difference statement is that the offspring fit the predicted ratio and any gap is chance.
Video: Watch: No preference: the null hypothesis and the expected counts
The dish from above with 16 and 4 written under its halves; the null hypothesis of no preference written beneath it; 20 divided by two sides; 10 and 10 written under the damp half and the dry half.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04Ba.mp4
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Here is the test dish again, seen from above: the damp half shaded, 16 pill bugs on the damp side and 4 on the dry side.
Two things could explain the lopsided count. The first is chance.
Each pill bug wanders on its own. So 20 pill bugs can stray from an even spread by chance alone.
The second is that the pill bugs chose the damp side.
In Unit 5 the null hypothesis was the no-difference statement: the counts fit the model, and any gap is due to chance.
For the dish, the null hypothesis is the same kind of statement: the pill bugs have no preference, and any gap from an even spread is due to chance.
The null hypothesis is never “the pill bugs prefer damp”. A preference is the real difference the test looks for.
So “the pill bugs prefer damp” is the alternative hypothesis: the claim that the counts really differ from an even spread.
No preference means the pill bugs spread evenly between the sides. So the expected count on each side is the total divided by the number of sides (or sections).
The expected count for one side, : the total number of pill bugs divided by the number of sides (or sections)
20 pill bugs are tipped into a dish with two sides, damp and dry; the null hypothesis is that they have no preference. Calculate the expected count on each side.
So no preference expects 10 pill bugs on the damp side and 10 on the dry side. The table below sets the observed count on each side beside the expected count.
Now consider a chamber with four sections instead: damp and dark, dry and dark, damp and light, dry and light.
With 32 pill bugs and no preference, the expected count in each section is the total divided by four: 8 pill bugs.
What you are expected to know State the null hypothesis for a choice chamber as no preference, and calculate the expected count on each side as the total divided by the number of sides.
A student tips 30 pill bugs into a chamber with a dark half and a light half.
Which of the following is the null hypothesis for the test?
- A. ✓ The pill bugs have no preference, and any gap from 15 and 15 is due to chance
- B. The pill bugs prefer the dark half, so the gap from 15 and 15 is realA preference is a real difference between the halves.
A claim of a real difference is the alternative hypothesis. - C. Exactly 15 pill bugs sit on each half, with no straying at allThe null hypothesis expects chance to make the counts stray from 15 and 15.
The null hypothesis never promises exact counts.
Why: The null hypothesis is the no-difference statement.
No preference means an even spread of the 30 pill bugs: 15 on each half.
So the null hypothesis says the pill bugs have no preference, and the gap from 15 and 15 is chance.
A student tips 20 pill bugs into a chamber with a damp half and a dry half, and writes the null hypothesis as: “The pill bugs prefer the damp half.”
Is the student's statement the null hypothesis?
- A. YesA preference is a real difference between the halves.
The null hypothesis claims no real difference. - B. ✓ No
Why: The null hypothesis is the no-difference statement.
A preference for the damp half is a real difference.
So the student has written the alternative hypothesis; the null hypothesis is that the pill bugs have no preference, and any gap from 10 and 10 is chance.
A student tips 24 pill bugs into a chamber with a warm half and a cool half. The null hypothesis is that the pill bugs have no preference.
Calculate the expected count on each half.
Answer: 12 pill bugs (tolerance ±0)
A student tips 36 pill bugs into a chamber with four sections: damp and dark, dry and dark, damp and light, dry and light. The null hypothesis is that the pill bugs have no preference.
Calculate the expected count in each section.
Answer: 9 pill bugs (tolerance ±0)
A student tips 28 pill bugs into a chamber with a dark half and a light half. The null hypothesis is that the pill bugs have no preference.
Calculate the expected count on each half.
Answer: 14 pill bugs (tolerance ±0)
29Did they choose? Chi-square and the verdict
In a fruit-fly cross, every class has an observed count and an expected count.
Which of the following is chi-square?
- A. The number of offspring whose class did not fit the ratio the model predictsChi-square is a sum of squared gaps divided by expected counts.
It is not a count of offspring. - B. ✓ For every class, the squared gap between o and e divided by e, added over the classes
- C. The gap between the largest observed count and the smallest observed countChi-square compares each observed count with its own expected count.
It does not compare one observed count with another.
Why: For each class, chi-square takes the gap between o and e, squares it, and divides by e.
Then the classes are added.
So chi-square is the squared gaps divided by the expected counts, summed.
A chi-square test compares two classes. The table below is the formula sheet's.
Which of the following is the critical value at p = 0.05?
- A. ✓ 3.84
- B. 5.995.99 is the p = 0.05 value for two degrees of freedom.
Two classes give one degree of freedom. - C. 6.636.63 is the p = 0.01 row's value for one degree of freedom.
The 0.05 row is the row to read.
Why: Two classes give one degree of freedom: two minus one.
Go down the column for one degree of freedom and along the 0.05 row.
The cell reads 3.84.
Chi-square for a set of counts comes out larger than the critical value.
Which of the following is the verdict?
- A. Accept the null hypothesisA chi-square test never accepts or proves a hypothesis.
The verdict is written as reject, or fail to reject. - B. Fail to reject the null hypothesisFail to reject is the verdict when chi-square is not larger than the critical value.
Here chi-square is larger. - C. ✓ Reject the null hypothesis
Why: Chi-square is larger than the critical value.
So the gap between the counts and the model is too large to be chance.
The verdict is: reject the null hypothesis.
Video: Watch: Did they choose? Chi-square and the verdict
The chi-square working appearing one line at a time: the damp side's term, the dry side's term, the sum 7.2; the table's cell for one degree of freedom at p = 0.05 ringed; 7.2 placed beyond 3.84 on the number line; the verdict written in its wording.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L04Bb.mp4
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The null hypothesis of no preference expects 10 and 10. The dish holds 16 and 4.
Chi-square puts one number on the gap between the observed counts and the expected counts. Here is its equation, exactly as your formula sheet prints it, with each symbol named beneath.
Chi-square, exactly as the AP formula sheet writes it.
is chi-square, the one number for the gap
is the observed count on one side
is the expected count for that side
means add the sides
16 pill bugs sit on the damp side and 4 on the dry side; the null hypothesis of no preference expects 10 and 10. Calculate chi-square.
Each side is squared and divided by its own expected count, just as each class was in Unit 5. The sum, 7.2, has no unit.
The verdict needs the critical value. Two sides are two classes, and two classes minus one gives one degree of freedom.
Here is the chi-square table your formula sheet prints: a table with a column for each number of degrees of freedom and a row for each p value.
Go down the column for one degree of freedom and along the p = 0.05 row. The critical value is 3.84.
7.2 is larger than 3.84. So reject the null hypothesis: the counts do not fit no preference, and the pill bugs chose.
Rejecting the null hypothesis says the lopsided count was not chance. It does not say why the pill bugs chose the damp side.
If chi-square is not larger than the critical value, the verdict is fail to reject the null hypothesis: the counts are consistent with no preference.
The verdict is always written as reject, or fail to reject. It is never written as accept, and a chi-square test never proves that the pill bugs have no preference.
What you are expected to know Calculate chi-square for the observed counts against the even-split expected counts, read the p = 0.05 row at the right degrees of freedom, and state the verdict as reject or fail to reject the null hypothesis.
A student tips pill bugs into a chamber with a dark half and a light half. After two minutes, 14 sit on the dark side and 6 on the light side. The null hypothesis of no preference expects 10 and 10; the table below sets the counts side by side.
Calculate chi-square for these counts.
Part 1. Calculate the dark side's term: the squared gap divided by its expected count.
Answer: 1.6 (tolerance ±0.005)
Part 2. Calculate the light side's term.
Answer: 1.6 (tolerance ±0.005)
Answer: 3.2 (tolerance ±0.005)
In another dark-and-light chamber, 13 pill bugs sat on the dark side and 7 on the light side, and chi-square against no preference is 1.8. The table below is the formula sheet's.
Which of the following is the verdict on the null hypothesis at p = 0.05?
- A. RejectChi-square, 1.8, is not larger than the critical value, 3.84.
A gap this size is small enough to be chance. - B. ✓ Fail to reject
Why: Two sides give one degree of freedom, and the 0.05 row reads 3.84.
1.8 is not larger than 3.84.
So fail to reject the null hypothesis: 13 and 7 is consistent with no preference.
A student tips 40 pill bugs into a chamber with four sections: damp and dark, dry and dark, damp and light, dry and light. After two minutes the counts are 17, 12, 7 and 4, written in the drawing below. The null hypothesis is that the pill bugs have no preference.
Calculate chi-square for these counts.
Answer: 9.8 (tolerance ±0.005)
In another four-section chamber with 40 pill bugs, the counts are 18, 10, 8 and 4, and chi-square against no preference is 10.4. The table below is the formula sheet's.
Which of the following is the verdict on the null hypothesis at p = 0.05?
- A. ✓ Reject
- B. Fail to rejectFour sections give three degrees of freedom, and the 0.05 row reads 7.81.
10.4 is larger than 7.81.
Why: Four sections are four classes, so the degrees of freedom are three.
The 0.05 row at three degrees of freedom reads 7.81.
10.4 is larger than 7.81.
So reject the null hypothesis: the counts do not fit no preference.
Here is the test dish again: the damp half shaded, 16 pill bugs on the damp side and 4 on the dry side.
The null hypothesis of no preference expects 10 and 10.
Chi-square comes to 7.2, and 7.2 is larger than 3.84. So reject the null hypothesis: the pill bugs chose.
53Mixed practice mixed practice
A student tips 50 pill bugs into a chamber with a light half and a dark half. After two minutes, 34 sit on the dark side and 16 on the light side. The null hypothesis is that the pill bugs have no preference.
Calculate chi-square for these counts, to three significant figures.
Answer: 6.48 (tolerance ±0.005)
In another light-and-dark chamber with 50 pill bugs, 36 sat on the dark side and 14 on the light side, and chi-square against no preference is 9.68. The table below is the formula sheet's.
Which of the following is the verdict on the null hypothesis at p = 0.05?
- A. ✓ Reject
- B. Fail to rejectTwo sides give one degree of freedom, and the 0.05 row reads 3.84.
9.68 is larger than 3.84.
Why: Two sides give one degree of freedom, and the 0.05 row reads 3.84.
9.68 is larger than 3.84.
So reject the null hypothesis: the counts do not fit no preference.
A student tips 48 pill bugs into a chamber with four sections: damp and dark, dry and dark, damp and light, dry and light.
Which of the following is the null hypothesis for the test?
- A. Exactly the same count sits in each section with no straying at allThe null hypothesis expects chance to make the counts stray from 12 in each section.
The null hypothesis never promises exact counts. - B. The pill bugs prefer the damp and dark section so the gap from an even spread is realA preference is a real difference between the sections.
A claim of a real difference is the alternative hypothesis. - C. ✓ The pill bugs have no preference, and any gap from 12 in each section is due to chance
Why: The null hypothesis is the no-difference statement.
No preference means an even spread of the 48 pill bugs: 12 in each section.
So the null hypothesis says the pill bugs have no preference, and the gap from 12 in each section is chance.
A student tips 24 pill bugs into a chamber with a warm half and a cool half. After two minutes, 15 sit on the cool side and 9 on the warm side. The null hypothesis is that the pill bugs have no preference.
Calculate chi-square for these counts.
Answer: 1.5 (tolerance ±0.005)
In another warm-and-cool chamber with 24 pill bugs, 14 sat on the cool side and 10 on the warm side, and chi-square against no preference is 0.67. The table below is the formula sheet's.
Which of the following is the verdict on the null hypothesis at p = 0.05?
- A. RejectChi-square, 0.67, is not larger than the critical value, 3.84.
A gap this size is small enough to be chance. - B. ✓ Fail to reject
Why: Two sides give one degree of freedom, and the 0.05 row reads 3.84.
0.67 is not larger than 3.84.
So fail to reject the null hypothesis: 14 and 10 is consistent with no preference.
A student tips 44 pill bugs into a chamber with four sections: damp and dark, dry and dark, damp and light, dry and light. The null hypothesis is that the pill bugs have no preference.
Calculate the expected count in each section.
Answer: 11 pill bugs (tolerance ±0)
A student tips 36 pill bugs into a chamber with a damp half and a dry half. After two minutes, 25 sit on the damp side and 11 on the dry side. The null hypothesis is that the pill bugs have no preference.
Calculate chi-square for these counts, to three significant figures.
Answer: 5.44 (tolerance ±0.005)
In another damp-and-dry chamber with 36 pill bugs, 26 sat on the damp side and 10 on the dry side, and chi-square against no preference is 7.11. The table below is the formula sheet's.
Which of the following is the verdict on the null hypothesis at p = 0.05?
- A. ✓ Reject
- B. Fail to rejectTwo sides give one degree of freedom, and the 0.05 row reads 3.84.
7.11 is larger than 3.84.
Why: Two sides give one degree of freedom, and the 0.05 row reads 3.84.
7.11 is larger than 3.84.
So reject the null hypothesis: the counts do not fit no preference.
A student tips 60 pill bugs into a chamber with a warm half and a cool half. After two minutes, 42 pill bugs sit on the cool side and 18 on the warm side. The table below is the formula sheet's.
(a) State the null hypothesis for this test. (1 pt)
- Award 1 point for: the pill bugs have no preference (an even spread, 30 and 30) and the difference from it is due to chance (a no-difference statement).
- Do not award: a statement that the pill bugs prefer the cool half (that is the alternative hypothesis).
(b) Calculate chi-square for these counts against the null hypothesis. (1 pt)
Answer: 9.6 (tolerance ±0.005)
Chi-square for 42 and 18 against 30 and 30 is 9.6.
- Award 1 point for: chi-square of 9.6, with expected counts of 30 and 30.
(c) State the verdict of the test at p = 0.05, in the correct wording, and justify it with the two numbers it rests on. (1 pt)
Chi-square, 9.6, is larger than 3.84.
So reject the null hypothesis: the counts do not fit no preference, and the pill bugs chose the cool side.
- Award 1 point for: reject the null hypothesis, because the calculated chi-square is larger than the critical value 3.84 (one degree of freedom, p = 0.05).
- Do not award a verdict written only as “accept the alternative hypothesis” or “the test proves…”. “Reject the null hypothesis, so the data support a preference for the cool side” earns the point.
APBIO-U08-L05 A call in the grass
Photo: Amaury Laporte, Wikimedia Commons, CC BY 2.0 (cropped and resized).
A prairie dog town sits quiet at noon. One prairie dog stands upright on a mound and gives a sharp bark. Every prairie dog in sight stops feeding and runs to the mouth of its burrow.
Nothing else changed: no hawk had reached them yet. What did the bark do?
Unit 8 · Ecology
1A bark is a signal
How can one organism make a stimulus for another organism?
One prairie dog barked. Every other prairie dog in sight detected the bark and ran to the mouth of its burrow.
The bark was a stimulus for the other prairie dogs. This time another organism made the stimulus.
A stimulus one organism makes for another can travel as light, as sound, as touch, as an electric pulse or as a chemical.
In Unit 4, cells of the adrenal glands released epinephrine into the blood. Heart cells detected the epinephrine and beat faster.
Which of the following was the chemical signal?
- A. The bloodThe blood carried the epinephrine from the adrenal glands to the heart.
The chemical signal is the molecule the blood carried. - B. ✓ The epinephrine molecule
- C. The faster heartbeatThe faster heartbeat is the heart cells’ response after they detected the epinephrine.
The chemical signal is the molecule they detected.
Why: A chemical signal is a molecule one cell releases that carries information to other cells.
The adrenal cells released the epinephrine.
The heart cells detected it.
So the epinephrine molecule is the chemical signal.
Video: Watch: A bark is a signal
One prairie dog upright on its mound gives a sharp bark; every other prairie dog in sight stops feeding and runs to the mouth of its burrow.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L05a.mp4
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In Unit 4 the chemical signal was a molecule. One cell released it, and other cells detected it.
One prairie dog produced the bark. Every other prairie dog in sight detected the bark.
Something one organism produces that another organism detects is called a .
So the bark is a signal.
Unit 4’s chemical signal was the same idea one scale down: one cell produced a molecule, and other cells detected it. Here one whole organism produces the signal, and other whole organisms detect it.
The prairie dog that barked is the sender. The prairie dogs that detected the bark are the receivers.
One organism produces a stimulus, another detects it, and the receiver’s behavior changes; the sender’s success in leaving offspring can depend on it.
The receivers’ behavior changed: they stopped feeding and ran to the mouths of their burrows.
Here are four more cases, each judged by the same two questions: did an organism produce it, and did another organism detect it?
For example, a male stickleback’s belly turns red, and a female stickleback sees it. The red belly is a signal, because one organism produced it and another organism detected it.
And a whale sings, and a second whale far away hears the song. The song is a signal, because one organism produced it and another organism detected it.
But a wave breaks on a beach, and a seal hears the crash. The crash is not a signal, because no organism produced it.
And the sun comes out, and its warmth reaches the seal’s back. The warmth is not a signal, because no organism produced it.
Here is a table of the four cases with their verdicts: whether an organism made each one, whether another organism detected it, and whether it is a signal.
What you are expected to know Identify the signal in a described exchange between organisms: the thing one organism produces and another organism detects.
A male cricket chirps. A female cricket walks toward the chirping.
Which of the following is the signal?
- A. The air that carries the chirpThe air carries the chirp from one cricket to the other.
The signal is the chirp the air carries. - B. The female cricket’s movement toward the maleWalking toward the chirp is what the female cricket does after she detects it.
The signal is the chirp she detected. - C. ✓ The chirp
Why: A signal is something one organism produces that another organism detects.
The male cricket produced the chirp.
The female cricket detected the chirp.
So the chirp is the signal.
Signals pass between organisms.
What is a signal between organisms?
- A. Any change in the surroundings that an organism detectsA change no organism produced, such as rain, is a stimulus but not a signal.
An organism produces a signal. - B. What an organism does after it detects a changeWhat the organism does afterwards is its response.
The signal is the thing it detected. - C. ✓ Something one organism produces that another organism detects
Why: A signal is something one organism produces that another organism detects.
Rain falls on a leaf and on the snail sitting on it.
Is the rain a signal?
- A. YesNo organism produced the rain.
A signal is something an organism produces. - B. ✓ No
Why: The snail detected the rain.
But no organism produced the rain.
So the rain is not a signal.
A firefly flashes its light. A second firefly sees the flash.
Is the flash a signal?
- A. ✓ Yes
- B. NoThe first firefly produced the flash, and the second firefly detected it.
That is what a signal is.
Why: The first firefly produced the flash.
The second firefly detected the flash.
So the flash is a signal.
A stork on its nest performs its courtship display. A second stork sees the display.
Is the display a signal?
- A. ✓ Yes
- B. NoThe first stork produced the display, and the second stork detected it.
That is what a signal is.
Why: The first stork produced the display.
The second stork detected the display.
So the display is a signal.
Thunder rolls across a field. A horse in the field hears it.
Is the thunder a signal?
- A. YesNo organism produced the thunder.
A signal is something an organism produces. - B. ✓ No
Why: The horse detected the thunder.
But no organism produced the thunder.
So the thunder is not a signal.
A flower’s petals are brightly colored. A bee sees the color and lands on the flower.
Is the flower’s color a signal?
- A. ✓ Yes
- B. NoThe flower, a living organism, produced the color of its petals.
The bee detected the color.
Why: The flower produced the color of its petals.
The bee detected the color.
So the color is a signal, even though a plant produced it.
A cold wind blows across a hillside. A sheep on the hillside turns its back to the wind.
Is the wind a signal?
- A. YesNo organism produced the wind.
A signal is something an organism produces. - B. ✓ No
Why: The sheep detected the wind.
But no organism produced the wind.
So the wind is not a signal.
31Quick quiz: signal mixed practice
A rabbit thumps its hind foot on the ground. A second rabbit in a nearby burrow hears the thump.
Which of the following is the signal?
- A. ✓ The thump
- B. The ground under the rabbit’s footThe ground is what the foot struck.
The signal is the thump the second rabbit detected.
Why: The first rabbit produced the thump.
The second rabbit detected the thump.
So the thump is the signal.
A stork chick in its nest makes begging movements. The parent stork sees the movements and feeds the chick.
Which of the following is the signal?
- A. The food the parent bringsThe food is what the parent brings after it detects the movements.
The signal is the movements the parent detected. - B. ✓ The chick’s begging movements
- C. The parent’s feeding of the chickFeeding the chick is what the parent does after it detects the begging movements.
The signal is the movements.
Why: The chick produced the begging movements.
The parent stork detected the movements.
So the begging movements are the signal.
Signals pass between organisms.
(a) State what a signal between organisms is. (1 pt)
- Award 1 point for: something one organism produces (or makes) that another organism detects.
35Five ways a signal can travel
Video: Watch: Five ways a signal can travel
Five short clips, ten seconds each: a female stickleback seeing a male’s red belly; a whale hearing another whale’s song; one monkey grooming another; an electric fish’s pulse reaching a second fish; ants following a chemical trail.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L05b.mp4
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Now consider how the bark reached the other prairie dogs. The bark traveled through the air as sound.
A signal can reach the receiver in five ways: as light, as sound, as touch, as an electric pulse or as a chemical.
Each way has a name. The name comes from what the receiver detects, not from how the sender made the signal.
The female stickleback sees the male’s red belly. A signal the receiver sees is called a .
The second whale hears the first whale’s song. A signal the receiver hears is called an .
One monkey grooms another monkey, and its fingers touch the groomed monkey’s skin. A signal the receiver detects as touch is called a . Tactile means by touch.
An electric fish sends a weak electric pulse through the water, and a second electric fish detects the pulse. A signal the receiver detects as an electric pulse is called an .
An ant walking back from food leaves a chemical on the ground, and other ants detect the chemical and follow the trail. A signal the receiver detects as a chemical is called a .
Unit 4 used the same words, chemical signal, for a molecule passed between cells. Here the chemical passes between whole organisms.
When organisms of the same species pass a chemical signal between them, that chemical is called a : a chemical carried from one animal to another of its kind. The ant’s trail chemical is a pheromone.
Here is a table comparing the five kinds of signal: what the receiver detects, and one example of each.
One sender can use two ways at once. A dog growls and bares its teeth at a second dog, so the second dog hears an audible signal and sees a visual signal together.
What you are expected to know Classify a signal as visual, audible, tactile, electrical or chemical from what the receiver detects: light, sound, touch, an electric pulse or a chemical.
A firefly flashes its light at night. A second firefly sees the flash.
Which kind of signal is the flash?
- A. ✓ A visual signal
- B. An electrical signalAn electrical signal is an electric pulse the receiver detects.
The second firefly detected light. - C. A chemical signalA chemical signal is a chemical the receiver detects.
The second firefly detected light, whatever the first firefly used to make it.
Why: The kind of signal is named from what the receiver detects.
The second firefly saw the flash: it detected light.
So the flash is a visual signal.
A bird sings at dawn. A second bird hears the song.
Which kind of signal is the song?
- A. A visual signalA visual signal is one the receiver sees.
The second bird heard the song. - B. ✓ An audible signal
- C. A tactile signalA tactile signal is one the receiver detects as touch.
The second bird heard the song.
Why: The kind of signal is named from what the receiver detects.
The second bird heard the song: it detected sound.
So the song is an audible signal.
One chimpanzee grooms another, picking through its fur with its fingers.
Which kind of signal is the grooming?
- A. ✓ A tactile signal
- B. A visual signalA visual signal is one the receiver sees.
The groomed chimpanzee detected the touch of the fingers in its fur. - C. A chemical signalA chemical signal is a chemical the receiver detects.
The groomed chimpanzee detected the touch of the fingers in its fur.
Why: The kind of signal is named from what the receiver detects.
The groomed chimpanzee detected the fingers in its fur: it detected touch.
So the grooming is a tactile signal.
An electric fish sends a weak electric pulse through the water. A second fish detects the pulse.
Which kind of signal is the pulse?
- A. A tactile signalA tactile signal is one the receiver detects as touch.
The second fish detected an electric pulse through the water, with nothing touching it. - B. An audible signalAn audible signal is one the receiver hears.
The second fish detected an electric pulse, not a sound. - C. ✓ An electrical signal
Why: The kind of signal is named from what the receiver detects.
The second fish detected an electric pulse.
So the pulse is an electrical signal.
A female moth releases a chemical into the night air. A male moth detects the chemical.
Which kind of signal is the chemical?
- A. A visual signalA visual signal is one the receiver sees.
The male moth detected a chemical in the air. - B. ✓ A chemical signal
- C. A tactile signalA tactile signal is one the receiver detects as touch.
The male moth detected a chemical in the air.
Why: The kind of signal is named from what the receiver detects.
The male moth detected a chemical.
So the chemical is a chemical signal.
It passed between two moths of the same species.
So it is also a pheromone.
A peacock fans its tail and gives a loud call. A peahen sees the tail and hears the call.
Which kinds of signal did the peacock send?
- A. Visual onlyThe peahen saw the tail, and she also heard the call.
The call is a second signal, one she heard. - B. Audible onlyThe peahen heard the call, and she also saw the tail.
The fanned tail is a second signal, one she saw. - C. ✓ Visual and audible
Why: The kind of signal is named from what the receiver detects.
The peahen saw the fanned tail: a visual signal.
The peahen heard the call: an audible signal.
So the peacock sent both at once.
Here is the prairie dog town again: one prairie dog upright on its mound, barking, and every other prairie dog in sight at the mouth of its burrow.
The bark was a signal. One prairie dog produced it, and every other prairie dog in sight detected it.
The bark traveled through the air as sound. So the bark was an audible signal.
59Quick quiz: visual, audible, tactile, electrical and chemical signals; pheromone mixed practice
A rabbit thumps its hind foot on the ground. A second rabbit in a nearby burrow hears the thump.
Which kind of signal is the thump?
- A. ✓ An audible signal
- B. A tactile signalThe rabbit’s foot touched the ground, not the second rabbit.
The second rabbit heard the thump: it detected sound. - C. A visual signalA visual signal is one the receiver sees.
The second rabbit heard the thump: it detected sound.
Why: The kind of signal is named from what the receiver detects.
The second rabbit heard the thump: it detected sound.
So the thump is an audible signal.
A student says: “A signal between organisms is always a sound.”
Is the student correct?
- A. YesA signal can travel as light, touch, an electric pulse or a chemical as well as sound.
A firefly’s flash is a signal with no sound. - B. ✓ No
Why: A signal is something one organism produces that another organism detects.
The receiver may see it, hear it, detect it as touch, detect it as an electric pulse or detect it as a chemical.
So a signal is not always a sound.
A deer raises its white tail as it bounds away. A second deer sees the raised tail.
Which kind of signal is the raised tail?
- A. An audible signalAn audible signal is one the receiver hears.
The second deer saw the raised tail: it detected light. - B. ✓ A visual signal
- C. A chemical signalA chemical signal is a chemical the receiver detects.
The second deer saw the raised tail: it detected light.
Why: The kind of signal is named from what the receiver detects.
The second deer saw the raised tail: it detected light.
So the raised tail is a visual signal.
Signals travel between organisms in five ways.
What is a visual signal?
- A. ✓ A signal the receiver sees
- B. A signal the receiver hearsA signal the receiver hears is an audible signal.
- C. A signal the receiver detects as touchA signal the receiver detects as touch is a tactile signal.
Why: A visual signal is a signal the receiver sees.
Signals travel between organisms in five ways.
What is an audible signal?
- A. A signal the receiver seesA signal the receiver sees is a visual signal.
- B. ✓ A signal the receiver hears
- C. A signal the receiver detects as a chemicalA signal the receiver detects as a chemical is a chemical signal.
Why: An audible signal is a signal the receiver hears.
Signals travel between organisms in five ways.
What is a tactile signal?
- A. A signal the receiver hearsA signal the receiver hears is an audible signal.
- B. A signal the receiver detects as an electric pulseA signal the receiver detects as an electric pulse is an electrical signal.
- C. ✓ A signal the receiver detects as touch
Why: A tactile signal is a signal the receiver detects as touch.
Signals travel between organisms in five ways.
What is an electrical signal?
- A. ✓ A signal the receiver detects as an electric pulse
- B. A signal the receiver detects as touchA signal the receiver detects as touch is a tactile signal.
- C. A signal the receiver seesA signal the receiver sees is a visual signal.
Why: An electrical signal is a signal the receiver detects as an electric pulse.
Signals travel between organisms in five ways.
What is a chemical signal between organisms?
- A. A signal the receiver hearsA signal the receiver hears is an audible signal.
- B. ✓ A signal the receiver detects as a chemical
- C. A signal the receiver seesA signal the receiver sees is a visual signal.
Why: A chemical signal between organisms is a signal the receiver detects as a chemical.
Some chemical signals have a name of their own.
What is a pheromone?
- A. A visual signal passed between organisms of the same speciesA visual signal is one the receiver sees.
A pheromone is a chemical. - B. An audible signal passed between organisms of the same speciesAn audible signal is one the receiver hears.
A pheromone is a chemical. - C. ✓ A chemical signal passed between organisms of the same species
Why: A pheromone is a chemical signal passed between organisms of the same species.
The ant’s trail chemical, detected by other ants, is one.
Some chemical signals have a name of their own.
(a) State what a pheromone is. (1 pt)
- Award 1 point for: a chemical (or chemical signal) passed between organisms of the same species.
Glossary
- signal (between organisms)
- Something one organism produces that another organism detects: a prairie dog’s bark, a flower’s color seen by a bee. The same word Unit 4 used for a molecule passed between cells, one scale up.
- visual signal
- A signal the receiver sees: it detects light. A male stickleback’s red belly seen by a female is one.
- audible signal
- A signal the receiver hears: it detects sound. A whale’s song heard by another whale is one.
- tactile signal
- A signal the receiver detects as touch. One monkey grooming another is one.
- electrical signal
- A signal the receiver detects as an electric pulse. An electric fish’s pulse through the water, detected by a second fish, is one.
- chemical signal (between organisms)
- A signal the receiver detects as a chemical. An ant’s trail chemical detected by other ants is one. Unit 4 used the same words for a molecule passed between cells.
- pheromone
- A chemical signal passed between organisms of the same species, such as the ant’s trail chemical. Every pheromone is a chemical signal.
APBIO-U08-L06 What the signal is for
Photo: Ellie Attebery, Wikimedia Commons, CC BY 2.0 (resized).
Two wolves stand in a pack. One walks stiff-legged toward the other, tail raised, staring. The second wolf drops its tail, flattens its ears and looks away. No blow is struck, and no food changes hands.
What did the first wolf’s posture do?
Unit 8 · Ecology
1Four jobs a signal does
What are signals for, and what sets them off? An animal’s signal does a job for the animal that sends it. Four common jobs are these.
It shows dominance, as the wolf’s stare did. It leads others to food, as a honeybee’s dance does.
It marks a territory, as a fox’s scent on a post does. It wins a mate, as a stork’s bill-clattering does.
Something set each signal off: a change outside the animal, such as a rival coming close, or a change inside the animal, such as an empty stomach.
Two questions follow every signal: which job it does for the sender, and what set it off.
A prairie dog town sits quiet at noon. One prairie dog stands on a mound and barks, and every prairie dog in sight stops feeding and runs to the mouth of its burrow.
Which of the following is the signal?
- A. ✓ The bark
- B. The moundThe mound is where the prairie dog stands.
No prairie dog produced the mound for another to detect. - C. The burrowThe burrow is where each prairie dog goes.
No prairie dog produced the burrow for another to detect.
Why: A signal is something one organism produces that another organism detects.
One prairie dog produced the bark.
The other prairie dogs detected it.
So the bark is the signal.
Video: Watch: Four jobs a signal does
Four short scenes, one for each job: the first wolf’s stare and the second wolf giving way; the worker honeybee’s dance and the other workers flying out to the flowers; the fox’s scent on the post and a second fox turning back; the stork’s bill-clattering and its mate landing beside it. The job is written under each scene as it ends.
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Here are the two wolves again. The first wolf walked stiff-legged toward the second, tail raised, staring, and the second wolf dropped its tail, flattened its ears and looked away.
The posture was a signal. The first wolf produced it, and the second wolf detected it.
Now ask what the signal did for the wolf that sent it. The second wolf gave way without a fight.
The posture did one job for the sender: it showed dominance. The first wolf stands above the second in the pack, and the second wolf gives way to it.
Now consider a worker honeybee back in the hive from a patch of flowers. She dances on the comb, and the other workers then fly out to the same flowers.
The dance did one job for the sender: it led the other workers to food.
Now consider a fox that leaves its scent on a fence post at the edge of the ground it hunts. Another fox smells the post and turns back.
The second fox turned back, so the sender kept that ground to itself. The ground an animal keeps other animals of its kind out of is its territory. So the scent did one job for the sender: it marked the fox’s territory.
Now consider a stork at its nest. It clatters its bill in a display, and a second stork lands beside it and stays as its mate.
The clattering did one job for the sender: it won the stork a mate.
Here are the four cases side by side, each with the job its signal did for the sender.
Four jobs an animal’s signal often does for its sender: it shows dominance, it leads others to food, it marks territory, or it wins a mate. A warning bark, like the prairie dog’s bark, does a fifth job: it warns the receivers of a predator.
The table below compares the four jobs: the job for the sender, one example, and who detects the signal.
What you are expected to know Identify what a described animal signal does for the sender: shows dominance, leads others to food, marks territory or wins a mate.
In a wolf pack, one wolf walks stiff-legged toward another, tail raised, staring. The second wolf drops its tail and looks away.
Which job did the posture do for the sender?
- A. ✓ Shows dominance
- B. Leads others to foodNo wolf fed after the posture.
The second wolf gave way to the first. - C. Marks territoryThe posture marked no ground; the second wolf stayed in the pack.
The second wolf gave way to the first. - D. Wins a mateThe second wolf did not stay as the sender’s mate.
The second wolf gave way to the first.
Why: The sender stared, and the second wolf gave way without a fight.
A signal after which a rival gives way shows dominance.
So the posture showed dominance.
A stork at its nest clatters its bill in a display. A second stork lands beside it and stays as its mate.
Which job did the clattering do for the sender?
- A. Shows dominanceThe second stork did not give way; it stayed beside the sender.
- B. Leads others to foodNo stork fed after the clattering; a second stork stayed as the sender’s mate.
- C. Marks territoryThe clattering kept no stork away; a second stork landed beside the sender and stayed.
- D. ✓ Wins a mate
Why: After the clattering, a second stork landed beside the sender and stayed as its mate.
A signal after which a mate joins the sender wins a mate.
So the clattering won the stork a mate.
A worker honeybee returns to the hive from a patch of flowers and dances on the comb. The other workers then fly out to the same flowers.
Which job did the dance do for the sender?
- A. Shows dominanceNo worker gave way to the dancer; the other workers flew out to the flowers.
- B. ✓ Leads others to food
- C. Marks territoryThe dance kept no bee out of any ground; the other workers flew out to the flowers.
- D. Wins a mateNo bee stayed as the dancer’s mate; the other workers flew out to the flowers.
Why: After the dance, the other workers flew out to the flowers the dancer had found.
A signal after which others go to the food leads others to food.
So the dance led the other workers to food.
Suppose a fox leaves its scent on a fence post at the edge of the ground it hunts. A second fox smells the post and turns back.
Which job did the scent do for the sender?
- A. Shows dominanceThe two foxes never came face to face, so neither gave way to the other; the second fox turned back at the post.
- B. Leads others to foodNo fox fed after smelling the post; the second fox turned back.
- C. ✓ Marks territory
- D. Wins a mateNo fox stayed as the sender’s mate; the second fox turned back at the post.
Why: The scent sat at the edge of the ground the sender hunts.
The second fox smelled it and turned back, so the sender kept that ground to itself.
A signal that keeps others out of the sender’s ground marks territory.
So the scent marked the fox’s territory.
Suppose an ant finds a pile of seeds and lays a scent trail from the seeds back to its nest. Other ants from the nest then follow the trail to the seeds.
Which job did the trail do for the sender?
- A. Shows dominanceNo ant gave way to the sender; the other ants followed the trail to the seeds.
- B. ✓ Leads others to food
- C. Marks territoryThe trail kept no ant out of any ground; the other ants followed it to the seeds.
- D. Wins a mateNo ant stayed as the sender’s mate; the other ants followed the trail to the seeds.
Why: After the trail was laid, the other ants followed it to the seeds.
A signal after which others go to the food leads others to food.
So the trail led the other ants to food.
Suppose two male deer come face to face in a clearing. One lowers its antlers toward the other and roars. The other deer backs away, and no fight follows.
Which job did the lowered antlers and the roar do for the sender?
- A. ✓ Shows dominance
- B. Leads others to foodNo deer fed after the roar; the other deer backed away.
- C. Marks territoryThe roar marked no ground; the other deer backed away from the sender itself.
- D. Wins a mateThe other deer did not stay as the sender’s mate; it backed away.
Why: After the lowered antlers and the roar, the other deer backed away without a fight.
A signal after which a rival gives way shows dominance.
So the antlers and the roar showed dominance.
Suppose a male firefly flashes its light at dusk. A female firefly flashes back and moves toward him, and the two pair.
Which job did the flash do for the sender?
- A. Shows dominanceNo firefly gave way to the sender; a female moved toward him and the two paired.
- B. Leads others to foodNo firefly fed after the flash; a female moved toward the sender and the two paired.
- C. Marks territoryThe flash kept no firefly away; a female moved toward the sender and the two paired.
- D. ✓ Wins a mate
Why: After the flash, a female firefly moved toward the sender and the two paired.
A signal after which a mate joins the sender wins a mate.
So the flash won the firefly a mate.
Suppose a male songbird sings every morning in spring from a post at the edge of his patch. When another male lands nearby, the singer flies at him and chases him off the patch.
Which job did the song do for the sender?
- A. Shows dominanceThe other male did not give way to the singer face to face; the singer chased him off the patch.
- B. Leads others to foodNo bird fed after the song; the singer chased the other male off the patch.
- C. ✓ Marks territory
- D. Wins a mateNo bird stayed as the singer’s mate; the singer chased the other male off the patch.
Why: The singer sang from the edge of his patch and chased another male off it.
A signal that keeps others out of the sender’s ground marks territory.
So the song marked the singer’s territory.
31The cue that set it off
Suppose a bat has caught nothing for two nights, and its stomach is empty.
Which kind of change is the empty stomach?
- A. ✓ An internal change
- B. An external changeAn external change happens in the animal’s surroundings.
The empty stomach is inside the bat’s own body.
Why: An internal change happens inside the animal’s own body.
The bat’s stomach is inside its body.
So the empty stomach is an internal change.
Video: Watch: The cue that set it off
The two wolves again: the second wolf walks up close, and that approach is drawn outside the first wolf’s body and labeled the cue, then the stare. Then the chick in its nest: an empty stomach drawn inside its body and labeled the cue, then the begging call.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L06b.mp4
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Here are the two wolves again. A moment before the stare, the second wolf had walked up close to the first.
That approach set the stare off. Whatever sets a signal off is called the signal’s cue.
The approach happened outside the first wolf’s body. A cue outside the sender’s body is an external cue, the same word ‘external’ the earlier lesson used for a change in an animal’s surroundings.
Now consider a chick in a nest. Its stomach is empty, and it calls; a parent then feeds it.
The empty stomach set the begging call off, and the empty stomach is inside the chick’s body. A cue inside the sender’s body is an internal cue.
Nothing outside the chick changed, yet the chick still sent a signal. An internal cue is still a cue.
Now consider a deer that snorts when a predator moves at the edge of the forest. The moving predator is outside the deer’s body, so the cue is external.
The state inside an animal can be the cue as well. An animal cornered by a predator has a racing heart and tense muscles. That state is called fight-or-flight. From that state the animal may growl or hiss: the cue is inside it.
Now consider a plant. Suppose a grazing animal bites its leaves, and the plant then releases a chemical that neighboring plants detect.
The bite came from outside the plant, so the cue is external.
To find the cue, ask where the change that set the signal off happened. Outside the sender’s body, the cue is external; inside the sender’s body, the cue is internal.
What you are expected to know Identify the cue, internal or external, that made an organism send a signal.
In a wolf pack, the second wolf walks up close to the first. The first wolf then walks stiff-legged toward it, tail raised, staring.
Which of the following set the stare off?
- A. ✓ The second wolf coming close
- B. The first wolf’s empty stomachNothing in the case says the first wolf was hungry.
The second wolf came close, and the stare followed. - C. The first wolf’s tail risingThe rising tail is part of the posture, the signal itself.
The second wolf came close, and the stare followed.
Why: The second wolf came close.
The stare followed that approach.
So the second wolf coming close set the stare off.
In a wolf pack, the second wolf walks up close to the first, and the first wolf stares at it.
Is the cue for the stare internal or external?
- A. InternalAn internal cue is inside the sender’s body.
The second wolf’s approach happened outside the first wolf’s body. - B. ✓ External
Why: The cue was the second wolf coming close.
That approach happened outside the first wolf’s body.
A cue outside the sender’s body is external.
Suppose a male songbird sings more each morning as the days lengthen in spring.
Is the cue for the singing internal or external?
- A. InternalAn internal cue is inside the sender’s body.
The lengthening day is in the bird’s surroundings. - B. ✓ External
Why: The cue was the lengthening day.
Day length changes in the bird’s surroundings, outside its body.
A cue outside the sender’s body is external.
Suppose a female moth releases a scent into the air once the eggs inside her are ready to lay. Males detect the scent.
Is the cue for the scent release internal or external?
- A. ✓ Internal
- B. ExternalAn external cue is in the sender’s surroundings.
The eggs becoming ready are inside the moth’s own body.
Why: The cue was the eggs inside the moth becoming ready to lay.
That change happened inside the moth’s body.
A cue inside the sender’s body is internal.
Suppose a dog growls when a stranger steps into its yard.
Is the cue for the growl internal or external?
- A. InternalAn internal cue is inside the sender’s body.
The stranger stepped into the yard, in the dog’s surroundings. - B. ✓ External
Why: The cue was the stranger stepping into the yard.
The stranger is outside the dog’s body.
A cue outside the sender’s body is external.
Suppose a lamb bleats when its body is short of water. Its mother then leads it to the trough.
Is the cue for the bleat internal or external?
- A. ✓ Internal
- B. ExternalAn external cue is in the sender’s surroundings.
The shortage of water is in the lamb’s own body.
Why: The cue was the shortage of water in the lamb’s body.
That change happened inside the lamb’s body.
A cue inside the sender’s body is internal.
Suppose a puppy whines when its stomach is empty. Its mother then feeds it.
Is the cue for the whine internal or external?
- A. ✓ Internal
- B. ExternalAn external cue is in the sender’s surroundings.
The empty stomach is inside the puppy’s own body.
Why: The cue was the puppy’s empty stomach.
The stomach is inside the puppy’s body.
A cue inside the sender’s body is internal.
Suppose a ground squirrel whistles when a hawk’s shadow crosses the ground beside it. The other squirrels dive into their burrows.
Is the cue for the whistle internal or external?
- A. InternalAn internal cue is inside the sender’s body.
The hawk’s shadow crossed the ground beside the squirrel, in its surroundings. - B. ✓ External
Why: The cue was the hawk’s shadow crossing the ground.
The shadow is outside the squirrel’s body.
A cue outside the sender’s body is external.
Suppose a male cricket begins to chirp as the light fades at dusk. A female cricket moves toward the chirping, and the two pair.
(a) Identify the job the chirping did for the sender. (1 pt)
- Award 1 point for: wins a mate (a female moved toward the sender and the two paired).
(b) Identify the cue that set the chirping off. (1 pt)
- Award 1 point for: the fading light (the light fading at dusk).
(c) Explain why the cue is external. (1 pt)
Frame The cue is external because …
The fading happened outside the cricket’s body.
A cue outside the sender’s body is an external cue.
- Award 1 point for: the fading light is a change in the cricket’s surroundings, outside its body, so the cue is external.
Here are the two wolves again, one staring with its tail raised, the other looking away with its tail down.
The stare was a dominance signal.
The second wolf’s approach was the cue that set it off, outside the first wolf’s body: an external cue.
APBIO-U08-L06B Who leaves offspring
Photo: Geoff Gallice, Wikimedia Commons, CC BY 2.0 (resized).
A male frog calls at a pond edge every night in spring. Some males call louder and longer than others. By June, some males have fathered many tadpoles and some have fathered none.
What did the calling change?
Unit 8 · Ecology
1What the receiver does next
How can a signal decide who breeds?
A signal changes what the receiver does.
After a male frog’s call, a female frog moves toward the caller. After a prairie dog’s bark, the other prairie dogs run to the mouths of their burrows.
A signal does not force every receiver into one fixed action. A signal makes one action more likely.
The loud callers draw more females. So the loud callers leave more offspring.
So a signaling behavior can give some individuals more offspring than others.
A male cricket chirps at night. A female cricket detects the chirping.
Which cricket is the receiver of the signal?
- A. The male cricketThe male cricket produced the chirping, so the male cricket is the sender.
- B. ✓ The female cricket
Why: The receiver is the organism that detects the signal.
The female cricket detected the chirping.
So the female cricket is the receiver.
Video: Watch: What the receiver does next
A male frog calls at the pond edge; a female frog in the reeds detects the call and moves toward him; then a prairie dog’s bark, and the other prairie dogs at the mouths of their burrows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L06Ba.mp4
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Here is the pond edge again on a spring night. A male frog calls, and a female frog in the reeds detects the call.
The male frog produced the call. So the male frog is the sender.
The female frog detected the call. So the female frog is the receiver.
Before the call, the female frog sat still in the reeds. After the call, the female frog moves toward the caller.
The call changed what the receiver did. A signal changes the receiver’s behavior.
Now consider ten female frogs in the reeds, all within earshot of the same call.
Six of the ten female frogs move toward the caller. Four of the ten stay where they are.
The call did not force every female frog to move toward the caller. The call made that move more likely.
Now consider a prairie dog town at noon. One prairie dog barks from its mound.
The other prairie dogs stop feeding and run to the mouths of their burrows. The bark changed what the receivers did.
To predict what a receiver does next, ask what job the signal does for its sender.
A mating call draws a mate toward the sender. A bark that warns of a predator sends the receivers to cover.
The table below compares five jobs a signal can do for its sender, with one example of each and what the receiver does next.
What you are expected to know Predict what a receiver does after a signal: name the job the signal does for its sender, then state the change in the receiver’s behavior.
Suppose a blackbird feeding on a lawn gives a sharp call when a cat walks onto the grass. The other blackbirds on the lawn detect the call.
Which change is the call most likely to make in the other blackbirds’ behavior?
- A. ✓ The other blackbirds fly up into the hedge
- B. The other blackbirds fly toward the calling blackbirdThe blackbird called when the cat appeared, so the call warns of a predator.
Receivers of a warning move to cover, not toward the sender. - C. The other blackbirds keep feeding on the lawnThe call is a signal: one blackbird produced it and the others detected it.
A signal changes what the receivers do.
Why: The blackbird called when the cat appeared, so the call warns of a predator.
A signal changes what the receivers do.
Receivers of a predator warning move to cover.
So the other blackbirds fly up into the hedge.
Suppose two male sea lions come face to face on a beach. The first male barks loudly. The second male, which had been moving toward the first, turns and moves away.
(a) Explain how the second male’s behavior demonstrates what a signal does to its receiver. (1 pt)
Frame The second male’s behavior demonstrates this because …
The first male produced the bark, and the second male detected it.
Before the bark, the second male was moving toward the first.
After the bark, the second male turned and moved away.
So the bark changed what the receiver did.
A signal changes the receiver’s behavior.
- Award 1 point for: the bark is a signal (one male produced it, the other detected it) AND the second male’s behavior changed after it (moving toward, then moving away), so a signal changes what the receiver does.
A student says: “A signal forces the receiver to act the same way every time.”
Is the student correct?
- A. YesSome receivers of one call move toward the caller, and some stay where they are.
The call changed what most receivers did; it forced none of them. - B. ✓ No
Why: A signal changes what the receiver does.
Some receivers of one call move toward the caller, and some stay where they are.
So a signal makes one action more likely.
A signal does not force the receiver to act the same way every time.
Suppose a hen finds spilled grain and gives a soft, repeated call. Her chicks, scattered across the yard, detect the call.
Which of the following is the change the call makes in the chicks’ behavior?
- A. The chicks move away from the henThe hen called when she found grain, so her call is a food signal.
Receivers of a food signal move to the food, beside the sender. - B. The chicks stop moving and freeze in the grassFreezing is what receivers of a predator warning do.
The hen called when she found grain, not when a predator appeared. - C. ✓ The chicks move toward the hen
Why: The hen called when she found grain.
So the call’s job is to lead the chicks to food.
Receivers of a food signal move to the food.
The grain is beside the hen, so the chicks move toward the hen.
28Who leaves offspring
Here is the pond edge again in spring. Twenty male frogs call from the edge every night.
Some of the males call louder and longer than others.
Now consider two of the twenty males. The first male calls loud and long every night.
The second male calls quietly and stops early each night.
Female frogs detect both calls. More females move toward the loud caller than toward the quiet caller.
Suppose 8 females reach the loud caller over the spring, and 2 females reach the quiet caller.
Each female that reaches a male mates with him. She then lays a clutch of eggs, and the eggs grow into tadpoles.
So by June the loud caller has fathered 8 clutches of tadpoles. The quiet caller has fathered 2 clutches.
The loud caller left more offspring than the quiet caller. His loud, long call drew more females, and each female laid a clutch.
So a signaling behavior can give some individuals more offspring than others.
No male frog chose to call louder so as to leave more offspring. The loud callers simply drew more females, and more tadpoles followed.
In Unit 7, a small male guppy fathered 30 young that survived to reproduce, and a large male guppy fathered 12.
Which male has the higher evolutionary fitness?
- A. ✓ The small male guppy
- B. The large male guppyEvolutionary fitness is counted in offspring that survive to reproduce, not in size.
The small male left 30 such offspring; the large male left 12.
Why: Evolutionary fitness is counted in offspring that survive to reproduce.
The small male left 30 and the large male left 12.
So the small male has the higher evolutionary fitness.
Video: Watch: Who leaves offspring
Two males call at the pond edge, one loud and long, one quiet and brief; females move toward the loud caller; the clutch counts appear under each male as the spring passes.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L06Bb.mp4
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Unit 7 counted the same thing. The number of offspring an individual leaves that themselves survive to reproduce is its evolutionary fitness.
Suppose the same share of each male’s tadpoles survives to reproduce. Then the loud caller leaves more offspring that survive to reproduce.
So the loud caller has the higher evolutionary fitness. “More offspring” in this topic and “higher evolutionary fitness” in Unit 7 are the same measure.
What you are expected to know Explain how a signaling behavior gives some individuals more offspring than others: the signal draws more mates to some senders, so those senders leave more offspring.
Suppose in a field of crickets, some males chirp for most of the night and others chirp for an hour. The table below shows how many females reach each kind of male over the summer.
Which crickets leave more offspring by the end of the summer?
- A. The males that chirp for an hour5 females reach each male that chirps for most of the night; 1 reaches each male that chirps for an hour.
More matings give more offspring. - B. ✓ The males that chirp for most of the night
- C. Both kinds of male leave the same numberEach female that reaches a male lays a clutch of eggs.
5 clutches against 1 clutch is not the same number.
Why: 5 females reach each male that chirps for most of the night.
1 female reaches each male that chirps for an hour.
Each female that reaches a male mates with him and lays eggs.
So the males that chirp for most of the night leave more offspring.
Suppose male songbirds of one kind sing at dawn in spring. Females move toward the males that sing the longest songs. Over one spring, each long-singing male fathers 3 broods of chicks, and each short-singing male fathers 1 brood or none.
(a) Explain how this case demonstrates that a signaling behavior can give some individuals more offspring than others. (1 pt)
Frame This case demonstrates this because …
Each male produced his song, and the females detected it.
The females moved toward the long-singing males, so the long-singing males won more mates.
Each mate raised a brood of chicks with him.
So the long-singing males left more offspring than the short-singing males.
The singing behavior gave some individuals more offspring than others.
- Award 1 point for: the song is a signal that drew more females (mates) to the long-singing males, so those males fathered more broods, so the signaling behavior gave some individuals more offspring than others.
In a pond of frogs, the loud callers father more tadpoles than the quiet callers. Every female at the pond mates with one male and lays one clutch. A student says: “The loud calls help the whole pond of frogs leave more offspring.”
Is the student correct?
- A. YesThe females that reached the loud callers would otherwise have reached other males.
The loud calls changed which males fathered the tadpoles, not how many tadpoles the pond produced. - B. ✓ No
Why: Each female lays one clutch whichever male she reaches.
The loud calls drew females to the loud callers and away from the quiet callers.
So the loud callers fathered more clutches and the quiet callers fewer.
The loud calls helped the loud callers, not the whole pond.
Suppose in one kind of bird, the females move toward the males with the brightest feathers.
Which males leave the most offspring?
- A. ✓ The males with the brightest feathers
- B. The males with the dullest feathersThe females move toward the brightest males, not the dullest.
A male that no female reaches leaves no offspring. - C. Every male leaves the same numberMore females reach the brightest males.
Each female that reaches a male mates with him, so the brightest males father more offspring.
Why: The bright feathers are a signal the females detect.
More females move toward the brightest males.
Each female that reaches a male mates with him and lays eggs.
So the males with the brightest feathers leave the most offspring.
Here is the pond edge again on a spring night: two males calling, one loud and long, one quiet and brief.
The females move toward the loud caller. By June the loud caller has fathered 8 clutches of tadpoles and the quiet caller 2.
So the loud caller’s call changed who left offspring.
53Mixed practice mixed practice
Suppose a raven finds a dead deer in the snow and gives a loud, repeated call. Other ravens far off detect the call.
Which change is the call most likely to make in the other ravens’ behavior?
- A. The other ravens fly away from the callerThe raven called when it found food, so the call is a food signal.
Receivers of a food signal move to the food. - B. The other ravens stay where they are and call backA signal changes what the receivers do.
The call was given at food, so the receivers move to the food. - C. ✓ The other ravens fly to the dead deer
Why: The raven called when it found the dead deer.
So the call’s job is to lead others to food.
Receivers of a food signal move to the food.
So the other ravens fly to the dead deer.
Suppose peahens choose among peacocks by the size of the tail fan each male displays. The table below shows how many peahens reach each kind of peacock over a season.
Which peacocks leave more offspring?
- A. The peacocks with small tail fans1 peahen reaches each peacock with a small tail fan; 4 reach each peacock with a large tail fan.
Each peahen that reaches a peacock lays a clutch of eggs. - B. ✓ The peacocks with large tail fans
- C. Both kinds of peacock leave the same number4 peahens reach each peacock with a large tail fan and 1 reaches each peacock with a small tail fan.
More matings give more offspring.
Why: 4 peahens reach each peacock with a large tail fan.
1 peahen reaches each peacock with a small tail fan.
Each peahen that reaches a peacock mates with him and lays eggs.
So the peacocks with large tail fans leave more offspring.
A student says: “A female frog that detects a male’s call is more likely to move toward him than a female that detects no call.”
Is the student correct?
- A. ✓ Yes
- B. NoThe call makes the move toward the caller more likely.
A female that detects no call has no caller to move toward.
Why: A signal changes what the receiver does.
The call makes a move toward the caller more likely.
A female that detects no call has no caller to move toward.
So the female that detects the call is more likely to move toward him.
Suppose a large male toad fathers 3 clutches of eggs in a spring and a small male toad fathers 7 clutches. The same share of each clutch survives to reproduce.
Which toad has the higher evolutionary fitness?
- A. The large male toadEvolutionary fitness is counted in offspring that survive to reproduce, not in body size.
The small toad fathered 7 clutches; the large toad fathered 3. - B. ✓ The small male toad
Why: Evolutionary fitness is counted in offspring that survive to reproduce.
The small male toad fathered 7 clutches and the large male 3.
The same share of each clutch survives, so the small male leaves more offspring that survive to reproduce.
So the small male toad has the higher evolutionary fitness.
Suppose, in the mating season, a male grasshopper rubs its legs against its wings and makes a buzzing song. A female grasshopper nearby detects the song.
Which of the following is the change the song makes in the female’s behavior?
- A. ✓ The female grasshopper moves toward the buzzing male
- B. The female grasshopper moves away from the buzzing maleA male’s song in the mating season is a mating signal.
Receivers of a mating signal move toward the sender. - C. The female grasshopper freezes in the grassFreezing is what receivers of a predator warning do.
The male sang in the mating season; no predator appeared.
Why: The male sang in the mating season, so the song’s job is to win a mate.
A signal changes what the receiver does.
Receivers of a mating signal move toward the sender.
So the female grasshopper moves toward the buzzing male.
In a pond, the quiet callers father few tadpoles. A student says: “A quiet male frog will call louder next spring because he needs more offspring.”
Is the student correct?
- A. YesA male frog does not change his call because he needs offspring.
The loud callers drew more females this spring, so more tadpoles followed. - B. ✓ No
Why: A male frog does not change his call because he needs offspring.
The loud callers drew more females this spring, so more tadpoles followed.
The behavior came first, and the count of offspring followed from it.
Suppose 30 male frogs call at a pond in spring. The 10 loudest callers each mate with 4 females. The other 20 callers each mate with 1 female or none. Each female lays one clutch of eggs, and the same share of every clutch survives to reproduce.
(a) Explain how the frogs’ case demonstrates that a signaling behavior can give some individuals more offspring than others. (1 pt)
Frame The frogs’ case demonstrates this because …
More females moved toward the loudest callers than toward the other callers.
Each female that reached a male mated with him and laid a clutch.
So each of the 10 loudest callers fathered 4 clutches, and the other callers fathered 1 or none.
The calling behavior gave the loudest callers more offspring than the others.
- Award 1 point for: the call is a signal that drew more females to the loudest callers, so those callers fathered more clutches (4 against 1 or none), so the signaling behavior gave some individuals more offspring than others.
(b) Justify the claim that the 10 loudest callers have the higher evolutionary fitness. (1 pt)
Frame The 10 loudest callers have the higher evolutionary fitness because …
Each of the loudest callers fathered 4 clutches, and each other caller fathered 1 or none.
The same share of every clutch survives to reproduce.
So the loudest callers leave more offspring that survive to reproduce.
- Award 1 point for: evolutionary fitness is counted in offspring that survive to reproduce, and the loudest callers fathered more clutches (4 against 1 or none) with the same share surviving, so they leave more such offspring.
APBIO-U08-L07 Born knowing it, or learned it
Photo: J.M.Garg, Wikimedia Commons, CC BY 3.0.
Here are two painted stork chicks in their nest. On the day they hatched, each chick gaped and called for food, with no chick to copy. Beside them is a rat in a maze with food at the far end. Given food at the end each day, the rat took seven days to learn the way.
Both animals end up fed. What is different about how each behavior got there?
Unit 8 · Ecology
1Born knowing it, or learned it
Where does a behavior come from, and why does it spread?
Some behaviors are present with no experience at all, not shaped by the animal’s own experience. The chicks’ begging is one of these.
Other behaviors are changed by the animal’s own experience. The rat’s route through the maze is one of these.
A behavior present without experience can still be set off by something outside the animal. Present without experience means the pattern was not learned, not that nothing triggers it.
Either kind of behavior spreads through a population for one reason. The animals that have it survive more or leave more offspring, so more of the next generation have it too.
That is natural selection, the process Unit 7 taught, acting on a behavior.
A hawk’s shadow passes over a mouse. The mouse darts into a hole.
Which is the stimulus?
- A. The mouse darting into the holeDarting into the hole is what the mouse does: the response.
The stimulus is the change the mouse detected. - B. ✓ The hawk’s shadow passing over the mouse
Why: The stimulus is the change the organism detects.
The mouse detected the shadow passing over it.
So the shadow is the stimulus, and the dart into the hole is the response.
Video: Watch: Born knowing it, or learned it
The stork chicks begging for food on their first day beside the rat in the maze on day 1 and on day 7; the two verdicts written beneath them.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L07a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L07a.mp4
Consider the chicks first. Each chick gapes and calls for food on the day it hatches, with no chick to copy.
The chick had no experience of begging before its first day. So experience did not build the behavior.
A behavior that is present without any experience, not shaped by the individual’s own experience, is called an . Innate means inborn: present from birth.
Now consider the rat. Here is the rat’s route through the maze on day 1 and its route on day 7.
On day 1 the rat turns into dead ends. On day 7 the rat walks straight to the food.
The rat’s own experience changed its behavior. Each day the rat found the food, and each day it took fewer wrong turns.
A behavior that is changed by the individual’s own experience is called a .
Here are the two cases side by side, each with its verdict.
For example, a duckling follows the first adult it sees after hatching. A duckling that sees a human first follows the human instead.
So following is a learned behavior, because the duckling’s own experience decided what it follows. Biologists call this fast, early learning imprinting.
But a woodlouse speeds up the first time it walks onto a warm surface, with no experience of warmth.
So speeding up is an innate behavior, because it is present without experience: nothing the woodlouse went through shaped it.
The warm surface set the speeding up off. An innate behavior can still be set off by something outside the animal: innate means the pattern was not learned, not that nothing triggers it.
The table below compares an innate behavior with a learned behavior: present at birth, shaped by the individual’s experience, whether it can vary between animals, and one example of each.
What you are expected to know Classify a behavior as an innate behavior (present without experience, not shaped by the individual’s own experience) or a learned behavior (changed by the individual’s own experience).
Three animals each do something.
Which of the following is an innate behavior?
- A. A horse that trots to the fence at the sound of a feed bucket, after being fed there many timesBeing fed there many times changed what the horse does at the sound.
A behavior changed by the individual’s experience is a learned behavior. - B. A parrot that says a word after hearing its owner say the word many timesHearing the word many times changed what the parrot says.
A behavior changed by the individual’s experience is a learned behavior. - C. ✓ A newborn kangaroo that climbs into its mother’s pouch minutes after birth, with no joey to copy
Why: The newborn kangaroo climbs into the pouch minutes after birth, with no joey to copy.
So experience did not build the behavior.
A behavior present without experience is an innate behavior.
A newborn baby grips a finger placed in its palm on its first day. Every newborn grips the same way.
Which kind of behavior is the grip?
- A. ✓ An innate behavior
- B. A learned behaviorThe baby grips on its first day, with no experience of gripping.
A behavior present without experience is an innate behavior.
Why: The baby grips on its first day, with no experience.
Nothing the baby went through shaped the grip.
So the grip is an innate behavior.
A dog sits when it hears the word sit, after weeks of being given food each time it sat.
Which kind of behavior is the sitting?
- A. An innate behaviorWeeks of food each time it sat changed what the dog does at the word.
A behavior changed by the individual’s own experience is a learned behavior. - B. ✓ A learned behavior
Why: The food each time it sat is the dog’s own experience.
That experience changed what the dog does at the word.
So the sitting is a learned behavior.
Suppose a bird pecks open the foil top of a milk bottle after watching another bird do it.
Which kind of behavior is the pecking?
- A. An innate behaviorWatching the other bird is the bird’s own experience, and it changed what the bird does.
A behavior changed by experience is a learned behavior. - B. ✓ A learned behavior
Why: The bird watched another bird open a bottle.
That experience changed what the bird does.
So the pecking is a learned behavior.
Suppose a spider raised alone spins a web of the same pattern as every spider of its kind.
Which kind of behavior is the web-spinning?
- A. ✓ An innate behavior
- B. A learned behaviorThe spider was raised alone, so no experience of other webs built the pattern.
A behavior present without experience is an innate behavior.
Why: The spider spins the pattern with no web to copy.
Nothing the spider went through shaped the pattern.
So the web-spinning is an innate behavior.
Suppose a moth flies toward a lamp on its first night out of the cocoon, like every moth of its kind.
Which kind of behavior is the flight toward the lamp?
- A. ✓ An innate behavior
- B. A learned behaviorThe moth flies to the lamp on its first night, with no experience of lamps.
A behavior present without experience is an innate behavior.
Why: The moth flies to the lamp on its first night, with no experience.
Nothing the moth went through shaped the flight.
So the flight toward the lamp is an innate behavior.
A prairie dog sounds an alarm call at human footsteps. After hearing footsteps many times with no harm following, it stops calling at them.
Which kind of behavior is the prairie dog’s silence at footsteps?
- A. An innate behaviorHearing footsteps many times with no harm is the prairie dog’s own experience, and it changed the calling.
A behavior changed by experience is a learned behavior. - B. ✓ A learned behavior
Why: The prairie dog heard footsteps many times with no harm.
That experience changed its calling.
So the silence is a learned behavior.
Suppose a gull chick pecks at a red spot on its parent’s bill on the day it hatches, and every chick of its kind pecks the same way. The chick pecks when the bill appears and stops when the bill is gone.
Which kind of behavior is the pecking?
- A. ✓ An innate behavior
- B. A learned behaviorThe bill sets the pecking off, but the chick’s experience did not build it.
A behavior present without experience is an innate behavior.
Why: The chick pecks on its first day, with no experience.
Nothing the chick went through shaped the pecking.
So the pecking is an innate behavior.
The parent’s bill sets the pecking off; a trigger from outside does not make a behavior learned.
33Quick quiz: innate behavior, learned behavior mixed practice
A sea turtle hatchling crawls straight to the sea on its first night, like every hatchling of its kind.
Which kind of behavior is the crawl to the sea?
- A. ✓ An innate behavior
- B. A learned behaviorThe hatchling crawls to the sea on its first night, with no experience of the sea.
A behavior present without experience is an innate behavior.
Why: The hatchling crawls to the sea on its first night, with no experience.
Nothing the hatchling went through shaped the crawl.
So the crawl is an innate behavior.
A pigeon pecks a button more and more often after being given a seed each time it pecked.
Which kind of behavior is the pecking at the button?
- A. An innate behaviorA seed each time it pecked is the pigeon’s own experience, and it changed how often the pigeon pecks.
A behavior changed by experience is a learned behavior. - B. ✓ A learned behavior
Why: The seed each time it pecked is the pigeon’s own experience.
That experience changed how often the pigeon pecks.
So the pecking at the button is a learned behavior.
A newborn calf stands and walks on its first day, with no experience of walking, like every calf.
Which kind of behavior is the walking?
- A. ✓ An innate behavior
- B. A learned behaviorThe calf walks on its first day, with no experience of walking.
A behavior present without experience is an innate behavior.
Why: The calf walks on its first day, with no experience.
Nothing the calf went through shaped the walking.
So the walking is an innate behavior.
An animal does something.
What is a learned behavior?
- A. ✓ A behavior changed by the individual’s own experience
- B. A behavior present without experience, so not learnedA behavior present without experience, not shaped by the individual’s own experience, is an innate behavior.
Why: A learned behavior is a behavior that the individual’s own experience has changed.
An animal does something.
What is an innate behavior?
- A. A behavior changed by the individual’s own experienceA behavior changed by the individual’s own experience is a learned behavior.
- B. ✓ A behavior present without experience, so not learned
- C. A behavior that nothing outside the animal can set offAn innate behavior can be set off by something outside the animal.
Innate means the pattern was not learned.
Why: An innate behavior is present without any experience: nothing the individual went through shaped it.
Animals do some things from their first day and other things only after experience.
(a) State what an innate behavior and a learned behavior are. (1 pt)
A learned behavior is changed by the individual’s own experience.
- Award 1 point for: innate = present without experience (not learned; performed with nothing to copy); learned = changed by the individual’s own experience. Accept ‘present without experience’ alone for innate.
40Why a behavior spreads
Unit 7 taught natural selection.
In natural selection, which phenotype becomes more common in later generations?
- A. The phenotype an individual picks up during its own lifeA phenotype picked up during life is not heritable.
Natural selection makes a phenotype more common only when it is inherited. - B. The phenotype of the largest and strongest individualsSize and strength are not counted.
The phenotype whose owners survive or reproduce more becomes more common. - C. ✓ The heritable phenotype whose owners survive or reproduce more
Why: In natural selection, individuals with one heritable phenotype survive or reproduce more.
So that phenotype becomes more common in later generations.
Unit 7 taught evolutionary fitness.
What does an animal’s evolutionary fitness count?
- A. ✓ The offspring it leaves that survive to reproduce
- B. The number of years it lives before it diesLifespan is not counted.
Evolutionary fitness counts the offspring an animal leaves that survive to reproduce. - C. Its size and strength compared with the othersSize and strength are not counted.
Evolutionary fitness counts the offspring an animal leaves that survive to reproduce.
Why: Evolutionary fitness is the number of offspring an animal leaves that themselves survive to reproduce.
Video: Watch: Why a behavior spreads
A field of ten fawns as a predator passes; the fawns that lie flat survive and the predator catches three that bolt; the next generation appears with six of ten lying flat.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L07b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L07b.mp4
Why does a behavior spread through a population? Suppose a field holds 10 fawns.
When a predator passes, 4 of the 10 fawns lie flat and still. The other 6 fawns bolt.
The predator sees the bolting fawns and catches 3 of the 6. The predator passes the 4 lying fawns without seeing them.
So 4 lying fawns and 3 bolting fawns survive. The fawns that lie flat survive more often.
Each survivor grows up and has fawns of its own. Each young fawn behaves as its parent did.
4 of the 7 survivors lay flat, so more than half of the next generation lie flat. Of the next 10 fawns, 6 lie flat and 4 bolt.
Lying flat became more common because the fawns that lay flat survived more and so left more offspring. That is the one reason a behavior spreads.
Unit 7 counted the offspring an animal leaves that survive to reproduce and called that number its evolutionary fitness. The fawns that lie flat have the higher evolutionary fitness.
This is natural selection, the process Unit 7 taught, acting on a behavior.
The four steps in the field of fawns:
- the fawns vary: some lie flat, some bolt
- each fawn behaves as its parent did
- the fawns that lie flat survive more
- so lying flat becomes more common
In this field, lying flat is an innate behavior: a fawn lies flat with no experience of predators.
A learned behavior spreads the same way. Now consider a troop of monkeys in which a few monkeys have learned to crack hard nuts with a stone.
The young of these monkeys grow up beside them and learn the trick from them. In a dry season the nut-crackers find food while the other monkeys go hungry.
The nut-crackers survive the dry season more often and raise more young, and their young crack nuts too. So nut-cracking becomes more common in the troop.
What matters is the behavior’s effect on survival and offspring. An innate behavior and a learned behavior spread for the same reason: the animals that have the behavior leave more offspring that have it too.
A behavior that lowers survival or the number of offspring spreads the other way. Fewer of the next generation have it.
What you are expected to know Explain why a behavior, innate or learned, that raises survival or the number of offspring becomes more common in a population.
Suppose that on an open hillside some young hares lie still when a fox comes near, and other young hares bolt across the open ground. The fox chases what it sees.
Which young hares survive the fox more often?
- A. The hares that boltThe fox chases what it sees, and a bolting hare is seen.
A hare lying still is passed by. - B. ✓ The hares that lie still
- C. Both kinds equallyThe fox sees the bolting hares and passes the still ones.
So the two kinds do not survive equally.
Why: The fox chases what it sees.
A bolting hare is seen; a hare lying still is passed by.
So the hares that lie still survive the fox more often.
Suppose that in a desert some beetles burrow into the sand in the midday heat and others stay on the surface. A beetle on the surface at midday overheats, and some of those beetles die. Each beetle’s young behave as their parent did. After several generations, most beetles in that desert burrow into the sand at midday.
(a) Explain how the beetles demonstrate that a behavior that raises survival becomes more common in a population. (1 pt)
Frame Burrowing became more common because …
A beetle that burrows into the sand stays cooler and survives.
So the beetles that burrowed survived more often.
The survivors grew up and had young.
Their young burrowed, as their parents did.
So each generation had a larger share of beetles that burrow.
- Award 1 point for: the beetles that burrow survive more and so leave more young that also burrow, so the share of the population that burrows rises.
A student looks at the hillside hares and says: “Lying still spread through the hares because it is good for the hare species as a whole.”
Is the student correct?
- A. YesA behavior spreads because the individual hares that have it survive more and leave more offspring.
Being good for the species is not what spreads it. - B. ✓ No
Why: The hares that lie still survive the fox more often.
So they leave more young, and their young lie still too.
Lying still spread because the hares that had it left more offspring, not because it helped the species as a whole.
Suppose a few crows in a city learn to drop nuts onto a road so that cars crack them. Young crows learn the trick from their parents. The crows with the trick find more food in winter and raise more young.
Over several generations, what happens to the share of the city’s crows that use the trick?
- A. ✓ The share rises
- B. The share fallsThe crows with the trick raise more young, and the young learn it.
So the share rises. - C. The share stays the sameThe trick-users raise more young than the other crows, and their young learn the trick.
So the share rises.
Why: The crows with the trick find more food and raise more young.
Their young learn the trick.
So each generation has a larger share of crows with the trick.
Here are the painted stork chicks and the rat again: the chicks begging for food on their first day, and the rat walking straight to the food on its seventh day.
The chicks’ begging is an innate behavior. The rat’s route is a learned behavior.
Both became common for one reason: the animals that had them survived more or left more offspring.
68Mixed practice mixed practice
Suppose that on an island some lizards dart into rock cracks when a hawk’s shadow passes, and other lizards stay in the open. The hawk takes the lizards it can see.
Which lizards leave more offspring over their lives?
- A. ✓ The lizards that dart into cracks
- B. The lizards that stay in the openThe hawk takes the lizards it can see, and a lizard in the open is seen.
A lizard in a crack survives to breed.
Why: The hawk takes the lizards it can see.
A lizard in a crack is hidden; a lizard in the open is taken.
So the lizards that dart into cracks survive longer and leave more offspring.
A squirrel opens the lid of a bird feeder after many failed tries over several days.
Which kind of behavior is opening the lid?
- A. An innate behaviorMany failed tries are the squirrel’s own experience, and they changed what the squirrel does at the feeder.
A behavior changed by experience is a learned behavior. - B. ✓ A learned behavior
Why: The squirrel’s many tries are its own experience.
That experience changed what the squirrel does at the feeder.
So opening the lid is a learned behavior.
Suppose darting into cracks has become common among the island’s lizards. A student says: “Darting became common because each lizard chose to protect its species.”
Is the student correct?
- A. YesDarting became common because the lizards that darted survived the hawk and left more offspring.
A lizard’s choice for its species plays no part. - B. ✓ No
Why: The lizards that dart into cracks survive the hawk more often.
So they leave more offspring, and their offspring dart too.
Darting became common because the darting lizards left more offspring, not because any lizard chose to protect the species.
Suppose a kitten arches its back and hisses the first time a dog comes near, with no experience of dogs, and every kitten does the same. A student says: “The dog set the hissing off, but the hissing is still an innate behavior, because the kitten needed no experience to do it.”
Is the student correct?
- A. ✓ Yes
- B. NoThe kitten hisses with no experience of dogs.
A trigger from outside does not make a behavior learned.
Why: The kitten hisses with no experience of dogs.
Nothing the kitten went through shaped the hissing.
So the hissing is an innate behavior.
The dog sets the hissing off; a trigger from outside does not make a behavior learned.
Suppose a cuckoo chick raised by birds of another species, with no cuckoo to copy, sings the cuckoo’s own song when it is grown.
Which kind of behavior is the song?
- A. ✓ An innate behavior
- B. A learned behaviorThe cuckoo had no cuckoo to copy, so experience did not build the song.
A behavior present without experience is an innate behavior.
Why: The cuckoo sings the song with no cuckoo to copy.
Nothing the cuckoo went through shaped the song.
So the song is an innate behavior.
Suppose some gulls at a harbor learn to feed on scraps thrown from a road with fast traffic, and young gulls learn it from their parents. Cars kill the gulls that feed at the road more often than the other gulls.
Over several generations, what happens to the share of the harbor’s gulls that feed at the road?
- A. The share risesCars kill the gulls that feed at the road more often, so those gulls leave fewer young.
The share falls. - B. ✓ The share falls
- C. The share stays the sameThe gulls that feed at the road leave fewer young than the other gulls.
So the share falls.
Why: Cars kill the gulls that feed at the road more often.
So those gulls leave fewer young, and fewer young learn the behavior.
So each generation has a smaller share of gulls that feed at the road.
Suppose that in a marsh, some newly hatched chicks of a ground-nesting bird crouch still when their parent gives an alarm call, and other chicks keep moving. The chicks that crouch do so on their first day, with no experience of alarm calls. A hawk hunting over the marsh catches the moving chicks more often. The chicks that survive grow up and have chicks of their own, and those chicks behave as their parents did.
(a) Identify the crouching as an innate behavior or a learned behavior. (1 pt)
- Award 1 point for: innate behavior (the chicks crouch on their first day, with no experience).
(b) Explain how the chicks demonstrate that a behavior that raises survival becomes more common in a population. (1 pt)
Frame Crouching becomes more common because …
The survivors grow up and have chicks of their own.
Their chicks crouch as their parents did.
So each generation has a larger share of chicks that crouch.
- Award 1 point for: the chicks that crouch survive more and so leave more offspring that also crouch, so the share of the population that crouches rises.
Glossary
- innate behavior
- A behavior that is present without any experience, not shaped by the individual’s own experience, like a woodlouse speeding up on a warm surface. Innate means inborn. Something outside the animal can still set it off: innate means the pattern was not learned.
- learned behavior
- A behavior that the individual's own experience has changed, like a rat that learns the way through a maze or a duckling that follows the first adult it sees.
APBIO-U08-L08 Safety in numbers
Photo: Bernard Dupont, Wikimedia Commons, CC BY-SA 4.0 (resized).
A meerkat stands upright on a termite mound. Below it, the rest of its group digs for beetles. The meerkat on the mound is not eating. When a hawk appears, it calls, and every meerkat dives for cover.
The meerkat on the mound gave up a meal to stand there. What did it get back?
Unit 8 · Ecology
1Standing guard while the others eat
Why would an animal spend its effort on others?
Standing guard, sharing a kill, hunting as a pack: each of these costs the animal something now and helps the others in its group.
The animal gets something back. The sentinel eats less this hour.
When its own turn to dig comes, another meerkat stands guard.
The group gains too. More eyes spot a hawk sooner. So hawks take fewer of the group. So the population survives.
Helping a close relative pays in one more way. Relatives share many alleles.
So a worker bee that raises the queen’s offspring spreads copies of its own alleles without breeding itself.
A hawk appears over a meerkat colony. One meerkat on a mound gives a sharp call. The other meerkats hear the call and dive for cover.
Which of the following is the signal?
- A. The hawkThe hawk is the change the sentinel detected.
The signal is the thing one organism produced and the others detected: the call. - B. ✓ The call
- C. The dive for coverThe dive is what the other meerkats did after they detected the call.
The signal is the call they detected.
Why: A signal is something one organism produces that another organism detects.
The sentinel produced the call.
The other meerkats detected the call.
So the call is the signal.
Video: Watch: Standing guard while the others eat
One meerkat upright on the mound, the others digging below; a hawk appears; the sentinel calls and every meerkat dives for cover.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L08a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L08a.mp4
Here is the mound again: one meerkat upright on top, the rest of the group digging for beetles below.
The meerkat on the mound is not digging. So for as long as it stands guard, it is not eating.
Standing upright in the open, the meerkat is easy for a hawk to see. So it puts itself at risk.
When the hawk appears, the meerkat on the mound calls. The others hear the call and reach cover before the hawk reaches them.
So standing guard costs this meerkat something now: its digging time, and its safety. And standing guard helps the others in its group.
Behavior that costs the individual now and helps the others in its group is called .
A meerkat that stands guard is called a sentinel. Sentinel is an old word for a guard.
Here are six cases, each judged by the same two questions: does it cost the individual now, and does it help the others in its group?
For example, a meerkat stands upright on a mound and watches for hawks while the others dig. This is cooperative behavior, because it costs the meerkat its digging time now and helps the others in its group.
But a meerkat digs up a beetle and eats it. This is not cooperative behavior, because it helps no other meerkat.
And a wild dog carries meat from a kill back to the den for pack members that missed the hunt. This is cooperative behavior, because it costs the wild dog meat now and helps the others in its pack.
But a wild dog eats the meat at the kill. This is not cooperative behavior, because it helps no other wild dog.
And a worker bee feeds the queen’s young in the hive. This is cooperative behavior, because it costs the worker its time and food now and helps the others in its hive.
And a lemur carries and grooms an infant that is not its own. This is cooperative behavior, because it costs the lemur its time and effort now and helps another lemur in its group.
Here is a table of the six cases with their verdicts: whether each one costs the individual now, whether it helps the others in its group, and whether it is cooperative behavior.
What you are expected to know Identify cooperative behavior in a described case: behavior that costs the individual now and helps the others in its group.
A meerkat stands upright on a mound and watches for hawks while the others of its group dig for beetles.
Which of the following makes the standing guard cooperative behavior?
- A. The whole group stands guard on the mound at the same momentOnly one meerkat stands guard at a time.
The test is a cost to the individual now and a help to the others. - B. ✓ It misses its digging while it stands, and its call sends the others to cover
- C. Another meerkat in the group detects the guard’s call and respondsSomething another animal detects is a signal.
Cooperative behavior is judged by its cost to the individual and its help to the group.
Why: The meerkat on the mound is not digging, so standing guard costs it its digging time now.
Its call lets the others reach cover, so it helps the others in its group.
So standing guard is cooperative behavior.
A ground squirrel stands up in the grass and watches for hawks while the others of its colony feed. When a hawk comes it whistles, and the others dive into their burrows.
Is the watching cooperative behavior?
- A. ✓ Yes
- B. NoWatching costs the ground squirrel its feeding time now.
The whistle it gives sends the others to their burrows in time.
Why: Watching costs the ground squirrel its feeding time now.
Watching helps the others in its colony: they hear the whistle and reach their burrows.
So the watching is cooperative behavior.
A lizard lies on a warm rock in the sun until its body warms.
Is the basking cooperative behavior?
- A. YesBasking warms the lizard itself and helps no other lizard.
Cooperative behavior helps the others in the group. - B. ✓ No
Why: Basking warms the lizard’s own body.
Basking helps no other lizard.
So the basking is not cooperative behavior.
Wolves that hunted carry meat back to pack members that missed the hunt.
Is the carrying of meat cooperative behavior?
- A. ✓ Yes
- B. NoCarrying the meat back costs the hunters meat they could have eaten now.
The meat feeds the others in the pack.
Why: Carrying the meat back costs the hunters meat they could have eaten now.
The meat feeds the others in the pack.
So the carrying of meat is cooperative behavior.
Ten ants carry one leaf back to their nest together. The leaf feeds the whole colony.
Is the carrying of the leaf cooperative behavior?
- A. ✓ Yes
- B. NoCarrying the leaf costs each ant its effort now.
The leaf feeds the others in the colony.
Why: Carrying the leaf costs each ant its effort now.
The leaf feeds the others in the colony.
So the carrying of the leaf is cooperative behavior.
A robin sings from a branch to keep other robins out of its patch of garden.
Is the singing cooperative behavior?
- A. YesThe song keeps the patch for the singer alone and helps no other robin.
Cooperative behavior helps the others in the group. - B. ✓ No
Why: The song keeps the patch of garden for the singer.
The song helps no other robin.
So the singing is not cooperative behavior.
A school of small fish swims together. When a predator comes near, each fish turns with the school instead of darting off along its own escape route. A predator catches fewer fish from a school that turns together, so the fish around it are less likely to be caught.
Is turning with the school cooperative behavior?
- A. ✓ Yes
- B. NoEach fish gives up its own escape route now, a cost to it.
Fewer of the fish around it are caught, so the turning helps the others in its group.
Why: Cooperative behavior costs the individual now and helps the others in its group.
Each fish gives up its own escape route now: a cost.
Fewer of the fish around it are caught: a help to the others.
So the turning is cooperative behavior.
34Quick quiz: cooperative behavior mixed practice
A bird in a flock gives a loud call when a fox appears. The call marks the caller out to the fox, and every other bird in the flock flies up out of reach.
Is the call cooperative behavior?
- A. ✓ Yes
- B. NoThe call marks the caller out to the fox, so it costs the caller its safety now.
The call helps the others in the flock fly up in time.
Why: The call marks the caller out to the fox, so it costs the caller its safety now.
The call helps the others in the flock fly up in time.
So the call is cooperative behavior.
A cat washes its own fur with its tongue.
Is the washing cooperative behavior?
- A. YesWashing its own fur helps no other cat.
Cooperative behavior helps the others in the group. - B. ✓ No
Why: Washing its own fur cleans the cat itself.
Washing helps no other cat.
So the washing is not cooperative behavior.
Animals in a group behave in many ways toward one another.
What is cooperative behavior?
- A. Behavior that one animal produces and another animal detectsSomething one animal produces and another detects is a signal.
Cooperative behavior is judged by its cost to the individual and its help to the group. - B. Behavior that warms or cools the individual’s own bodyA behavior that changes the individual’s own body temperature helps only the individual.
Cooperative behavior helps the others in its group. - C. ✓ Behavior that costs the individual now and helps the others in its group
Why: Cooperative behavior is behavior that costs the individual now and helps the others in its group.
Animals in a group behave in many ways toward one another.
(a) State what cooperative behavior is. (1 pt)
- Award 1 point for: behavior that costs the individual (now) and helps (benefits) others in its group.
39What the sentinel gets back
Video: Watch: What the sentinel gets back
The sentinel climbs down and digs for beetles; another meerkat climbs the mound and stands guard over it; the two scenes side by side, the same meerkat watching and then eating under guard.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L08b.mp4
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Now consider the sentinel an hour later. Its turn on the mound is over, so it climbs down and digs for beetles with the others.
Another meerkat climbs the mound and stands guard in its place.
So the meerkat that stood guard now eats, while another meerkat watches for hawks over it.
The sentinel gave up an hour of eating. It gets back an hour of eating under guard.
That is the individual’s benefit: the helper is helped in its turn. Cooperative behavior costs the individual now, and another member of the group usually does the same for it later.
A meerkat that eats under guard, day after day, is more likely to stay alive and to leave offspring. That is the test the whole topic uses: alive, or offspring.
Now consider the wild dog again. It carried meat back to the den for a pack member that missed the hunt, so it gave up meat it could have eaten.
Suppose on a later hunt this wild dog stays at the den with the pups. The pack members that hunted carry meat back to it.
So the wild dog gave up meat on one day and was fed on another. The helper was helped in its turn.
What you are expected to know State the benefit to the individual doing a described cooperative behavior: what another member of the group later does for it.
A meerkat stands guard on the mound for an hour while the rest of its group digs for beetles. Then another meerkat takes its place on the mound.
What does the first meerkat get back for its hour on the mound?
- A. ✓ An hour of digging and eating while another meerkat stands guard
- B. The beetles the others dug up while it watchedThe others eat the beetles they dig up themselves.
What the first meerkat gets is its own turn to dig, under another meerkat’s guard. - C. The first place in the burrow when the hawk comesEvery meerkat dives for whatever cover is nearest when the hawk comes.
What the first meerkat gets is its own turn to dig, under another meerkat’s guard.
Why: After its hour on the mound, the first meerkat climbs down and digs.
Another meerkat stands guard over it.
So the first meerkat gets an hour of eating under guard.
In a flock of small birds, one bird stops feeding to watch for hawks while the others feed. The birds take turns at watching. A student says: “The watcher only loses. Watching costs it feeding time and gets it nothing.”
Is the student correct?
- A. YesLater another bird watches while the watcher feeds.
So the watcher gets feeding time under another bird’s guard. - B. ✓ No
Why: Watching costs the bird its feeding time now.
Later another bird watches while the watcher feeds.
So the watcher gets feeding time under guard.
The watcher does not only lose: it is helped in its turn.
A wolf that hunted carries meat back to the den for a pack member that stayed with the pups. Suppose on a later hunt this wolf is the one that stays at the den.
Which of the following does the wolf get back?
- A. A larger share of every future killNothing in the pack’s sharing gives one wolf a larger share for good.
The wolf gets back meat carried to it on the day it stays at the den. - B. Its own territory away from the packA wolf that leaves the pack is fed by no one.
What the wolf gets back is meat carried to it on the day it stays at the den. - C. ✓ Meat carried back to the den by the wolves that hunted
Why: The wolf gave up meat on the day it carried meat to the den.
On the day it stays at the den, the wolves that hunted carry meat back to it.
So the wolf is fed in its turn.
54Many eyes
Video: Watch: Many eyes
One antelope grazing alone, a lion creeping close unseen; then a herd, two heads always up, the lion seen far off and the whole herd bolting; then a wolf pack sharing meat at the den.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L08c.mp4
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Now consider a herd of antelope grazing on a plain, and one antelope grazing alone. A lion creeps toward each.
Each antelope lifts its head from the grass now and then to look about. The lone antelope has one pair of eyes, so between its looks the lion creeps closer unseen.
The herd has twenty pairs of eyes. At any moment some of those eyes are up, so one antelope sees the lion while it is still far off.
That antelope bolts. The whole herd bolts with it, and the lion catches nothing.
So the herd loses fewer of its members to lions than the same twenty antelope would lose grazing alone.
A school of fish does the same job another way. When the school turns as one, the predator cannot pick out a single fish, so it catches fewer fish.
A flock of birds does it a third way. When one bird finds food, the others see it feeding and join it, so more of the flock eats.
Fewer members lost and more members fed means more of the group stay alive to breed. So the population survives.
Cooperative behavior contributes to the survival of the population.
What you are expected to know Explain how cooperative behavior raises the survival of the population: more eyes see a predator sooner, a school confuses a predator, a flock shares the food one bird finds.
Thirty zebras graze together in one herd. Another thirty zebras graze alone, scattered across the same plain. Lions hunt on the plain all year.
Which group loses more zebras to the lions in a year?
- A. The herdThe herd has thirty pairs of eyes, and some are always up.
A lion is seen far off, and the whole herd bolts. - B. ✓ The scattered zebras
- C. The two groups lose the same numberA zebra alone has one pair of eyes, down in the grass most of the time.
A zebra in the herd is warned by the eyes that are up.
Why: A zebra grazing alone has one pair of eyes, down in the grass most of the time.
Between its looks a lion creeps close unseen.
In the herd some of thirty pairs of eyes are always up, so a lion is seen far off.
So the scattered zebras lose more.
Suppose forty wild goats graze together in one herd on a mountainside, and another forty graze alone, scattered across the same slopes. Wolves hunt on the mountain all year. Each goat lifts its head from the grass now and then to look about.
(a) Explain how grazing in a herd changes the number of goats the wolves catch in a year. (2 pt)
Frame Grazing in a herd changes the number caught because …
Between its looks a wolf creeps close unseen.
In the herd forty pairs of eyes are watching, and some are up at any moment.
So one goat sees the wolf while it is still far off, and the whole herd flees.
So the wolves catch fewer goats from the herd than from the scattered goats.
- Award 1 point for: a goat alone has one pair of eyes (looks up only now and then), so a wolf can creep close unseen.
- Award 1 point for: in the herd many pairs of eyes watch and some are always up, so the wolf is seen while still far off and the herd flees, so fewer are caught (more survive).
Suppose one flock of forty starlings shares each patch of food a bird finds: when one bird starts to eat, the others see it and join it. A second group of forty starlings feeds with each bird searching alone.
Which group has more birds fed by nightfall?
- A. ✓ The flock that shares each patch
- B. The birds that search aloneA bird searching alone eats only from the patches it finds itself.
In the flock, one bird’s find feeds many. - C. The two groups feed the same numberIn the flock, one find feeds many birds.
Alone, one find feeds one bird.
Why: In the flock, one bird’s find is seen by the others, and they join it.
So one patch of food feeds many birds.
Alone, each bird eats only from the patches it finds itself.
So the flock that shares has more birds fed.
69Helping a relative
In Unit 5, a gene for flower color came in two versions, one giving purple flowers and one giving white flowers.
What is each version of the gene called?
- A. A chromosomeA chromosome is one long DNA molecule carrying many genes.
Each version of one gene is an allele. - B. A gameteA gamete is a sex cell, a sperm or an egg.
Each version of one gene is an allele. - C. ✓ An allele
Why: When a gene comes in more than one version, each version is called an allele.
The purple version and the white version are two alleles of the flower-color gene.
In Unit 7, two guppies were compared for their evolutionary fitness.
What does evolutionary fitness count?
- A. ✓ Offspring that survive to reproduce
- B. Years the individual livesA long life with no offspring that reproduce is an evolutionary fitness of zero.
Evolutionary fitness counts offspring that survive to reproduce. - C. How large and strong the individual growsSize and strength are not counted.
Evolutionary fitness counts offspring that survive to reproduce.
Why: Evolutionary fitness is the number of offspring an individual leaves that themselves survive to reproduce.
Video: Watch: Helping a relative
A hive: the queen laying, workers feeding the young; the queen’s alleles drawn shaded and the same shading appearing in the workers and in a new queen; the new queen breeding and the shaded alleles passing on.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L08d.mp4
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Now consider a honeybee hive. It holds one queen and many thousands of workers.
The queen lays all the eggs. The workers are sterile: they cannot breed.
Yet the workers feed the queen and raise her young. Every one of those young is the queen’s offspring.
A worker leaves no offspring of its own. So its evolutionary fitness, counted the Unit 7 way, is zero.
Yet the helping stayed common in hives for millions of years. To see why, look at whose young the worker raises.
The workers are the queen’s daughters. The young they raise are the queen’s offspring too: the workers’ own sisters and brothers.
A daughter gets half of her alleles from her mother. So the worker and each of the queen’s new offspring got half of their alleles from the same queen.
So the worker and the queen’s new offspring share many of the same alleles.
Some of the queen’s new offspring are new queens. A new queen flies off, breeds, and passes her alleles to her own offspring.
Many of those alleles are alleles the worker carries too. So copies of the worker’s alleles pass to the next generation without the worker breeding.
Here is the hive as a diagram. A shaded square is an allele the queen carries; the worker and the new queen both carry shaded squares, and the new queen’s offspring carry them on.
Kin is an old word for relatives.
Selection that favors helping close relatives to breed, because the relatives share the helper’s alleles, is called .
Kin selection is not for the good of the species. The worker’s help spreads alleles the worker itself carries.
Workers whose alleles made them raise the queen’s young left more copies of those alleles, through their sisters, than workers that did not. So the helping stayed common.
That is natural selection, the process Unit 7 taught, working through relatives.
What you are expected to know Explain why helping a close relative to reproduce can spread an individual’s own alleles: the relative shares many of those alleles and passes them on.
A honeybee hive holds one queen and thousands of workers.
Do the workers breed?
- A. YesThe workers are sterile.
The queen lays every egg in the hive. - B. ✓ No
Why: The workers are sterile: they cannot breed.
The queen lays all the eggs.
Worker bees raise the queen’s offspring rather than offspring of their own.
Why can raising the queen’s offspring spread copies of a worker’s alleles?
- A. Feeding the young passes the worker’s alleles into themFood carries no alleles.
The young share the worker’s alleles because they and the worker came from the same queen. - B. Raising the young changes the alleles the worker itself carriesAn individual’s alleles do not change during its life.
The young share the worker’s alleles because they and the worker came from the same queen. - C. ✓ The queen’s offspring share many alleles with the worker, and some of them breed
Why: The worker and the queen’s new offspring came from the same queen, so they share many alleles.
Some of those offspring are new queens, and a new queen breeds.
So copies of the worker’s alleles pass on through her.
A student says: “Worker bees raise the queen’s young for the good of the species. Kin selection is selection for what helps the species.”
Is the student correct?
- A. YesThe worker’s help spreads alleles the worker itself carries, through relatives that share them.
Kin selection works through shared alleles, not for the species as a whole. - B. ✓ No
Why: The queen’s young share many alleles with the worker.
When those young breed, copies of the worker’s alleles pass on.
Workers whose alleles made them help left more copies, so the helping stayed common.
Kin selection works through shared alleles, not for the good of the species.
Here is the mound again: the sentinel upright on top, and the rest of the group digging for beetles below.
The sentinel eats less now. When its own turn to dig comes, another meerkat stands guard over it.
The group loses fewer of its members to hawks. So the population survives.
If the meerkats it guards are its close relatives, its help spreads copies of its own alleles too.
97Quick quiz: kin selection mixed practice
A bird feeds the chicks of an unrelated pair nesting in the next tree.
Can kin selection explain this helping?
- A. YesThe chicks are not the helper’s relatives, so they share no more of its alleles than any other bird does.
Kin selection works only through relatives. - B. ✓ No
Why: The chicks are not the helper’s relatives.
So they share no more of the helper’s alleles than any other bird does.
Their breeding spreads no extra copies of the helper’s alleles.
So kin selection cannot explain the helping.
A young bird stays at its parents’ nest for a year and feeds the new chicks there, its own brothers and sisters.
Can kin selection explain this helping?
- A. ✓ Yes
- B. NoThe chicks are the helper’s brothers and sisters, so they share many of its alleles.
Kin selection favors helping relatives that share the helper’s alleles.
Why: The chicks are the helper’s brothers and sisters.
So they share many of the helper’s alleles.
When they grow up and breed, copies of the helper’s alleles pass on.
So kin selection can explain the helping.
In an ant colony, sterile workers feed the queen’s larvae, their own sisters.
Can kin selection explain this helping?
- A. ✓ Yes
- B. NoThe larvae are the workers’ sisters, so they share many of the workers’ alleles.
Some larvae become new queens and breed.
Why: The larvae are the workers’ sisters.
So they share many of the workers’ alleles.
Some larvae become new queens and breed, passing those alleles on.
So kin selection can explain the helping.
Some animals help their relatives to breed.
What is kin selection?
- A. Selection that favors helping any member of the group, related or notAn unrelated member of the group shares no more of the helper’s alleles than a stranger.
Kin selection works through alleles shared with close relatives. - B. Selection that favors the strongest individual in a groupStrength is not what kin selection favors.
Kin selection favors helping relatives that share the helper’s alleles. - C. ✓ Selection that favors helping relatives that share the helper’s alleles
Why: Kin selection is selection that favors helping close relatives to breed, because the relatives share many of the helper’s alleles and pass copies of them on.
Some animals help their relatives to breed.
(a) State what kin selection is. (1 pt)
- Award 1 point for: selection favoring help to (close) relatives because they share the helper’s alleles (so the helper’s alleles are passed on).
103Mixed practice mixed practice
A shoal of two hundred herring swims as one school, and a seal hunts it. A student says: “A herring is safer in the school, because when the school turns as one the seal loses sight of any single fish.”
Is the student correct?
- A. No: a herring in the school is as easy for the seal to pick out as a herring aloneWhen the school turns as one, the seal loses sight of any single fish.
So it catches fewer herring from the school than from herring swimming alone. - B. ✓ Yes: when the school turns as one, the seal loses sight of any single fish, so it catches fewer
Why: The seal catches a herring by picking one out.
When the school turns as one, the seal loses sight of any single fish.
So the seal catches fewer herring from the school.
So a herring in the school is safer than one swimming alone.
A badger digs a burrow and sleeps in it alone.
Is the digging cooperative behavior?
- A. YesThe burrow shelters the badger itself and no other badger.
Cooperative behavior helps the others in the group. - B. ✓ No
Why: The burrow shelters the badger itself.
The digging helps no other badger.
So the digging is not cooperative behavior.
Suppose a group of monkeys in which one monkey watches for eagles from a treetop while the others feed on the ground beneath it.
Which of the following would let kin selection favor the watching?
- A. ✓ The monkeys feeding beneath the treetop are the watcher’s close relatives
- B. The monkeys feeding beneath the treetop are the largest in the forestSize is not what kin selection favors.
Kin selection favors helping relatives that share the helper’s alleles. - C. The monkeys feeding beneath the treetop are unrelated to the watcherUnrelated monkeys share no more of the watcher’s alleles than any other monkey does.
Kin selection works through relatives.
Why: Kin selection favors helping close relatives, because they share many of the helper’s alleles.
If the monkeys the watcher guards are its close relatives, they share many of its alleles.
When they breed, copies of the watcher’s alleles pass on.
So kin selection would favor the watching.
A student says: “A worker bee leaves zero offspring, so it leaves zero copies of its alleles in the next generation.”
Is the student correct?
- A. YesThe queen’s young share many of the worker’s alleles, and the new queens among them breed.
So copies of the worker’s alleles do reach the next generation. - B. ✓ No
Why: The worker and the queen’s young came from the same queen, so they share many alleles.
The new queens among those young breed.
So copies of the worker’s alleles reach the next generation without the worker breeding.
Three animals behave in the ways listed.
Which of the following is cooperative behavior?
- A. ✓ A hyena carries meat back to cubs at the den that did not hunt
- B. A tortoise pulls its head into its shell when a shadow passesPulling its head in protects the tortoise itself and helps no other tortoise.
- C. A frog calls at night to attract a mate to itselfThe call brings a mate to the caller itself and helps no other frog.
Why: Carrying meat back costs the hyena meat it could have eaten now.
The meat feeds the others in its group.
So the carrying of meat is cooperative behavior.
In a colony of ground squirrels, one squirrel gives an alarm call when a hawk appears. The squirrels around it are its sisters and their young. A student says: “The call can spread copies of the caller’s own alleles, because the squirrels it saves share many of its alleles and go on to breed.”
Is the student correct?
- A. No: a squirrel spreads copies of its alleles by breeding itself, not by callingThe saved squirrels are the caller’s sisters and their young, so they share many of its alleles.
When they breed, copies of the caller’s alleles pass on. - B. ✓ Yes: the saved squirrels share many of the caller’s alleles, and when they breed those alleles pass on
Why: The squirrels the call saves are the caller’s sisters and their young.
So they share many of the caller’s alleles.
When they breed, copies of the caller’s alleles pass on.
So the call spreads copies of the caller’s alleles without the caller breeding.
Wild dogs hunt in packs. After a hunt, the dogs that hunted carry meat back to the den for pack members that were not at the hunt, such as the dogs guarding the pups.
(a) Explain how the carrying of meat demonstrates cooperative behavior. (1 pt)
Frame The carrying of meat is cooperative behavior because …
The meat feeds the pack members that were not at the hunt.
So the behavior costs the individual now and helps the others in its group.
- Award 1 point for: it costs the carrier now (meat it could have eaten) AND it helps others in its pack (feeds the dogs that missed the hunt).
(b) Explain how the carrying of meat raises the number of pack members that survive to breed. (1 pt)
Frame More members of the pack survive to breed because …
So the guards stay alive, and the pups they guard stay alive.
On a later day the carrier stays at the den and is fed in its turn.
So the carrying keeps more of the pack alive to breed.
- Award 1 point for: the pack members that did not hunt (the guards, the pups) are fed and stay alive, so more of the pack survive to breed. Accept with or without the carrier being fed in its turn.
Glossary
- cooperative behavior
- Behavior that costs the individual now and helps the others in its group: a meerkat standing guard while the others feed; wolves carrying meat back to pack members that missed the hunt. The helper is usually helped in its turn, and the group loses fewer members, so the population survives.
- kin selection
- Selection that favors helping close relatives to breed, because the relatives share many of the helper’s alleles: a worker bee raises the queen’s offspring, which carry many of the worker’s own alleles, so copies of those alleles pass on without the worker breeding. Kin means relatives. Not selection for the good of the species.
APBIO-U08-L09 Capstone: a change in the environment, a change in behavior, a change in the population
Photo: Megan McCarty, Wikimedia Commons, CC BY 3.0 (resized).
Suppose a field of moths flies at night. Each female releases a pheromone into the air, and each male smells it and flies toward her. A farmer sprays the field with a chemical. The chemical blocks the receptors the males use to smell the females’ pheromone.
The following spring there are far fewer moths. Trace the chain from the spray to the count.
Unit 8 · Ecology
1The chain from the spray to the count
How does one change in the surroundings end up changing a population?
The chain has four links, each from this topic:
- the chemical is the change;
- the blocked receptors stop the males detecting the signal;
- the males cannot find females, so fewer matings and fewer offspring;
- the next generation is smaller.
A chain question asks for the end of the chain and the links, in order.
Here is the field again. Suppose a biologist counts the moths in it: 400 the summer before the spray, and 60 the spring after it.
The chemical arrived in the moths’ surroundings. A change in an organism’s surroundings is an external change, and this one is where the chain starts.
Before the spray, each female moth released a pheromone into the night air, and each male moth detected it. The pheromone is a chemical signal: a chemical one organism produces and another organism of the same species detects.
A male frog calls at a pond edge on a spring night. A female frog detects the call.
Which change does the call make in the female’s behavior?
- A. ✓ The female moves toward the caller
- B. The female stays where she isA signal makes one action more likely.
For a mating call, that action is the move toward the caller. - C. The female calls backA female frog that detects a male’s call moves toward him.
The calling is the male’s signal.
Why: A signal changes what the receiver does.
The female frog detected the male’s call.
So the female moves toward the caller.
After a signal, the receiver’s behavior changes. Here the male moth is the receiver: he detected the pheromone and flew toward the female.
The flight toward the female is a behavioral response: something the whole moth does. Because the male moves toward the stimulus, the flight is also a taxis.
Now the chemical blocks the males’ receptors. A male no longer detects the pheromone, so his behavior no longer changes: he does not fly toward the female.
So the change in the surroundings became a change in behavior. The males cannot find females.
At a pond, some male frogs call louder and longer than others, and those males draw more females.
Compared with the quiet callers, what do the loud callers leave?
- A. Fewer offspringThe loud callers draw more females, so they mate more often.
More matings leave more offspring, not fewer. - B. The same number of offspringThe loud callers draw more females than the quiet callers.
More matings leave more offspring. - C. ✓ More offspring
Why: The loud callers draw more females.
So the loud callers mate more often.
More matings leave more offspring.
So the loud callers leave more offspring than the quiet callers.
A male that finds a female mates, and the pair leave offspring. A male that cannot find a female leaves none.
Across the field, fewer males find females. So the change in behavior became a change in offspring: fewer matings, so fewer offspring.
The moths in the field are one population: all the moths of one species living in that field. Fewer offspring this year means fewer moths counted next spring: 60 instead of 400.
A change that alters a behavior alters who survives and breeds, and so alters the population.
In a herd of deer, some fawns freeze when a predator passes and others flee, and each fawn inherits which of the two it does from its parents. The fawns that freeze survive more often than the fawns that flee.
What happens to freezing in the next generation of fawns?
- A. ✓ Freezing becomes more common
- B. Freezing becomes less commonThe fawns that freeze survive more often, so more of them live to breed.
Their offspring make up more of the next generation. - C. Freezing stays as common as beforeThe fawns that freeze survive more often than the fawns that flee.
So the freezing fawns and the fleeing fawns do not breed in equal numbers.
Why: The fawns that freeze survive more often.
So more of the freezing fawns live to breed.
More of the next generation are their offspring, and they freeze too.
So freezing becomes more common.
Video: Watch: The chain from the spray to the count
The strip drawn link by link as each is said: the chemical sprayed on the field; the males that cannot detect the pheromone or find females; fewer matings and fewer offspring; fewer moths the next spring.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L09a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L09a.mp4
Now imagine that some males had receptors the chemical did not block. Those males would still find females, so they would leave more offspring than the rest.
So more of the next generation would be their offspring. A behavior that raises the number of offspring becomes more common in a population: that is natural selection, the process Unit 7 taught.
Here is the whole chain again, with the word this topic uses written under each link.
The table below compares the four links: what happened in the sprayed field, and the word this topic uses for it.
A chain like this one is ecology at work: living things interacting with one another and with their surroundings, here at the level of a population.
To answer a chain question, first predict the end: the count rises or falls. Then justify the prediction by stating each link, in order, from the change in the surroundings to the population.
What you are expected to know Predict and justify how a described change in the environment changes an organism’s behavior and, through survival or offspring, the population.
Suppose a female beetle releases a chemical into the air, and a male beetle of the same species detects it.
Which word from this topic names the chemical?
- A. A tactile signalA tactile signal is one the receiver detects as touch.
The male beetle detected a chemical in the air. - B. ✓ A pheromone
- C. An audible signalAn audible signal is one the receiver hears.
The male beetle detected a chemical in the air.
Why: The male beetle detected a chemical, so the chemical is a chemical signal.
It passed between two beetles of the same species.
A chemical signal passed between organisms of the same species is a pheromone.
Suppose a snail in dry air crawls faster and turns more often, with no fixed direction, until it reaches damp air and slows down.
Which word from this topic names the crawl?
- A. A taxisA taxis is a movement toward or away from the stimulus.
The snail crawled with no fixed direction. - B. ✓ A kinesis
- C. A phototropismA phototropism is a shoot growing toward the light.
The snail is an animal that crawled, with no fixed direction.
Why: The snail changed its speed and its turning, with no fixed direction.
A movement that changes speed or turning with no fixed direction is a kinesis.
A potted plant sits on a shelf under a lamp. Over a week its shoot grows toward the lamp.
Which word from this topic names the growth toward the lamp?
- A. A kinesisA kinesis is an animal changing its speed or turning.
The shoot grew in one direction, toward the light. - B. PhotoperiodismPhotoperiodism is a plant timing its flowering by the length of day or night.
The shoot grew toward the light; nothing flowered. - C. ✓ Phototropism
Why: The shoot grew toward the light.
A shoot growing toward the light is phototropism.
Suppose a chick hatches in a box on its own. On its first day it pecks at seeds, and it has seen no other bird peck.
Which word from this topic names the pecking?
- A. A learned behaviorA learned behavior is changed by the individual’s experience.
The chick pecked on its first day, with no experience of pecking. - B. A physiological responseA physiological response is a change inside the organism’s body.
Pecking is something the whole chick does. - C. ✓ An innate behavior
Why: The chick pecked without any experience of pecking.
A behavior present without experience is an innate behavior.
A male bird spreads its tail feathers, and a female bird sees them.
Which kind of signal is the spread tail?
- A. ✓ A visual signal
- B. An electrical signalAn electrical signal is an electric pulse the receiver detects.
The female bird saw the tail: she detected light. - C. A tactile signalA tactile signal is one the receiver detects as touch.
The female bird saw the tail: she detected light.
Why: The kind of signal is named from what the receiver detects.
The female bird saw the tail: she detected light.
So the spread tail is a visual signal.
A meerkat stands upright on a mound and eats less, while the rest of its group digs for food. When a hawk appears it calls, and the group takes cover.
Which word from this topic names the standing guard?
- A. A kinesisA kinesis is an animal changing its speed or turning.
The meerkat stood still and watched. - B. ✓ Cooperative behavior
- C. A physiological responseA physiological response is a change inside the organism’s body.
Standing guard is something the whole meerkat does.
Why: Standing guard costs the meerkat food now.
It helps the others in its group, who feed and take cover in time.
Behavior that costs the individual now and helps others in its group is cooperative behavior.
Worker honeybees do not breed. They raise the queen’s offspring, who share many of the workers’ alleles.
Which idea from this topic explains why raising the queen’s offspring spreads the workers’ own alleles?
- A. Learned behaviorWhether the raising is learned or innate does not decide whose alleles spread.
The queen’s offspring share many of the workers’ alleles. - B. A pheromoneA pheromone is a chemical signal passed between organisms of the same species.
The question is whose alleles the raised offspring carry. - C. ✓ Kin selection
Why: The queen’s offspring share many of the workers’ alleles.
So when the workers raise those offspring, copies of the workers’ own alleles spread.
Helping a close relative reproduce to spread shared alleles is kin selection.
A student tips 20 snails into a box with a warm tile on one side and a cool tile on the other, and after ten minutes counts the snails on each tile.
Which of the following is the dependent variable?
- A. The temperature of the tilesThe temperature of the tiles is the one condition the student deliberately set different on the two sides.
That is the independent variable. - B. The number of snails tipped inThe student keeps the number tipped in at 20.
A condition kept the same is not the result measured. - C. ✓ The count of snails on each tile
Why: The dependent variable is the result the student measures.
The student counts the snails on each tile after ten minutes.
So the count on each tile is the dependent variable.
A student tips 20 snails into a box with a warm tile on one side and a cool tile on the other. After ten minutes 14 snails sit on the cool tile and 6 on the warm tile.
Which of the following is the null hypothesis for this test?
- A. The snails prefer the cool tile, so the gap from 10 and 10 is realA preference is a real difference between the tiles.
A claim of a real difference is the alternative hypothesis. - B. ✓ The snails have no preference, and any gap from 10 on each tile is due to chance
- C. The snails prefer the warm tile, so the gap from 10 and 10 is realA preference for either tile is a real difference between the tiles.
A claim of a real difference is the alternative hypothesis, not the null.
Why: The null hypothesis for a choice box is no preference.
With no preference, the 20 snails split evenly: 10 expected on each tile.
Chi-square then measures how far 14 and 6 sit from 10 and 10.
A runner’s heart beats faster during a race.
Which kind of response is the faster heartbeat?
- A. ✓ A physiological response
- B. A behavioral responseA behavioral response is something the whole organism does.
The heartbeat is a change inside the runner’s body.
Why: The faster heartbeat is a change inside the runner’s body.
A response that is a change inside the organism’s body is a physiological response.
A biologist counts every oak tree of one species in one forest.
Which of the following has the biologist counted?
- A. A communityA community is all the populations of different species living in one place.
The biologist counted one species only. - B. An ecosystemAn ecosystem is the community together with the non-living surroundings.
The biologist counted one species only. - C. ✓ A population
Why: The biologist counted all the oaks of one species living in one place.
All the organisms of one species living in one place are a population.
Suppose a new road opens beside a pond where male frogs call each spring, and traffic noise now covers the calls at night.
Which of the following is the change in the frogs’ surroundings?
- A. The number of tadpoles at the pondThe number of tadpoles is the end of the chain, part of the population.
The change in the surroundings is what started the chain. - B. ✓ The traffic noise
- C. The females’ movement toward the callersThe females’ movement is the receivers’ behavior.
The change in the surroundings is what happened around the frogs.
Why: The traffic noise arrived in the frogs’ surroundings, outside their bodies.
A change in an organism’s surroundings is an external change.
So the traffic noise is the change in the surroundings.
Traffic noise covers the calls of the male frogs at a pond. Before the road, a female that detected a call moved toward the caller.
Do as many females move toward the callers now?
- A. YesThe noise covers the calls, so fewer females detect one.
A female that detects no call does not move toward a caller. - B. ✓ No
Why: A signal changes what the receiver does.
The noise covers the calls, so fewer females detect one.
A female that detects no call does not move toward a caller.
So fewer females move toward the callers.
Traffic noise covers the male frogs’ calls at a pond through the breeding season.
What happens to the number of tadpoles at the pond?
- A. ✓ Fewer tadpoles
- B. The same number of tadpolesFewer females reach the callers, so fewer pairs mate.
Fewer matings leave fewer offspring. - C. More tadpolesFewer females reach the callers, so fewer pairs mate.
Fewer matings leave fewer offspring, not more.
Why: Fewer females reach the calling males.
So fewer pairs mate.
Fewer matings leave fewer offspring, and the offspring are the tadpoles.
So there are fewer tadpoles.
A road opens beside a pond, and traffic noise covers the male frogs’ calls through one breeding season.
Predict the number of young frogs at the pond the following spring.
- A. ✓ Smaller than before the road
- B. The same as before the roadFewer tadpoles hatched, so fewer young frogs grow up.
The next generation is smaller. - C. Larger than before the roadFewer tadpoles hatched, so fewer young frogs grow up.
The next generation is smaller, not larger.
Why: Fewer tadpoles hatched this year.
So fewer young frogs grow up to join the population.
So the number of young frogs at the pond the following spring is smaller than before the road.
A student reads about the pond beside the new road and says: “Noise only changes what the frogs hear, so the number of frogs stays the same.”
Is the student correct?
- A. YesThe noise covers the calls, so fewer females move toward the callers.
Fewer matings follow, so fewer tadpoles hatch and the next generation is smaller. - B. ✓ No
Why: The noise covers the calls.
So fewer females detect one.
So fewer females move toward the callers.
So fewer pairs mate, and fewer tadpoles hatch.
A change that alters a behavior alters who breeds, and so alters the population: the number of frogs falls.
Suppose the town closes the road at night the next year, and the pond is quiet again during the frogs’ calling season.
Will more tadpoles hatch than in the noisy year?
- A. ✓ Yes
- B. NoWith the pond quiet, the females detect the calls again and move toward the callers.
More matings leave more offspring than in the noisy year.
Why: With the pond quiet, the females detect the calls again.
So more females move toward the callers, and more pairs mate.
More matings leave more offspring.
So more tadpoles hatch than in the noisy year.
Here is the sprayed field again, and the count the following spring: 60 moths where there had been 400.
The chemical blocked the signal, so the males could not find females.
Fewer matings followed, so there were fewer offspring, and the next year’s count fell.
47Practice: predict and justify the chain mixed practice
Female moths of one species release a pheromone into the night air, and the males detect it and fly toward them. Suppose a farmer sprays a field of these moths with a chemical that breaks the pheromone down in the air before it reaches the males. The males’ receptors work as before.
(a) Predict how the number of moths counted in the field changes the following spring. (1 pt)
- Award 1 point for: fewer moths (a smaller count) the following spring.
(b) Justify your prediction by stating each link, in order, from the spray to the count. (1 pt)
Frame The count changes because …
So the males no longer detect the females’ signal, and they do not fly toward the females.
Fewer males find females, so there are fewer matings and fewer offspring.
So the next generation is smaller, and the count the following spring falls.
- Award 1 point for: the males cannot detect the pheromone (the signal), so they do not find females, so there are fewer matings and fewer offspring, so the next generation is smaller — the links in that order. Accept with or without the word signal.
Suppose newly hatched turtles dig out of a nest on a beach at night. On a dark beach each hatchling crawls toward the brightest part of the horizon, which is the sea, and reaches the water within minutes. A row of hotels is then built along the beach, and their lights stay on all night. The lights are brighter than the horizon over the sea.
(a) Predict what happens to the number of hatchlings from this beach that reach the sea. (1 pt)
- Award 1 point for: fewer hatchlings reach the sea.
(b) Justify your prediction by stating each link from the lights to the number of hatchlings that reach the sea. (1 pt)
Each hatchling crawls toward the brightest part of the horizon.
The lights are now brighter than the sea, so the hatchlings crawl toward the hotels instead of the sea.
So fewer hatchlings reach the sea.
- Award 1 point for: the hatchlings crawl toward the brightest part of the horizon, which is now the lights, so they crawl away from the sea, so fewer reach it.
(c) Explain how this change in the hatchlings’ behavior changes the size of the turtle population. (1 pt)
So fewer of this year’s hatchlings survive to breed.
Fewer breeding turtles leave fewer offspring.
So the turtle population is smaller in later years.
- Award 1 point for: hatchlings that do not reach the sea die, so fewer survive to breed, so fewer offspring and a smaller population.
APBIO-U08-P81 Practice questions: Topic 8.1
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one choice-chamber test one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table from the formula sheet is printed beside it.
Video: Watch first: Responses to the environment, summed up
A stimulus and a response at the scale of the whole organism; behavioral or physiological; taxis, kinesis and the plant’s two versions; a signal as a stimulus made for another organism, its five channels and four common jobs; innate or learned; cooperation and kin; the choice-chamber test and its chi-square.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T81-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T81-summary.mp4
A student reads that ecologists study the badgers of one wood and says: “Ecology is the study of wild animals in their natural surroundings.”
Which of the following corrects the student’s statement?
- A. Ecology studies wild animals in the laboratory as well as in the wild, but only animalsEcology is not limited to animals: plants, fungi and microbes interact too.
The student’s error is the ‘animals’, not the place. - B. Ecology is the care of wild places, so it is a job rather than a branch of biologyCaring for wild places is something people do.
Ecology is a branch of biology: the study of how living things interact. - C. ✓ Ecology studies how living things of every kind interact with one another and with their surroundings
- D. Ecology studies the organs and cells inside wild animals rather than the animals’ surroundingsThe organs and cells inside one animal are the biology of one organism.
Ecology asks how living things interact with one another and their surroundings.
Why: Ecology is the study of how living things interact with one another and with their surroundings.
Badgers, the plants they eat and the soil they dig all belong to it.
The student keeps only the animals and drops the interactions.
So the correction names every living thing and their interactions.
Researchers let field mice choose between an insulated shelter and an open chamber for 30 minutes, at two air temperatures. The table gives the results.
Which of the following is a physiological response by the mice to the cold air?
- A. The 18 minutes the mice spent in the insulated shelterMoving into the shelter is something the whole mouse does.
That is a behavioral response. - B. ✓ The rise in the oxygen the mice used
- C. The fall in the air temperature to 8 °CThe cold air is the change the mice detected: the stimulus, not a response.
- D. The mice’s body temperature staying near 37 °CA body temperature that stays near 37 °C is not a change inside the mouse.
The change that kept it steady was the rise in oxygen use.
Why: A physiological response is a change inside the organism’s body.
In the cold the mice used more oxygen: 2.4 mL per gram per hour against 1.1.
Using more oxygen is a change inside the mouse’s body.
So the rise in oxygen use is the physiological response.
Four small animals each respond to a change around them.
Which of the following movements is a taxis?
- A. ✓ Mosquito larvae swim straight down whenever a shadow falls on the water above them
- B. Earthworms turn more often on warm soil than on cool soil, in no fixed directionThe earthworms change how often they turn, with no fixed direction.
A change in turning with no fixed direction is a kinesis. - C. Pond snails crawl faster in water with little oxygen and slower in oxygen-rich water, whichever way they are headingThe snails change speed with the oxygen, with no fixed direction.
A change in speed with no fixed direction is a kinesis. - D. Water fleas dart about more in warm water than in cool water, heading nowhere in particularThe water fleas move more in the warmth, with no fixed direction.
A change in how much an animal moves, with no fixed direction, is a kinesis.
Why: A taxis is movement toward or away from the stimulus.
The larvae swim straight down whenever a shadow falls.
The shadow sets the direction of their swim: away from it.
So the swim is a taxis.
In a kind of wasp, some young females stay at the nest they hatched in and spend their lives feeding their mother’s later young, their own sisters, instead of laying eggs of their own. A student says: “The helpers stay because raising sisters is good for the wasp species.”
Which of the following corrects the student’s reason?
- A. The helpers stay because feeding the young passes their own alleles into the young with the foodFood carries no alleles.
A sister carries copies of the helper’s alleles because the two share the same parents. - B. The helpers stay because staying at the nest costs them nothingStaying costs the helper the eggs she would have laid.
The helping is kept because of what it does for her sisters’ survival, and so for her alleles. - C. The helpers stay because their own alleles change to match their sisters’ while they feed themAn individual’s alleles are fixed when it hatches.
Feeding changes which relatives survive, not the helper’s own alleles. - D. ✓ The helpers stay because their sisters share many of their alleles, and the fed sisters survive to breed
Why: Kin selection favors helping relatives that share the helper’s alleles.
Sisters share many of their alleles.
The fed sisters survive and breed.
So copies of the helper’s alleles pass on through her sisters’ young, though she lays no eggs herself.
Two males of one kind of electric fish live in one shoal. A worm sinks to the river bed between them, and both turn toward it. Each fish gives off a steady train of weak electric pulses, and a researcher records the pulses before and while the two face each other. The table gives the record.
Which of the following best describes what the larger male’s change in pulses did for it?
- A. ✓ It showed dominance
- B. It marked the river bed around the worm as its own patchA territory mark is left on an area for fish that come later.
The pulses settled which of two fish gave way over one worm. - C. It led the smaller male to the wormThe smaller male backed away from the worm; nothing led it there.
- D. It won the smaller male as a mateBoth fish are males, and the smaller one backed away and hid rather than staying.
Why: A signal that settles which of two animals gives way shows dominance.
The larger male raised its pulses from 58 to 92 per second and approached.
The smaller male dropped its pulses to 15 per second, backed away and hid.
So the larger male’s pulses showed dominance.
When an ant of one kind that forages at night finds food, it returns to the nest and its nestmates then leave for the food. Researchers place a barrier between a returning ant and its nestmates and count the nestmates that leave for the food each hour. The table gives the results and what each barrier lets through.
Which kind of signal do the returning ants use to bring their nestmates to the food?
- A. A visual signalSight passed through the clear glass and the wire mesh, and 3 to 5 nestmates left.
Seeing the returning ant was not enough. - B. An audible signalSound passed through both meshes, and 5 or 6 nestmates left.
Hearing the returning ant was not enough. - C. ✓ A tactile signal
- D. A chemical signalChemicals passed through both meshes, and 5 or 6 nestmates left.
Smelling the returning ant was not enough.
Why: A tactile signal is one the receiver detects as touch.
With no barrier, 64 nestmates left each hour.
Every barrier let sight, sound or chemicals through, and every barrier cut the count to 3 to 6.
Touch is the one thing every barrier stopped, so the signal is tactile.
Suppose caterpillars feed on a bean seedling for an hour. Researchers then remove the caterpillars and place the damaged seedling near an undamaged one in four ways, and after 24 hours measure a defensive chemical in the undamaged seedling’s leaves. The table gives the results.
Which conclusion do the results support?
- A. The damaged seedling passes a chemical through the shared soil to the undamaged seedlingThe two seedlings in the shared soil tray, with their shoots kept apart, gave 5.0 micrograms per gram, the same as the undamaged pair.
Nothing passed through the soil. - B. ✓ The damaged seedling gives off a chemical into the air that the undamaged seedling detects
- C. The caterpillars must touch the undamaged seedling before it makes the defensive chemicalThe caterpillars were gone before the undamaged seedling was placed, and it still made 18.4 micrograms per gram in open air.
No caterpillar touched it. - D. An undamaged seedling makes the defensive chemical whenever another seedling stands near itAn undamaged seedling next to an undamaged neighbor made 5.1 micrograms per gram, the low value.
Only a damaged neighbor in open air raised it.
Why: A signal is something one organism produces that another detects.
Beside a damaged seedling in open air, the undamaged one made 18.4 micrograms per gram.
A charcoal filter or an air barrier cut that to about 5.
So the damaged seedling gives off an airborne chemical that the other detects.
Researchers play one recorded sound to each of several feeding flocks of finches and record the share of each flock that flies into the cover of a shrub within 30 seconds. The table gives the results.
Which conclusion do the results support?
- A. The finches fly into cover when they hear any loud soundThe same three notes in reverse order, as loud as before, sent only 0.20 of the flock into cover.
Loudness alone did not move the birds. - B. The finches fly into cover after they see a hawk, whatever sound they hearNo hawk appeared in any trial; the researchers played sounds only.
The call alone sent 0.72 of the flock into cover. - C. ✓ The sharp three-note call, in its usual order, sends the flock into cover
- D. The finches fly into cover after the low two-note feeding callThe feeding call sent 0.17 of the flock into cover, about the same as no sound at all.
Why: A signal changes what the receiver does.
With no sound, 0.15 of a flock flew into cover; with the feeding call, 0.17.
With the sharp three-note call in its usual order, 0.72 flew into cover; reversed, 0.20.
So the call in its usual order changed the flock’s behavior.
Suppose male bowerbirds of one kind each build a bower of sticks on the ground and display in front of it to visiting females. Some males decorate the bower with blue objects. Researchers watch 10 males of each kind through a season and count the females that visit and mate with each. The table gives the means.
Which claim do the results support?
- A. Decorating the bower keeps the male alive longerThe table records visits and matings, not how long the males live.
A signal that draws mates changes who reproduces. - B. Every male mates with about the same number of females, whatever his bowerThe males with decorated bowers mated with 2.9 females each on average; the others with 0.4.
The matings differ seven-fold. - C. Females that visit a decorated bower lay more eggsThe table counts how many females mated with each male, not how many eggs each female laid.
More matings, not larger clutches, is what the decorated males gained. - D. ✓ Decorated males mate with more females, so they leave more offspring
Why: A signaling behavior can give some individuals more offspring than others.
A decorated bower drew 6.8 female visits per male against 2.1.
The decorated males mated with 2.9 females each on average; the undecorated males with 0.4.
So the decorated males leave more offspring.
A parasitic wasp attacks caterpillars of one kind for the first time in their lives, and each caterpillar does one of two things: it thrashes, or it drops off the leaf on a thread of silk. Every caterpillar does its behavior perfectly on this first attack, with no caterpillar to copy, and its young later do as it did. Researchers record what happens to the caterpillars of each kind. The table gives the results.
Which of the following predicts what happens to dropping over the generations, and why?
- A. Thrashing becomes more common, because dropping costs the caterpillar its place on the leaf and its mealDropping does cost the caterpillar its place on the leaf, but 75 % of the droppers survive to adulthood against 20 % of the thrashers.
Survival decides which behavior spreads. - B. ✓ Dropping becomes more common, because more droppers survive to breed and their young drop too
- C. Dropping becomes more common, because the young caterpillars copy the droppers they see surviveEach caterpillar drops or thrashes on its first attack with no caterpillar to copy, so the behavior is innate.
Dropping spreads because more droppers survive to breed. - D. Dropping becomes more common, because the parasitized caterpillars change their alleles to droppingAlleles are inherited; a wasp’s attack does not rewrite them.
The droppers were born droppers, and more of them live to pass the behavior on.
Why: The caterpillars drop or thrash on their first attack with no experience, so the behavior is innate.
75 % of the droppers survive to become adults; 20 % of the thrashers do.
So more droppers live to breed, and their young drop as they did.
So dropping becomes more common.
A student tests whether brine shrimp gather in the light. He shades one end of a narrow trough of sea water and lights the other end, tips 26 brine shrimp into the middle, and after five minutes counts the shrimp at each end. The counts are drawn below.
(a) Identify the independent variable in the student’s test. (1 pt)
Frame The independent variable is …, because it is the condition the student …
Hint Which one condition did the student make different between the two ends of the trough?
- Award 1 point for: light or shade at an end as the independent variable.
Slip Naming the number of shrimp at each end. That is what was counted: the dependent variable.
(b) Identify the dependent variable in the student’s test. (1 pt)
Frame The dependent variable is …, the quantity the student …
Hint What did the student count to see the effect?
- Award 1 point for: the number of shrimp at each end as the dependent variable.
Slip Naming the 26 shrimp tipped in. That number was the same for the whole trough: a condition kept the same.
(c) State the null hypothesis for the test. (1 pt)
Frame The null hypothesis is that the shrimp have …, so any gap between the two counts is due to …
Hint What would the two counts look like if the light made no difference to the shrimp?
- Award 1 point for: no preference (an even split), any difference due to chance.
Slip Writing “the shrimp prefer the light”. That is the claim the test might support; the null hypothesis is no preference.
(d) Calculate the expected count at each end under the null hypothesis. (1 pt)
Frame Under no preference the expected count at each end is … ÷ … = …
Hint Under the null hypothesis, how do 26 shrimp share two ends?
Answer: 13 shrimp (tolerance ±0)
- Award 1 point for: 13 shrimp at each end.
(e) Calculate chi-square for the observed counts against the expected counts, to three significant figures. (1 pt)
Frame For the lit end, = …; for the shaded end, … ; so chi-square = …
Hint The formula sheet gives chi-square; o is each count you were given, and e is your answer to part (d).
Answer: 5.54 (tolerance ±0.005)
Shaded end: .
So .
- Award 1 point for: chi-square = 5.54 (accept 5.53 to 5.55).
(f) Determine whether the null hypothesis is rejected at p = 0.05, and justify the verdict with the two numbers it rests on. (1 pt)
Frame Two ends give … degree(s) of freedom, and the p = 0.05 row reads … there; chi-square is … than that, so the verdict is …
Hint How many degrees of freedom do two ends give? Read the p = 0.05 row in that column and compare your chi-square with it.
The p = 0.05 row reads 3.84 at one degree of freedom.
5.54 is larger than 3.84.
So the null hypothesis is rejected: the counts support a preference for the lit end.
- Award 1 point for: reject the null hypothesis, because 5.54 is larger than 3.84 at one degree of freedom.
Slip Reading the table at two degrees of freedom because there are two ends. Degrees of freedom are one fewer than the number of classes.
A student studies a roundworm of one kind that lives in soil and eats soil bacteria. The bacteria give off a chemical, diacetyl, into the air spaces of the soil. The student puts 100 worms in the middle of a dish, a drop of diacetyl at one edge and a drop of water at the opposite edge, and after 60 minutes counts the worms near each drop and in the middle. The table gives the counts.
(a) Describe what the counts show about the worms’ response to diacetyl. (1 pt)
- Award 1 point for: the worms gather near the diacetyl (most of the 100 end up there), so they detect it and respond to it.
Slip Saying the diacetyl pulled the worms across the dish. The worms moved; the chemical was the change they detected.
(b) Explain how the response you described in part (a) helps a worm survive. (1 pt)
So where the diacetyl is, the bacteria are.
A worm that gathers where the diacetyl is ends up among its food.
So it eats, and it survives.
- Award 1 point for: diacetyl comes from the bacteria the worm eats, so gathering where it is brings the worm to its food and it survives.
Slip Saying the worm likes the smell. The point is what the response does for survival: it brings the worm to food.
(c) A classmate says: “The counts show the worms crawled straight toward the diacetyl, so the movement is a taxis.” Evaluate the classmate’s claim. (1 pt)
A taxis is movement toward the stimulus; a kinesis is a change in speed or turning with no fixed direction.
Worms that slow down and turn less wherever diacetyl is would also pile up near the drop.
So the same counts fit a kinesis, and only the worms’ paths, not their end positions, can settle which it is.
- Award 1 point for: the judgement (the claim is not supported by the counts alone) AND the ground (a kinesis, slowing or turning less near the chemical, would give the same end counts; the paths would have to be watched to tell).
Slip Agreeing because most worms ended up near the diacetyl. Where the animals finish is the same for a taxis and a kinesis; how they move differs.
(d) The student spreads diacetyl evenly over the whole surface of a second dish and tips in 100 worms. Predict where the worms are after 60 minutes, and justify your prediction. (1 pt)
Diacetyl is everywhere, so no part of the dish differs from another in the chemical.
With no difference to detect, the worms have nothing to gather toward, so they spread by chance.
If instead the worms slow down wherever diacetyl is strong, they slow down everywhere, so they stay near the middle, where they were tipped in.
- Award 1 point for EITHER prediction with its matching reason: an even (chance) spread, because the diacetyl is the same everywhere so there is no difference for the worms to detect; OR the worms stay near the middle, because a worm that slows wherever diacetyl is strong slows everywhere, so it stays near where it was tipped in. A prediction paired with the other prediction’s reason does not earn the point.
Slip Predicting that the worms all gather at one edge. With no difference between the parts of the dish, there is nothing to choose between.
APBIO-U08-T81 End-of-topic test: Responses to the Environment
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table from the formula sheet is printed beside it.
An ecologist records where in a meadow the voles feed and how far each vole keeps from the owls’ perch.
Which feature of the study makes it a study in ecology?
- A. ✓ It asks how the voles interact with the owls and with the meadow
- B. It is carried out on wild animals in the open rather than in a laboratoryWhere a study happens does not make it ecology.
Ecology is the study of how living things interact with one another and with their surroundings. - C. It counts whole animals rather than measuring their cellsCounting whole animals is a method, and many branches use it.
Ecology is about the interactions among living things and with their surroundings. - D. It watches the voles without handling themWatching without handling is a method, and many branches use it.
What makes the study ecology is the interactions it asks about.
Why: Ecology is the study of how living things interact with one another and with their surroundings.
The study records how the voles keep away from the owls and where in the meadow they feed.
Those are interactions among living things and with their surroundings.
So the study is ecology.
A student puts tadpoles into containers of water they cannot see out of and counts each tadpole’s tail movements. Half the containers hold plain water; half hold water taken from a tank that had held crayfish, which eat tadpoles. No crayfish is present. The table gives the results.
Which of the following is the stimulus the tadpoles detected?
- A. The opaque walls of the containerThe container was the same for every tadpole, so it cannot be the change that set the two groups apart.
- B. The crayfish in the neighboring tankNo crayfish was present, and the tadpoles could see nothing outside the container.
What reached the tadpoles was the water. - C. ✓ The chemical in the water that had held crayfish
- D. The tadpoles’ slower tail movementsThe slower tail movement is what the tadpoles did about the change.
That is the response, and the stimulus is the change they detected.
Why: A stimulus is the change the organism detects.
The only difference between the groups was the water: plain, or from a tank that had held crayfish.
The tadpoles in the crayfish water moved their tails less.
So the stimulus was the chemical in that water.
A student counts the snails on one desert rock face four times in a day, on the open rock and in the shaded cracks, and records the surface temperature each time. The table gives the counts.
Which additional observation would show that a behavioral response by the snails caused the change in the counts?
- A. The snails in the cracks have a body temperature below the surface temperature at middayA body temperature is a state inside the snail, and a difference in it is not something the snail does.
A behavioral response is something the whole snail does. - B. ✓ Snails the student had marked with a dot of paint crawl into the cracks as the rock warms
- C. More young snails hatch in the cool months than in the hot monthsHatching in the cool months changes how many snails exist, and it happens over months.
The counts shift within one day, so the same snails must have moved. - D. Snails on the hot open rock make proteins that protect their cells from heatMaking protective proteins is a change inside the snail’s body: a physiological response.
A behavioral response is something the whole snail does.
Why: A behavioral response is something the whole organism does.
The counts alone could change because snails moved or because different snails were counted.
Watching marked snails crawl into the cracks as the rock warms shows the whole snail moving in answer to the heat.
So it shows a behavioral response.
On one winter morning a hare’s blood glucose falls, and snow covers the grass it feeds on. Both changes send the hare to dig for food.
Which of the following describes the two changes?
- A. ✓ The fall in blood glucose is internal, the snow is external, and both are stimuli
- B. Only the snow is a stimulus, because the hare can see it and cannot see its blood glucoseA stimulus is a change the organism detects, and detecting is more than seeing.
The hare detects its falling blood glucose from inside. - C. Only the fall in blood glucose is a stimulus, because it is inside the hare’s bodyThe snow is a change in the hare’s surroundings that it detects.
An external change is a stimulus too. - D. Both changes are external, because the snow caused the fall in blood glucoseWhatever caused it, the fall in blood glucose happens inside the hare’s body.
A change inside the body is internal.
Why: A stimulus is a change the organism detects, inside its body or outside it.
Blood glucose is a state of the hare’s own body, so its fall is internal.
Snow lies in the hare’s surroundings, so the snow is external.
The hare detects both and digs, so both are stimuli.
A student puts mites into a dish with a damp half and a dry half and films them for ten minutes.
Which of the following observations would show that the mites’ movement is a kinesis?
- A. Most of the mites are in the damp half by the end of the ten minutesEnding up in the damp half fits a taxis and a kinesis alike.
Where the animals finish does not tell the two apart; how they move does. - B. A mite tipped straight onto the damp half stays thereA mite that stays where conditions suit it fits a taxis and a kinesis alike.
How the mites move in dry air tells the two apart. - C. Every mite is still moving at ten minutesMoving at all tells neither the speed nor the direction.
A kinesis is a change in speed or turning with the stimulus, in no fixed direction. - D. ✓ The mites crawl faster in dry air than in damp air, in no fixed direction
Why: A kinesis is a change in speed or turning with no fixed direction.
Faster in dry air and slower in damp air is a change of speed with the stimulus.
No fixed direction rules out a taxis.
So that observation would show a kinesis.
A grower keeps four potted plants of one kind and watches each for a month.
Which of the following observations is photoperiodism?
- A. A plant on a shelf leans toward the nearest windowLeaning toward the window is a direction of growth toward light.
That is phototropism. - B. A plant grows taller in a warm room than in a cool oneGrowing taller in warmth is a response to temperature, and it is neither a direction of growth nor a timing by day length.
- C. ✓ A plant forms flower buds once the nights pass twelve hours
- D. A plant flowers earlier in a warm spring than in a cold oneFlowering earlier in warmth is timing by temperature.
Photoperiodism is timing by the length of the day or the night.
Why: Photoperiodism is a plant timing its flowering by the length of the day or the night.
The plant forms flower buds once the nights pass twelve hours.
The plant is timing its flowering by the length of the night.
So that observation is photoperiodism.
A student follows one desert lizard through a day, recording the air temperature in the open, where the lizard is, and its body temperature. The table gives the record.
Which of the following best describes what the move under the shrub at 13:00 does for the lizard?
- A. The move warms the lizard’s body further, so it can keep hunting through middayUnder the shrub the lizard’s body stayed at 39 °C while the open air reached 44 °C.
The move stopped the warming; it did not add to it. - B. ✓ The move stops the lizard’s body warming when the air is hottest, so the lizard survives
- C. The move changes nothing, because the lizard’s body follows the air temperature wherever it sitsAt 13:00 the air in the open was 44 °C and the lizard’s body was 39 °C.
Under the shrub, its body did not follow the air. - D. The move gives the lizard the temperature it decided it wantedA response helps because of what it does to the animal’s survival, not because the animal wants it.
The shade kept the lizard’s body from overheating.
Why: In the open the lizard’s body warms with the air.
At 13:00 the air in the open reached 44 °C, hot enough to kill the lizard.
Under the shrub, its body stayed at 39 °C.
So moving into shade stopped its body warming further, and the lizard survived.
Suppose that when a ladybug attacks a pea aphid, the aphid releases a chemical into the air. Researchers blow air over groups of unharmed aphids feeding on a leaf for one minute and count the aphids that drop off the leaf. The table gives the results.
Which of the following is the signal in this exchange between aphids?
- A. The attack by the ladybugThe ladybug’s attack is not an exchange between aphids.
The signal here is what one aphid produces and another aphid detects: the chemical. - B. The leaf the aphids are feeding onThe leaf was there in every trial and was made by no aphid.
A signal is something one organism produces and another detects. - C. The unharmed aphids dropping off the leafDropping off the leaf is what the receiving aphids did after they detected the chemical.
That is the response, and the signal is the chemical. - D. ✓ The chemical the attacked aphid releases
Why: A signal is something one organism produces that another organism detects.
The attacked aphid produced the chemical.
Air carrying that chemical made 85 % of the unharmed aphids drop; ordinary air and air from unharmed aphids made about 5 % drop.
So the chemical is the signal.
Suppose that on a mudflat, a male fiddler crab stands at its burrow and waves its one large claw. Researchers record how many of 20 female crabs approach under four conditions. The table gives the results.
Which conclusion about the male’s signal do the results support?
- A. ✓ A visual signal: a large claw in motion drew the females
- B. A chemical signal: a scent from the large claw drew the femalesA scent would reach the females from a still claw as well as a waving one.
The still large claw drew only 3 of 20 females. - C. A tactile signal: the male’s touch drew the femalesThe females approached from a distance, before any male touched them.
A tactile signal needs contact. - D. An audible signal: the sound of the waving claw drew the femalesThe male that had lost its large claw waved just as often, and only 4 of 20 females approached.
A sound would not depend on the claw’s size.
Why: A visual signal is one the receiver sees.
A waving large claw drew 16 of 20 females.
A still large claw drew 3, and a waving small claw drew 4.
So the females answered what they saw, a large claw in motion, and the signal is visual.
A researcher follows one female deer for two weeks, recording a reproductive hormone in her blood, the number of scent marks she leaves on plants each hour, and the number of males that approach her. The table gives the record.
Which of the following is most likely the cue for the female’s scent-marking?
- A. The scent marks on the plantsThe scent marks are the signal the female sent, and a signal is not the cue that set it off.
- B. The males approaching herThe males approached after the marking rose, and their approaches rose and fell with it.
The approaches are the receivers’ response to the marks. - C. ✓ The rise in the hormone inside her body
- D. The length of the daysThe record has no day lengths, and the marking rose and fell within thirteen days.
The change that rose and fell with the marking was the hormone in her blood.
Why: A cue can come from inside the organism or from its surroundings.
The hormone in the female’s blood rose to day 10 and fell by day 13.
Her scent-marking rose and fell with it.
So the cue was internal: the rise in the hormone inside her.
On an island, some woodpeckers of one kind use a twig to pry insects from bark. Researchers raise 50 young birds with experienced adults and 50 apart from all adults, record which young use the tool at six months, and follow every bird to breeding age. The table gives the results.
Which conclusion do the results support?
- A. Tool use is innate, because every young bird does it whether or not it saw an adultOnly 14 % of the young raised apart used the tool.
A behavior shaped by whether the bird saw adults is learned. - B. The young raised with adults survived better because the adults cared for them, whatever they learnedWithin the group raised with adults, the tool users survived at 88 % and the non-users at 41 %.
Tool use, not the adults’ care, went with survival. - C. Because tool use is learned, selection cannot make it more commonSelection acts on any behavior that raises survival or offspring, innate or learned.
The young that use the tool survive to breed more often, so selection favors it. - D. ✓ Tool use is learned from adults, and the young that use it survive to breed more often
Why: A learned behavior is shaped by the individual’s experience.
92 % of the young raised with adults used the tool, 14 % of those raised apart.
So the young learned it from adults.
In both groups the tool users survived to breed about twice as often as the rest.
Shore crickets carry either an R allele or an r allele. When a hawk’s shadow passes, an R cricket darts into a crevice; an r cricket stays in the open. A researcher stocks three enclosures at an R allele frequency of 0.50: hawks with open crevices, hawks with the crevices filled in, and no hawks with open crevices. After one breeding season she measures the R allele frequency in the eggs laid. The table gives the results.
Which of the following explains why the R allele frequency rose only in the enclosure with hawks and open crevices?
- A. The hawk’s shadow made the r crickets develop the R response before they bredAn allele is inherited, and a passing shadow does not change which alleles a cricket carries.
Only the R carriers darted, and only they survived more often. - B. ✓ The crickets that darted into crevices survived to breed more often, so more of the eggs carried R
- C. Darting into a crevice is good for the crickets as a whole, so every cricket in that enclosure took it upA behavior spreads when its carriers leave more offspring, not because it is good for the group.
Only the R carriers darted, and only they survived to breed more often. - D. The crickets decided to hide from the hawks, so every cricket in that enclosure became an R carrierA behavior spreads because of what it does to survival and offspring, not because the animals decide.
Only the R carriers darted, and the frequency reached 0.71, not 1.
Why: A behavior that raises survival spreads when its carriers leave more offspring.
With hawks and open crevices, R crickets hid and r crickets stayed exposed.
So hawks caught more r crickets, and more R crickets survived to lay eggs.
With crevices filled, or no hawks, hiding saved no one.
In a troop of baboons, one baboon spends part of the morning feeding time picking ticks from another baboon’s fur.
Which additional observation would show that the grooming is cooperative behavior: that it costs the groomer now and helps another baboon?
- A. ✓ The groomer eats less that morning, and the groomed baboon carries fewer ticks afterwards
- B. The groomer swallows every tick it picks, and the groomed baboon scratches as often afterwards as beforeA groomer that feeds on the ticks pays no cost now.
A baboon that scratches as often as before was not helped. - C. Several baboons in the troop groom at the same momentDoing something at the same moment is not the test.
Cooperative behavior costs the individual now and helps the others. - D. The groomed baboon sits still and lifts its arm as soon as the grooming startsSitting still and lifting an arm shows the groomed baboon responding to the touch.
It shows no cost to the groomer and no help to another.
Why: Cooperative behavior costs the individual now and helps the others in its group.
Eating less that morning is a cost to the groomer now.
Fewer ticks on the other baboon is a help to another in its group.
So that observation shows both halves, and the grooming is cooperative.
Suppose that in a roost of vampire bats, a bat that has fed well sometimes brings up part of its blood meal for a roost-mate that found nothing to feed on. Over a year, researchers record for each of 20 bats how many nights it shared and how often a roost-mate fed it when it came back with nothing. The table gives the results.
Which conclusion do the results support?
- A. Sharing a meal is a one-way gift that the giver never gets backThe bats that shared were fed on 8 of every 10 hungry nights.
What they gave came back to them. - B. A bat is fed on a hungry night by chance, whether or not it has sharedBats that shared were fed on 8 of 10 hungry nights; bats that never shared on 1 of 10.
Being fed followed having shared. - C. ✓ A bat that shares its meal is fed in turn when it comes back with nothing
- D. A bat that never shares is fed more often, because it keeps its own mealsBats that never shared were fed on 1 of every 10 hungry nights, the least.
Keeping its own meals left a bat unfed when it failed to feed.
Why: Cooperative behavior costs the individual now and helps others in its group.
A sharer gives up part of its meal that night.
Sharers were fed on 8 of every 10 hungry nights; non-sharers on 1 of 10.
So a sharer gets help back when its own hungry night comes.
A researcher watches hawks attack small birds of one kind on a plain: 30 attacks on a flock of 50 birds feeding together, and 30 attacks on 50 birds of the same kind feeding scattered, each alone. The table gives the number of birds caught.
Which of the following explains why the hawks caught fewer birds from the flock?
- A. The birds in the flock are larger and stronger than the scattered birdsThe two groups are birds of one kind feeding on one plain.
What differs is that one group feeds together and the other apart. - B. The hawk cannot see a flock of birds against the groundFifty birds together are easier to see than one bird alone.
The hawk sees the flock; the flock sees the hawk sooner. - C. Each bird in the flock decides to protect the others, so it stays to warn themThe flock survives because of what feeding together does, not because each bird decides anything.
Fifty pairs of eyes notice a hawk before one pair does. - D. ✓ More eyes in the flock spot the hawk sooner, so the birds fly up before it strikes
Why: In a group, more eyes watch for a predator.
One of fifty birds spots the hawk sooner than a lone bird does.
When it flies up, the others fly up with it, before the hawk strikes.
So the hawk caught 2 from the flock and 11 from the scattered birds.
In a wood, young female birds of one kind spend their first year feeding the chicks at their parents’ nest, their own younger brothers and sisters. Researchers remove the helpers from half the nests. The table gives the chicks that left each set of nests.
Which of the following explains how the helping spreads copies of the helpers’ own alleles?
- A. Each chick a helper feeds carries a copy of an allele taken from the helper’s bodyFood carries no alleles.
A chick shares the helper’s alleles because the two have the same parents. - B. ✓ The chicks share many of each helper’s alleles, and more of them survive to breed
- C. The helper is fed by these chicks in later years, so she leaves more offspring herselfThe table counts chicks that left the nest, not what the helper receives later.
The alleles spread through the chicks themselves. - D. More chicks in the wood is good for the species, so the species keeps the helpingKin selection is about shared alleles, not the species.
The chicks carry copies of the helper’s alleles because they are her brothers and sisters.
Why: Kin selection favors helping relatives that share the helper’s alleles.
The chicks are the helper’s brothers and sisters, so they carry many copies of her alleles.
Nests with helpers sent out 62 chicks; nests without, 36.
More of those chicks survive to breed, so the copies spread.
A student tips 22 millipedes into a tray with a damp cloth on one half and a dry cloth on the other. After two minutes 14 millipedes sit on the damp half and 8 on the dry half, as drawn below. The null hypothesis is that the millipedes have no preference.
Which of the following is chi-square for these counts?
- A. 0.550.55 adds the gaps 3 and 3 and divides by 11, without squaring them.
Each gap must be squared before it is divided by e. - B. 0.820.82 is one half’s term alone, .
Chi-square adds the term for every half. - C. ✓ 1.64
- D. 1818 adds the two squared gaps, 9 and 9, and stops.
Each squared gap must be divided by its expected count before the terms are added.
Why: Under no preference, e = 22 ÷ 2 = 11 on each half.
The damp half’s term is .
The dry half’s term is .
So .
A student tests whether ground beetles of one kind gather under cover. She lays leaf litter on one half of a shallow tray and bare sand on the other half, tips 16 beetles into the middle, and after five minutes counts the beetles on each half. She also sets up a second tray with leaf litter on both halves, 16 beetles, counted the same way. The counts from the first tray are drawn below.
(a) Identify the independent variable and the dependent variable in the student’s test. (1 pt)
The dependent variable is the number of beetles counted on each half, the quantity she measured.
- Award 1 point for: the cover on a half (leaf litter or bare sand) as the independent variable AND the number of beetles on each half as the dependent variable.
Slip Naming the number of beetles tipped in as a variable. Sixteen went into every tray; that is a condition kept the same.
(b) State the null hypothesis for the first tray. (1 pt)
- Award 1 point for: the beetles have no preference (an even split), and any difference from it is due to chance. Accept with or without: 8 and 8.
Slip Writing “the beetles prefer leaf litter” as the null hypothesis. The null is no preference; the preference is what the test may reject it in favor of.
(c) Justify the student’s decision to set up the second tray with leaf litter on both halves. (1 pt)
The second tray shows how 16 beetles spread when nothing is chosen.
So a lopsided count in the first tray can be put down to the cover, and not to the beetles bunching up on their own.
- Award 1 point for: the second tray has no difference to choose between, so it shows the spread with no preference; a lopsided count in the first tray can then be credited to the cover. Accept: it is the control, with what it shows (the spread with no difference to choose between).
Slip Calling the second tray a repeat that makes the result more reliable. It differs from the first tray in one way: it offers no choice.
(d) Calculate chi-square for the counts in the first tray against the null hypothesis. (1 pt)
Answer: 9.0 (tolerance ±0.005)
Leaf litter: .
Bare sand: .
So .
- Award 1 point for: chi-square = 9.0 (accept 9).
(e) Determine whether the null hypothesis is rejected at p = 0.05, and justify the verdict with the two numbers it rests on. (1 pt)
The p = 0.05 row reads 3.84 at one degree of freedom.
9.0 is larger than 3.84.
So the null hypothesis is rejected: a gap this large is too large to be chance, and the counts support a preference for the leaf litter.
- Award 1 point for: reject the null hypothesis, because chi-square (9.0) is larger than the critical value (3.84) at one degree of freedom. Accept: reject, so the data support a preference.
Slip Writing “accept the preference” as the verdict. The verdict is about the null hypothesis: reject it, or fail to reject it.
In a clear stream, the males of one kind of small fish turn bright red on the belly each spring and swim in short darts in front of the females. A female swims toward the male with the reddest belly and lays her eggs in his nest, where he fertilizes them. Fish of this kind mate only at a nest, and the males’ nests lie several meters apart along the stream bed. In the clear water a female can see a red belly from ten meters away. A quarry then opens upstream, and from that spring on the stream is cloudy with fine clay through the whole breeding season; a female can now see a male’s belly only from within half a meter.
(a) Describe the job the red belly does for the male that shows it. (1 pt)
- Award 1 point for: the red belly wins a mate (draws a female to the male’s nest).
Slip Saying the red belly marks the male’s territory. A territory mark keeps rivals out; this display draws females in.
(b) Explain how the red belly gives some males more offspring than others. (1 pt)
So the reddest males are reached by more females.
More females lay eggs in those males’ nests, and those males fertilize them.
So the reddest males leave more offspring than the duller males.
- Award 1 point for: females move toward the reddest males, so those males mate with more females and leave more offspring.
Slip Stopping at “the red belly attracts females”. The point needs the step from attracting females to more eggs fertilized, and so more offspring.
(c) Predict how the number of young fish in the stream changes in the springs after the quarry opens. (1 pt)
- Award 1 point for: the number of young fish falls (is smaller than before the quarry).
Slip Predicting no change because the males still turn red. The signal is sent as before, but it no longer reaches the receivers.
(d) Justify your prediction. (1 pt)
So the red belly reaches fewer females.
Fewer females swim to a male’s nest, so fewer pairs mate.
Fewer eggs are laid and fertilized, so fewer young fish hatch.
So the number of young fish in the stream falls.
- Award 1 point for: the chain in order — the cloudy water stops the signal reaching the females → fewer females reach a nest and mate → fewer eggs fertilized → fewer young fish. Accept the chain with the mating step and the egg step written as one link.
Slip Jumping from “cloudy water” to “fewer fish”. Each link has to be stated: the signal fails to reach the receiver, so the receiver’s behavior changes, so the matings fall, so the young fall.
APBIO-U08-L10 What the energy is used for
Photo: David Perez, Wikimedia Commons, CC BY 3.0 (resized).
Here is a field mouse in January. It eats seeds every day. It does not grow, and it has no litter.
The same mouse in June eats the same seeds. It grows heavier, and it raises a litter. Where did January’s energy go?
Unit 8 · Ecology
1Four things the energy is used for
Every living thing needs a continuous input of energy.
Which two sources can that input come from?
- A. ✓ Light, or chemical compounds such as the sugar in food
- B. Warmth from the surroundings, or lightWarmth is not an input a living thing can use.
A warm animal still needs food, and a warm plant still needs light. - C. Water, or chemical compounds such as the sugar in foodWater carries no chemical energy a living thing can release.
The input comes from light or from chemical compounds.
Why: A green plant in the light takes energy in as light.
A mouse eating seeds takes energy in as food.
So the input comes from light, or from chemical compounds such as the sugar in food.
Video: Watch: Four things the energy is used for
The field mouse in January; the chemical energy of its seeds arriving, and four labeled arrows leaving it: keeping its parts in order, growing, reproducing, keeping its inside steady; the heat leaving at every step.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L10a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L10a.mp4
What does an organism use its energy for, and what decides whether any is left over?
Every organism uses energy for four things: keeping its parts in order, growing, reproducing and keeping its inside steady.
In January the mouse uses almost all of its energy for staying warm and for moving to find seeds. So nothing is left for growth or a litter.
When energy in beats energy out, the organism stores the surplus as fat or starch. Then it grows, and it breeds more.
When energy out beats energy in, the organism loses mass and breeds less. If the loss goes on, it dies.
The energy the mouse uses is not destroyed. It leaves the mouse as heat, as the second law of thermodynamics in Unit 3 said it must.
Here is the mouse in January. Every day it eats seeds.
The chemical energy in those seeds is the mouse’s energy in.
The mouse uses that energy for four things:
- Keeping its parts in order. Its cells rebuild worn-out molecules and pump ions across their membranes. Unit 3 said keeping that order needs energy.
- Growing. The mouse builds new tissue from the energy and the matter in its food.
- Reproducing. The mouse builds a litter of young inside its body, then feeds them with milk.
- Keeping its inside steady. The mouse’s body stays warm while the January air is cold.
Keeping the conditions inside the body steady while the outside changes is homeostasis, the word Unit 1 gave it. Staying warm in cold air is one case of homeostasis.
A small animal loses heat quickly. So in January the mouse uses most of its energy replacing the heat it loses.
The energy the mouse uses does not vanish. At every step, some of it leaves the mouse as heat.
That heat warms the air around the mouse. The mouse cannot get the heat back, so it must keep eating.
What you are expected to know State the four things an organism uses energy for: keeping its parts in order, growing, reproducing and keeping its inside steady.
A mouse shivers on a cold night.
Which of the four things the mouse uses energy for is this?
- A. Keeping its parts in orderShivering rebuilds nothing.
Shivering makes heat, and the heat keeps the mouse’s body temperature steady. - B. GrowingShivering adds no new tissue.
Shivering makes heat, and the heat keeps the mouse’s body temperature steady. - C. ReproducingShivering makes no young.
Shivering makes heat, and the heat keeps the mouse’s body temperature steady. - D. ✓ Keeping its inside steady
Why: Shivering muscles release heat.
The heat keeps the mouse’s body temperature steady while the air is cold.
So shivering is keeping its inside steady.
A tadpole gets longer and heavier each week.
Which of the four things the tadpole uses energy for is this?
- A. Keeping its parts in orderGetting heavier adds new tissue; rebuilding worn-out molecules adds none.
- B. ✓ Growing
- C. ReproducingA tadpole makes no young.
Getting longer and heavier is building new tissue. - D. Keeping its inside steadyGetting heavier is not a steady condition.
Getting longer and heavier is building new tissue.
Why: The tadpole builds new tissue each week.
Building new tissue is growing.
A bird lays a clutch of eggs.
Which of the four things the bird uses energy for is this?
- A. Keeping its parts in orderLaying eggs rebuilds nothing in the bird’s own body.
The eggs are the bird’s young. - B. GrowingLaying eggs adds no tissue to the bird’s own body.
The eggs are the bird’s young. - C. ✓ Reproducing
- D. Keeping its inside steadyLaying eggs keeps no condition inside the bird steady.
The eggs are the bird’s young.
Why: The eggs are the bird’s young.
Building and laying them is reproducing.
A liver cell rebuilds a worn-out protein.
Which of the four things the cell uses energy for is this?
- A. ✓ Keeping its parts in order
- B. GrowingRebuilding a worn-out protein adds no new tissue.
It replaces a part the cell already had. - C. ReproducingRebuilding a protein makes no young.
It replaces a part the cell already had. - D. Keeping its inside steadyRebuilding a protein keeps no condition steady.
It replaces a part the cell already had.
Why: The cell replaces a part it already had.
Replacing worn-out parts is keeping its parts in order.
A student says: “The mouse used up the energy in its seeds, so that energy is gone.”
Is the student correct?
- A. Yes: the mouse used the energy up, so it is goneThe energy left the mouse as heat.
The heat is still energy; it has spread into the air. - B. ✓ No: the energy left the mouse as heat
Why: The mouse used the energy for staying warm and moving.
At every step, some of that energy left the mouse as heat.
The heat is still energy, spread into the air.
So the energy is not gone: it left the mouse as heat.
23When energy in beats energy out
Video: Watch: When energy in beats energy out
The two budget bars, January over June, drawn to one scale; June’s energy out shorter, the surplus filling as fat, new tissue and a litter; the table filling row by row.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L10b.mp4
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Now consider the mouse in June. Suppose it takes in 60 kJ per day from its seeds, the same as in January.
In June the air is warm. So the mouse uses far less energy keeping its inside warm.
In January the mouse used 55 kJ per day. In June it uses 40 kJ per day.
In June, energy in beats energy out. The 20 kJ per day that the mouse does not use is its surplus.
The mouse stores part of the surplus as fat. A plant with a surplus stores it as starch.
The mouse builds the rest of the surplus into new tissue. So the mouse grows heavier.
A mouse with a surplus also has energy to spare for a litter. So it produces more offspring.
In January the surplus was only 5 kJ per day. That 5 kJ per day went into a little fat, and nothing was left for growth or a litter.
When energy in equals energy out, there is no surplus. The organism stores nothing new, its mass stays the same, and it produces few offspring or none.
The table below compares two of the cases. For each it gives what is stored, what happens to the organism’s mass, and how many offspring it produces.
What you are expected to know Predict what happens to an organism whose energy in beats its energy out: it stores energy as fat or starch, grows, and produces more offspring.
Suppose a trout takes in more energy from its food each day than it uses.
What happens to the trout’s mass?
- A. It loses massA trout loses mass only when it uses more energy than it takes in.
This trout takes in more than it uses. - B. It stays the sameIts mass stays the same only when energy in equals energy out.
This trout takes in more than it uses. - C. ✓ It gains mass
Why: The trout takes in more energy than it uses.
The energy it does not use is a surplus.
The trout stores the surplus as fat and builds new tissue.
So the trout gains mass.
Suppose a female deer eats well all autumn and stores fat.
Compared with a thin female deer, how many offspring is she likely to produce?
- A. FewerFewer offspring follow when energy out beats energy in.
This deer has stored a surplus. - B. The same numberThe same number would follow if the two deer had the same surplus.
This deer has stored a surplus; the thin deer has not. - C. ✓ More
Why: The well-fed deer took in more energy than she used.
She stored the surplus as fat.
Stored energy is energy to spare for young.
So she is likely to produce more offspring than the thin deer.
Suppose a potato plant makes more sugar in photosynthesis each day than it respires.
What does the plant do with the sugar left over after respiration?
- A. ✓ It stores the sugar as starch
- B. It releases the sugar’s energy as heatHeat leaves from the sugar the plant respires.
The sugar it does not respire is a surplus, and a plant stores its surplus as starch.
Why: The plant makes more sugar than it respires.
The sugar it does not respire is a surplus.
A plant stores its surplus as starch.
So the plant stores that sugar as starch.
39When energy out beats energy in
Video: Watch: When energy out beats energy in
A thin elk in deep snow; the mouse’s hard-January bar, the energy in shorter than the energy out, the dashed part taken from its own fat; the table’s third row filling.
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Now consider a mouse in a hard January. Suppose it gets only 45 kJ per day from its seeds.
It still uses 55 kJ per day staying warm and moving. So energy out beats energy in by 10 kJ per day.
The mouse takes the missing 10 kJ per day from its own store: it breaks down its fat. So it loses mass.
With no surplus, the mouse has nothing to spare for a litter. So it produces fewer offspring, or none.
A short loss is normal: every animal uses more energy than it takes in while it sleeps and eats nothing.
It replaces the loss when it wakes and eats.
If the loss continues for weeks, the mouse’s fat is gone. Then the mouse breaks down its muscle, and in the end it dies.
In one study of elk, biologists measured each female’s body fat at the end of autumn. Females with less than 6 % body fat were far less likely to become pregnant than females with more than 10 % body fat.
A female elk with too little stored energy does not become pregnant. So a thin female produces fewer offspring.
The table below compares all three cases. For each it gives what is stored, what happens to the organism’s mass, and how many offspring it produces.
What you are expected to know Predict what happens to an organism whose energy out beats its energy in: it loses mass, produces fewer offspring and, if the loss continues, dies.
Suppose a hedgehog uses more energy each day than it takes in from its food.
What happens to the hedgehog’s mass?
- A. ✓ It loses mass
- B. It stays the sameIts mass stays the same only when energy in equals energy out.
This hedgehog uses more than it takes in. - C. It gains massA hedgehog gains mass only when it takes in more energy than it uses.
This hedgehog uses more than it takes in.
Why: The hedgehog uses more energy than it takes in.
It takes the missing energy from its own fat.
So the hedgehog loses mass.
The two bars below show one animal’s energy in and energy out for a day, drawn to one scale.
What happens to the animal’s mass?
- A. It loses massThe energy out bar is shorter than the energy in bar.
The animal uses less than it takes in. - B. It stays the sameThe two bars are not the same length.
The animal takes in more than it uses. - C. ✓ It gains mass
Why: The energy in bar is longer than the energy out bar.
So the animal takes in more energy than it uses.
It stores the surplus and builds new tissue.
So the animal gains mass.
The two bars below show one animal’s energy in and energy out for a day, drawn to one scale.
What happens to the animal’s mass?
- A. ✓ It loses mass
- B. It stays the sameThe two bars are not the same length.
The animal uses more than it takes in. - C. It gains massThe energy out bar is longer than the energy in bar.
The animal uses more than it takes in.
Why: The energy out bar is longer than the energy in bar.
So the animal uses more energy than it takes in.
It takes the missing energy from its own stores.
So the animal loses mass.
A student says: “An animal that uses more energy than it takes in for one night will die.”
Is the student correct?
- A. Yes: one night’s loss is enough to kill itEvery animal uses more energy than it takes in while it sleeps and eats nothing.
It takes the difference from its fat and replaces it when it eats again. - B. ✓ No: a short loss comes from its fat and is replaced
Why: While it sleeps, the animal eats nothing but still uses energy.
It takes the difference from its fat, a small loss.
When it wakes, it eats and replaces the loss.
Death follows only a loss that continues for a long time.
Suppose a female elk enters winter thin, with little body fat, and deep snow makes food scarce.
Compared with a fat female elk, how likely is she to produce a calf in spring?
- A. ✓ Less likely
- B. As likelyThe thin elk has less stored energy than the fat elk.
Energy to spare for a calf comes from stored energy. - C. More likelyMore offspring follow when energy in beats energy out.
The thin elk has little stored energy and little food.
Why: The thin elk has little stored energy.
Scarce food means her energy out beats her energy in all winter.
She has no energy to spare for a calf.
So she is less likely than the fat elk to produce a calf.
Here is the mouse again, in January and in June.
In January, 55 of its 60 kJ per day go on staying warm and moving. Nothing is left for growth or a litter.
In June, 40 of its 60 kJ per day go on staying warm and moving. The other 20 kJ per day go into new tissue and a litter.
60Mixed practice mixed practice
Suppose a robin in a cold, wet spring takes in less energy each day than it uses.
How many young is it likely to raise?
- A. ✓ Fewer than usual
- B. As many as usualAs many as usual would follow if the robin took in as much as it used.
This robin uses more than it takes in. - C. More than usualMore young follow when energy in beats energy out.
This robin uses more than it takes in.
Why: The robin uses more energy than it takes in.
It has no surplus to spare for young.
So it raises fewer young than usual.
A caterpillar sheds its skin and gets bigger each week.
Which of the four things the caterpillar uses energy for is this?
- A. Keeping its parts in orderGetting bigger adds new tissue; rebuilding worn-out molecules adds none.
- B. ✓ Growing
- C. ReproducingA caterpillar makes no young.
Getting bigger is building new tissue. - D. Keeping its inside steadyGetting bigger is not a steady condition.
Getting bigger is building new tissue.
Why: The caterpillar builds new tissue each week.
Building new tissue is growing.
Suppose a sunflower makes more sugar each day than it respires.
What happens to the sunflower’s mass?
- A. It loses massA plant loses mass only when it respires more sugar than it makes.
This sunflower makes more than it respires. - B. It stays the sameIts mass stays the same only when the sugar made equals the sugar respired.
This sunflower makes more than it respires. - C. ✓ It gains mass
Why: The sunflower makes more sugar than it respires.
The sugar it does not respire is a surplus.
The plant stores the surplus as starch and builds new tissue.
So the sunflower gains mass.
A student says: “A mouse that stays the same mass all winter has taken in exactly as much energy as it used.”
Is the student correct?
- A. ✓ Yes: its energy in equals its energy out
- B. No: a mouse of steady mass has used no energyA mouse of steady mass still uses energy staying warm and moving.
That energy leaves as heat, and it equals the energy taken in.
Why: A mouse that stays the same mass takes in as much energy as it uses.
It still uses energy staying warm and moving.
That energy leaves the mouse as heat.
So its energy in equals its energy out.
Suppose a deer uses more energy than it takes in for six weeks of deep snow.
What happens to the deer’s fat over the six weeks?
- A. ✓ The deer breaks down its fat
- B. The deer’s fat stays the sameIts fat stays the same only when energy in equals energy out.
This deer uses more than it takes in. - C. The deer stores more fatA deer stores more fat only when energy in beats energy out.
This deer uses more than it takes in.
Why: The deer uses more energy than it takes in.
It takes the missing energy from its own store.
So the deer breaks down its fat.
A person’s body stays at 37 °C on a cold day.
Which of the four things the person uses energy for is this?
- A. Keeping its parts in orderHolding a temperature rebuilds no worn-out part.
Body temperature is a condition inside the body, kept steady. - B. GrowingHolding a temperature adds no new tissue.
Body temperature is a condition inside the body, kept steady. - C. ReproducingHolding a temperature makes no young.
Body temperature is a condition inside the body, kept steady. - D. ✓ Keeping its inside steady
Why: Body temperature is a condition inside the body.
The person’s body keeps it steady while the outside is cold.
So this is keeping its inside steady.
Suppose a hare’s food becomes scarce in a hard winter. For three months the hare takes in 25 % less energy each day than it did in autumn. It uses the same amount of energy each day as it did in autumn.
(a) Predict two changes in the hare over the three months. (2 pt)
Frame Over the three months the hare …
It also produces fewer offspring, or none.
- Award 1 point for: the hare loses mass (or breaks down its fat, or its stores).
- Award 1 point for: the hare produces fewer offspring, or none (or, if the loss continues long enough, it dies).
(b) Justify your predictions using the hare’s energy in and energy out. (1 pt)
Frame The hare changes in these ways because …
The hare takes the missing energy from its own fat, so it loses mass.
With no surplus, it has no energy to spare for young, so it produces fewer offspring.
- Award 1 point for: energy out exceeds energy in, so the hare draws on its own stores (loses mass) and has no surplus for offspring.
APBIO-U08-L10B Gain or loss: the energy budget
Photo: David Perez, Wikimedia Commons, CC BY 3.0 (resized).
Here is the mouse again. In January it takes in 60 kJ per day from its seeds, and it uses 55 kJ per day staying warm and moving. In June it takes in 60 kJ per day, and it uses 40 kJ per day.
How much is left each day, and what happens to it?
Unit 8 · Ecology
1Net energy: energy in minus energy out
An organism’s energy out is greater than its energy in for a whole season.
What happens to the organism’s mass over the season?
- A. The organism gains massAn organism gains mass when its energy in is greater than its energy out.
Here energy out is the larger. - B. The organism’s mass stays the sameAn organism’s mass stays the same when energy in equals energy out.
Here energy out is the larger. - C. ✓ The organism loses mass
Why: Energy out is greater than energy in.
So the organism takes the missing energy from its own stores.
It breaks down its fat, so it loses mass.
Video: Watch: Net energy: energy in minus energy out
The mouse’s two budgets as two-bar drawings, energy in over energy out; the equation written on its own line; the working for January appearing line by line to +5 kJ per day and a little fat drawn; the working for June to +20 kJ per day and a litter drawn; then a hard January, the out bar longer, the working to −10 kJ per day and the fat store shrinking.
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How do you work out whether an organism is gaining or losing?
You subtract the energy it uses from the energy it takes in. What is left is its net energy, in kilojoules per day.
net energy: what is left over each day, in kJ per day
energy in: the chemical energy in the food eaten each day, in kJ per day
energy out: the energy used each day on the four things, in kJ per day
In January the mouse’s net energy is +5 kJ per day. That small surplus goes into a little fat.
In June the mouse’s net energy is +20 kJ per day. That larger surplus is enough for growth and a litter.
A negative net energy means the organism uses more energy than it takes in. It takes the difference from its own stores and loses mass.
Write down the values, write down the equation, substitute, and state the answer with its unit and its sign. That is the whole routine.
Here is the mouse again in January. Suppose it takes in 60 kJ per day from its seeds, and it uses 55 kJ per day staying warm and moving.
The energy in is the chemical energy in the seeds the mouse eats each day.
The energy out is the energy the mouse uses each day on the four things: keeping its parts in order, growing, reproducing and keeping its inside steady.
The drawing sets the two bars to one scale. The energy in bar is a little longer, and the stretch beyond the end of the energy out bar is what is left over.
That leftover is the net energy. You find it by subtracting the energy out from the energy in, as the equation below says.
net energy: what is left over each day, in kJ per day
energy in: the chemical energy in the food eaten each day, in kJ per day
energy out: the energy used each day on the four things, in kJ per day
Both values are in kilojoules per day. So the net energy comes out in kilojoules per day too, as the line below shows.
kilojoules per day minus kilojoules per day leaves kilojoules per day: the net energy carries the same unit as the two values it comes from
In January the mouse takes in 60 kJ per day and uses 55 kJ per day. Calculate its net energy.
The net energy is positive. So energy in beats energy out, and the mouse stores the 5 kJ per day as a little fat.
Now consider the mouse in June. Suppose it still takes in 60 kJ per day, but in the warm air it uses only 40 kJ per day.
In June the mouse takes in 60 kJ per day and uses 40 kJ per day. Calculate its net energy.
The net energy is positive again, and it is four times January’s. So the mouse builds new tissue and raises a litter.
The larger the positive net energy, the more an organism can grow and the more offspring it can raise.
Now consider a mouse in a hard January. Suppose it gets only 45 kJ per day from its seeds, and it still uses 55 kJ per day staying warm and moving.
This time the energy out bar is the longer one. The stretch beyond the end of the energy in bar is energy the mouse uses but does not take in.
In a hard January the mouse takes in 45 kJ per day and uses 55 kJ per day. Calculate its net energy.
The net energy is negative. So energy out beats energy in, and the mouse takes the missing 10 kJ per day from its own fat.
The mouse breaks down its fat to make up the difference. So it loses mass.
The sign of the net energy tells you which way the organism’s mass goes. The table below sets a positive, a zero and a negative net energy beside what each one means for the mass.
A net energy of zero is the boundary case: energy in equals energy out, so the organism stores nothing new and its mass stays the same.
What you are expected to know Calculate an organism’s net energy from its energy in and its energy out, in kilojoules per day, and state from its sign whether the organism gains or loses mass.
An organism’s energy in and energy out for one day are both known, in kilojoules per day.
Which of the following gives the organism’s net energy?
- A. Energy out minus energy inEnergy out minus energy in gives the right size with the wrong sign.
A surplus would come out negative. - B. ✓ Energy in minus energy out
- C. Energy in plus energy outAdding the two values does not say which is larger.
The net energy is what is left after the energy used is taken away.
Why: The net energy is what is left over each day.
What is left over is the energy taken in with the energy used taken away.
So net energy is energy in minus energy out.
Suppose a house sparrow in spring takes in 110 kJ per day from seeds and insects, and it uses 95 kJ per day flying, keeping warm and feeding its chicks.
Substitute the values into the equation and calculate the sparrow’s net energy, with its sign.
Part 1. Write down the values in the question. What is the sparrow’s energy in?
Answer: 110 kJ per day (tolerance ±0)
Part 2. What is the sparrow’s energy out?
Answer: 95 kJ per day (tolerance ±0)
Answer: 15 kJ per day (tolerance ±0)
A house sparrow in spring takes in 110 kJ per day and uses 95 kJ per day. Its net energy is +15 kJ per day.
What happens to the sparrow’s mass?
- A. ✓ The sparrow gains mass
- B. The sparrow’s mass stays the sameA net energy of zero would keep the mass the same.
Here the net energy is +15 kJ per day, not zero. - C. The sparrow loses massA negative net energy means a loss of mass.
Here the net energy is positive.
Why: The net energy is positive.
So energy in beats energy out.
The sparrow stores the surplus or builds it into new tissue, so it gains mass.
Suppose a hedgehog in late autumn takes in 260 kJ per day from its food, and it uses 330 kJ per day keeping warm and searching for food.
Calculate the hedgehog’s net energy, with its sign.
Answer: -70 kJ per day (tolerance ±0)
A hedgehog in late autumn takes in 260 kJ per day from its food and uses 330 kJ per day. Its net energy is −70 kJ per day.
What happens to the hedgehog’s mass?
- A. The hedgehog gains massA positive net energy means a gain of mass.
Here the net energy is negative. - B. The hedgehog’s mass stays the sameA net energy of zero would keep the mass the same.
Here the net energy is −70 kJ per day, not zero. - C. ✓ The hedgehog loses mass
Why: The net energy is negative.
So energy out beats energy in.
The hedgehog takes the missing energy from its own fat, so it loses mass.
Here is the mouse again, with its two budgets drawn to one scale.
In January its net energy is +5 kJ per day, stored as a little fat.
In June its net energy is +20 kJ per day, enough for growth and a litter.
38Numeric practice: net energy mixed practice
Suppose a rabbit takes in 1 200 kJ per day from grass, and it uses 950 kJ per day.
Calculate the rabbit’s net energy, with its sign.
Answer: 250 kJ per day (tolerance ±0)
The two bars below show a toad’s energy in and energy out for one day, drawn to one scale.
What happens to the toad’s mass?
- A. ✓ The toad gains mass
- B. The toad’s mass stays the sameThe two bars are not the same length.
The energy in bar is the longer one. - C. The toad loses massThe energy out bar is shorter than the energy in bar.
The toad uses less energy than it takes in.
Why: The energy in bar is longer than the energy out bar.
So energy in beats energy out, and the net energy is positive.
The toad stores the surplus, so it gains mass.
Suppose a lizard in summer takes in 16 kJ per day from insects, and it uses 9 kJ per day.
Calculate the lizard’s net energy, with its sign.
Answer: 7 kJ per day (tolerance ±0)
Suppose a gray squirrel in autumn uses 520 kJ per day, and it takes in 640 kJ per day from acorns.
Calculate the squirrel’s net energy, with its sign.
Answer: 120 kJ per day (tolerance ±0)
Suppose a vole in a cold week takes in 48 kJ per day from roots and seeds, and it uses 61 kJ per day.
Calculate the vole’s net energy, with its sign.
Answer: -13 kJ per day (tolerance ±0)
Suppose a deer in deep snow takes in 14 000 kJ per day from twigs and bark, and it uses 21 000 kJ per day keeping warm and wading through the snow.
Calculate the deer’s net energy, with its sign.
Answer: -7000 kJ per day (tolerance ±0)
The two bars below show a snail’s energy in and energy out for one day, drawn to one scale.
What happens to the snail’s mass?
- A. The snail gains massA gain of mass needs the energy in bar to be the longer one.
Here the two bars are the same length. - B. ✓ The snail’s mass stays the same
- C. The snail loses massA loss of mass needs the energy out bar to be the longer one.
Here the two bars are the same length.
Why: The two bars are the same length.
So energy in equals energy out, and the net energy is zero.
The snail stores nothing new, so its mass stays the same.
APBIO-U08-L11 Warm from inside, warm from outside
Photo: Nicholas Stadie, Wikimedia Commons, CC BY-SA 3.0 (cropped and resized).
Here are a lizard and a mouse of the same mass, in the same desert at dawn. The mouse is already warm. The lizard is cold, and it crawls onto a sunlit rock.
By noon both animals are warm. But only one of them used up food to get warm. Which one, and where did the other one’s warmth come from?
Unit 8 · Ecology
1Warm from inside, or warm from outside
In Unit 3, a newborn baby’s brown fat kept it warm.
What does it mean to say a mammal is endothermic?
- A. Taking the body’s temperature from the surroundingsA fish, a frog or a lizard takes its temperature from its surroundings.
That is the opposite of endothermic. - B. ✓ Keeping the body warm from within, using heat the animal’s own cells release
- C. Keeping the body cool by giving heat to the surroundingsEndo means inside and therm means heat.
The heat is made inside, and it keeps the body warm.
Why: Endo means inside and therm means heat.
An endothermic animal keeps its body warm from within, using heat its own cells release.
Video: Watch: Warm from inside, or warm from outside
The lizard and the mouse at dawn, a thermometer beside each; the mouse’s reading already high, the lizard’s rising only as the sun reaches its rock; the two names arriving with the table.
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Where does an animal’s warmth come from, and does the animal have to eat for it?
The mouse keeps its body warm with heat from its own respiration. Its body temperature stays steady while the desert air changes.
The lizard’s body temperature follows its surroundings. So the lizard warms by what it does: onto the rock at dawn, under the rock at noon, in a heap with other lizards on a cold night.
The mouse’s heat is made by respiring food. Food respired for heat is not built into new tissue.
So the mouse must eat more per gram of body than a lizard of the same size.
Here is the mouse at dawn. The desert air is cold, and the mouse is warm.
All night the mouse’s cells respired food. Some of the energy released by that respiration spread out as heat.
That heat kept the mouse’s body warm. Unit 3 called the mouse endothermic: it keeps its body warm from within, using heat its own cells release.
An animal that keeps its body warm with heat its own body makes is called an . Endo means inside, and therm means heat.
An endotherm is Unit 3’s endothermic animal, named as a noun. The mouse is an endotherm.
Now consider the lizard at dawn. The air is cold, and the lizard is cold too: its body is at the temperature of the sand it lies on.
The lizard crawls onto the sunlit rock. Heat passes from the sun and the warm rock into the lizard’s body, and the lizard warms.
An animal whose body temperature follows its surroundings is called an . Ecto means outside: the heat that warms it comes from outside its body.
You may have heard a lizard called cold-blooded. At noon the lizard on its rock is warmer than you are.
So cold-blooded is the wrong word. An ectotherm’s body is cold at dawn and hot at noon, because its body temperature follows its surroundings.
Mammals and birds are endotherms. Reptiles, amphibians, most fish and insects are ectotherms.
For example, a mouse at a cold dawn is warm. This is an endotherm, because its own body makes the heat that keeps it warm.
But a lizard at the same cold dawn is cold until the sun warms it. This is an ectotherm, because its body temperature follows its surroundings.
And a snake on the same rock is cold until the sun warms it. This is an ectotherm, because its body temperature follows its surroundings.
But a rabbit at the same cold dawn is warm. This is an endotherm, because its own body makes the heat that keeps it warm.
And a fox on a frosty morning is warm. This is an endotherm, because its own body makes the heat that keeps it warm.
But a goldfish in a cold pond is as cold as the water. This is an ectotherm, because its body temperature follows its surroundings.
The table below shows the six cases with their verdicts: where each animal’s heat comes from, and which name it takes.
The table below compares an endotherm with an ectotherm: where the heat comes from, what the body temperature does, and one example of each.
What you are expected to know Classify an animal as an endotherm (it keeps its body warm with heat its own body makes) or an ectotherm (its body temperature follows its surroundings).
Two animals sit in the same cold air.
Which of the following makes an animal an endotherm?
- A. ✓ Its own body makes the heat that keeps it warm
- B. Its body takes heat from the sun and a warm rockHeat from the sun and a rock is heat from outside the body.
An endotherm’s heat is made inside its body. - C. Its body is always warmer than the airAn ectotherm on a hot rock can be warmer than the air.
What makes an endotherm is where its heat is made: inside its body.
Why: An endotherm’s own body makes heat in respiration.
That heat keeps its body warm while the air is cold.
So an animal is an endotherm because its own body makes the heat that keeps it warm.
A sparrow sits on a frosty branch.
Which kind of animal is this?
- A. ✓ An endotherm
- B. An ectothermA sparrow is a bird.
A bird’s own body makes the heat that keeps it warm.
Why: A sparrow is a bird.
A bird’s own body makes the heat that keeps it warm.
So a sparrow is an endotherm.
A trout swims in a cold stream.
Which kind of animal is this?
- A. An endothermA trout is a fish.
A fish’s body temperature follows the water around it. - B. ✓ An ectotherm
Why: A trout is a fish.
A fish’s body temperature follows the water around it.
So a trout is an ectotherm.
A frog sits at the edge of a cold pond.
Which kind of animal is this?
- A. An endothermA frog is an amphibian.
An amphibian’s body temperature follows its surroundings. - B. ✓ An ectotherm
Why: A frog is an amphibian.
An amphibian’s body temperature follows its surroundings.
So a frog is an ectotherm.
A bat flies through the cool evening air.
Which kind of animal is this?
- A. ✓ An endotherm
- B. An ectothermA bat is a mammal.
A mammal’s own body makes the heat that keeps it warm.
Why: A bat is a mammal.
A mammal’s own body makes the heat that keeps it warm.
So a bat is an endotherm.
A deer stands in falling snow.
Which kind of animal is this?
- A. ✓ An endotherm
- B. An ectothermA deer is a mammal.
A mammal’s own body makes the heat that keeps it warm.
Why: A deer is a mammal.
A mammal’s own body makes the heat that keeps it warm.
So a deer is an endotherm.
A grasshopper waits in the grass at dawn.
Which kind of animal is this?
- A. An endothermA grasshopper is an insect.
An insect’s body temperature follows its surroundings. - B. ✓ An ectotherm
Why: A grasshopper is an insect.
An insect’s body temperature follows its surroundings.
So a grasshopper is an ectotherm.
A student says: “A lizard has cold blood, so it is cold all day.”
Is the student correct?
- A. Yes: a lizard is cold all day longAt noon a lizard on a sunlit rock is warmer than you are.
Its body temperature follows its surroundings: cold at dawn, hot at noon. - B. ✓ No: a lizard is cold at dawn and hot at noon
Why: A lizard’s body temperature follows its surroundings.
At dawn the sand is cold, so the lizard is cold.
At noon the rock is hot, so the lizard is hot.
So the lizard is not cold all day: its body is as warm as its surroundings.
37Quick quiz: endotherm, ectotherm mixed practice
A goat stands in the snow.
Which kind of animal is this?
- A. ✓ An endotherm
- B. An ectothermA goat is a mammal.
A mammal’s own body makes the heat that keeps it warm.
Why: A goat is a mammal.
A mammal’s own body makes the heat that keeps it warm.
So a goat is an endotherm.
A salamander rests under a damp log.
Which kind of animal is this?
- A. An endothermA salamander is an amphibian.
An amphibian’s body temperature follows its surroundings. - B. ✓ An ectotherm
Why: A salamander is an amphibian.
An amphibian’s body temperature follows its surroundings.
So a salamander is an ectotherm.
Two words name where an animal’s warmth comes from.
What is an ectotherm?
- A. An animal whose body is cold all dayAn ectotherm on a hot rock at noon is hot.
Its body follows its surroundings, cold or hot. - B. ✓ An animal whose body temperature follows its surroundings
- C. An animal that keeps its body warm with heat its own body makesAn animal that keeps its body warm with heat its own body makes is an endotherm.
An ectotherm’s heat comes from outside.
Why: Ecto means outside.
An ectotherm’s body temperature follows its surroundings, because the heat that warms it comes from outside its body.
Two words name where an animal’s warmth comes from.
What is an endotherm?
- A. ✓ An animal that keeps its body warm with heat its own body makes
- B. An animal whose body temperature follows its surroundingsAn animal whose body temperature follows its surroundings is an ectotherm.
An endotherm makes its own heat. - C. An animal that is always warmer than the air around itAn ectotherm on a hot rock can be warmer than the air.
An endotherm is named for where its heat is made: inside its body.
Why: Endo means inside and therm means heat.
An endotherm keeps its body warm with heat its own body makes.
Two words name where an animal’s warmth comes from.
(a) State what an endotherm is and what an ectotherm is. (2 pt)
An ectotherm’s body temperature follows its surroundings.
- Award 1 point for: an endotherm keeps its body warm with heat its own body makes (or heat from its own respiration).
- Award 1 point for: an ectotherm’s body temperature follows its surroundings (or it warms from outside).
43Onto the rock, then under it
In topic 8.1, a lizard lying in full sun walked into the shade of a bush.
Which kind of response was the walk?
- A. ✓ A behavioral response
- B. A physiological responseA physiological response is a change inside the organism’s body.
The walk is something the whole lizard does.
Why: The whole lizard walked.
A response that the whole organism does is a behavioral response.
Video: Watch: Onto the rock, then under it
The rock through the day: the sun rising, the lizard climbing onto the rock at dawn and sliding under it at noon; a heap of lizards on a cold night.
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Here is the lizard at dawn again, cold on the cold sand.
Its body makes little heat of its own. So to get warm, the lizard must move to where heat is.
At dawn the lizard crawls onto the sunlit rock. Heat passes from the sun and the warm rock into its body, so its body temperature rises.
By noon the rock is hot. A lizard that stayed on top would overheat.
So at noon the lizard moves under the rock, into the shade. Heat passes from its body into the cooler shaded air, so its body temperature falls.
On a cold night, some ectotherms huddle together in a heap. An animal in the heap has less of its skin open to the cold air, so it loses heat more slowly than an animal alone.
Climbing into the sun, moving into the shade and huddling are all things the whole lizard does. So an ectotherm controls its body temperature by what it does: by behavioral responses.
Unit 3 said an enzyme that gets too hot loses its shape and stops working, and a cold enzyme works slowly. The moves between sun and shade keep the lizard’s body in the range where its enzymes work well.
What you are expected to know Describe how an ectotherm controls its body temperature by behavior: moving into sun or shade, or huddling with others.
Suppose a lizard sits on top of a desert rock at noon, and the rock is hot.
Where does the lizard go?
- A. Up to the top of the rock in full sunIn full sun at noon, more heat passes into the lizard, and it overheats.
The lizard moves to where heat passes out of it: the shade. - B. Nowhere: it sweats where it sitsSweating is a physiological response of mammals such as people.
A lizard controls its temperature by moving: at noon, into the shade. - C. ✓ Under the rock into the shade
Why: At noon the rock is hot, so heat passes into the lizard faster than it can lose it.
Under the rock the shaded air is cooler, so heat passes out of the lizard’s body.
So the lizard moves under the rock, into the shade.
Suppose a cold night falls on a group of lizards.
How are the lizards most likely to be found?
- A. Spread out, each lizard aloneA lizard alone has all of its skin open to the cold air.
In a heap, less of each lizard’s skin is open to the air. - B. ✓ Together in a heap
Why: Each lizard in a heap has less of its skin open to the cold air.
So each lizard loses heat more slowly than a lizard alone.
So the lizards are most likely to be found together in a heap.
Suppose a snake lies on cold sand at dawn.
Which of the following warms the snake?
- A. ShiveringShivering muscles make heat inside an endotherm such as a mouse.
A snake warms from outside, by moving to where heat is. - B. Moving into the shade of a bushShade is cooler than the sun, so heat passes out of a body there.
The snake needs heat to pass in: the sunlit rock. - C. ✓ Crawling onto a sun-warmed rock
Why: The snake is an ectotherm, so its body temperature follows its surroundings.
On a sun-warmed rock, heat passes from the rock and the sun into the snake.
So crawling onto the rock warms it.
A student says: “The lizard on the sunlit rock is warm because its body is making heat, the way the mouse’s body does.”
Is the student correct?
- A. Yes: the lizard’s own body made the heatThe lizard’s heat passed in from the sun and the rock.
Its own body makes little heat. - B. ✓ No: the heat passed in from the sun and the rock
Why: The lizard was cold at dawn, on the cold sand.
It warmed only after it crawled onto the sunlit rock.
Heat passed from the sun and the rock into its body.
So the lizard’s warmth came from outside its body, not from its own body making heat.
59Why the endotherm must eat more
In Unit 3, a cell broke down glucose. About 33 % of the energy released ended up in ATP.
What happened to the rest of the energy?
- A. ✓ It spread out as heat that can do no more work
- B. It was stored in the cell as new fatThe cell stored none of that energy as fat.
The energy that did not reach ATP spread out as heat. - C. It was destroyed, so the total energy fellEnergy is never created or destroyed.
The energy that did not reach ATP spread out as heat.
Why: At every energy transfer, some of the energy spreads out as heat that can do no more work.
That is the second law of thermodynamics.
So the energy that did not reach ATP spread out as heat.
Video: Watch: Why the endotherm must eat more
The mouse and the lizard at noon, both warm; two food piles per gram of body, the mouse’s far bigger; the caterpillar’s and the squirrel’s bars, about 18 % and about 2 % built into new tissue.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L11c.mp4
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Here are the mouse and the lizard again, side by side at noon. Both animals are warm, and the two have the same mass.
The mouse’s warmth came from respiration. When its cells respire food, some of the energy released spreads out as heat, as the second law of thermodynamics in Unit 3 says.
In the mouse, that heat is what keeps its body warm. So the mouse must respire food all the time, day and night, just to stay warm.
Food that the mouse respires for heat leaves it as heat. None of that food is built into new tissue.
The lizard’s warmth came from the sun and the rock. The lizard respired no food to get warm.
So the lizard needs less food per gram of body than the mouse. More of what the lizard eats can be built into new tissue.
The drawing below shows the food each animal needs per gram of body in a day. The mouse’s heap is far bigger.
Unit 2 said a small animal loses heat faster per gram of body than a large one. So a small endotherm such as a mouse must respire the most food per gram of all.
Biologists measured how much of the food energy an animal eats ends up in its new tissue.
A caterpillar eating leaves built about 18 % of the food energy into new tissue. The other 82 % left the caterpillar, as heat and as waste.
A squirrel eating acorns built about 2 % of the food energy into new tissue. The other 98 % left the squirrel, most of it as heat from respiration.
The squirrel is no worse at digesting than the caterpillar. Both animals digest their food well.
The difference comes after digestion. The squirrel is an endotherm, so its cells respire most of the food energy to make heat, and that heat leaves its body.
So an endotherm needs more food per gram of body than an ectotherm of the same size. The extra food is respired to make the heat that keeps it warm.
What you are expected to know Explain why an endotherm needs more food per gram of body than an ectotherm of the same size: it respires food to make the heat that keeps it warm, and that food is not built into new tissue.
Suppose a robin and a frog have the same mass and live in the same meadow.
Which animal eats more food per gram of body each day?
- A. ✓ The robin
- B. The frogA frog is an ectotherm: it respires no food to stay warm.
The robin respires food for heat, so it must eat more.
Why: The robin is an endotherm, so its own respiration makes the heat that keeps it warm.
The food it respires for heat is not built into new tissue.
The frog is an ectotherm, so it respires no food to stay warm.
So the robin eats more food per gram.
Suppose a shrew and a lizard have the same mass and live in the same dry field. Each day the shrew eats more food per gram of body than the lizard does.
(a) Explain why the shrew needs more food per gram of body than the lizard. (2 pt)
Frame The shrew needs more food because …
Its own respiration makes the heat that keeps its body warm.
The food it respires for heat is not built into new tissue.
The lizard is an ectotherm, so its warmth comes from its surroundings and it respires no food to stay warm.
So the shrew must eat more food per gram of body than the lizard.
- Award 1 point for: the shrew is an endotherm whose heat is made by respiring food (or: the lizard is an ectotherm that respires no food to stay warm).
- Award 1 point for: the food respired for heat is not built into new tissue (or leaves the body as heat), so the shrew needs more food.
A student says: “A squirrel builds so little of its food into new tissue because it is worse at digesting than a caterpillar is.”
Is the student correct?
- A. Yes: the squirrel digests its food worseThe squirrel digests its food as well as the caterpillar does.
The squirrel respires most of the food energy for heat, so little is left to build into tissue. - B. ✓ No: the squirrel respires most of the food energy for heat
Why: The squirrel digests its food well.
The squirrel is an endotherm, so its cells respire most of the food energy to make heat.
That heat leaves its body.
So little of the food energy is left to build into new tissue; digestion is not the reason.
Suppose a vole and a snake of the same mass each eat 50 g of food in a week.
Which animal builds more of that food into new tissue?
- A. The voleThe vole is an endotherm: it respires most of its food energy to make heat.
The snake respires no food to stay warm, so more of its food becomes tissue. - B. ✓ The snake
Why: The snake is an ectotherm, so it respires no food to stay warm.
The vole is an endotherm, so it respires most of the food energy to make the heat that keeps it warm.
So more of the snake’s 50 g is left to build into new tissue.
Here are the lizard and the mouse again, on the desert at dawn.
The mouse is an endotherm. Its own respiration warms it, so it must eat more per gram of body.
The lizard is an ectotherm. It warms on the sunlit rock, and it respires no food to do it.
84Mixed practice mixed practice
Suppose several snakes spend a cold night in one heap in a rock crevice.
Why do the snakes lie in a heap rather than apart?
- A. The snakes’ muscles make heat for the heap, as a mouse’s muscles doA snake is an ectotherm.
Its muscles make no heat to keep its body warm. - B. The heap keeps the snakes cooler than the night air around themThe heap slows the loss of heat.
The snakes stay warmer than a snake alone, not cooler. - C. ✓ Less of each snake’s skin is open to the cold air, so each loses heat more slowly
Why: Each snake is an ectotherm, so it loses heat to the cold air.
In a heap, less of each snake’s skin is open to the air.
So each snake loses heat more slowly than a snake alone.
A student says: “A trout is an ectotherm, so its body is at the temperature of the water it swims in.”
Is the student correct?
- A. ✓ Yes: an ectotherm’s body follows its surroundings
- B. No: a trout’s body is always colder than the waterAn ectotherm’s body temperature follows its surroundings.
The trout’s body is at the water’s temperature, not below it.
Why: An ectotherm’s body temperature follows its surroundings.
The trout’s surroundings are the water.
So the trout’s body is at the water’s temperature.
A hen scratches in the snow on a frosty morning.
Which kind of animal is this?
- A. ✓ An endotherm
- B. An ectothermA hen is a bird.
A bird’s own body makes the heat that keeps it warm.
Why: A hen is a bird.
A bird’s own body makes the heat that keeps it warm.
So a hen is an endotherm.
A student says: “A lizard moving into the shade at noon is a physiological response.”
Is the student correct?
- A. Yes: moving into the shade is physiologicalMoving into the shade is something the whole lizard does.
That is a behavioral response. - B. ✓ No: the whole lizard moves, so it is behavioral
Why: The whole lizard moves.
A response the whole organism does is a behavioral response.
So moving into the shade is a behavioral response, not a physiological one.
Suppose a mouse and a grasshopper have the same mass and live in the same barn.
Which animal eats more food per gram of body each day?
- A. ✓ The mouse
- B. The grasshopperA grasshopper is an ectotherm: it respires no food to stay warm.
The mouse respires food for heat, so it must eat more.
Why: The mouse is an endotherm, so its own respiration makes the heat that keeps it warm.
The food it respires for heat is not built into new tissue.
The grasshopper is an ectotherm, so it respires no food to stay warm.
So the mouse eats more food per gram.
Animals fall into two groups by where their warmth comes from.
Which of the following groups are endotherms?
- A. Reptiles and amphibiansA reptile or an amphibian takes its body temperature from its surroundings.
Reptiles and amphibians are ectotherms. - B. Fish and insectsA fish or an insect takes its body temperature from its surroundings.
Fish and insects are ectotherms. - C. ✓ Mammals and birds
Why: Mammals and birds keep their bodies warm with heat their own bodies make.
So mammals and birds are endotherms.
Suppose a biologist keeps a grasshopper and a mouse of the same mass in the same warm room for a week. Each day she weighs the food each animal eats. The mouse eats far more food per gram of body than the grasshopper does.
(a) Explain how this result demonstrates that an endotherm’s warmth comes from respiring food. (2 pt)
Frame The result demonstrates this because …
The mouse’s own respiration makes the heat that keeps it warm, so it must respire food all week just to stay warm.
The grasshopper’s body temperature follows the warm room, so it respires no food to stay warm.
So the mouse must eat far more food per gram of body, and the extra food is the food it respires for heat.
- Award 1 point for: the mouse (an endotherm) respires food to make the heat that keeps it warm, while the grasshopper (an ectotherm) respires no food to stay warm.
- Award 1 point for: the extra food the mouse eats is the food respired for heat, so the difference in food eaten is the heat made from food.
(b) Predict how the mouse’s food intake per gram of body would change if the biologist kept the room cold instead. (1 pt)
Frame In a cold room the mouse’s food intake per gram of body would …
- Award 1 point for: the mouse’s food intake per gram would rise (increase).
(c) Justify your prediction. (1 pt)
Frame The intake changes this way because …
Its own respiration must replace that heat to keep its body warm.
So the mouse must respire more food, and it must eat more.
- Award 1 point for: in cold air the mouse loses heat faster, so it must respire more food to replace the heat and keep its body warm.
Glossary
- endotherm
- An animal that keeps its body warm with heat its own body makes in respiration: a mouse, a sparrow.
- ectotherm
- An animal whose body temperature follows its surroundings, so it warms and cools by moving into sun or shade: a lizard, a trout.
APBIO-U08-L12 Breeding when the energy is there
Photo: Whitney Cranshaw, Colorado State University, Wikimedia Commons, CC BY 3.0 US (resized).
Here is a rose shoot in early summer, crowded with aphids. The large adult at the right is giving birth. Aphids can give birth like this without mating.
In July every aphid on the shoot is a female, and each gives birth to daughters without mating, dozens a week. On the same rose in October there are males, females and eggs.
The rose did not change species. Why did the aphids change how they breed?
Unit 8 · Ecology
1Matching breeding to the food on offer
A field mouse in June takes in more energy from its seeds than it uses staying warm and moving. The mouse has a surplus.
Which of the following describes the number of offspring the mouse produces?
- A. Fewer offspring than a mouse with no surplusFewer offspring is what a mouse produces when its energy out beats its energy in.
This mouse has energy to spare. - B. The same number as a mouse with no surplusThe surplus is energy the mouse does not need for staying warm and moving.
The mouse puts part of it into a litter. - C. ✓ More offspring than a mouse with no surplus
Why: The mouse’s energy in beats its energy out.
The surplus is energy to spare.
Part of the surplus goes into a litter.
So the mouse produces more offspring.
In Unit 5, a new mouse began when a sperm from the father fused with an egg from the mother.
Which of the following describes sexual reproduction?
- A. ✓ Two gametes fuse, one from each parent, into one new organism
- B. A skin cell divides by mitosis so that a cut healsMitosis makes two body cells with the same chromosomes.
It makes no new organism. - C. One parent produces offspring with no gametes fusingReproduction from one parent with no gametes fusing is the other way of reproducing.
Sexual reproduction needs two gametes, one from each parent.
Why: In sexual reproduction two gametes fuse, one from each parent.
Each parent contributes one chromosome set to the new organism.
Video: Watch: Matching breeding to the food on offer
The rose shoot in July: one female, then dozens of daughters with no male among them. The season turns on screen: the leaves yellow, males and females mate, eggs sit on the stem through the winter and hatch as the new leaves open. Then three quick cases: deer mating in autumn, a carrot plant’s two summers, an insect that lays no eggs through a dry season.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L12a.mp4
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How does a species match its breeding to the energy on offer? The aphid gives one answer: it changes how it reproduces as the food changes.
In July, with food everywhere, one female makes daughters on her own, dozens a week. Each daughter is a copy of her mother.
In October, as the food grows scarce, the aphids mate. The females lay eggs that wait out the winter.
Other species time their breeding rather than switch it:
- Deer in cool climates mate in autumn, so the fawns arrive with the spring grass.
- A carrot plant stores food in its first summer and flowers in its second.
- Some insects stop breeding through a poor season.
Switching or timing is a response to conditions, not a decision.
The photograph shows aphids on a rose shoot. The large adult at the right is giving birth, and aphids can give birth like this without mating.
Every aphid on the shoot does the same in July. One parent, the mother, makes each daughter, and no gametes fuse.
No sperm joins an egg, so every chromosome the daughter carries came from her mother. The daughter is genetically identical to her mother.
Reproduction from one parent, with no gametes fusing, that gives offspring genetically identical to the parent is called .
A- means without. Asexual reproduction is reproduction without gametes fusing, and so without a second parent.
Sexual reproduction, from Unit 5, is the other way: two gametes fuse, one from each parent. So the offspring carries chromosomes from both parents and is genetically identical to neither.
In July the food is everywhere. A female that reproduces asexually needs no mate, so she turns her surplus into daughters at once.
Each daughter does the same. So while the food lasts, one female’s line fills the shoot in a few weeks.
In October the rose’s leaves yellow, and less sap flows in its stems. The aphids’ energy in falls.
Now the aphids reproduce sexually: males and females mate, and the females lay eggs on the stem.
An egg uses almost no energy through the winter. When the rose grows new leaves in spring, the eggs hatch, and the food is there.
So the aphid switches how it reproduces as the energy on offer changes: asexual reproduction while food is plentiful, sexual reproduction with eggs as the food grows scarce.
Other species do not switch. They time one way of breeding so that the young arrive when the food does.
Deer in cool climates mate in autumn. Their fawns are born in late spring, when new grass is growing.
A carrot plant stores food in its thick root through its first summer. It flowers and sets seed only in its second summer, on the energy it stored.
Some insects stop breeding through a poor season. The adults live on, but they lay no eggs until the food returns.
Timing and pausing differ in what sets the date. A species that times its breeding mates in the same season every year, and its young arrive with the food.
A species that pauses breeds whenever the food is there. It stops when the food fails and starts again when the food returns, however long that takes.
The table below compares the four strategies: what the species does, what the timing matches, and one example of each.
None of these is a decision. For the aphid, less food and shorter days are the stimulus, and the switch to mating and eggs is the response.
Aphids that switched left eggs that survived the winter, so the switch stayed common. That is natural selection, the process Unit 7 taught.
What you are expected to know Describe how a species switches its reproduction to match the energy on offer: asexual reproduction while food is plentiful, sexual reproduction as the food grows scarce.
What you are expected to know Describe how a species times its reproduction to match the energy on offer: mating timed so the young arrive with the food, breeding on stored energy, or a pause through a poor season.
In October the aphids switch from asexual reproduction to mating and laying eggs.
Which of the following changes are the aphids responding to?
- A. The food on offer is rising as the rose grows new leavesNew leaves grow in spring, and that is when the eggs hatch.
In October the leaves yellow and less sap flows. - B. ✓ The food on offer is falling as the rose’s leaves yellow
- C. The shoot has become too crowded with aphidsIn October the rose’s leaves yellow and less sap flows.
The aphids answer the fall in food.
Why: In October the rose’s leaves yellow and less sap flows in its stems.
So the food on offer falls.
The aphids answer the fall in food by mating and laying eggs that wait for spring.
Suppose a bird nests in a forest. It lays its eggs in April, so its chicks hatch in May, when caterpillars are most plentiful.
Which of the following strategies is the bird using?
- A. Switches from asexual to sexual reproduction as food fallsThe bird breeds one way only, from a mate and an egg.
Switching means asexual reproduction while food is plentiful, then sexual. - B. ✓ Times its mating so the young arrive when the food does
- C. Stores energy in one season and breeds on it in the nextStoring first means breeding on energy saved in an earlier season.
The bird’s timing puts the chicks with the caterpillars. - D. Pauses breeding through a poor season, then breeds againPausing means laying no eggs through a poor season.
The bird lays every spring.
Why: The chicks hatch in May, when caterpillars are most plentiful.
So the young arrive when the food does.
The bird times its mating to the food on offer.
Suppose a tiny animal lives in a pond. In May the pond is thick with algae; every one of the animals is a female, and each gives birth to daughters without mating. In September the algae are gone; the animals mate and lay eggs that sink to the mud.
Which of the following strategies is the pond animal using?
- A. ✓ Switches from asexual to sexual reproduction as food falls
- B. Times its mating so the young arrive when the food doesTiming means one way of breeding, placed in one season.
This animal breeds two ways: without mating in May, by mating in September. - C. Stores energy in one season and breeds on it in the nextStoring first means breeding on energy saved in an earlier season.
This animal breeds in May on the algae there and then. - D. Pauses breeding through a poor season, then breeds againPausing means no breeding through the poor season.
This animal breeds in September, by mating.
Why: In May each female gives birth without mating: asexual reproduction while the algae are plentiful.
In September the animals mate and lay eggs: sexual reproduction as the food grows scarce.
So the animal switches as the food changes.
Suppose a beetle lives where the dry season is long. The adults live through the dry months but lay no eggs until the rains bring fresh leaves.
Which of the following strategies is the beetle using?
- A. Switches from asexual to sexual reproduction as food fallsSwitching means asexual reproduction while food is plentiful, then sexual.
The beetle breeds one way; it stops and restarts. - B. Times its mating so the young arrive when the food doesTiming means mating in one season every year.
The beetle stops laying until the food returns, however long the dry season lasts. - C. Stores energy in one season and breeds on it in the nextStoring first means breeding on energy saved in an earlier season.
The beetle breeds when fresh leaves arrive, not on a store. - D. ✓ Pauses breeding through a poor season, then breeds again
Why: Through the dry months the beetle lays no eggs.
When the rains bring fresh leaves, it lays again.
So the beetle pauses breeding through the poor season.
Suppose a plant grows only leaves in its first summer and stores sugar in a thick root. In its second summer it flowers, sets seed and dies.
Which of the following strategies is the plant using?
- A. Switches from asexual to sexual reproduction as food fallsSwitching means asexual reproduction while food is plentiful, then sexual.
The plant reproduces once, by flowering and seed. - B. Times its mating so the young arrive when the food doesTiming means placing the breeding so the young arrive with the food.
The plant’s point is the year of storing that comes first. - C. ✓ Stores energy in one season and breeds on it in the next
- D. Pauses breeding through a poor season, then breeds againPausing means stopping through a poor season and restarting.
The plant has not started: it is storing for its one flowering.
Why: In its first summer the plant stores sugar in its root.
In its second summer it flowers and sets seed on that store.
So the plant stores energy in one season and breeds on it in the next.
A student watches the aphids in October and says: “The aphids know winter is coming, so they decide to mate.”
Is the student correct?
- A. Yes: the aphids know winter is coming, so they decide to mateLess food and shorter days are the stimulus.
Mating and eggs are the response, kept by selection; the aphids decide nothing. - B. ✓ No: shorter days and less food are the stimulus; mating is the response
Why: Less food and shorter days are the stimulus.
The switch to mating and eggs is the response.
Aphids that switched left eggs that survived the winter, so the switch stayed common.
The aphids know and decide nothing.
Suppose a moth’s caterpillars eat oak leaves. The moths mate and lay their eggs in autumn. The eggs sit on the twigs through the winter and hatch in April, as the oak’s new leaves open.
(a) Explain how this case demonstrates that a species matches its reproduction to the energy on offer. (1 pt)
Frame The case demonstrates this because …
The moths mate in autumn, and the eggs wait through the winter.
The eggs hatch in April, as the new leaves open.
So the young arrive when the food is there: the moth times its breeding to the energy on offer.
- Award 1 point for: the eggs hatch in April, when the new oak leaves (the caterpillars’ food) are on offer, so the moth’s breeding is timed to the energy available.
Here is the rose again, in July and in October.
In July, with food everywhere, one female reproduces asexually: daughters without mating, dozens a week.
In October, as the food grows scarce, the insects reproduce sexually: they mate and lay eggs that wait for spring.
43Quick quiz: asexual reproduction mixed practice
A strawberry plant sends a runner out along the soil. The runner roots and grows into a new plant with the same chromosomes as the parent plant.
Is this asexual reproduction?
- A. ✓ Yes
- B. NoOne parent made the new plant, no gametes fused, and the new plant is genetically identical to the parent.
That is asexual reproduction.
Why: One parent plant made the new plant.
No gametes fused.
The new plant carries the same chromosomes as the parent.
So this is asexual reproduction.
A hen’s egg is fertilized by a rooster’s sperm. The egg hatches into a chick.
Is this asexual reproduction?
- A. YesTwo gametes fused, the sperm and the egg, one from each parent.
That is sexual reproduction. - B. ✓ No
Why: The sperm and the egg are two gametes, one from each parent.
They fused into the chick’s first cell.
So this is sexual reproduction, not asexual.
A pea plant’s pollen fertilizes an egg in a flower on another pea plant. The seed grows into a new plant.
Is this asexual reproduction?
- A. YesThe pollen’s gamete fused with the egg, one from each parent plant.
That is sexual reproduction. - B. ✓ No
Why: The pollen carries a gamete from one plant, and the egg is a gamete of the other plant.
The two gametes fused into the seed’s first cell.
So this is sexual reproduction, not asexual.
A bacterium copies its DNA and splits into two bacteria, each carrying the same DNA.
Is this asexual reproduction?
- A. ✓ Yes
- B. NoOne bacterium made two, no gametes fused, and each new bacterium carries the same DNA.
That is asexual reproduction.
Why: One parent bacterium made the two new bacteria.
No gametes fused.
Each new bacterium carries the same DNA as the parent.
So this is asexual reproduction.
An organism reproduces.
Which of the following is asexual reproduction?
- A. Two gametes fuse, one from each parent, into one new organismTwo gametes fusing, one from each parent, is sexual reproduction.
- B. A skin cell divides by mitosis so that a cut in the skin healsMitosis in a skin cell makes two body cells of the same organism.
No new organism is produced. - C. ✓ One parent makes genetically identical offspring, no gametes fusing
Why: Asexual reproduction is reproduction from one parent.
No gametes fuse.
The offspring are genetically identical to the parent.
An organism reproduces.
(a) State what asexual reproduction is. (1 pt)
- Award 1 point for: one parent, no gametes fusing, offspring genetically identical to the parent (any two of the three).
Glossary
- asexual reproduction
- Reproduction from one parent, with no gametes fusing, that gives offspring genetically identical to the parent. The pair to sexual reproduction, in which two gametes fuse, one from each parent.
APBIO-U08-L13 From a pond to a biome, and the two things that pass through it
Here is one pond. Its frogs live in it. So does every other species in the pond. Around them lie the pond’s water, mud and light. And across the continent’s climate zone lie thousands of ponds like it.
Sunlight falls on the pond every day and never comes back. The carbon in a frog was in the air last year and will be again. Why does the energy leave while the carbon goes round?
Unit 8 · Ecology
1From the frogs of one pond to a biome
In the first week of the course you sorted an oak forest into levels. All the oaks of one kind living in that forest made one group.
Which level is that group?
- A. ✓ A population
- B. A communityA community is all the populations living together in one place: the oaks with the deer and the beetles.
All the oaks of one kind are one species only. - C. An ecosystemAn ecosystem is the community together with its non-living surroundings: the soil, the rain and the sunlight.
All the oaks of one kind are living things of one species only.
Why: All the oaks of one kind in one forest are all the organisms of one species living in one place.
All the organisms of one species living in one place are a population.
Video: Watch: From the frogs of one pond to a biome
The four nested boxes drawn out from the frogs of one pond: the frogs, every species in the pond, the species with the water, mud and light, and every pond like it across the climate zone; each box named as it appears.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L13a.mp4
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How is life sorted into levels, and what passes through all of them?
The frogs of one pond are a population. Every species in the pond is its community.
The community with its water, light and mud is the ecosystem. All the ecosystems of one kind across a climate zone, grouped together, make one level higher still.
Through every level, energy passes once. It enters as sunlight, moves from organism to organism, and leaves as heat, never to return.
Matter does the opposite. Its atoms are used again and again, passing between organisms and their surroundings and back.
One sentence carries every cycle that follows: energy flows through, and matter goes round.
Here is the pond, and here are all the frogs of one kind living in it. The first week gave that group its name: a population, all the organisms of one species living in one place.
Now add every other species in the pond: the pondweed, the snails, the dragonflies and the bacteria in the mud. All the populations living together in one place are a community.
Now add the pond’s water, its mud and the light falling on it. The community together with the non-living parts of its surroundings is an ecosystem.
Now consider every pond like this one across the whole climate zone. Thousands of ponds share the same warm summers, the same cold winters and the same rain.
Because the ponds share one climate, they hold much the same kinds of living things. So the whole region is one kind of ecosystem, repeated thousands of times.
All the ecosystems of one kind across a region with one climate, grouped together, are called a .
Ecologists group biomes by climate: how warm a region is, and how much rain falls on it.
So a biome’s edge is where the climate changes, not where a country’s border lies. One country can hold several biomes, and one biome can stretch across many countries.
In the first week the levels went population, community, ecosystem, biosphere. The biome is one more level, between the ecosystem and the biosphere.
The table below lists the four levels above the organism, from smallest to largest. For each it gives what the level adds and what it is at the pond.
What you are expected to know State the four levels above the organism in order, population, community, ecosystem and biome, and what each level adds: the other species, the non-living surroundings, every ecosystem of one kind across a climate zone.
Two ecosystems lie 500 km apart.
Which of the following puts the two ecosystems in the same biome?
- A. They lie inside the borders of the same countryA country’s border is a line people drew.
A biome is grouped by climate and the living things the climate supports. - B. ✓ They share one climate
- C. They lie on the banks of the same riverOne river can flow from mountains to a hot plain, through more than one climate.
A biome is grouped by climate.
Why: A biome is all the ecosystems of one kind across a region with one climate.
Two ecosystems that share one climate hold the same kinds of living things.
So the two ecosystems belong to the same biome.
The levels above the organism can be listed in order of size.
Which list puts the levels in order from smallest to largest?
- A. community, population, ecosystem, biomeA community is made of populations.
So the population comes before the community. - B. population, ecosystem, community, biomeAn ecosystem is a community plus its non-living surroundings.
So the community comes before the ecosystem. - C. ✓ population, community, ecosystem, biome
Why: Organisms of one species in one place make a population.
The populations of one place make a community.
The community with its non-living surroundings makes an ecosystem.
All the ecosystems of one kind across a climate zone make a biome.
Suppose you take the species of one river together with the river’s water, gravel and sunlight.
Which level is that?
- A. A populationA population is one species only.
This includes many species and the water, gravel and sunlight. - B. A communityA community is living things only.
This includes the water, gravel and sunlight. - C. ✓ An ecosystem
- D. A biomeA biome holds every ecosystem of one kind across a climate zone.
This is one river.
Why: The species of the river are its community.
The water, gravel and sunlight are the non-living parts of its surroundings.
The community together with the non-living parts of its surroundings is an ecosystem.
Suppose you count all the trout living in one river.
Which level is that group?
- A. ✓ A population
- B. A communityA community holds every species living in the river.
The trout are one species. - C. An ecosystemAn ecosystem holds the living things and the non-living surroundings.
The trout are living things of one species. - D. A biomeA biome holds every ecosystem of one kind across a climate zone.
The trout are one species in one river.
Why: The trout are all the organisms of one species living in one place.
All the organisms of one species in one place are a population.
Suppose you take every river and lake of one mild, rainy region, with their living things, water and light, together.
Which level is that?
- A. A populationA population is one species in one place.
This is every river and lake of a region. - B. A communityA community is the living things of one place.
This is every river and lake of a region, with their water and light. - C. An ecosystemAn ecosystem is one river or one lake with its surroundings.
This is every river and lake of a region. - D. ✓ A biome
Why: Each river or lake with its surroundings is one ecosystem.
The region has one climate, so its rivers and lakes are ecosystems of one kind.
All the ecosystems of one kind across a region with one climate are a biome.
Suppose you list the trout, the mayflies, the reeds and every other species living in one river.
Which level is that group?
- A. A populationA population is one species only.
This group holds many species. - B. ✓ A community
- C. An ecosystemAn ecosystem also holds the non-living surroundings: the water, the gravel and the light.
This group holds living things only. - D. A biomeA biome holds every ecosystem of one kind across a climate zone.
This group is the living things of one river.
Why: The group is every population living together in one place, living things only.
All the populations living together in one place are a community.
A student says: “Every country is one biome, so Brazil is one biome and Canada is another.”
Is the student correct?
- A. Yes: each country is one biome, so Brazil is one biome and Canada is anotherA country’s border is a line people drew, and the climate does not change at it.
Brazil holds hot wet forest and dry grassland: more than one biome. - B. ✓ No: Brazil holds hot wet forest and dry grassland, two biomes inside one country
Why: A biome is all the ecosystems of one kind across a region with one climate.
A country’s border is a line people drew; the climate does not change at it.
One country can hold several biomes.
So a country is not a biome.
28Quick quiz: biome mixed practice
Take all the hot, wet forests across one region of the tropics, with their living things, rain and soil, together.
Is that a biome?
- A. ✓ Yes
- B. NoThe forests share one climate and hold the same kinds of living things.
Every ecosystem of one kind across a region with one climate is a biome.
Why: Each forest with its living things, rain and soil is one ecosystem.
The forests share one hot, wet climate, so they are ecosystems of one kind.
All the ecosystems of one kind across a region with one climate are a biome.
Take one forest’s trees, birds and insects, together with the forest’s rain and soil.
Is that a biome?
- A. YesOne forest with its living things and its non-living surroundings is one ecosystem.
A biome is every ecosystem of one kind across a whole climate zone. - B. ✓ No
Why: The trees, birds and insects are the forest’s community.
The community with its rain and soil is one ecosystem.
A biome is all the ecosystems of one kind across a region with one climate, not one forest.
Ecologists sort life into levels above the organism.
What is a biome?
- A. All the organisms of one species living in one placeAll the organisms of one species living in one place are a population.
- B. The living things of one place together with their non-living surroundingsThe living things of one place with their non-living surroundings are an ecosystem.
- C. ✓ All the ecosystems of one kind across a region with one climate
Why: A biome is all the ecosystems of one kind across a region with one climate, grouped together.
Ecologists sort life into levels above the organism.
(a) State what a biome is. (1 pt)
- Award 1 point for: all the ecosystems of one kind (or with the same kinds of living things) across a region with one climate.
33Energy flows through; matter goes round
Why does the sunlight’s energy leave the pond for good, while the carbon in the frog goes round?
A frog uses the chemical energy in its food to jump.
Which of the following does the second law of thermodynamics say happens at that energy transfer?
- A. All of the energy becomes the movement of the jumpNo energy transfer is fully efficient.
Some of the energy spreads out as heat. - B. ✓ Some of the energy spreads out as heat
- C. The energy used for the jump is destroyedEnergy is never created or destroyed, as the first law of thermodynamics says.
The energy that does not become movement spreads out as heat.
Why: The second law of thermodynamics says that in every energy transfer some energy spreads out as heat.
That heat can no longer do work.
So only part of the food’s energy becomes the jump.
Video: Watch: Energy flows through; matter goes round
A single arrow of energy entering the pond as sunlight, passing from pondweed to snail to frog, and leaving as heat; beneath it a loop of carbon atoms turning from the air to the pondweed to the frog to the mud and back to the air.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L13b.mp4
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Consider the pond’s energy first. Almost all of it enters as sunlight falling on the pondweed.
The pondweed stores part of the sunlight’s energy in sugar. A snail eats the pondweed, and a frog eats the snail, so the energy passes from organism to organism.
Each organism uses that energy to live. At every use, some of the energy spreads out as heat, as the second law of thermodynamics says it must.
That heat warms the water and the air, and then it spreads away. No organism in the pond can take it back.
So the pond must take in fresh sunlight every day. Energy passes through the pond once.
Now consider the pond’s matter. Its atoms sit in stores: the air, the water, the mud and the living things themselves.
Photosynthesis, eating, breathing out and decay move the atoms from store to store. The moves make no new atom and destroy none.
Ecologists call each store a reservoir, and each way the atoms move a process. Atoms move between reservoirs by processes; none is made or lost, so a fall in one reservoir is a rise in another.
So the pond needs no fresh atoms. The same atoms are used again and again: matter goes round.
Here are five cases, each judged by one question: does it pass through the pond once, or does it go round?
For example, take the energy in one sugar molecule in a pondweed leaf, once a tadpole has eaten the leaf and used the sugar. That energy passes through once, because it spread out as heat and never comes back.
But take the carbon atom in that same sugar molecule, once the tadpole has breathed it out as carbon dioxide. That carbon atom goes round, because a pondweed leaf takes the carbon dioxide in and uses the atom again.
And take a water molecule in the pond, once the sun has evaporated it and the rain has brought it back. That water molecule goes round, because a pondweed root takes it in and uses it again.
But take the energy in a fallen leaf, once bacteria and fungi in the mud have broken the leaf down and used the energy. That energy passes through once, because it spread out as heat and never comes back.
And take the atoms of a dead frog, once bacteria and fungi in the mud have broken the body down. Those atoms go round, because they pass into the mud, water and air, and pondweed takes them in again.
Here is a table of the five cases with their verdicts: whether each passes through the pond once or goes round.
Energy passes through once: it enters as sunlight, moves from organism to organism, and leaves as heat, never to return. Matter goes round: its atoms are used again and again.
The table below compares energy with matter at the pond. For each it gives how it enters, how it moves, how it leaves, and whether it comes back.
Here is the pond drawn both ways: one straight arrow of energy passing through, and one loop of carbon atoms turning.
The path of an element’s atoms round and round between living things and their non-living surroundings is called a .
Bio means the living things. Geo means the earth: its rock, water and air.
Chemical means the element’s atoms. So a biogeochemical cycle is an element’s atoms going round between living things and the earth.
The carbon atom’s path and the water molecule’s path are both biogeochemical cycles. The sunlight’s energy has no cycle: it passes through once.
What you are expected to know Classify what happens to energy and to matter in an ecosystem: energy enters as sunlight, passes from organism to organism and leaves as heat, never to return. The atoms of matter are used again and again, in biogeochemical cycles.
A rabbit eats a grass blade and uses the sugar’s energy to hop.
Does that energy pass through the meadow once, or go round?
- A. ✓ Passes through once
- B. Goes roundThe rabbit used the sugar’s energy to hop.
At every use, some energy spreads out as heat, and the heat never comes back.
Why: The rabbit used the sugar’s energy to hop.
That energy spread out as heat.
Heat that has spread out can do no more work, and no organism takes it back.
So the energy passes through the meadow once.
A rabbit breathes out carbon dioxide over a meadow.
Does a carbon atom in that carbon dioxide pass through the meadow once, or go round?
- A. Passes through onceA grass blade takes the carbon dioxide in and builds the carbon atom into new sugar.
The same atom is used again. - B. ✓ Goes round
Why: The rabbit breathes the carbon atom out as carbon dioxide.
A grass blade takes that carbon dioxide in and builds the carbon atom into sugar.
So the same carbon atom is used again: it goes round.
A hawk’s muscles release energy as heat while the hawk flies over a meadow.
Does that energy pass through the meadow once, or go round?
- A. ✓ Passes through once
- B. Goes roundHeat that has spread into the air can do no more work.
No grass blade, rabbit or hawk takes it back.
Why: The hawk’s muscles release the energy as heat.
The heat spreads into the air.
Heat that has spread out can do no more work, and no organism takes it back.
So the energy passes through the meadow once.
A cow drinks water from a stream and later breathes some of it out.
Do those water molecules pass through the meadow once, or go round?
- A. Pass through onceThe water the cow breathes out rises into the air and falls again as rain.
The same molecules are used again. - B. ✓ Go round
Why: The cow breathes the water molecules out into the air.
The rain brings them back to the stream and the soil.
A grass root or another cow takes them in again.
So the same water molecules are used again: they go round.
A dead leaf lies on the soil of a meadow.
Does a nitrogen atom in the leaf pass through the meadow once, or go round?
- A. Passes through onceBacteria and fungi break the leaf down, and the nitrogen atom passes into the soil.
A grass root takes it in again. - B. ✓ Goes round
Why: Bacteria and fungi in the soil break the leaf down.
The nitrogen atom passes into the soil.
A grass root takes it in and builds it into new leaf.
So the same nitrogen atom is used again: it goes round.
Fungi on a meadow’s soil break a fallen branch down and use its energy.
Does that energy pass through the meadow once, or go round?
- A. ✓ Passes through once
- B. Goes roundThe fungi used the branch’s energy to live.
At every use, some energy spreads out as heat, and the heat never comes back.
Why: The fungi used the branch’s energy to live.
That energy spread out as heat.
Heat that has spread out can do no more work, and no organism takes it back.
So the energy passes through the meadow once.
A student says: “A pond must take in fresh sunlight every day, but it can use the same carbon atoms year after year.”
Is the student correct?
- A. ✓ Yes: the pond’s energy leaves as heat, while its carbon comes back to the pondweed again
- B. No: a pond recycles the sunlight’s energy as it recycles its carbon atoms, year after yearEnergy leaves the pond as heat at every use, and that heat never comes back.
Only the pond’s atoms are used again.
Why: Every time an organism uses energy, some spreads out as heat.
That heat leaves the pond and never comes back, so fresh sunlight must enter every day.
The carbon atoms pass back to the air and are taken in again.
So the same atoms are used year after year.
A forest takes in sunlight every day. It takes in almost no new carbon atoms: the same carbon atoms pass from the air to the leaves to the animals and back to the air.
(a) Explain why the forest needs fresh sunlight every day but needs no fresh carbon atoms. (2 pt)
Frame The forest needs fresh sunlight every day because …
Every time an organism uses energy, some spreads out as heat that can do no more work.
That heat leaves the forest and never comes back.
The carbon atoms do not leave for good: a deer breathes them out as carbon dioxide, and the leaves take them in again.
So the same carbon atoms are used again and again.
- Award 1 point for: energy leaves the forest as heat at every transfer (the second law of thermodynamics) and the heat cannot be used again, so fresh energy must enter as sunlight.
- Award 1 point for: the carbon atoms pass back to the air (as carbon dioxide) and are taken in again by the leaves, so the same atoms are used again and again.
Here is the pond again: its frogs, every species living in it, the species with the pond’s water, mud and light, and every pond like it across the climate zone.
Those four are the population, the community, the ecosystem and the biome.
The sunlight’s energy leaves the pond as heat and never returns. Its carbon atoms are used again and again.
72Quick quiz: biogeochemical cycle mixed practice
Sunlight’s energy passes into a leaf, from the leaf into a caterpillar, and out of the caterpillar as heat.
Is this path a biogeochemical cycle?
- A. YesA biogeochemical cycle follows an element’s atoms, and the atoms come back to where they started.
Energy is not made of atoms, and the heat does not come back. - B. ✓ No
Why: A biogeochemical cycle follows an element’s atoms, and those atoms come back to their start.
Energy is not an element’s atoms.
The heat leaves and never comes back.
So the energy’s path is not a biogeochemical cycle.
Carbon atoms pass from the air into a leaf, from the leaf into a caterpillar, and back into the air when the caterpillar breathes out.
Is this path a biogeochemical cycle?
- A. ✓ Yes
- B. NoThe carbon atoms pass through living things and back to the air, where a leaf takes them in again.
The same atoms go round.
Why: The path follows an element’s atoms, carbon.
The atoms pass between living things and the air and come back to where they started.
An element’s atoms going round between living things and their non-living surroundings is a biogeochemical cycle.
Ecologists follow the atoms of an element through an ecosystem.
What is a biogeochemical cycle?
- A. The path of heat from a pond’s warm water up into the cold air above itHeat leaving the pond is energy passing through once.
A cycle comes back to its start, and only the atoms of matter come back. - B. The path of one organism’s food from its mouth through its gut to its cellsFood passing through one organism is one step inside a cycle, not the cycle itself.
A biogeochemical cycle follows an element’s atoms through living things and the earth. - C. ✓ The path of an element’s atoms round and round between living things and their surroundings
Why: Bio means the living things, geo means the earth, and chemical means the element’s atoms.
A biogeochemical cycle is an element’s atoms going round between living things and their non-living surroundings.
Ecologists follow the atoms of an element through an ecosystem.
(a) State what a biogeochemical cycle is. (1 pt)
- Award 1 point for: the path of an element’s atoms (matter) round between living things and their non-living surroundings (air, water, soil or rock), used again and again.
Glossary
- biome
- All the ecosystems of one kind across a region with one climate, grouped together. Ecologists group biomes by climate (how warm the region is and how much rain falls), never by a country's borders.
- biogeochemical cycle
- The path of an element's atoms round and round between living things (bio) and the earth's rock, water and air (geo), the same atoms used again and again. Carbon and water go round in biogeochemical cycles; energy has no cycle, it passes through once and leaves as heat.
APBIO-U08-L14 Boxes and arrows: how every cycle is drawn
Here is a blank diagram: four boxes, and arrows between them. It will be the same diagram for water, for carbon, for nitrogen and for phosphorus.
What do the boxes stand for, and what do the arrows do?
Unit 8 · Ecology
1A box is a reservoir, an arrow is a process
Sunlight falls on a pond every day. Carbon atoms sit in the pond’s frogs, its plants and its water.
Which of these is used again and again, going round between the pond’s living things and their surroundings?
- A. The energy of the sunlightEvery transfer of energy gives off heat, and the heat leaves the pond for good.
The sunlight’s energy passes through once. - B. ✓ The carbon atoms
Why: The sunlight’s energy leaves the pond as heat and does not come back.
The carbon atoms move from the water to the plants, to the frogs, and back.
So the carbon atoms are used again and again: a biogeochemical cycle.
Video: Watch: A box is a reservoir, an arrow is a process
The blank template building itself: four boxes appear, then the arrows between them; the boxes take the names of the carbon example and each arrow takes the name of the process that moves the carbon.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L14a.mp4
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How is a cycle drawn so that four cycles need one picture?
Each box is a place where the matter sits: the air, the water, the rock, the plants, the animals, the microbes in the soil.
Each arrow is something that moves the matter from one place to another.
Nothing makes or destroys an atom on the way. So when one box loses atoms, another box gains the same number.
Every cycle drawn from here on is this diagram with the boxes and arrows named.
Here is the blank diagram again: four boxes, and arrows between them.
Consider carbon as the example. Carbon atoms sit in the air, as carbon dioxide.
Carbon atoms also sit in plants, in animals and in the soil. Each of those places holds a store of carbon.
A place where the matter of a cycle sits is called a . A reservoir is an everyday word for a store of water behind a dam; here it means any store of the matter.
So the air is a reservoir of carbon. Each box on the diagram is one reservoir.
Now the arrows. Photosynthesis takes carbon dioxide out of the air and builds its carbon into a plant’s sugar.
So photosynthesis moves carbon from the air to the plants. Something that moves matter from one reservoir to another is called a process.
Each arrow on the diagram is one process. The arrow starts at the reservoir the matter leaves and ends at the reservoir the matter enters.
Here is the diagram with its carbon reservoirs and its processes named.
Respiration moves carbon from the plants and the animals back to the air. Eating moves carbon from the plants to the animals.
Dead remains and waste move carbon from the plants and the animals to the soil. Microbes in the soil respire that carbon, and it goes back to the air.
Every cycle drawn as boxes and arrows is drawn this way: the air at the top, the rock and soil at the bottom, the living things in the middle.
What you are expected to know Identify, on a drawn cycle, which parts are reservoirs (the boxes, where the matter sits) and which are processes (the arrows, which move matter from one reservoir to another).
In the carbon diagram, a box and an arrow each stand for one thing.
Which of the following does a box stand for?
- A. ✓ The air, where carbon sits as carbon dioxide
- B. Photosynthesis, which moves carbon from the air into the plantsPhotosynthesis moves carbon from one place to another.
Something that moves the matter is a process, and a process is drawn as an arrow.
Why: The air is a place where carbon sits.
A place where the matter sits is a reservoir.
A reservoir is drawn as a box.
The drawing shows a cycle as boxes and arrows. A black pointer labeled this part marks one part.
Which of the following is the marked part?
- A. A reservoirA reservoir is drawn as a box.
The marked part is a line with an arrowhead. - B. ✓ A process
Why: The marked part is an arrow from one box to another.
An arrow is a process: it moves matter from one reservoir to another.
Rain falls from a cloud onto a field.
On a drawn cycle, which of the following is the rain falling?
- A. A reservoirA reservoir is a place where the matter sits.
The falling rain moves water from the cloud to the field. - B. ✓ A process
Why: The falling rain moves water from the cloud to the field.
Something that moves matter from one reservoir to another is a process.
A lake holds water all year.
On a drawn cycle, which of the following is the lake?
- A. ✓ A reservoir
- B. A processA process moves matter from one place to another.
The lake is a place where the water sits.
Why: The lake is a place where the water sits.
A place where the matter sits is a reservoir.
The drawing shows a cycle as boxes and arrows. A black pointer labeled this part marks one part.
Which of the following is the marked part?
- A. ✓ A reservoir
- B. A processA process is drawn as an arrow.
The marked part is a box.
Why: The marked part is a box.
A box is a reservoir: a place where the matter of the cycle sits.
A cow eats grass.
On a drawn cycle of carbon, which of the following is the eating?
- A. A reservoirA reservoir is a place where the carbon sits.
Eating moves carbon from the grass into the cow. - B. ✓ A process
Why: Eating moves carbon from the grass into the cow.
Something that moves matter from one reservoir to another is a process.
A river carries dissolved salts down to the sea.
On a drawn cycle, which of the following is the river carrying the salts?
- A. A reservoirA reservoir is a place where the matter sits.
The river moves the salts from the land to the sea. - B. ✓ A process
Why: The river carries the salts from the land to the sea.
Something that moves matter from one reservoir to another is a process.
29Quick quiz: reservoir, process mixed practice
The ocean holds most of the planet’s water.
Is the ocean a reservoir of water?
- A. ✓ Yes
- B. NoThe ocean is a place where water sits.
A place where the matter sits is a reservoir.
Why: The ocean is a place where water sits.
So the ocean is a reservoir of water.
A fox breathes out carbon dioxide.
Is the fox breathing out a process of the carbon cycle?
- A. ✓ Yes
- B. NoBreathing out moves carbon from the fox to the air.
Something that moves matter from one reservoir to another is a process.
Why: Breathing out moves carbon from the fox to the air.
So breathing out is a process.
A granite cliff holds phosphorus in its rock.
Is the cliff a process of the phosphorus cycle?
- A. YesA process moves matter from one reservoir to another.
The cliff is a place where the phosphorus sits. - B. ✓ No
Why: The cliff is a place where the phosphorus sits.
A place where the matter sits is a reservoir, not a process.
Every cycle has reservoirs and processes.
What is a reservoir?
- A. ✓ A place where the matter of the cycle sits
- B. What moves the matter from one place to anotherWhat moves the matter from one place to another is a process.
- C. The element whose atoms the cycle followsThe element is what moves; a reservoir is a place where it sits.
Why: A reservoir is a place where the matter of the cycle sits: the air, the water, the rock, the living things.
A mudflat’s nitrogen cycle has reservoirs and processes.
Which of the following is a process of that cycle?
- A. The nitrogen dissolved in the water over the mudflatThe dissolved nitrogen sits in one place, the water.
A place where the matter sits is a reservoir. - B. The cordgrass growing on the mudflatThe cordgrass is a place where nitrogen sits, built into its leaves and roots.
A place where the matter sits is a reservoir. - C. ✓ The cordgrass roots taking nitrogen in from the mud
Why: The roots take nitrogen from the mud into the cordgrass.
So the nitrogen moves from one reservoir, the mud, to another, the cordgrass.
Something that moves matter from one reservoir to another is a process.
Every cycle has reservoirs and processes.
(a) State what a reservoir and a process of a cycle are. (1 pt)
A process is something that moves the matter from one reservoir to another.
- Award 1 point for: a reservoir is a place where the matter sits (a store) and a process moves matter from one reservoir to another.
36Living boxes and non-living boxes
Video: Watch: Living boxes and non-living boxes
The named carbon diagram; the plants box and the animals box each gain a second outline while the air box and the soil box keep one; six cases sorted, alive or not, into the two kinds of reservoir.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L14b.mp4
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Look at the four carbon reservoirs again. Two of them are alive, and two of them are not.
The plants are alive. The animals are alive.
The air is not alive. The soil is not alive.
A reservoir that is not alive, like the air, the water or the rock, is called an reservoir. Bio- means life, and a- means not.
A reservoir that is alive, like the plants, the animals or the microbes in the soil, is called a biotic reservoir.
On every drawn cycle, a biotic reservoir has a double outline. An abiotic reservoir has a single outline.
Here are six cases, each judged by one question: is this reservoir alive?
For example, the air is not alive. So the air is an abiotic reservoir.
But a forest is alive. So a forest is a biotic reservoir.
And a herd of deer is alive. So a herd of deer is a biotic reservoir.
But the ocean’s water is not alive. So the ocean’s water is an abiotic reservoir.
And the fish in the ocean are alive. So the fish in the ocean are a biotic reservoir, even though they live inside an abiotic one.
But a mountain’s rock is not alive. So a mountain’s rock is an abiotic reservoir.
Here is a table of the six cases with their verdicts: whether each one is alive, and which kind of reservoir it is.
What you are expected to know Classify a reservoir as abiotic (not alive: air, water, rock) or biotic (alive: plants, animals, the microbes in the soil).
A cycle diagram of a river has a box for the river’s water.
Which outline does the water box carry?
- A. ✓ A single outline
- B. A double outlineA double outline marks a biotic reservoir.
Water is not alive.
Why: The river’s water is not alive, so the water is an abiotic reservoir.
An abiotic reservoir is drawn with a single outline.
A cycle diagram of a forest has a box for the trees.
Which outline does the trees box carry?
- A. A single outlineA single outline marks an abiotic reservoir.
Trees are alive. - B. ✓ A double outline
Why: The trees are alive, so the trees are a biotic reservoir.
A biotic reservoir is drawn with a double outline.
A lake’s water holds dissolved carbon dioxide.
Is the lake’s water a biotic reservoir?
- A. YesWater is not alive.
A reservoir that is not alive is abiotic. - B. ✓ No
Why: The lake’s water is not alive.
A reservoir that is not alive is an abiotic reservoir.
The grass of a meadow holds carbon in its leaves.
Is the grass a biotic reservoir?
- A. ✓ Yes
- B. NoGrass is alive.
A reservoir that is alive is biotic.
Why: The grass is alive.
A reservoir that is alive is a biotic reservoir.
The bacteria in a pond’s mud hold nitrogen.
Are the bacteria a biotic reservoir?
- A. ✓ Yes
- B. NoBacteria are alive.
A reservoir that is alive is biotic, even when it sits inside the mud.
Why: The bacteria are alive.
A reservoir that is alive is a biotic reservoir.
A cliff’s rock holds phosphorus.
Is the cliff’s rock a biotic reservoir?
- A. YesRock is not alive.
A reservoir that is not alive is abiotic. - B. ✓ No
Why: The cliff’s rock is not alive.
A reservoir that is not alive is an abiotic reservoir.
The air above a city holds carbon dioxide.
Is the air a biotic reservoir?
- A. YesAir is not alive.
A reservoir that is not alive is abiotic. - B. ✓ No
Why: The air is not alive.
A reservoir that is not alive is an abiotic reservoir.
The earthworms in a garden’s soil hold carbon.
Are the earthworms a biotic reservoir?
- A. ✓ Yes
- B. NoEarthworms are alive.
A reservoir that is alive is biotic, even when it lives in the soil.
Why: The earthworms are alive.
A reservoir that is alive is a biotic reservoir.
The drawing shows a cycle of carbon in a lake. One arrow runs from the water plants to the fish.
Which process does that arrow stand for?
- A. PhotosynthesisPhotosynthesis moves carbon from the lake’s water into the water plants.
That arrow runs from the water to the plants. - B. RespirationRespiration moves carbon from the plants or the fish back to the lake’s water.
Those arrows end at the water box. - C. ✓ Eating
Why: The arrow starts at the water plants and ends at the fish.
The fish eat the water plants, and eating moves carbon from the plants into the fish.
So the arrow stands for eating.
62Quick quiz: abiotic, biotic mixed practice
A reservoir of a cycle can be one of two kinds.
What does abiotic mean?
- A. AliveAlive is biotic.
A- means not, so abiotic means not alive. - B. ✓ Not alive
- C. Made of carbonAbiotic says whether the reservoir is alive, not what it is made of.
Why: Bio- means life and a- means not.
An abiotic reservoir is one that is not alive: air, water, rock.
A cycle diagram of an orchard names three reservoirs.
Which of the following is a biotic reservoir?
- A. The orchard’s soilSoil is not alive.
A reservoir that is not alive is abiotic. - B. The air above the orchardAir is not alive.
A reservoir that is not alive is abiotic. - C. ✓ The orchard’s pear trees
Why: The pear trees are alive.
A reservoir that is alive is a biotic reservoir.
A reservoir of a cycle can be one of two kinds.
(a) State what an abiotic reservoir and a biotic reservoir are. (1 pt)
A biotic reservoir is a store that is alive, such as the plants, the animals or the microbes in the soil.
- Award 1 point for: abiotic means not alive (air, water, rock) and biotic means alive (plants, animals, microbes).
66No atom is lost
A young tree doubles in mass over a summer.
Where did the atoms the tree added come from?
- A. The tree made the atoms from the energy of sunlightSunlight is energy, not matter.
Energy cannot become a carbon atom. - B. ✓ The tree took the atoms in from the air, the water and the soil
- C. The tree made the atoms inside its own cellsThe tree cannot make atoms, and neither can any other organism.
Why: The tree cannot make atoms.
Every atom it adds came from outside the tree: carbon dioxide from the air, water, and substances dissolved in the soil water.
Video: Watch: No atom is lost
The carbon diagram with the atoms drawn as dots in every box; one dot at a time hops along the photosynthesis arrow from the air to the plants; the air’s count falls as the plants’ count rises, and the total under the diagram stays the same.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L14c.mp4
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Now consider the carbon diagram with the atoms counted. Suppose the four reservoirs hold 20 carbon atoms in all: 8 in the air, 4 in the plants, 3 in the animals, 5 in the soil.
Over one day, photosynthesis moves 3 of the air’s carbon atoms into the plants’ sugar.
Now the air holds 5 carbon atoms and the plants hold 7. The animals still hold 3 and the soil still holds 5.
Count the dots in the second drawing: 20 carbon atoms, the same total as before.
No organism can make an atom, as Unit 1 said. No process on the diagram destroys an atom either.
So every atom that leaves one reservoir arrives in another. A fall in one reservoir is a rise in another, by the same number of atoms.
No process makes or destroys matter; processes only move it from reservoir to reservoir. Scientists call this the conservation of matter, and every biogeochemical cycle shows it.
Now consider a dead leaf on the soil. Its carbon atoms are not used up when the leaf rots.
Microbes in the soil respire the leaf’s carbon. So the carbon goes back to the air as carbon dioxide.
Because no atom is lost, every atom that leaves the air comes back to the air by some path. That is why the arrows of a cycle join up into a loop.
What you are expected to know Explain why a biogeochemical cycle shows the conservation of matter: atoms move from one reservoir to another, no process makes or destroys one, so a fall in one reservoir is a rise in another.
A deer respires all night. Its respiration moves carbon atoms out of the deer.
Which carbon reservoir gains those atoms?
- A. ✓ The air
- B. The plantsCarbon reaches the plants by photosynthesis, from the air.
Respiration releases carbon dioxide into the air. - C. The soilCarbon reaches the soil as dead remains and waste.
Respiration releases carbon dioxide into the air.
Why: Respiration releases the deer’s carbon as carbon dioxide.
The carbon dioxide goes into the air.
So the air gains the atoms the deer lost.
A salmon respires all night in a river. Its respiration moves carbon atoms out of the salmon.
(a) Explain why the number of carbon atoms entering the river water is the same as the number the salmon lost. (1 pt)
Frame The same number enter the water because …
Every carbon atom that leaves the salmon arrives somewhere.
Respiration releases those atoms into the water as carbon dioxide.
So the water gains one atom for every atom the salmon lost.
- Award 1 point for: no process makes or destroys an atom, so every atom that leaves the salmon enters the water (the fall in one reservoir equals the rise in the other).
A student says: “When a rabbit dies and its body rots away, the matter in it is used up and gone.”
Is the student correct?
- A. Yes: the rotted matter is used up and goneThe rabbit’s atoms are not destroyed.
Microbes respire them to the air, and the rest go into the soil. - B. ✓ No: its atoms move into the air and the soil
Why: No process destroys an atom.
The microbes in the soil respire the rabbit’s carbon, so that carbon goes to the air.
The rest of the rabbit’s matter goes into the soil.
So the rabbit’s matter is moved to other reservoirs, not used up.
A log burns to ash. Most of its carbon leaves the log as carbon dioxide.
What happens to the air’s store of carbon?
- A. It fallsThe carbon dioxide from the log goes into the air.
The air gains carbon. - B. It stays the sameThe log’s carbon atoms are not destroyed.
They go into the air as carbon dioxide. - C. ✓ It rises
Why: Burning releases the log’s carbon as carbon dioxide.
The carbon dioxide goes into the air.
So the air’s store of carbon rises by the carbon the log lost.
Here is the blank template again, four boxes and the arrows between them.
The boxes are reservoirs, living or non-living. The arrows are processes.
A fall in one box is a rise in another, because no atom is lost on the way.
87Mixed practice mixed practice
Suppose a pond’s water plants take in carbon by photosynthesis all day, and no process returns carbon to the water.
What happens to the store of carbon in the pond’s water over the day?
- A. ✓ It falls
- B. It stays the samePhotosynthesis moves carbon out of the water and into the plants.
The water loses that carbon. - C. It risesPhotosynthesis moves carbon out of the water, not into it.
Why: Photosynthesis moves carbon from the water into the plants.
Nothing returns carbon to the water.
So the water’s store of carbon falls by the carbon the plants gained.
A bed of seaweed holds carbon in its fronds.
Which kind of reservoir is the seaweed?
- A. ✓ A biotic reservoir
- B. An abiotic reservoirAn abiotic reservoir is not alive.
Seaweed is alive.
Why: Seaweed is alive.
A reservoir that is alive is a biotic reservoir.
The drawing shows a cycle as boxes and arrows. A black pointer labeled this part marks one part.
Which of the following is the marked part?
- A. A reservoirA reservoir is drawn as a box.
The marked part is a line with an arrowhead. - B. ✓ A process
Why: The marked part is an arrow from one box to another.
An arrow is a process: it moves matter from one reservoir to another.
A student says: “The arrows of a cycle join up into a loop because no atom is lost: every atom that leaves a reservoir arrives in another.”
Is the student correct?
- A. ✓ Yes: no atom is lost, so every atom arrives in another reservoir
- B. No: living things make new atoms to close the loopNo organism can make an atom.
The loop closes because every atom that leaves a reservoir arrives in another.
Why: No organism can make an atom.
No process destroys one.
So every atom that leaves a reservoir arrives in another and comes back by some path.
That is why the arrows join up into a loop.
The drawing shows a cycle of carbon in a lake. The fish box and the water box carry different outlines.
Which of the following is the reason the fish box carries the outline it does?
- A. The fish hold the most carbon of the lake’s reservoirsHow much carbon a reservoir holds is not drawn in its outline.
The outline shows whether the reservoir is alive. - B. The fish move about the lakeWhether a reservoir moves about is not drawn in its outline.
The outline shows whether the reservoir is alive. - C. ✓ The fish are alive
Why: A double outline marks a biotic reservoir.
A biotic reservoir is alive.
The fish are alive, so their box carries a double outline.
Wind blows dust from a dry field into the sea.
On a drawn cycle, which of the following is the wind carrying the dust?
- A. A reservoirA reservoir is a place where the matter sits.
The wind moves the dust from the field to the sea. - B. ✓ A process
Why: The wind moves the dust from the field to the sea.
Something that moves matter from one reservoir to another is a process.
Suppose the air above a forest holds 10 000 kg of carbon in spring. Over the summer, photosynthesis moves 500 kg of carbon from the air into the forest’s plants, and no other process moves carbon into or out of either reservoir.
(a) Explain how the changes in the forest’s air reservoir and plant reservoir demonstrate the conservation of matter. (2 pt)
Frame The changes demonstrate the conservation of matter because …
So the air reservoir falls by 500 kg of carbon.
No process makes or destroys an atom of that carbon on the way.
So the plant reservoir rises by the same 500 kg of carbon.
A fall in one reservoir is a rise in another by the same amount, which is the conservation of matter.
- Award 1 point for: the air reservoir falls by 500 kg of carbon and the plant reservoir rises by 500 kg of carbon (the same amount).
- Award 1 point for: no atom is made or destroyed, so the fall in one reservoir equals the rise in the other (conservation of matter).
Glossary
- reservoir and process (of a cycle)
- A reservoir is a place where the matter of a cycle sits: the air, the water, the rock, the plants, the animals, the microbes in the soil. A process is something that moves the matter from one reservoir to another: photosynthesis moves carbon from the air to the plants. On a drawn cycle, a box is a reservoir and an arrow is a process.
- abiotic and biotic
- An abiotic reservoir is not alive: the air, the water, the rock (a- means not, bio- means life). A biotic reservoir is alive: the plants, the animals, the microbes in the soil. On every drawn cycle, a biotic reservoir has a double outline and an abiotic reservoir a single outline.
APBIO-U08-L15 Where the rain came from
A raindrop sits on a leaf. Yesterday it was in the sea.
Tomorrow it will be inside the plant, and by the weekend it will be back in the air. Which moves took it round, and what are they called?
Unit 8 · Ecology
1Name the boxes and arrows for water
Unit 1 studied the properties of water.
What is water’s heat of vaporization?
- A. ✓ The energy needed to turn one gram of liquid water into a gas
- B. The temperature at which liquid water boilsThe boiling point is a temperature, measured in degrees.
Heat of vaporization is an amount of energy, per gram. - C. The energy needed to warm one gram of water by 1 °CThe energy that warms one gram of water by 1 °C is water’s specific heat.
Heat of vaporization turns a gram of liquid into a gas.
Why: Heat of vaporization is the energy needed to turn one gram of a liquid into a gas.
Water’s heat of vaporization is high: each gram that evaporates carries away a lot of energy.
Video: Watch: Name the boxes and arrows for water
The cycle template with its boxes named for water; each arrow drawn bold in turn as its process is named: evaporation, condensation, precipitation, transpiration.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L15a.mp4
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How does water move between its stores?
Water sits in four reservoirs: the oceans, the surface water on land, the atmosphere and living organisms.
Four processes carry water from one reservoir to the next. Once the four processes have names, one molecule can be followed all the way round.
Here is the cycle template from the last lesson, with each box named for water. The living organisms box has the double outline of a biotic reservoir; the other three reservoirs are abiotic.
The sun warms the surface of the sea. The fastest water molecules break free of the liquid and leave as water vapor, a gas.
Water leaving a liquid surface as vapor is called .
Each gram of water that evaporates carries away a lot of energy. That energy is water’s heat of vaporization, and the sun supplies it.
Lakes and rivers lose water to the air the same way. The template draws the evaporation arrow from the oceans, where most of the world’s evaporation happens.
Far above the ground, the air is colder. The vapor cools, and the water molecules join into tiny droplets of liquid water.
Those droplets, floating in the air, are a cloud.
Vapor turning back into liquid droplets is called .
Condensation happens inside the atmosphere. So on the template it is drawn as an arrow that loops out of the atmosphere box and back into it.
The droplets in a cloud join into drops heavy enough to fall. They fall as rain, or they freeze and fall as snow.
Water falling from a cloud as rain or snow is called .
Rain falls on the sea and on the land. So two precipitation arrows leave the atmosphere: one to the oceans and one to the surface water.
On land, some rain flows in rivers to the sea. Some soaks into the soil, and a plant’s roots absorb it.
Water travels up through the plant to its leaves. There it evaporates through tiny holes and leaves the leaf as vapor.
Water leaving a plant’s leaves as vapor is called .
Here is the template with every box and every arrow named.
The four reservoirs and the processes that join them are called the . Another name for it is the hydrologic cycle.
What you are expected to know Name the water cycle’s four reservoirs and four processes on the cycle template: the oceans, surface water, the atmosphere and living organisms; evaporation, condensation, precipitation and transpiration.
On the water cycle template below, no box is named. One box is marked with a question mark.
Which reservoir is the marked box?
- A. The oceansThe oceans are the bottom-left box.
The evaporation arrow leaves it. - B. Surface waterSurface water is the bottom-right box.
The river flow arrow leaves it. - C. The atmosphereThe atmosphere is the top box.
The condensation loop sits on it. - D. ✓ Living organisms
Why: The marked box has a double outline, so it is the biotic reservoir.
The absorbed-by-roots arrow enters it, and the transpiration arrow leaves it.
So the marked box is living organisms.
On the water cycle template below, no box is named. One box is marked with a question mark.
Which reservoir is the marked box?
- A. The oceansThe oceans are the bottom-left box.
The evaporation arrow leaves it. - B. Surface waterSurface water is the bottom-right box.
The river flow arrow leaves it. - C. ✓ The atmosphere
- D. Living organismsLiving organisms are the middle box, with a double outline.
The transpiration arrow leaves it.
Why: The marked box is at the top of the template.
Evaporation and transpiration enter it, and the condensation loop leaves it and comes back.
So the marked box is the atmosphere.
On the water cycle template below, no box is named. One box is marked with a question mark.
Which reservoir is the marked box?
- A. ✓ The oceans
- B. Surface waterSurface water is the bottom-right box.
The river flow arrow leaves it. - C. The atmosphereThe atmosphere is the top box.
The condensation loop sits on it. - D. Living organismsLiving organisms are the middle box, with a double outline.
The transpiration arrow leaves it.
Why: The marked box is at the bottom left.
The river flow arrow enters it, and rivers end in the sea.
So the marked box is the oceans.
On the water cycle template below, no box is named. One box is marked with a question mark.
Which reservoir is the marked box?
- A. The oceansThe oceans are the bottom-left box.
The evaporation arrow leaves it. - B. ✓ Surface water
- C. The atmosphereThe atmosphere is the top box.
The condensation loop sits on it. - D. Living organismsLiving organisms are the middle box, with a double outline.
The transpiration arrow leaves it.
Why: The marked box is at the bottom right.
The absorbed-by-roots arrow leaves it, and rivers flow from it to the sea.
So the marked box is surface water: lakes, rivers and soil.
The water cycle template below has its boxes named and its arrows unnamed. One arrow is drawn bold.
Which process is the bold arrow?
- A. EvaporationEvaporation leaves a liquid surface such as the sea.
The bold arrow leaves living organisms. - B. CondensationCondensation happens inside the atmosphere, so its arrow loops back into the atmosphere box.
- C. PrecipitationPrecipitation falls out of the atmosphere.
The bold arrow goes up into the atmosphere. - D. ✓ Transpiration
Why: The bold arrow leaves the living organisms box and enters the atmosphere.
Water leaving a plant’s leaves as vapor is transpiration.
The water cycle template below has its boxes named and its arrows unnamed. One arrow is drawn bold.
Which process is the bold arrow?
- A. ✓ Evaporation
- B. CondensationCondensation happens inside the atmosphere, so its arrow loops back into the atmosphere box.
- C. PrecipitationPrecipitation falls out of the atmosphere.
The bold arrow goes up into the atmosphere. - D. TranspirationTranspiration leaves living organisms.
The bold arrow leaves the oceans.
Why: The bold arrow leaves the oceans and enters the atmosphere.
Water leaving the sea’s surface as vapor is evaporation.
The water cycle template below has its boxes named and its arrows unnamed. One arrow is drawn bold.
Which process is the bold arrow?
- A. EvaporationEvaporation carries water from a liquid surface into the atmosphere.
The bold arrow starts and ends in the atmosphere. - B. ✓ Condensation
- C. PrecipitationPrecipitation carries water out of the atmosphere.
The bold arrow starts and ends in the atmosphere. - D. TranspirationTranspiration carries water from living organisms into the atmosphere.
The bold arrow starts and ends in the atmosphere.
Why: The bold arrow loops out of the atmosphere box and back into it.
The one process that happens inside the atmosphere is vapor turning into droplets: condensation.
The water cycle template below has its boxes named and its arrows unnamed. Two arrows are drawn bold.
Which process are the two bold arrows?
- A. EvaporationEvaporation goes up into the atmosphere.
The bold arrows come down out of it. - B. CondensationCondensation happens inside the atmosphere, so its arrow loops back into the atmosphere box.
- C. ✓ Precipitation
- D. TranspirationTranspiration goes up from living organisms into the atmosphere.
The bold arrows come down out of it.
Why: The bold arrows leave the atmosphere and enter the oceans and surface water.
Water falling from a cloud as rain or snow is precipitation.
33Quick quiz: evaporation, condensation, precipitation, transpiration, water cycle mixed practice
Water moves between its reservoirs by named processes.
What is evaporation?
- A. ✓ Water leaving a liquid surface as vapor
- B. Vapor turning back into liquid dropletsVapor turning back into liquid droplets is condensation.
- C. Water leaving a plant’s leaves as vaporWater leaving a plant’s leaves as vapor is transpiration.
Why: Evaporation is water leaving a liquid surface, such as the sea, as vapor.
Water moves between its reservoirs by named processes.
What is condensation?
- A. Water falling from a cloud as rain or snowWater falling from a cloud as rain or snow is precipitation.
- B. ✓ Vapor turning back into liquid droplets
- C. Water leaving a liquid surface as vaporWater leaving a liquid surface as vapor is evaporation.
Why: Condensation is water vapor turning back into liquid droplets; the droplets are a cloud.
Water moves between its reservoirs by named processes.
What is precipitation?
- A. Vapor turning back into liquid dropletsVapor turning back into liquid droplets is condensation.
- B. Water leaving a plant’s leaves as vaporWater leaving a plant’s leaves as vapor is transpiration.
- C. ✓ Water falling from a cloud as rain or snow
Why: Precipitation is water falling from a cloud as rain or snow.
Water moves between its reservoirs by named processes.
What is transpiration?
- A. ✓ Water leaving a plant’s leaves as vapor
- B. Roots absorbing water from the soilRoots absorbing water carries water into a plant.
Transpiration carries water out of a plant. - C. Water leaving a liquid surface as vaporWater leaving a liquid surface, such as the sea, as vapor is evaporation.
Why: Transpiration is water leaving a plant’s leaves as vapor.
Water moves between its reservoirs by named processes.
What is the water cycle?
- A. Water’s path from the sea into a cloudThe path from the sea into a cloud is two processes, evaporation and condensation, not the whole cycle.
- B. Water’s turning from vapor into rainVapor turning into cloud and rain is condensation and precipitation, two of the processes, not the whole cycle.
- C. ✓ Water’s four reservoirs and the processes joining them
Why: The water cycle is the four reservoirs of water and the processes that move water between them.
Water moves between its reservoirs by named processes.
(a) State what condensation is. (1 pt)
- Award 1 point for: water vapor (a gas) turning into liquid water droplets. Accept with or without: the droplets form a cloud.
40Follow one molecule all the way round
Video: Watch: Follow one molecule all the way round
One molecule leaving the sea, joining a cloud, falling on a field, absorbed by a root, leaving the leaf, and falling back into the sea, each arrow named as it is crossed.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L15b.mp4
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Now follow one water molecule all the way round. Suppose the molecule sits in the ocean, just below the surface.
Here is its path drawn bold on the template. Each arrow it crosses is one process, and the name is on the arrow.
The molecule’s path, one arrow at a time:
- Evaporation. The sun warms the sea surface. The molecule breaks free and rises into the atmosphere as vapor.
- Condensation. Far above the ground, the vapor cools. The molecule joins a droplet in a cloud.
- Precipitation. The drop grows heavy and falls as rain onto a field. The molecule soaks into the soil.
- Absorbed by a root. A wheat plant’s root absorbs the molecule from the soil. The molecule travels up to a leaf.
- Transpiration. The molecule leaves the leaf as vapor and rises into the atmosphere.
- Condensation again. The molecule joins a new cloud.
- Precipitation again. The cloud drifts over the sea. The rain carries the molecule back into the ocean.
The molecule is back where it began, in the ocean. On the way it passed through all four processes: evaporation, condensation, precipitation and transpiration.
Every molecule can be followed like this. Start in a named box, cross one arrow at a time, name the process on each arrow, and stop when you are back in the box you started in.
What you are expected to know Trace a water molecule from any named reservoir through the water cycle back to where it began, naming each process in order.
Now trace one yourself, one arrow at a time. Suppose a water molecule is in a hill lake, and a river carries it down to the sea.
A river has carried a water molecule from a hill lake into the sea. The sun warms the sea’s surface, and the molecule rises into the atmosphere as vapor.
Which process carried the molecule from the sea into the atmosphere?
- A. ✓ Evaporation
- B. PrecipitationPrecipitation carries water down out of the atmosphere.
The molecule went up into the atmosphere. - C. TranspirationTranspiration carries water out of a plant’s leaves.
The molecule left the sea.
Why: The molecule left the sea’s surface as vapor.
Water leaving a liquid surface as vapor is evaporation.
A water molecule from the sea is now in the atmosphere. It cools and joins a droplet in a cloud.
Which process is that?
- A. EvaporationEvaporation turns liquid into vapor.
The molecule turned from vapor into a liquid droplet. - B. ✓ Condensation
Why: The molecule turned from vapor into part of a liquid droplet.
Vapor turning back into liquid droplets is condensation.
A water molecule in a cloud droplet over the hills grows heavy and falls as rain into a hill lake.
Which process carried the molecule into the lake?
- A. CondensationCondensation made the droplet inside the cloud.
The fall from the cloud into the lake is a different process. - B. ✓ Precipitation
- C. TranspirationTranspiration carries water out of a plant’s leaves.
The molecule fell from a cloud.
Why: The molecule fell from a cloud as rain.
Water falling from a cloud as rain or snow is precipitation.
A water molecule rose from the sea into the atmosphere, joined a cloud, and fell as rain into a hill lake.
In order, which processes did the molecule pass through?
- A. ✓ Evaporation, condensation, precipitation
- B. Condensation, evaporation, precipitationThe molecule left the sea first, and a molecule leaves a liquid surface by evaporation.
Condensation came second, in the atmosphere. - C. Evaporation, precipitation, condensationThe droplet formed before the fall.
Condensation makes the droplet; precipitation is the fall.
Why: The molecule left the sea by evaporation.
In the atmosphere it joined a droplet by condensation.
The droplet fell as rain into the lake by precipitation.
Suppose a water molecule is in a droplet in a cloud over a forest. It falls, an oak root absorbs it from the soil, and later it returns to a cloud.
In order, which processes did the molecule pass through?
- A. ✓ Precipitation, transpiration, condensation
- B. Condensation, precipitation, evaporationThe molecule started in a droplet, so condensation had already happened.
Water leaves a leaf by transpiration. - C. Transpiration, precipitation, condensationThe molecule started in a cloud, so the first arrow it crossed was the fall as rain: precipitation.
Why: The molecule fell from the cloud by precipitation.
A root absorbed it, and it left the tree’s leaves as vapor by transpiration.
In the atmosphere it joined a new droplet by condensation.
Suppose a water molecule is inside a leaf of a maple tree. Over the next week the molecule leaves the tree and comes back into a plant.
(a) Trace the molecule from inside the leaf until it is back inside a plant, naming each process it passes through in order. (2 pt)
Frame From inside the leaf, the molecule …
In the atmosphere the vapor cools, and the molecule joins a cloud droplet by condensation.
The droplet falls as rain onto the soil by precipitation.
A root absorbs the molecule from the soil, so the molecule is back inside a plant.
- Award 1 point for: transpiration, then condensation, then precipitation, named in that order. Accept any correct chain of named processes from the leaf to the soil or surface water, for example with evaporation from a puddle, condensation and precipitation added before a root absorbs the water.
- Award 1 point for: a root absorbing the water from the soil (or from surface water) returns it to a plant.
Here is the raindrop again, sitting on its leaf. Yesterday it was in the sea.
Evaporation lifted it from the sea into the atmosphere. Condensation made it part of a cloud, and precipitation dropped it on the leaf.
Tomorrow it will drip into the soil, and a root will absorb it into the plant. By the weekend, transpiration will send it back into the air.
58Mixed practice mixed practice
On a clear night, drops of water form on the grass.
Which process is that?
- A. EvaporationEvaporation turns liquid water into vapor.
The vapor here turned into liquid drops. - B. PrecipitationPrecipitation is water falling from a cloud.
These drops formed on the grass, out of the air. - C. ✓ Condensation
Why: Water vapor in the air turned into liquid drops.
Vapor turning back into liquid is condensation.
A puddle on a road is gone by the afternoon of a sunny day.
Which process moved the puddle’s water?
- A. ✓ Evaporation
- B. PrecipitationPrecipitation brings water down out of a cloud.
The puddle’s water left the road. - C. TranspirationTranspiration is water leaving a plant’s leaves.
A puddle is not a plant.
Why: The sun warmed the puddle, and its water left the surface as vapor.
Water leaving a liquid surface as vapor is evaporation.
Transpiration carries water out of a plant’s leaves.
Which reservoir does transpiration carry the water into?
- A. ✓ The atmosphere
- B. The oceansThe oceans receive water from rain and rivers.
Vapor from a leaf rises into the air. - C. Surface waterSurface water receives rain and rivers.
Vapor from a leaf rises into the air.
Why: Water leaves the leaf as vapor.
The vapor rises into the air, so it enters the atmosphere.
Suppose a water molecule falls as snow onto a hillside. The snow melts, a river carries the water to the sea, and in summer the molecule leaves the sea into the air.
In order, which two processes did the molecule pass through?
- A. Evaporation, then precipitationThe molecule fell as snow first.
Water falling from a cloud as snow is precipitation. - B. ✓ Precipitation, then evaporation
- C. Precipitation, then transpirationThe molecule left the sea, not a plant’s leaves.
Water leaving a liquid surface as vapor is evaporation.
Why: Snow falling from a cloud is precipitation.
Then the molecule left the sea’s surface as vapor: evaporation.
A student says: “Transpiration is rain falling onto a plant’s leaves.”
Is the student correct?
- A. Yes: transpiration is rain falling on the leavesRain falling from a cloud onto a leaf is precipitation.
Transpiration is water leaving the leaf as vapor. - B. ✓ No: transpiration is water leaving the leaves as vapor
Why: Transpiration is water leaving a plant’s leaves as vapor, into the atmosphere.
Rain falling onto the leaves is precipitation, coming out of the atmosphere.
The student has the direction reversed: transpiration carries water up out of the plant.
A student says: “The water in a cloud over the sea may have left a leaf in a forest the day before.”
Is the student correct?
- A. ✓ Yes: transpired vapor rises, and the air carries it
- B. No: a cloud over the sea holds only sea waterTranspiration carries water from leaves into the atmosphere, and the air moves.
A cloud’s water can come from any reservoir.
Why: Water left the leaf as vapor by transpiration.
The vapor rose into the atmosphere, and the air carried it out over the sea.
There it condensed into the cloud, so the student is correct.
A student says: “The water a plant’s roots absorb stays inside the plant for good.”
Is the student correct?
- A. Yes: absorbed water stays in the plantAlmost all the water a plant absorbs leaves its leaves as vapor by transpiration.
So the water returns to the atmosphere. - B. ✓ No: almost all of it leaves the leaves as vapor
Why: A root absorbs water from the soil.
The water travels up to the leaves and leaves them as vapor by transpiration.
So almost all of the water the plant absorbs returns to the atmosphere; it is not used up.
Snow lies on a fence post on a hill. Two days ago, the same water was in a lake in the valley.
(a) Identify, in order, the processes that carried the water from the lake to the fence post. (1 pt)
- Award 1 point for: evaporation, condensation, precipitation, in that order.
(b) The snow melts, a root absorbs the water, and the water later leaves a leaf as vapor. Explain how this shows that no water molecule is lost from the water cycle. (1 pt)
Frame No water molecule is lost because …
No atom is made or destroyed on the way.
So the plant’s store of water falls, and the atmosphere’s store rises by the same amount.
- Award 1 point for: the water moves into another reservoir (the atmosphere); none is destroyed, so a fall in one reservoir is a rise in another.
Glossary
- evaporation
- Water leaving a liquid surface, such as the sea, as water vapor, a gas. On the cycle template it is the arrow from the oceans up to the atmosphere.
- condensation
- Water vapor turning back into tiny droplets of liquid water; the droplets are a cloud. It happens inside the atmosphere, so on the template it is drawn as a loop out of the atmosphere box and back into it.
- precipitation
- Water falling from a cloud as rain or snow. On the cycle template it is the arrow from the atmosphere down to the oceans or to surface water.
- transpiration
- Water leaving a plant’s leaves as vapor through tiny holes. On the cycle template it is the arrow from living organisms up to the atmosphere.
- water cycle (hydrologic cycle)
- Water’s four reservoirs (the oceans, surface water, the atmosphere and living organisms) and the processes that move water between them (evaporation, condensation, precipitation and transpiration). Another name for it is the hydrologic cycle.
APBIO-U08-L16 Where your carbon has been
Here is a carbon atom in the sugar of a leaf. Last month it was carbon dioxide in the air.
Next month it may be in a caterpillar, in a bird, or back in the air. Which arrows can it take, and which arrow did a felled forest change?
Unit 8 · Ecology
1Four arrows touch the air
Unit 2 wrote photosynthesis as a word equation.
Which of the following is the word equation for photosynthesis?
- A. glucose + oxygen → carbon dioxide + water, and energy is releasedGlucose and oxygen on the left is aerobic cellular respiration’s equation.
Photosynthesis builds glucose; it does not use it up. - B. ✓ carbon dioxide + water → glucose + oxygen, using the energy of light
- C. carbon dioxide + oxygen → glucose + water, using the energy of lightPhotosynthesis uses water, not oxygen.
Oxygen is what photosynthesis gives off.
Why: Photosynthesis takes in carbon dioxide and water.
It builds glucose and gives off oxygen.
The energy comes from light: carbon dioxide + water → glucose + oxygen, using the energy of light.
Every cycle drawn as boxes and arrows uses one template.
Which of the following does an arrow stand for?
- A. ✓ A process that moves matter from one reservoir to another
- B. A place where the matter of the cycle sitsA place where the matter sits is a reservoir, and a reservoir is drawn as a box.
Why: An arrow is a process.
A process moves matter from one reservoir to another.
Video: Watch: Four arrows touch the air
The cycle template filling in for carbon: the air, the living things, the dead material and the fossil fuel take their names; then the four arrows that touch the air are drawn and named one at a time, photosynthesis in, cellular respiration, decomposition and combustion out.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L16a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L16a.mp4
How does carbon leave the air, and how does it get back?
At the largest scale, four processes move carbon into or out of the air: photosynthesis, cellular respiration, the rotting of dead material, and burning. Unit 2 gave the first two as word equations.
Photosynthesis takes carbon dioxide out of the air and builds its carbon into carbohydrate. Cellular respiration sends that carbon back to the air as carbon dioxide.
Rotting is respiration too: the fungi and microbes that eat a dead log respire its carbon. So a rotting log returns its carbon to the air.
Burning wood or fossil fuel returns carbon to the air too. Fossil fuel’s carbon was stored underground long ago.
Here is the cycle template, filled in for carbon.
Carbon sits in four reservoirs:
- the air, as carbon dioxide
- the living things, plants and animals, in their sugar, starch, protein and fat
- dead material in the soil
- fossil fuel, coal, oil and natural gas, deep in the rock
In this drawing every living thing, plant or animal, sits in one box. The air, the dead material and the fossil fuel are not alive, so their boxes have a single outline.
Start with the arrow from Unit 2. In sunlight, a leaf takes carbon dioxide in from the air and builds its carbon into glucose.
That is photosynthesis, in Unit 2’s words: carbon dioxide + water → glucose + oxygen, using the energy of light.
So photosynthesis is the arrow from the air into the living things. It is the only arrow that takes carbon out of the air.
Now consider a caterpillar eating the leaf. The carbon goes from the leaf into the caterpillar.
The leaf and the caterpillar are both living things. So the carbon stays inside the living-things box.
The caterpillar’s cells respire the glucose. That is cellular respiration, in Unit 2’s words: glucose + oxygen → carbon dioxide + water, and energy is released.
The carbon dioxide leaves the caterpillar and goes into the air. So cellular respiration is the arrow from the living things back to the air.
Every living thing respires, plants included. So the cellular respiration arrow carries carbon from every plant and animal, day and night.
Now consider a log that has fallen in the forest. Fungi and microbes in the soil eat the dead wood.
Their cells respire the log’s carbon, by the word equation of cellular respiration. So carbon dioxide leaves the rotting log and goes into the air.
Fungi and microbes breaking dead material down, and respiring its carbon back to the air as carbon dioxide, is called .
Decomposition is the arrow from the dead material to the air. The rotting log’s carbon is not destroyed: it goes into the air, respired by the fungi and microbes that eat the log.
On the drawing, cellular respiration is the arrow from the living things to the air. Decomposition is the arrow from the dead material to the air, even though it is the decomposers’ cells that respire.
Now consider a log on a fire, or coal burning in a power station. The fuel’s carbon joins oxygen from the air, and the carbon leaves as carbon dioxide.
A fuel burning, its carbon leaving as carbon dioxide, is called . Combustion is the arrow from the fossil fuel, or from a burning plant, to the air.
Coal, oil and natural gas are fossil fuels: the remains of living things buried millions of years ago, before fungi and microbes could decompose them.
So burning fossil fuel returns carbon to the air that was stored underground for millions of years.
Count the arrows that touch the air box: four. Photosynthesis takes carbon out of the air; cellular respiration, decomposition and combustion put it back.
The other two arrows, death and burial, move carbon between the boxes on the land. Neither arrow touches the air.
Carbon moving round between the air, the living things, the dead material and the fossil fuel by these arrows is called the .
The table below compares the four processes that touch the air: where each moves carbon from, where it moves carbon to, and what does the moving.
Here are four cases, each judged by one question: which process moved the carbon?
For example, a fern in sunlight takes carbon dioxide in from the air. This is photosynthesis, because the carbon went from the air into a living thing.
But a fox breathes carbon dioxide out into the air. This is cellular respiration, because the fox’s own living cells respired the carbon.
And a rotting apple gives off carbon dioxide into the air. This is decomposition, because fungi and microbes respired the carbon of something dead.
But a gas cooker’s flame gives off carbon dioxide into the air. This is combustion, because the carbon left a burning fuel, with no living cell respiring it.
The table below shows the four cases with their verdicts: which process moved the carbon in each case.
What you are expected to know Identify, on the carbon cycle drawn as boxes and arrows, the four processes that touch the air: photosynthesis takes carbon into the living things; cellular respiration, decomposition and combustion return it to the air.
The drawing below shows the carbon cycle with its boxes named and its arrows unnamed. One arrow is drawn bold.
Which named arrow of the carbon cycle is the bold arrow?
- A. ✓ Photosynthesis
- B. Cellular respirationCellular respiration goes from the living things up to the air.
The bold arrow goes from the air down to the living things. - C. DecompositionDecomposition goes from the dead material up to the air.
The bold arrow starts at the air. - D. CombustionCombustion goes from the fossil fuel up to the air.
The bold arrow starts at the air.
Why: The bold arrow starts at the air and ends at the living things.
Photosynthesis takes carbon from the air into the living things.
So the bold arrow is photosynthesis.
The drawing below shows the carbon cycle with its boxes named and its arrows unnamed. One arrow is drawn bold.
Which named arrow of the carbon cycle is the bold arrow?
- A. PhotosynthesisPhotosynthesis goes from the air down to the living things.
The bold arrow starts at the dead material. - B. Cellular respirationCellular respiration goes from the living things up to the air.
The bold arrow starts at the dead material. - C. ✓ Decomposition
- D. CombustionCombustion goes from the fossil fuel up to the air.
The bold arrow starts at the dead material.
Why: The bold arrow starts at the dead material and ends at the air.
Decomposition returns dead material’s carbon to the air.
So the bold arrow is decomposition.
The drawing below shows the carbon cycle with its boxes named and its arrows unnamed. One arrow is drawn bold.
Which named arrow of the carbon cycle is the bold arrow?
- A. PhotosynthesisPhotosynthesis goes from the air down to the living things.
The bold arrow starts at the fossil fuel. - B. Cellular respirationCellular respiration goes from the living things up to the air.
The bold arrow starts at the fossil fuel. - C. DecompositionDecomposition goes from the dead material up to the air.
The bold arrow starts at the fossil fuel. - D. ✓ Combustion
Why: The bold arrow starts at the fossil fuel and ends at the air.
Combustion returns a burning fuel’s carbon to the air.
So the bold arrow is combustion.
The drawing below shows the carbon cycle with its boxes named and its arrows unnamed. One arrow is drawn bold.
Which named arrow of the carbon cycle is the bold arrow?
- A. PhotosynthesisPhotosynthesis goes from the air down to the living things.
The bold arrow goes from the living things up to the air. - B. ✓ Cellular respiration
- C. DecompositionDecomposition starts at the dead material.
The bold arrow starts at the living things. - D. CombustionCombustion starts at the fossil fuel.
The bold arrow starts at the living things.
Why: The bold arrow starts at the living things and ends at the air.
Cellular respiration returns living things’ carbon to the air.
So the bold arrow is cellular respiration.
A wildfire burns through a field of dry grass.
Which process moves the grass’s carbon into the air?
- A. PhotosynthesisPhotosynthesis takes carbon out of the air.
The burning grass puts carbon into the air. - B. Cellular respirationCellular respiration is done by living cells.
Fire, not the grass’s cells, sends this carbon into the air. - C. DecompositionDecomposition is done by fungi and microbes eating dead material.
Here the grass burns. - D. ✓ Combustion
Why: The grass is a fuel, and it burns.
Its carbon joins oxygen and leaves as carbon dioxide.
A fuel burning is combustion.
A dead fish lies on the bottom of a pond. Bacteria in the mud feed on it, and carbon dioxide leaves the mud.
Which named arrow of the carbon cycle moves the fish’s carbon out of the mud?
- A. PhotosynthesisPhotosynthesis takes carbon dioxide in.
Here carbon dioxide is given off. - B. Cellular respirationThe bacteria do respire the fish’s carbon.
On the drawing, carbon leaving dead material is the decomposition arrow; cellular respiration leaves the living things. - C. ✓ Decomposition
- D. CombustionCombustion is a fuel burning.
Nothing burns under water.
Why: The fish is dead material.
Bacteria eat the dead fish and respire its carbon.
Microbes respiring dead material’s carbon to carbon dioxide is decomposition.
A student says: “When a log rots away, its carbon is destroyed. The log is simply gone.”
Is the student correct?
- A. Yes: the log’s carbon is destroyedNo process destroys an atom.
The fungi and microbes that eat the log respire its carbon into the air. - B. ✓ No: fungi and microbes respire its carbon into the air
Why: Fungi and microbes eat the rotting log.
Their cells respire the log’s carbon.
The carbon leaves as carbon dioxide and goes into the air.
So the log’s carbon is moved to the air, not destroyed.
48Quick quiz: decomposition, combustion, carbon cycle mixed practice
A cow breathes out carbon dioxide.
Is the cow breathing out decomposition?
- A. YesDecomposition is fungi and microbes respiring dead material’s carbon.
The cow is alive, and its own cells respired this carbon. - B. ✓ No
Why: The cow is a living thing.
Its own cells respired the glucose.
A living thing’s cells respiring is cellular respiration, not decomposition.
A candle burns down over an evening.
Is the candle burning combustion?
- A. ✓ Yes
- B. NoThe candle’s wax is a fuel.
The wax burns, and its carbon leaves as carbon dioxide.
Why: The wax is a fuel, and it burns.
Its carbon joins oxygen and leaves as carbon dioxide.
A fuel burning is combustion.
Carbon moves between the air, the living things, the dead material and the fossil fuel.
What is the carbon cycle?
- A. The energy of sunlight passing through an ecosystem once and leaving as heatEnergy passes through once; it is not carbon and it does not go round.
- B. The word equation for photosynthesis, carbon dioxide and water making glucoseThe word equation for photosynthesis describes one arrow of the cycle, not the whole cycle.
- C. ✓ Carbon moving round between the air, living things, dead material and fossil fuel
Why: The carbon cycle is carbon moving round between its reservoirs: the air, the living things, the dead material and the fossil fuel.
Dead material returns its carbon to the air.
What is decomposition?
- A. ✓ Fungi and microbes respiring dead material’s carbon to the air
- B. A fuel burning and its carbon leaving as carbon dioxideA fuel burning is combustion.
- C. A plant taking carbon dioxide in from the airA plant taking carbon dioxide in is photosynthesis.
Why: Decomposition is fungi and microbes breaking dead material down.
They respire its carbon back to the air as carbon dioxide.
A fuel returns its carbon to the air.
What is combustion?
- A. Fungi and microbes respiring dead material’s carbon to the airFungi and microbes respiring dead material’s carbon is decomposition.
- B. ✓ A fuel burning and its carbon leaving as carbon dioxide
- C. A living thing’s cells respiring glucoseA living thing’s cells respiring glucose is cellular respiration.
Why: Combustion is a fuel burning.
The fuel’s carbon joins oxygen and leaves as carbon dioxide.
Dead material and fuel both return their carbon to the air.
(a) State what decomposition and combustion are. (1 pt)
Combustion is a fuel burning so that its carbon leaves as carbon dioxide.
- Award 1 point for: decomposition is fungi and microbes breaking down dead material and respiring its carbon to carbon dioxide; combustion is a fuel burning and its carbon leaving as carbon dioxide.
55Change one arrow, watch the air
A gardener burns a heap of dry leaves. The leaves’ carbon leaves them as carbon dioxide.
What happens to the air’s store of carbon?
- A. It fallsThe carbon dioxide from the leaves goes into the air.
The air gains carbon. - B. It stays the sameNo process destroys an atom.
The leaves’ carbon goes into the air as carbon dioxide. - C. ✓ It rises
Why: Burning releases the leaves’ carbon as carbon dioxide.
The carbon dioxide goes into the air.
So the air’s store of carbon rises by the carbon the leaves lost.
Video: Watch: Change one arrow, watch the air
The carbon cycle with the atoms counted in every box; the combustion arrow thickens, atoms leave the fossil fuel box along it, and the air box fills as its count rises while the total under the drawing stays the same.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L16b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L16b.mp4
Speed up one arrow, or slow one down, and the air’s store of carbon changes.
Clear a forest, and the plants take less carbon out of the air. Burn more fuel, and the fires put more carbon in.
Look at the air box on the carbon cycle. One arrow takes carbon out of it: photosynthesis.
Three arrows put carbon into it: cellular respiration, decomposition and combustion.
Suppose that over one year, those three arrows put as much carbon into the air as photosynthesis takes out. Then the air’s store of carbon stays steady.
Now consider people burning more coal, oil and gas each year. The combustion arrow carries more carbon into the air each year.
Here is the carbon cycle drawn twice, with the carbon atoms counted.
Left, a steady year: the air holds 12 carbon atoms. Right, a year of burning more fuel: the combustion arrow is drawn bold.
In the year of burning more fuel, the fossil fuel box loses 3 carbon atoms and the air box gains 3. The air now holds 15 carbon atoms.
More carbon comes into the air than photosynthesis takes out. So the air’s store of carbon dioxide rises.
No process made an atom on the way. The carbon the air gained is the carbon the fossil fuel lost, as the drawing shows.
Now consider farmers clearing a forest for fields. They cut the trees down, and far fewer plants photosynthesize there now.
Here is the carbon cycle with the photosynthesis arrow dashed. In this drawing, a dashed arrow is one that has slowed.
Photosynthesis now takes less carbon out of the air. Cellular respiration, decomposition and combustion still put carbon in.
So more carbon goes into the air than comes out. The air’s store of carbon dioxide rises.
The felled trees’ carbon does not vanish. The farmers burn some of the felled trees, and combustion puts their carbon into the air.
The rest of the felled trees rot, and decomposition puts their carbon into the air. So clearing a forest raises the air’s carbon dioxide twice over: less taken out, and the trees’ carbon put in.
The rule works the other way too. Suppose a country plants a new forest, and its young trees photosynthesize.
Photosynthesis now takes more carbon out of the air. So more carbon leaves the air than comes in, and the air’s store of carbon dioxide falls.
When the air’s store of carbon dioxide rises, Earth’s climate changes.
What you are expected to know Predict how a change in one carbon-cycle process changes the carbon dioxide in the air: speed up an arrow into the air, or slow the arrow out of it, and the air’s store rises.
Suppose loggers clear a wide forest for farmland. They cut every tree down.
What happens to the carbon dioxide in the air over the following years?
- A. It fallsPhotosynthesis takes carbon out of the air.
With the trees gone, less carbon is taken out. - B. It stays the sameThe trees took carbon out of the air.
With the trees gone, the arrows into the air are no longer matched. - C. ✓ It rises
Why: The felled trees no longer photosynthesize.
So less carbon is taken out of the air.
Cellular respiration, decomposition and combustion still put carbon in.
So the air’s carbon dioxide rises.
Suppose a city drains a marsh to build houses. The reeds are cut, half are burned and the rest are left in heaps to rot.
(a) Explain how the change affects the carbon dioxide in the air above the city. (2 pt)
Frame After the marsh is drained, the carbon dioxide in the air …
The cut reeds no longer photosynthesize, so less carbon is taken out of the air.
The burned reeds’ carbon goes into the air by combustion.
The rotting reeds’ carbon goes into the air by decomposition.
So more carbon goes into the air than comes out, and the air’s store of carbon dioxide rises.
- Award 1 point for: the cut reeds no longer photosynthesize, so less carbon is taken out of the air (and the air’s carbon dioxide rises).
- Award 1 point for: the reeds’ carbon returns to the air by combustion (the burned reeds) and by decomposition (the rotting reeds).
A student says: “When a forest is felled, the trees’ carbon vanishes with the trees. So clearing a forest adds no carbon dioxide to the air.”
Is the student correct?
- A. Yes: the felled trees’ carbon vanishesNo process destroys an atom.
The trees’ carbon goes into the air when the felled trees burn or rot. - B. ✓ No: the trees’ carbon goes into the air when they burn or rot
Why: No atom is destroyed.
Burned trees return their carbon to the air by combustion.
Rotting trees return their carbon to the air by decomposition.
So the trees’ carbon goes into the air; it does not vanish.
Suppose a country closes half its coal power stations, while everything else stays the same.
What happens to the carbon dioxide in the air?
- A. ✓ It falls
- B. It stays the sameBurning coal is combustion, which puts carbon into the air.
With half the stations closed, less carbon is put in. - C. It risesCombustion puts carbon into the air.
Less coal burned means less carbon put in, not more.
Why: Burning coal is combustion.
With half the coal power stations closed, combustion puts less carbon into the air.
Photosynthesis takes out as much as before.
So the air’s carbon dioxide falls.
Suppose ash from a volcano darkens the sky over a lake for a whole summer, so the lake’s algae photosynthesize far less.
What happens to the carbon dioxide in the air above the lake?
- A. It fallsPhotosynthesis takes carbon out of the air.
With less photosynthesis, less carbon is taken out. - B. It stays the sameCellular respiration, decomposition and combustion still put carbon in.
Photosynthesis takes out less than before. - C. ✓ It rises
Why: The algae photosynthesize less.
So less carbon is taken out of the air.
Cellular respiration, decomposition and combustion still put carbon in.
So the air’s carbon dioxide rises.
Here is the carbon atom again, in the sugar of the leaf.
Photosynthesis took it in from the air. Cellular respiration, decomposition or combustion will send it back.
If farmers clear the forest, photosynthesis takes less carbon in, and the air’s store of carbon dioxide rises.
87Mixed practice mixed practice
A dead branch lies on the forest floor. Fungi feed on it, and carbon dioxide leaves the branch.
Which named arrow of the carbon cycle moves the branch’s carbon into the air?
- A. PhotosynthesisPhotosynthesis takes carbon dioxide in.
Here carbon dioxide is given off. - B. Cellular respirationThe fungi do respire the branch’s carbon.
On the drawing, carbon leaving dead material is the decomposition arrow; cellular respiration leaves the living things. - C. ✓ Decomposition
- D. CombustionCombustion is a fuel burning.
The branch rots; nothing burns.
Why: The branch is dead material.
Fungi eat the dead branch and respire its carbon.
Fungi respiring dead material’s carbon to carbon dioxide is decomposition.
A student says: “Clearing a forest lowers the carbon dioxide in the air, because fewer trees are left to respire.”
Is the student correct?
- A. Yes: fewer trees respiring lowers the air’s carbon dioxideThe trees took far more carbon out of the air by photosynthesis than they put in by respiration.
With the trees gone, less carbon is taken out. - B. ✓ No: the trees took out more by photosynthesis than they put in
Why: A tree respires, but it photosynthesizes more than it respires.
So a forest takes carbon out of the air on balance.
With the trees gone, less carbon is taken out.
So the air’s carbon dioxide rises, not falls.
Three of the four processes on the carbon cycle put carbon into the air.
Which process takes carbon out of the air?
- A. ✓ Photosynthesis
- B. Cellular respirationCellular respiration puts carbon into the air as carbon dioxide.
- C. DecompositionDecomposition puts dead material’s carbon into the air as carbon dioxide.
- D. CombustionCombustion puts a fuel’s carbon into the air as carbon dioxide.
Why: Photosynthesis takes carbon dioxide in from the air and builds its carbon into glucose.
It is the one arrow out of the air box.
A family burns dried peat in a stove all winter.
Which process moves the peat’s carbon into the air?
- A. PhotosynthesisPhotosynthesis takes carbon out of the air.
The burning peat puts carbon into the air. - B. Cellular respirationCellular respiration is done by living cells.
Fire, not living cells, sends this carbon into the air. - C. DecompositionDecomposition is done by fungi and microbes eating dead material.
Here the peat burns. - D. ✓ Combustion
Why: The peat is a fuel, and it burns.
Its carbon joins oxygen and leaves as carbon dioxide.
A fuel burning is combustion.
Suppose a warm, wet year makes the fungi and microbes in a forest’s soil decompose the dead leaves twice as fast as usual, while everything else stays the same.
What happens to the carbon dioxide in the air above the forest?
- A. It fallsDecomposition puts carbon into the air, not out of it.
Faster decomposition means more carbon put in. - B. It stays the sameThe decomposition arrow now carries more carbon into the air.
Photosynthesis takes out no more than before. - C. ✓ It rises
Why: Decomposition returns dead leaves’ carbon to the air.
Faster decomposition puts more carbon into the air.
Photosynthesis takes out no more than before.
So the air’s carbon dioxide rises.
A student says: “Decomposition is a kind of respiration.”
Is the student correct?
- A. ✓ Yes: fungi and microbes respire the dead material’s carbon
- B. No: decomposition is rotting, not respirationThe fungi and microbes that eat dead material respire its carbon.
That respiration is what returns the carbon to the air.
Why: Fungi and microbes eat dead material.
Their cells respire its carbon, by the word equation of cellular respiration.
So decomposition is respiration, done by the fungi and microbes that eat dead material.
Suppose a farmer drains a peat bog. Peat is dead plant material that lay under water for thousands of years, where fungi and microbes could not feed on it. Once the peat is dry and open to the air, fungi and microbes feed on it.
(a) Predict what happens to the carbon dioxide in the air above the bog over the following years, and explain why. (2 pt)
Fungi and microbes now feed on the peat and respire its carbon, so carbon dioxide leaves the peat by decomposition.
Carbon that the wet peat held for thousands of years goes into the air.
- Award 1 point for: the carbon dioxide in the air rises.
- Award 1 point for: fungi and microbes now feed on the peat, so decomposition (the arrow from the dead material to the air) puts the peat’s carbon into the air as carbon dioxide.
(b) Explain how the drained bog demonstrates that decomposition is a kind of respiration. (2 pt)
Frame The drained bog demonstrates this because …
Their cells respire the peat’s carbon: glucose + oxygen → carbon dioxide + water, and energy is released.
The carbon dioxide leaves them and goes into the air.
So the peat’s carbon returns to the air the way a respiring animal’s carbon does: by cells respiring it.
- Award 1 point for: the fungi and microbes that feed on the dead peat respire its carbon (the word equation of cellular respiration).
- Award 1 point for: the carbon leaves them as carbon dioxide and goes into the air, so the peat’s carbon returns to the air by respiration.
Glossary
- decomposition
- Fungi and microbes breaking dead material down and respiring its carbon back to the air as carbon dioxide. On the carbon cycle it is the arrow from the dead material to the air.
- combustion
- A fuel burning: its carbon joins oxygen and leaves as carbon dioxide. On the carbon cycle it is the arrow from the fossil fuel, or from a burning plant, to the air.
- carbon cycle
- Carbon moving round between its reservoirs, the air, the living things, the dead material and the fossil fuel. Four processes touch the air: photosynthesis takes carbon out, and cellular respiration, decomposition and combustion put it back.
APBIO-U08-L17 Surrounded by nitrogen, and unable to use it
Photo: Rasbak, Wikimedia Commons, CC BY-SA 3.0 (resized).
Here is a young cabbage plant short of nitrogen: its lower leaves have yellowed and died, and it has stopped growing. A maize plant short of nitrogen goes the same way, and stands yellow-leaved and stunted in its field. The air around it is about 78 % nitrogen gas. Its proteins and its DNA all need nitrogen.
Why is the plant starving in a sea of the element it needs?
Unit 8 · Ecology
1Starving in a sea of nitrogen
Unit 1 matched each element to the classes of molecule that need it.
Which classes of molecule need nitrogen?
- A. Phospholipids onlyThe phosphate head of a phospholipid needs phosphorus.
Nitrogen is in every protein and every nucleic acid. - B. Carbohydrates and fatsA carbohydrate is built from carbon, hydrogen and oxygen.
A fat is built from the same three elements. - C. ✓ Proteins and nucleic acids
Why: Every amino acid carries an amine group, –NH₂, so every protein contains nitrogen.
Every nitrogenous base contains nitrogen, so every nucleic acid contains nitrogen.
Why can a plant not use the air’s nitrogen, and who makes it usable?
The two nitrogen atoms in nitrogen gas are joined by a bond that no plant enzyme can break.
So a plant takes up nitrogen only as ammonium or nitrate, two nitrogen compounds dissolved in the soil water.
Bacteria break that bond. Some of these bacteria live free in the soil, and some live in swellings on the roots of bean plants.
Add the nitrogen compounds the bacteria made to the soil, and the maize plant’s leaves green up.
Here is the maize plant again, yellow-leaved and stunted in its field. Suppose the soil water under it holds very little ammonium and very little nitrate.
The plant needs nitrogen for every new protein and every new nucleic acid. With no nitrogen coming in, it makes no new protein, so it stops growing.
Chlorophyll, the green pigment in a leaf, contains nitrogen too. With no nitrogen coming in, the plant makes no new chlorophyll, and the leaves turn yellow.
Yet the air around the plant is about 78 % nitrogen gas. Every breath of wind through the leaves is full of it.
Nitrogen gas is two nitrogen atoms bonded together: N₂.
Unit 1 drew the bond in a hydrogen molecule, H₂.
What is a covalent bond?
- A. ✓ A pair of electrons shared between two atoms
- B. An electron handed over completely from one atom to the otherAn atom that hands its electron over completely is not sharing.
That pair of atoms is not covalently bonded.
Why: In a covalent bond the two atoms share a pair of electrons.
The shared pair holds the two atoms together.
Video: Watch: Starving in a sea of nitrogen
The maize plant among nitrogen molecules it cannot use; one molecule enlarged, its two atoms sharing three pairs of electrons; then a bean root with its nodules, and the bacteria inside turning nitrogen gas into ammonia and ammonium.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L17a.mp4
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In hydrogen gas, H₂, the two atoms share one pair of electrons: one covalent bond.
In nitrogen gas, N₂, the two atoms share three pairs of electrons. Three shared pairs between the same two atoms is called a triple covalent bond.
Three shared pairs hold the two nitrogen atoms together far more strongly than one pair would.
To use the nitrogen, a plant would have to pull the two atoms apart. No enzyme in a plant can break the triple covalent bond.
So the plant’s roots take up nitrogen only as two compounds dissolved in the soil water: ammonium, and nitrate.
In each of those compounds the nitrogen atom is already on its own, bonded to hydrogen or to oxygen. So the plant’s enzymes can build it into an amino acid.
Now the question turns round. Where do the ammonium and the nitrate in the soil water come from?
Some bacteria carry an enzyme that breaks the triple covalent bond. Some of them live free in the soil, and some live inside swellings on the roots of bean plants, called root nodules.
These bacteria pull the two nitrogen atoms apart. They build each nitrogen atom into a compound a plant can take up: ammonia, and from it ammonium.
Turning nitrogen gas into ammonia and ammonium is called . The nitrogen is ‘fixed’: taken out of the air and held in a compound.
A bean plant with nodules gets its fixed nitrogen straight from the bacteria in its roots. So it grows green in soil where a maize plant goes yellow.
A farmer’s nitrogen fertilizer is fixed nitrogen, made from ammonia. In the soil it becomes ammonium and nitrate, and the maize plant’s roots take those up.
The plant makes new chlorophyll and new protein, and the leaves green up.
What you are expected to know Explain why a plant cannot use the nitrogen gas that makes up most of the air, and needs it fixed into ammonia or ammonium first.
A maize plant’s roots sit in soil water. The air around its leaves is about 78 % nitrogen gas.
In which form does the plant take up its nitrogen?
- A. As nitrogen gas, through its leavesThe two atoms in nitrogen gas are held by a triple covalent bond.
No plant enzyme breaks it. - B. ✓ As ammonium or nitrate, through its roots
- C. As protein, through its rootsSoil water carries small dissolved compounds, not protein.
A plant builds its own protein from the nitrogen it takes up.
Why: No plant enzyme breaks the triple covalent bond in nitrogen gas.
Ammonium and nitrate hold nitrogen atoms the plant’s enzymes can use.
Both compounds are dissolved in the soil water, so the roots take them up.
Suppose a tomato plant grows in a pot of washed sand that holds almost no ammonium or nitrate. The air around it is about 78 % nitrogen gas. Its leaves turn yellow.
(a) Explain why the plant stays short of nitrogen in air that is about 78 % nitrogen gas. (1 pt)
Frame The plant stays short of nitrogen because …
No enzyme in the plant can break that bond.
So the plant cannot pull the two nitrogen atoms apart to build them into protein or nucleic acid.
It takes up nitrogen only as ammonium or nitrate, and the sand holds almost none of either.
- Award 1 point for: the two nitrogen atoms in N₂ are held by a triple covalent bond that no plant enzyme can break, so the plant takes up nitrogen only as ammonium or nitrate dissolved around its roots.
A student says: “A plant breathes in nitrogen gas through its leaves and builds it into protein.”
Is the student correct?
- A. Yes: leaves take in nitrogen gas for proteinNo plant enzyme breaks the triple covalent bond in nitrogen gas.
A plant takes up nitrogen as ammonium or nitrate through its roots. - B. ✓ No: the roots take up ammonium or nitrate instead
Why: The two nitrogen atoms in nitrogen gas are held by a triple covalent bond.
No plant enzyme breaks it, so a plant builds no protein from nitrogen gas.
A plant takes up its nitrogen as ammonium or nitrate from the soil water.
A farmer grows clover in a field and adds no nitrogen fertilizer. The clover grows green, and its roots carry nodules.
Which of the following supplies the clover with nitrogen it can use?
- A. The clover’s leaves, taking in nitrogen gas from the airNo plant enzyme breaks the triple covalent bond in nitrogen gas.
Leaves take in no usable nitrogen from the air. - B. The clover’s roots, taking up nitrogen gas from the soilNitrogen gas dissolved in soil water is still N₂, with its triple covalent bond.
A root takes up nitrogen as ammonium or nitrate. - C. ✓ Bacteria in the roots’ nodules, fixing nitrogen gas
Why: Bacteria in the clover’s root nodules break the triple covalent bond.
They turn nitrogen gas into ammonia and ammonium.
So the clover gets fixed nitrogen from its own roots, with no fertilizer.
33Quick quiz: nitrogen fixation mixed practice
Bacteria in a bean’s root nodule turn nitrogen gas into ammonia.
Is this nitrogen fixation?
- A. ✓ Yes
- B. NoThe bacteria start from nitrogen gas and make ammonia.
Turning nitrogen gas into ammonia is nitrogen fixation.
Why: The starting substance is nitrogen gas, and the product is ammonia.
Turning nitrogen gas into ammonia and ammonium is nitrogen fixation.
A maize root takes up nitrate from the soil water.
Is this nitrogen fixation?
- A. YesNitrate is already a compound, with its nitrogen atom on its own.
Nitrogen fixation starts from nitrogen gas. - B. ✓ No
Why: The root moves nitrate that is already fixed from the soil water into the plant.
No nitrogen gas is turned into a compound, so this is not nitrogen fixation.
A leaf cell builds ammonium into an amino acid.
Is this nitrogen fixation?
- A. YesThe ammonium is already fixed nitrogen.
Nitrogen fixation is the earlier step, from nitrogen gas to ammonia and ammonium. - B. ✓ No
Why: The leaf cell starts from ammonium, which is already a compound.
No nitrogen gas is turned into a compound, so this is not nitrogen fixation.
Bacteria living free in the soil turn nitrogen gas into ammonium.
Is this nitrogen fixation?
- A. ✓ Yes
- B. NoThe bacteria start from nitrogen gas and end with ammonium.
Turning nitrogen gas into ammonia and ammonium is nitrogen fixation, wherever the bacteria live.
Why: The starting substance is nitrogen gas, and the product is ammonium.
Turning nitrogen gas into ammonia and ammonium is nitrogen fixation, in the soil as in a nodule.
Bacteria in the soil and in root nodules carry out nitrogen fixation.
What is nitrogen fixation?
- A. A root taking up nitrate from the soil waterA root taking up nitrate moves nitrogen that is already fixed.
Nitrogen fixation starts from nitrogen gas. - B. A plant building nitrogen into proteinBuilding nitrogen into protein uses nitrogen that is already fixed.
Nitrogen fixation turns nitrogen gas into ammonia and ammonium. - C. ✓ Turning nitrogen gas into ammonia and ammonium
Why: Nitrogen fixation is bacteria turning nitrogen gas into ammonia and then ammonium: nitrogen taken out of the air and held in a compound.
Some bacteria live free in the soil, and some live in the root nodules of bean plants.
(a) State what nitrogen fixation is. (1 pt)
- Award 1 point for: (bacteria) turning nitrogen gas into ammonia and ammonium (a compound a plant can take up).
40Nitrogen gas to ammonia to ammonium
Unit 1 wrote the hydrogen ion as H⁺.
What is a hydrogen ion?
- A. A hydrogen atom that has gained an extra electronA hydrogen atom that gains an electron carries a negative charge.
The hydrogen ion H⁺ carries a positive charge. - B. ✓ A hydrogen atom that has given up its electron
- C. Two hydrogen atoms bonded togetherTwo hydrogen atoms bonded together are a hydrogen molecule, H₂.
Why: A hydrogen atom has one proton and one electron.
When it gives up its electron, one positive proton is left: the hydrogen ion, H⁺.
Video: Watch: Nitrogen gas to ammonia to ammonium
The chain written step by step: N₂ splits and each nitrogen atom takes three hydrogen atoms, NH₃; a hydrogen ion from the soil solution joins, making an ammonium ion, NH₄⁺; the formulas appear one arrow at a time.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L17b.mp4
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Now consider the bacteria at work free in the soil. Nitrogen gas from the air reaches them, and they break its triple covalent bond.
The bacteria bond each nitrogen atom to three hydrogen atoms. One nitrogen atom bonded to three hydrogen atoms is called , and its formula is NH₃.
Here is nitrogen gas becoming ammonia, written in formulas.
Read the arrow as ‘becomes’: nitrogen gas becomes ammonia
Ammonia dissolves in the soil water. Soil water with everything dissolved in it is called the soil solution.
Among the dissolved substances are hydrogen ions, H⁺. An ammonia molecule picks up one hydrogen ion from the soil solution.
The nitrogen atom is now bonded to four hydrogen atoms. The hydrogen ion brought its positive charge with it, so the new particle carries a charge of +1.
One nitrogen atom bonded to four hydrogen atoms, carrying a +1 charge, is called . The ammonium ion has the formula NH₄⁺.
Here is ammonia becoming ammonium, written in formulas: ammonia plus a hydrogen ion becomes ammonium.
ammonia plus a hydrogen ion becomes ammonium
Unit 1 showed the same move on an amino acid: the amine group –NH₂ took on a hydrogen ion and became –NH₃⁺.
So the whole of nitrogen fixation is one chain: nitrogen gas to ammonia to ammonium.
nitrogen fixation: nitrogen gas to ammonia to ammonium
Here is a table of the three forms of nitrogen in the chain: the name, the formula, the charge, and where each one is found.
Nitrogen fixation ends at ammonium. The nitrate in soil water is made from ammonium afterward, by other soil bacteria.
What you are expected to know Describe nitrogen fixation as a chain: bacteria convert nitrogen gas (N₂) to ammonia (NH₃), which picks up a hydrogen ion from the soil solution to become ammonium (NH₄⁺).
Nitrogen gas reaches the soil and the root nodules of a bean plant.
Which organisms turn the nitrogen gas into ammonia?
- A. The bean plant’s root cellsNo plant enzyme breaks the triple covalent bond in nitrogen gas.
The root cells house the bacteria that do. - B. FungiFungi feed on dead material, and no fungus breaks the triple covalent bond in nitrogen gas.
- C. ✓ Bacteria
Why: Bacteria carry the enzyme that breaks the triple covalent bond in nitrogen gas.
They bond each nitrogen atom to three hydrogen atoms, making ammonia.
Nitrogen fixation, written in formulas, is a chain of two arrows, and it begins N₂ → ___ .
Which of the following fills the blank?
- A. ✓ NH₃, ammonia
- B. NH₄⁺, ammoniumAmmonium is the second product.
The bacteria make ammonia first, and the ammonia then picks up a hydrogen ion.
Why: The bacteria break the triple covalent bond in N₂ and bond each nitrogen atom to three hydrogen atoms.
One nitrogen atom with three hydrogen atoms is ammonia, NH₃.
The chain of nitrogen fixation has two steps and ends at one product.
Which of the following is the last product of the chain?
- A. NH₃, ammoniaAmmonia is the first product.
It then picks up a hydrogen ion from the soil solution and becomes ammonium. - B. H⁺, hydrogen ionThe ammonia picks up the hydrogen ion from the soil solution.
Ammonia plus a hydrogen ion is ammonium, the chain’s last product. - C. ✓ NH₄⁺, ammonium
Why: The bacteria turn nitrogen gas into ammonia.
The ammonia picks up one hydrogen ion, H⁺, from the soil solution.
So the chain ends at ammonium, NH₄⁺.
A student says: “Nitrogen-fixing bacteria turn nitrogen gas into nitrate.”
Is the student correct?
- A. Yes: nitrogen-fixing bacteria turn the gas into nitrateNitrogen fixation makes ammonia, then ammonium.
Nitrate is made from ammonium afterward, by other soil bacteria. - B. ✓ No: fixation ends at ammonium; nitrate is made later
Why: Nitrogen-fixing bacteria turn nitrogen gas into ammonia.
The ammonia picks up a hydrogen ion and becomes ammonium.
Nitrogen fixation ends there: it makes ammonia and ammonium, never nitrate.
Here is the yellow maize plant again, standing in air that is about 78 % nitrogen gas.
No plant enzyme breaks the triple covalent bond, so the plant takes up only ammonium and nitrate from the soil water.
Bacteria fix the gas to ammonia and then ammonium. With fixed nitrogen reaching its roots, the maize plant’s leaves green up.
63Quick quiz: ammonia (NH₃), ammonium (NH₄⁺) mixed practice
A particle is drawn below: one nitrogen atom, N, with hydrogen atoms, H, bonded to it.
Which is the drawn particle?
- A. ✓ Ammonia, NH₃
- B. Ammonium, NH₄⁺The drawn particle has three hydrogen atoms and no charge sign.
Ammonium has four hydrogen atoms and a +1 charge.
Why: The drawing shows one nitrogen atom bonded to three hydrogen atoms, with no charge.
That is ammonia, NH₃.
A particle is drawn below: one nitrogen atom, N, with hydrogen atoms, H, bonded to it.
Which is the drawn particle?
- A. Ammonia, NH₃The drawn particle has four hydrogen atoms and a plus sign.
Ammonia has three hydrogen atoms and no charge. - B. ✓ Ammonium, NH₄⁺
Why: The drawing shows one nitrogen atom bonded to four hydrogen atoms, with a +1 charge.
That is ammonium, NH₄⁺.
Three forms of nitrogen appear in nitrogen fixation.
Which of the following is the formula of ammonium?
- A. N₂N₂ is nitrogen gas: two nitrogen atoms and no hydrogen.
- B. NH₃NH₃ is ammonia: one nitrogen atom with three hydrogen atoms and no charge.
- C. ✓ NH₄⁺
Why: Ammonium is one nitrogen atom bonded to four hydrogen atoms, carrying a charge of +1.
Its formula is NH₄⁺.
Ammonium is one nitrogen atom bonded to four hydrogen atoms.
What charge does ammonium carry?
- A. A charge of −1Ammonium formed when ammonia picked up a hydrogen ion, and a hydrogen ion carries a positive charge.
- B. No chargeAmmonia carries no charge.
Ammonium has picked up a hydrogen ion, and the hydrogen ion brought a positive charge. - C. ✓ A charge of +1
Why: Ammonium is one nitrogen atom bonded to four hydrogen atoms, NH₄⁺.
The fourth hydrogen arrived as a hydrogen ion, so ammonium carries a charge of +1.
Ammonia is one nitrogen atom bonded to hydrogen atoms.
How many hydrogen atoms are bonded to the nitrogen atom in ammonia?
- A. TwoAmmonia is NH₃: the 3 counts its hydrogen atoms.
- B. ✓ Three
- C. FourFour hydrogen atoms on one nitrogen atom is ammonium, NH₄⁺.
Ammonia is NH₃.
Why: Ammonia is one nitrogen atom bonded to three hydrogen atoms, NH₃, with no charge.
A hydrogen ion joins an ammonia molecule.
What does the ammonia molecule become?
- A. Nitrogen gasNitrogen gas is two nitrogen atoms bonded together, the starting substance of fixation.
An ammonia molecule holds one nitrogen atom. - B. ✓ Ammonium
- C. NitrateNitrate holds oxygen atoms, not four hydrogen atoms.
Ammonia plus a hydrogen ion is ammonium.
Why: The ammonia molecule, NH₃, gains one hydrogen ion, H⁺.
One nitrogen atom with four hydrogen atoms and a +1 charge is ammonium, NH₄⁺.
Ammonia made by nitrogen-fixing bacteria dissolves in the soil water.
(a) State how an ammonia molecule becomes an ammonium ion. (1 pt)
- Award 1 point for: ammonia picks up (acquires) a hydrogen ion (H⁺) from the soil solution (soil water) and becomes ammonium.
71Mixed practice mixed practice
Nitrogen gas, N₂, is two nitrogen atoms bonded together.
Which of the following bonds holds the two atoms together?
- A. A single covalent bondOne shared pair is the bond in hydrogen gas, H₂.
The two nitrogen atoms share three pairs. - B. A double covalent bondThe two nitrogen atoms share three pairs of electrons, not two.
- C. ✓ A triple covalent bond
Why: The two nitrogen atoms share three pairs of electrons.
Three shared pairs between two atoms is a triple covalent bond.
A gardener waters a yellowing pot plant with a solution of ammonium. Over the next weeks the leaves green up.
Which of the following explains the change?
- A. ✓ The roots take up the ammonium and the plant builds it into new chlorophyll
- B. The ammonium breaks the triple covalent bond in the air’s nitrogen gasAmmonium is already fixed nitrogen.
The plant takes it up through its roots and builds it into new molecules. - C. The leaves take in the ammonium from the air around themAmmonium dissolves in the water in the pot.
The roots take it up, not the leaves.
Why: Ammonium is fixed nitrogen, dissolved in the water round the roots.
The roots take it up.
With nitrogen coming in, the plant builds new chlorophyll and new protein, so the leaves green up.
A pea plant grows green in a field with no fertilizer. A maize plant beside it, in the same soil, turns yellow. The pea’s roots carry nodules and the maize’s roots carry no nodules.
Which of the following explains the difference?
- A. The pea’s leaves take in nitrogen gas from the airNo plant enzyme breaks the triple covalent bond in nitrogen gas.
The pea’s leaves take in no usable nitrogen. - B. ✓ Bacteria in the pea’s nodules fix nitrogen gas for the pea
- C. The pea plant needs no nitrogen for its proteinsEvery plant needs nitrogen for its proteins and nucleic acids.
The pea gets fixed nitrogen from the bacteria in its nodules.
Why: Bacteria in the pea’s root nodules turn nitrogen gas into ammonia and ammonium.
So the pea gets fixed nitrogen from its own roots.
The maize has no nodules, and the soil water holds too little ammonium and nitrate for it.
Ammonia made by bacteria in the soil becomes ammonium.
Where does the hydrogen ion that joins the ammonia come from?
- A. The airThe air around the roots holds nitrogen gas and oxygen, not dissolved hydrogen ions.
- B. The nitrogen gasNitrogen gas is two nitrogen atoms; it holds no hydrogen.
- C. ✓ The soil solution
Why: Soil water with its dissolved substances is the soil solution.
Hydrogen ions are among the dissolved substances, and ammonia picks up one of them.
A student says: “Nitrogen fixation happens in the roots of a bean plant, where the nodule bacteria live.”
Is the student correct?
- A. ✓ Yes: the bacteria in the root nodules do the fixing
- B. No: the leaves take the gas straight from the airNo plant cell fixes nitrogen.
Bacteria in the root nodules break the triple covalent bond.
Why: Only bacteria carry the enzyme that breaks the triple covalent bond in nitrogen gas.
In a bean plant those bacteria live in the root nodules.
So nitrogen fixation happens in the roots, and a leaf does no fixing.
A maize root sits in soil water.
Which of the following does the root take up as the plant’s nitrogen supply?
- A. Nitrogen gasNitrogen gas dissolved in the soil water is still N₂, with its triple covalent bond.
No plant enzyme breaks it. - B. ✓ Ammonium
- C. ProteinSoil water carries small dissolved compounds, not protein.
The plant builds its own protein.
Why: Ammonium is fixed nitrogen dissolved in the soil water.
Its nitrogen atom is on its own, so the plant’s enzymes can use it.
The root takes up ammonium, and nitrate, as the plant’s nitrogen supply.
On a gravel bar by a river, young alder trees grow green while birch seedlings beside them turn yellow. The alders’ roots carry nodules; the birches’ roots carry none. No fertilizer reaches the bar.
(a) Explain how the two kinds of tree demonstrate that a plant needs its nitrogen fixed before it can use it. (2 pt)
Frame The two kinds of tree demonstrate this because …
No plant enzyme breaks the triple covalent bond in nitrogen gas, so neither tree uses the gas itself.
Bacteria in the alders’ root nodules turn nitrogen gas into ammonia and then ammonium, so the alders get fixed nitrogen and stay green.
The birches have no nodules, and little fixed nitrogen reaches their roots in the gravel, so they turn yellow.
- Award 1 point for: both trees stand in nitrogen gas, but no plant enzyme breaks its triple covalent bond, so neither tree uses the gas directly.
- Award 1 point for: bacteria in the alders’ nodules fix nitrogen gas into ammonia and ammonium, which the alders use; the birches have no nodules and the gravel holds too little fixed nitrogen, so they turn yellow.
Glossary
- nitrogen fixation
- Bacteria turning nitrogen gas (N₂) into ammonia (NH₃) and then ammonium (NH₄⁺): nitrogen taken out of the air and held in a compound a plant can take up. Some of the bacteria live free in the soil; some live in the root nodules of plants such as beans.
- ammonia (NH₃)
- One nitrogen atom bonded to three hydrogen atoms, with no charge. The first product of nitrogen fixation; in the soil solution it picks up a hydrogen ion and becomes ammonium.
- ammonium (NH₄⁺)
- One nitrogen atom bonded to four hydrogen atoms, carrying a charge of +1. Ammonia becomes ammonium by picking up a hydrogen ion from the soil solution. A plant's roots take up ammonium (and nitrate) from the soil water.
APBIO-U08-L17B Four more steps, by what goes in and what comes out
Now consider the ammonium that bacteria living free in the soil made. A month later, some of it is in the maize plant’s leaves as protein.
Some of it has become nitrate in the soil water. Some has gone back to the air as nitrogen gas. Which step did each?
Unit 8 · Ecology
1Name the four steps by what goes in and what comes out
The last lesson followed nitrogen fixation from the air into the soil water.
Which chain is nitrogen fixation?
- A. ✓ Nitrogen gas to ammonia, then to ammonium
- B. Nitrogen gas straight to nitrateNitrogen fixation ends at ammonium.
Other soil bacteria make the nitrate from ammonium afterward. - C. Nitrate to nitrogen gasNitrogen fixation starts with nitrogen gas from the air and ends at ammonium, in the soil water.
Why: Bacteria turn nitrogen gas into ammonia.
The ammonia picks up a hydrogen ion from the soil solution and becomes ammonium.
Nitrogen gas to ammonia to ammonium is the whole of nitrogen fixation.
Every cycle drawn as boxes and arrows uses the same template.
What is each arrow on the template?
- A. A reservoirA box is a reservoir, a place where the matter sits.
An arrow moves matter from one reservoir to another. - B. ✓ A process
Why: Each arrow on the template is a process.
The arrow starts at the reservoir the matter leaves and ends at the reservoir the matter enters.
Video: Watch: Name the four steps by what goes in and what comes out
The cycle template named for nitrogen; each of the four arrows lights in turn as its step is named, what goes in at its tail and what comes out at its head.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L17Ba.mp4
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How does nitrogen move on once it is fixed?
Four more steps carry it round. Plants do one of them, and microbes in the soil do the other three.
Each step is named by what goes in and what comes out.
One step builds ammonium or nitrate into a plant’s own molecules. One step turns dead remains and waste back into ammonium.
One step turns ammonium into nitrate ions. One step turns nitrate back into nitrogen gas, and the loop to the air is closed.
Here is the cycle template again, with its boxes named for nitrogen. The one arrow already named is nitrogen fixation, from the last lesson.
On every arrow, the box at the tail is what goes in, and the box at the head is what comes out. Nitrogen gas goes into fixation, and ammonium comes out.
The living organisms box has the double outline of a biotic reservoir. The air, the ammonium and the nitrate are abiotic reservoirs.
The air box is the largest nitrogen reservoir of all: about 78 % of the air is nitrogen gas.
Two forms of nitrogen sit dissolved in the soil water, so the template gives each its own box: the ammonium that fixation made, and nitrate.
Start with the ammonium in the soil water. Suppose the maize plant’s roots take some of it up.
The plant bonds each nitrogen atom into its own new molecules: protein, and DNA.
So the arrow from the ammonium box to the living organisms box carries nitrogen into the plant. A root takes up nitrate the same way, so a second arrow enters from the nitrate box.
A plant taking up ammonium or nitrate and building the nitrogen into its own molecules is called .
Ammonium or nitrate goes in. The plant’s own protein and DNA come out.
assimilation: what goes in, and what comes out
The animals in the living organisms box get their nitrogen by eating plants, or by eating other animals.
Now consider a maize leaf that has died and fallen onto the soil. Its protein still holds the nitrogen atoms the plant built in.
Microbes in the soil decompose the dead leaf. As they break its protein apart, they release the nitrogen atoms as ammonium, back into the soil water.
The microbes do the same with animal waste, and with dead animals.
Soil microbes turning the nitrogen in dead remains and waste back into ammonium is called .
Dead remains and waste go in. Ammonium comes out.
ammonification: what goes in, and what comes out
On the template, the ammonification arrow leaves the living organisms box and enters the ammonium box. It sits beside the assimilation arrow, pointing the other way.
Now consider the ammonium that no root took up. Other bacteria in the soil change it.
These bacteria turn the ammonium first into nitrite, and then into nitrate.
One nitrogen atom bonded to three oxygen atoms, carrying a charge of −1, is called .
Soil bacteria turning ammonium into nitrite and then nitrate is called .
Ammonium goes in. Nitrate comes out.
nitrification: what goes in, and what comes out
Nitrification changes one dissolved form into another, inside the soil water. So on the template its arrow joins the ammonium box to the nitrate box along the bottom.
Now consider nitrate in a layer of soil with no oxygen in it.
Unit 3 named the bacteria that live there. They have no oxygen to take the electrons at the end of their electron transport chain.
So they pass the electrons to nitrate instead. Unit 3 called this anaerobic respiration.
The nitrate that takes the electrons turns, step by step, back into nitrogen gas. The gas leaves the soil and joins the air.
Soil bacteria turning nitrate back into nitrogen gas is called .
Nitrate goes in. Nitrogen gas comes out, and the air gets its nitrogen back.
denitrification: what goes in, and what comes out
On the template, the denitrification arrow leaves the nitrate box and enters the air box. It closes the loop: every nitrogen atom that fixation took out of the air can return to it.
Here is the template with every arrow named.
The table below lists the five steps: what goes in, what comes out, and who does it.
Two of the names look alike. Nitrification puts nitrogen into nitrate.
Denitrification takes it out again, as nitrogen gas: de- means undoing.
What you are expected to know Identify each of the four steps after nitrogen fixation from what goes in and what comes out: assimilation, ammonification, nitrification and denitrification.
On the nitrogen cycle template below, only nitrogen fixation is named. One arrow is drawn bold.
Which step is the bold arrow?
- A. AssimilationAssimilation carries nitrogen into living organisms.
The bold arrow joins the two soil water boxes. - B. AmmonificationAmmonification ends at ammonium.
The bold arrow starts at ammonium. - C. ✓ Nitrification
- D. DenitrificationDenitrification ends in the air.
The bold arrow ends at nitrate.
Why: The bold arrow leaves the ammonium box and enters the nitrate box.
Ammonium goes in and nitrate comes out.
That step is nitrification.
On the nitrogen cycle template below, only nitrogen fixation is named. Two arrows are drawn bold.
Which step are the two bold arrows?
- A. ✓ Assimilation
- B. AmmonificationAmmonification leaves living organisms.
The bold arrows enter living organisms. - C. NitrificationNitrification joins the ammonium box to the nitrate box.
The bold arrows leave those boxes and go up. - D. DenitrificationDenitrification ends in the air.
The bold arrows end at living organisms.
Why: The bold arrows leave the ammonium box and the nitrate box and enter the living organisms box.
Ammonium or nitrate goes in, and a plant’s own molecules come out.
That step is assimilation.
On the nitrogen cycle template below, only nitrogen fixation is named. One arrow is drawn bold.
Which step is the bold arrow?
- A. AssimilationAssimilation ends at living organisms.
The bold arrow ends at the air. - B. AmmonificationAmmonification ends at ammonium.
The bold arrow ends at the air. - C. NitrificationNitrification ends at nitrate.
The bold arrow starts at nitrate. - D. ✓ Denitrification
Why: The bold arrow leaves the nitrate box and enters the air box.
Nitrate goes in and nitrogen gas comes out.
That step is denitrification.
On the nitrogen cycle template below, only nitrogen fixation is named. One arrow is drawn bold.
Which step is the bold arrow?
- A. AssimilationAssimilation enters living organisms.
The bold arrow leaves living organisms. - B. ✓ Ammonification
- C. NitrificationNitrification starts at ammonium.
The bold arrow ends at ammonium. - D. DenitrificationDenitrification ends in the air.
The bold arrow ends at ammonium.
Why: The bold arrow leaves the living organisms box and enters the ammonium box.
Dead remains and waste go in, and ammonium comes out.
That step is ammonification.
Now name each step from its drawing alone: what goes in on the left, what comes out on the right.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. NitrificationNitrification ends at nitrate.
Here nitrate goes in. - B. ✓ Denitrification
- C. Nitrogen fixationNitrogen fixation starts with nitrogen gas.
Here nitrogen gas comes out.
Why: Nitrate goes in and nitrogen gas comes out.
Bacteria turning nitrate back into nitrogen gas is denitrification.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. AssimilationAssimilation builds nitrogen into a plant’s molecules.
Here dead leaves and waste are broken down. - B. NitrificationNitrification starts with ammonium.
Here ammonium comes out. - C. ✓ Ammonification
Why: Dead leaves and animal waste go in, and ammonium comes out.
Soil microbes turning dead remains and waste back into ammonium is ammonification.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. ✓ Assimilation
- B. AmmonificationAmmonification ends at ammonium.
Here ammonium goes in. - C. NitrificationNitrification ends at nitrate.
Here a plant’s protein comes out.
Why: Ammonium goes in and a plant’s protein comes out.
A plant building ammonium into its own molecules is assimilation.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. ✓ Nitrification
- B. DenitrificationDenitrification ends at nitrogen gas.
Here nitrate comes out. - C. AssimilationAssimilation ends inside a plant.
Here nitrate comes out, in the soil water.
Why: Ammonium goes in and nitrate comes out.
Soil bacteria turning ammonium into nitrate is nitrification.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. DenitrificationDenitrification ends at nitrogen gas.
Here nitrogen gas goes in. - B. ✓ Nitrogen fixation
- C. NitrificationNitrification starts with ammonium.
Here ammonium comes out.
Why: Nitrogen gas goes in and ammonium comes out.
Bacteria turning nitrogen gas into ammonia and then ammonium is nitrogen fixation.
One step of the nitrogen cycle is drawn below: what goes in, and what comes out.
Which step is it?
- A. NitrificationNitrification ends at nitrate.
Here nitrate goes in. - B. DenitrificationDenitrification ends at nitrogen gas.
Here a plant’s DNA comes out. - C. ✓ Assimilation
Why: Nitrate goes in and a wheat plant’s DNA comes out.
A plant building nitrate into its own molecules is assimilation.
Suppose a scientist measures the gases leaving a patch of soil, and finds nitrogen gas among them.
Which step of the nitrogen cycle is releasing the nitrogen gas?
- A. Nitrogen fixationNitrogen fixation takes nitrogen gas out of the air.
Here nitrogen gas is leaving the soil. - B. NitrificationNitrification ends at nitrate, which stays dissolved in the soil water.
- C. ✓ Denitrification
Why: The nitrogen gas is coming out of the soil.
The one step whose product is nitrogen gas is denitrification: nitrate goes in, and nitrogen gas comes out.
A student says: “Denitrification closes the loop: it sends nitrogen back to the air as nitrogen gas.”
Is the student correct?
- A. ✓ Yes: nitrate goes in, and nitrogen gas comes out
- B. No: nitrification is the step that returns nitrogen to the airNitrification turns ammonium into nitrate, which stays in the soil water.
Denitrification turns nitrate into nitrogen gas.
Why: Denitrification takes nitrate in and sends nitrogen gas out.
The gas leaves the soil and joins the air.
So denitrification is the step that returns nitrogen to the air and closes the loop.
Here is the month-old ammonium again, in the soil water under the maize plant.
Assimilation built some of it into the maize plant’s protein. Nitrification turned some of it into nitrate.
Denitrification sent some of that nitrate back to the air as nitrogen gas.
61Quick quiz: assimilation, ammonification, nitrification, denitrification, nitrate mixed practice
Four steps carry nitrogen on after nitrogen fixation.
What is assimilation?
- A. Soil bacteria turning ammonium into nitrateSoil bacteria turning ammonium into nitrate is nitrification.
- B. ✓ A plant building ammonium or nitrate into its own molecules
- C. Soil microbes turning dead remains back into ammoniumSoil microbes turning dead remains back into ammonium is ammonification.
Why: Assimilation is a plant taking up ammonium or nitrate and building the nitrogen into its own protein and DNA.
Four steps carry nitrogen on after nitrogen fixation.
What is ammonification?
- A. Bacteria turning nitrogen gas into ammonia and then ammoniumBacteria turning nitrogen gas into ammonia and then ammonium is nitrogen fixation.
- B. A plant building ammonium into its own moleculesA plant building ammonium into its own molecules is assimilation.
- C. ✓ Soil microbes turning dead remains and waste back into ammonium
Why: Ammonification is soil microbes turning the nitrogen in dead remains and waste back into ammonium.
Four steps carry nitrogen on after nitrogen fixation.
What is nitrification?
- A. ✓ Soil bacteria turning ammonium into nitrite and then nitrate
- B. Soil bacteria turning nitrate back into nitrogen gasSoil bacteria turning nitrate back into nitrogen gas is denitrification.
- C. Soil microbes turning waste back into ammoniumSoil microbes turning waste back into ammonium is ammonification.
Why: Nitrification is soil bacteria turning ammonium into nitrite and then nitrate.
Four steps carry nitrogen on after nitrogen fixation.
What is denitrification?
- A. Soil bacteria turning nitrite into nitrateTurning nitrite into nitrate is the second half of nitrification.
- B. ✓ Soil bacteria turning nitrate back into nitrogen gas
- C. Soil microbes turning dead remains back into ammoniumSoil microbes turning dead remains back into ammonium is ammonification.
Why: Denitrification is soil bacteria turning nitrate back into nitrogen gas, which returns to the air.
Plants take up two forms of nitrogen from the soil water.
What is nitrate?
- A. One nitrogen atom bonded to four hydrogen atoms, with a +1 chargeOne nitrogen atom bonded to four hydrogen atoms, with a +1 charge, is ammonium.
- B. Two nitrogen atoms joined by a triple covalent bondTwo nitrogen atoms joined by a triple covalent bond is nitrogen gas.
- C. ✓ One nitrogen atom bonded to three oxygen atoms, with a −1 charge
Why: Nitrate is one nitrogen atom bonded to three oxygen atoms, carrying a charge of −1: NO₃⁻.
Four steps carry nitrogen on after nitrogen fixation.
(a) State what denitrification is, naming what goes in and what comes out. (1 pt)
- Award 1 point for: nitrate turned into nitrogen gas (by bacteria in the soil). Accept with or without: the gas returning to the air.
Glossary
- assimilation
- A plant taking up ammonium or nitrate from the soil water and building the nitrogen into its own molecules, such as protein and DNA. On the cycle template it is the arrows from the ammonium box and the nitrate box up to living organisms.
- ammonification
- Microbes in the soil turning the nitrogen in dead remains and waste back into ammonium in the soil water. On the cycle template it is the arrow from living organisms down to the ammonium box.
- nitrate (NO₃⁻)
- One nitrogen atom bonded to three oxygen atoms, carrying a charge of −1. Soil bacteria make it from ammonium by nitrification; a plant's roots take it up along with ammonium.
- nitrification
- Soil bacteria turning ammonium into nitrite and then into nitrate, inside the soil water. On the cycle template it is the arrow along the bottom from the ammonium box to the nitrate box.
- denitrification
- Soil bacteria turning nitrate back into nitrogen gas, which leaves the soil and returns to the air. On the cycle template it is the arrow from the nitrate box up to the air box.
APBIO-U08-L18 The cycle that starts in rock
Photo: Tom Koerner, U.S. Fish and Wildlife Service, Wikimedia Commons, public domain (resized).
A deer’s antler lies on a hillside, shed last winter. Its phosphate was in grass a year ago, in soil water before that, and in the hill’s rock before anything alive touched it.
Nothing in this story went through the air. Where does phosphorus come from, and how does it get back?
Unit 8 · Ecology
1Follow phosphorus round its cycle
Unit 1 drew a phospholipid’s head.
What is a phosphate group?
- A. ✓ A phosphorus atom bonded to four oxygen atoms, carrying a negative charge
- B. A nitrogen atom bonded to three hydrogen atoms, carrying no chargeA nitrogen atom bonded to three hydrogen atoms is ammonia.
- C. A chain of three carbon atoms, each carrying an oxygen atomA chain of three carbon atoms, each carrying an oxygen, is glycerol, the backbone of a phospholipid.
Why: A phosphate group is a phosphorus atom bonded to four oxygen atoms.
In a phospholipid’s head it carries a full negative charge.
Unit 1 matched each element to the classes of molecule that need it.
Which classes of molecule are built with phosphorus?
- A. Carbohydrates and proteinsCarbohydrates need only carbon, hydrogen and oxygen.
Proteins add nitrogen and sulfur.
Neither class needs phosphorus. - B. ✓ Phospholipids and nucleic acids
- C. Proteins and nucleic acidsProteins and nucleic acids are the two classes that need nitrogen.
A protein needs no phosphorus.
Why: Every nucleotide of a nucleic acid carries a phosphate group.
Every phospholipid’s head carries a phosphate group.
So phosphorus is used to build phospholipids and nucleic acids.
How does a cycle work with no store in the air?
Phosphorus sits in rock, in the soil and its water, in plants and in animals. No store of phosphorus sits in the air.
Rock slowly crumbles and lets its phosphorus into the soil water. Roots absorb it, animals eat the plants, and dead remains and waste carry it back to the soil.
Video: Watch: Follow phosphorus round its cycle
The cycle template with no box at the top; rock crumbling and its phosphate dissolving into the soil water; a root absorbing it; a deer eating the grass; the shed antler breaking down into the soil.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L18a.mp4
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Here is the cycle template again, with its boxes named for phosphorus.
The top box is missing. No store of phosphorus sits in the air.
Plants and animals have the double outline of a biotic reservoir. Rock and the soil are abiotic reservoirs.
In nature, phosphorus is found as an ion: one phosphorus atom bonded to four oxygen atoms, carrying three negative charges.
That ion is called a (PO₄³⁻). It is the same phosphate group as in a phospholipid’s head and in every nucleotide.
Rock holds phosphate in its minerals. Rock is the slow store: phosphate leaves it only as the rock itself crumbles.
Rain, frost and the acids in soil slowly crumble the surface of rock.
The crumbling frees the phosphate ions. They dissolve in the soil water.
Rock crumbling and releasing what it holds is called . On the template it is the arrow from rock to the soil.
A plant’s roots absorb the soil water, and the dissolved phosphate ions with it.
The plant builds the phosphate into its molecules: into every nucleotide of its DNA, into every phospholipid of its membranes, and into its ATP.
An animal eats the plant, and the plant’s phosphate becomes the animal’s.
The animal builds phosphate into its own DNA and membranes. It also builds phosphate into its bones: the hard part of bone is calcium phosphate.
When a leaf falls or an animal dies, the remains lie on the soil. An animal’s waste falls on the soil too.
Microbes in the soil break the remains and the waste down: decomposition. The phosphate they held dissolves back into the soil water.
Rain washes some phosphate out of the soil into rivers, and rivers carry it to the sea.
In the sea the phosphate settles to the sea floor. Over millions of years the sediment hardens into new rock.
That arrow is drawn dashed. It is the slowest process on the template.
Here is the template with every box and every arrow named.
Rock, the soil, plants and animals, and the processes that carry phosphate between them, are called the .
Look for an arrow into the air. There is none.
Of the four cycles, the phosphorus cycle is the only one with no store in the air.
Now follow one phosphate ion all the way round. Suppose the ion sits in the hill’s rock, under the grass.
Here is its path drawn bold on the template. Each arrow it crosses is one process, and the name is on the arrow.
The ion’s path, one arrow at a time:
- Weathering. Frost cracks the rock face, and rain dissolves the ion into the soil water.
- Absorbed by roots. A grass root absorbs the soil water. The grass builds the ion into a nucleotide of its DNA.
- Eating. A deer eats the grass. The deer builds the ion into the bone of its growing antler.
- Dead remains. In winter the deer sheds the antler onto the hillside. Over the years, microbes break the antler down, and the ion is back in the soil water.
The ion is back in the soil, ready for the next root. It will be rock again only after a trip to the sea and millions of years on the sea floor.
Every phosphate ion can be followed like this. Start in a named box, cross one arrow at a time, and name the process on each arrow.
What you are expected to know Trace a phosphate ion through the phosphorus cycle on the template, naming each process in order: weathering, absorbed by roots, eating, dead remains and waste.
Now trace one yourself, one arrow at a time. Suppose a phosphate ion is dissolved in the soil water of a meadow.
A phosphate ion is dissolved in the soil water of a meadow. A clover plant’s root absorbs the water, and the ion is now inside the clover.
Which process carried the ion into the clover?
- A. WeatheringWeathering carries phosphate from rock into the soil.
This ion left the soil water and entered a plant. - B. ✓ Absorbed by roots
- C. EatingEating carries phosphate from a plant into an animal.
This ion entered a plant.
Why: The ion left the soil water and entered the clover through its root.
The arrow from the soil to plants is absorbed by roots.
A rabbit eats the clover, and the phosphate ion is now in the rabbit. In the autumn the rabbit dies, and microbes break down its body.
Which reservoir does the ion enter when the body breaks down?
- A. ✓ The soil
- B. RockPhosphate reaches rock only by settling on the sea floor over millions of years.
The rabbit’s remains lie on the soil. - C. The airNo arrow on the phosphorus template leads into the air.
Phosphate is never a gas.
Why: The rabbit’s remains lie on the soil.
Microbes break them down, and the phosphate dissolves into the soil water.
So the ion enters the soil.
A phosphate ion in a meadow’s soil water was absorbed by a clover root, eaten with the clover by a rabbit, and returned to the soil when the rabbit’s body broke down.
In order, which processes did the ion pass through?
- A. Weathering, eating, dead remains and wasteThe ion started in the soil water, not in rock.
Its first arrow was into the clover: absorbed by roots. - B. Eating, absorbed by roots, dead remains and wasteThe ion entered the clover before the rabbit ate the clover.
Absorbed by roots came first, then eating. - C. ✓ Absorbed by roots, eating, dead remains and waste
Why: The clover root absorbed the ion: absorbed by roots.
The rabbit ate the clover: eating.
The rabbit’s body broke down on the soil: dead remains and waste.
On the phosphorus cycle template below, no arrow is named. One arrow is drawn bold.
Which process is the bold arrow?
- A. ✓ Weathering
- B. Absorbed by rootsAbsorbed by roots leaves the soil and enters plants.
The bold arrow leaves rock. - C. EatingEating leaves plants and enters animals.
The bold arrow joins two abiotic boxes. - D. Dead remains and wasteDead remains and waste leave animals and enter the soil.
The bold arrow leaves rock.
Why: The bold arrow leaves rock and enters the soil.
Rock crumbling and releasing its phosphate into the soil is weathering.
On the phosphorus cycle template below, no box is named. One box is marked with a question mark.
Which reservoir is the marked box?
- A. RockRock is the box the weathering arrow leaves.
The marked box is the one the weathering arrow enters. - B. ✓ The soil
- C. PlantsThe plants box has a double outline.
The marked box has a single outline. - D. AnimalsThe animals box has a double outline.
The marked box has a single outline.
Why: The marked box has a single outline, so it is an abiotic reservoir.
The weathering arrow enters it, and the absorbed-by-roots arrow leaves it.
So the marked box is the soil.
Water, carbon, nitrogen and phosphorus each have a cycle drawn on the template.
Which of the four cycles has no reservoir in the air?
- A. The water cycleWater sits in the atmosphere as water vapor.
The water template has a box at the top. - B. The carbon cycleCarbon sits in the air as carbon dioxide.
The carbon template has a box at the top. - C. The nitrogen cycleNitrogen sits in the air as nitrogen gas.
The nitrogen template has a box at the top. - D. ✓ The phosphorus cycle
Why: Phosphate is never a gas.
No process carries it into the air, so the phosphorus template has no top box.
Suppose a phosphate ion is in a granite boulder on a mountainside. Centuries later it is in the bone of a goat grazing lower down the mountainside.
In order, which processes did the ion pass through?
- A. Weathering, eating, absorbed by rootsA goat cannot eat rock.
The ion entered a plant before the goat ate the plant. - B. ✓ Weathering, absorbed by roots, eating
- C. Absorbed by roots, weathering, eatingThe ion started in rock, and roots absorb from the soil water, not from rock.
Weathering came first.
Why: Weathering freed the ion from the boulder into the soil water.
A plant’s root absorbed it: absorbed by roots.
The goat ate the plant: eating.
44Quick quiz: phosphate, weathering, phosphorus cycle mixed practice
A clover root has just absorbed a phosphate ion from the soil water.
Is the ion still in the soil reservoir?
- A. YesThe absorbed-by-roots arrow moved the ion out of the soil box and into the plants box.
- B. ✓ No
Why: Absorbed by roots is the arrow from the soil to the plants.
The ion has crossed that arrow.
So it is now in the plants reservoir, not the soil.
The phosphorus cycle has four reservoirs on its template.
Is rock one of them?
- A. ✓ Yes
- B. NoRock is the bottom-left box: the slow store of phosphate.
Why: Rock holds phosphate in its minerals.
Weathering releases that phosphate, so rock is a reservoir of the phosphorus cycle.
A fallen branch lies rotting on a forest floor.
Is the phosphate in the branch still in the phosphorus cycle?
- A. ✓ Yes
- B. NoMicrobes break the branch down, and its phosphate dissolves into the soil water.
No atom leaves the cycle.
Why: The branch is dead remains lying on the soil.
Microbes break it down, and its phosphate goes into the soil water.
The phosphate is on the arrow from plants to the soil, still in the cycle.
Phosphate moves between its reservoirs by named processes.
What is weathering?
- A. Roots absorbing soil water and the phosphate dissolved in itRoots absorbing soil water is the arrow from the soil to plants.
- B. ✓ Rock crumbling and releasing the phosphate it holds
- C. Microbes breaking down dead remainsMicrobes breaking down dead remains is decomposition.
Why: Weathering is rock slowly crumbling and releasing what it holds; its phosphate dissolves into the soil water.
Phosphorus moves between its reservoirs as one ion, the phosphate ion.
Which of the following is the formula of the phosphate ion?
- A. ✓ PO₄³⁻
- B. PPhosphorus is never found on its own in nature.
In rock, water and living things it is bonded to four oxygen atoms. - C. NO₃⁻NO₃⁻ is the nitrate ion, a form of nitrogen.
Why: The phosphate ion is one phosphorus atom bonded to four oxygen atoms, carrying three negative charges.
Written as a formula, that is PO₄³⁻.
Phosphate moves between its reservoirs by named processes.
What is the phosphorus cycle?
- A. Phosphate’s path from rock into the soil waterThe path from rock into the soil is one process, weathering, not the whole cycle.
- B. Phosphate’s path from a plant into an animalThe path from a plant into an animal is one process, eating, not the whole cycle.
- C. ✓ Phosphorus’s reservoirs and the processes joining them
Why: The phosphorus cycle is phosphorus’s reservoirs, rock, the soil, plants and animals, and the processes that carry phosphate between them.
Phosphate moves between its reservoirs by named processes.
(a) State what weathering is. (1 pt)
- Award 1 point for: rock crumbling (breaking down) and releasing its phosphate (into the soil or water). Accept with or without: rain, frost or acids as the cause.
52Speed one arrow up
In a cycle, a process moves atoms out of one reservoir and into another.
What happens to the size of the reservoir the atoms enter?
- A. It fallsThe reservoir the atoms leave falls.
The reservoir they enter gains those atoms. - B. It stays the sameThe atoms arrive in the reservoir, and no process destroys them.
So the reservoir gains atoms. - C. ✓ It rises
Why: No process makes or destroys an atom.
Every atom that leaves one reservoir arrives in another.
So the reservoir the atoms enter rises by that number.
Video: Watch: Speed one arrow up
A field beside a lake; rain washing fertilizer off the field; a new arrow appearing from the soil box into a lake box, and the lake box filling.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L18b.mp4
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Now consider a field of maize beside a lake. The farmer spreads fertilizer on the field to feed the crop.
Fertilizer is rich in phosphate ions and in nitrate ions (NO₃⁻), two of the ions that roots absorb.
Heavy rain falls before the roots can absorb the fertilizer. The rain washes some of it off the field and into the lake.
On the template, that is a new arrow: from the field’s soil into the lake’s water.
The lake’s water is a reservoir of phosphate. The new arrow feeds it, and no arrow out of the lake gets any faster.
So the lake’s store of phosphate rises. Its store of nitrate rises the same way.
The algae in the lake absorb phosphate and nitrate from the water to build new cells. With more of both to build with, the algae multiply.
The same rule works for every reservoir on every template. A reservoir is fed by the arrows that enter it and drained by the arrows that leave it.
When a process into a reservoir gets faster, or a new process starts feeding it, the reservoir rises. When a process out of a reservoir gets faster, the reservoir falls.
Now consider the same field at harvest. The farmer cuts the whole maize crop and carts it away.
The plants’ phosphate leaves the field with the crop. The arrow from the plants back to the field’s soil, dead remains, is cut.
Roots still absorb phosphate from the soil every summer. Nothing returns it.
So the field’s soil loses phosphate year by year. That is why the farmer adds fertilizer.
What you are expected to know Predict how a reservoir changes when a process into it or out of it gets faster, or a new one starts. A faster or new arrow in raises the reservoir; a faster arrow out lowers it.
Suppose a town’s sewage pipe empties into a slow river. Sewage carries phosphate. No process takes phosphate out of the river’s water any faster than before.
What happens to the river’s store of phosphate?
- A. ✓ It rises
- B. It stays the sameThe pipe is a new arrow into the river’s water.
No arrow out of the river got faster, so more phosphate enters than leaves. - C. It fallsThe pipe brings phosphate in.
Nothing takes phosphate out any faster, so the store cannot fall.
Why: The sewage pipe is a new arrow into the river’s water.
No arrow out of the river got faster.
So more phosphate enters the river than leaves it, and the store rises.
Suppose hundreds of cormorants begin to roost every night in the trees of a small island in a lake. Their droppings fall into the lake’s water. No process takes phosphate out of the lake’s water any faster than before.
(a) Explain how the roosting cormorants change the lake’s store of phosphate. (1 pt)
Frame The cormorants change the store because …
Droppings carry phosphate, and they fall into the water every night.
No process carries phosphate out of the lake any faster than before.
So more phosphate enters the lake’s water than leaves it, and the store rises.
- Award 1 point for: a new (or faster) process brings phosphate into the lake while no process takes it out any faster, so more enters than leaves and the store rises.
A student says: “The phosphate that rain washes off a field into a lake is lost. It has left the phosphorus cycle.”
Is the student correct?
- A. Yes: the phosphate washed into the lake has left the phosphorus cycleThe phosphate moved from the field’s soil into the lake’s water, another reservoir.
No process destroys an atom. - B. ✓ No: the phosphate has moved into the lake’s water, another reservoir of the cycle
Why: Rain carried the phosphate from the field’s soil into the lake’s water.
The lake’s water is a reservoir of phosphate.
No process destroys an atom, so the phosphate is still in the cycle.
The field’s store fell, and the lake’s store rose by the same amount.
Suppose a forest has a warm, wet year. The microbes in its soil break down the fallen leaves faster than usual. Nothing else in the forest changes.
What happens to the soil’s store of phosphate during that year?
- A. It fallsDecomposition carries phosphate into the soil water, not out of it.
A faster arrow in raises the store. - B. It stays the sameDecomposition got faster, and no arrow out of the soil got faster.
So more phosphate enters the soil than leaves it. - C. ✓ It rises
Why: Decomposition frees phosphate from dead leaves into the soil water.
In the warm, wet year that process got faster.
No process out of the soil got faster.
So the soil’s store of phosphate rises.
73One cycle leans on another
Unit 3 wrote photosynthesis as a word equation.
Besides carbon dioxide, which substance does photosynthesis take in?
- A. ✓ Water
- B. OxygenOxygen is what photosynthesis gives out, not what it takes in.
- C. GlucoseGlucose is what photosynthesis makes, not what it takes in.
Why: Photosynthesis: carbon dioxide + water → glucose + oxygen.
It takes in carbon dioxide and water.
Video: Watch: One cycle leans on another
Rain falling on rock, dissolving phosphate and carrying it through the soil to a root and down a river; the rain stopping and the phosphate staying in the rock; the four templates side by side.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L18c.mp4
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Look at the weathering arrow again. Rain crumbles the rock, and the freed phosphate dissolves in the rainwater.
Soil water carries the phosphate to the roots. Rivers carry the rest to the sea.
Every one of those moves is water moving. Rain, water moving through the soil, and rivers are all arrows of the water cycle.
Now imagine the rain stops for good. No rain crumbles the rock, and no soil water carries phosphate to a root.
The phosphate stays in the rock. Without the water cycle, the phosphorus cycle stops.
So the phosphorus cycle depends on the water cycle. One cycle’s arrow, rain, carries another cycle’s atoms, phosphate.
The four cycles are drawn on four templates, but they share arrows. Where one cycle’s arrow carries another cycle’s atoms, the two cycles depend on each other.
The table below compares the four cycles: the non-living store each one’s living things draw on, the arrow into living things, the arrow back out, and whether the cycle has a store in the air.
Only the phosphorus row says no. Every other cycle has a gas at the top of its template.
What you are expected to know Explain, with one example, how two cycles depend on each other: rain and rivers, arrows of the water cycle, carry phosphate from weathered rock to roots and to the sea, so the phosphorus cycle needs the water cycle.
Weathered rock releases phosphate, and the phosphate ends up in a river.
Which cycle’s arrows carry the phosphate from the rock into the river?
- A. ✓ The water cycle’s
- B. The carbon cycle’sThe carbon cycle’s arrows are photosynthesis, respiration, decomposition and combustion.
None of them carries phosphate into a river. - C. The nitrogen cycle’sThe nitrogen cycle’s arrows move nitrogen between its forms.
None of them carries phosphate into a river.
Why: Rain dissolves the phosphate from the crumbling rock.
Soil water and rivers carry the dissolved phosphate downhill.
Rain, water moving through the soil, and rivers are arrows of the water cycle.
Suppose a long drought stops all rain on a rocky hillside for a year.
(a) Explain why the phosphorus cycle on the hillside slows during the drought. (1 pt)
Frame The phosphorus cycle slows because …
With no rain, no phosphate is freed from the rock.
With no soil water, no phosphate is carried to the roots.
So the arrows from rock to the soil and from the soil to plants both slow.
- Award 1 point for: water (rain, soil water) carries phosphate from rock into the soil and to roots, so with no rain those processes slow (or stop).
A student says: “Each cycle works on its own. An arrow of the water cycle carries only water, and no other cycle’s atoms.”
Is the student correct?
- A. Yes: an arrow of the water cycle carries only water and no other cycle’s atomsRain and rivers are arrows of the water cycle.
They carry dissolved phosphate, the phosphorus cycle’s atoms. - B. ✓ No: an arrow of the water cycle can carry another cycle’s atoms dissolved in it
Why: Rain crumbles rock and dissolves its phosphate.
Soil water and rivers carry that phosphate to roots and to the sea.
So the water cycle’s arrows carry the phosphorus cycle’s atoms: the cycles depend on each other.
Photosynthesis is the arrow from the air into plants on the carbon template. Its word equation has two substances going in.
The atoms of which two cycles does the photosynthesis arrow carry into the plant?
- A. The carbon cycle and the nitrogen cyclePhotosynthesis takes in no nitrogen.
Nitrogen enters a plant through its roots, as ammonium or nitrate. - B. The carbon cycle and the phosphorus cyclePhotosynthesis takes in no phosphate.
Phosphate enters a plant through its roots, dissolved in soil water. - C. ✓ The carbon cycle and the water cycle
Why: Photosynthesis takes in carbon dioxide from the air: the carbon cycle’s arrow.
It also takes in water the roots absorbed: the water cycle’s atoms.
So the one arrow carries the atoms of the carbon cycle and the water cycle.
Here is the antler again, where the deer shed it.
Weathering freed its phosphate from the hill’s rock, and rain carried the phosphate into the soil water. The grass absorbed it, and the deer ate the grass.
Microbes will break the antler down, and its phosphate will go back into the soil water for the next root, unless rain washes it off the hill into a lake first.
Nothing in the story went through the air.
94Mixed practice mixed practice
A quarry blasts open fresh rock faces on a hillside, and rain now reaches rock that was buried.
What happens to the soil’s store of phosphate downhill from the quarry?
- A. It fallsWeathering carries phosphate into the soil, not out of it.
More rock crumbling means a faster arrow in. - B. It stays the sameRain now crumbles more rock, so weathering is faster.
A faster arrow in raises the store. - C. ✓ It rises
Why: Rain reaches more rock, so weathering gets faster.
Weathering is the arrow from rock into the soil.
A faster arrow in raises the soil’s store of phosphate.
One reservoir of the phosphorus cycle holds its phosphate for millions of years.
Which reservoir is that?
- A. ✓ Rock
- B. The soilRoots absorb phosphate from the soil water within a season.
- C. PlantsA plant holds its phosphate until it is eaten or dies.
Why: Phosphate leaves rock only as the rock itself crumbles.
Rock is the slow store of the phosphorus cycle.
Rain soaks into a field’s soil. A wheat root takes up nitrate dissolved in that soil water.
Is the water soaking through the soil carrying the nitrogen cycle’s atoms as well as the water cycle’s?
- A. ✓ Yes
- B. NoWater moving through the soil is an arrow of the water cycle.
The nitrate dissolved in it moves with it, so the same water carries the nitrogen cycle’s atoms.
Why: Water moving through the soil is an arrow of the water cycle.
The dissolved nitrate moves with the water to the root.
So the same arrow carries the nitrogen cycle’s atoms.
Suppose a phosphate ion in the soil water is absorbed by an oak’s root and built into an acorn. A squirrel eats the acorn.
Which reservoir is the ion in now?
- A. PlantsThe acorn was the oak’s, but the squirrel has eaten it.
The ion is now inside the squirrel. - B. ✓ Animals
- C. The soilThe ion left the soil water when the oak’s root absorbed it.
Nothing has yet carried it back to the soil.
Why: The oak built the ion into an acorn, so the ion was in the plants reservoir.
The squirrel ate the acorn: eating.
Eating carries phosphate from plants to animals.
So the ion is now in the animals reservoir.
Every autumn a gardener rakes up all the leaves that fall under an apple tree and carts them away. The tree’s roots absorb phosphate from the soil every summer.
Over the years, what happens to the soil’s store of phosphate under the tree?
- A. It risesThe roots take phosphate out of the soil every summer.
The leaves that would return it are carted away. - B. It stays the sameThe arrow out, absorbed by roots, still works.
The arrow back, dead remains, is cut, so the store cannot stay the same. - C. ✓ It falls
Why: The roots absorb phosphate from the soil every summer: an arrow out.
The fallen leaves would return it, but the gardener carts them away.
So phosphate leaves the soil and nothing returns it, and the store falls.
On the phosphorus cycle template below, no arrow is named. One arrow is drawn bold.
Which process is the bold arrow?
- A. WeatheringWeathering runs from rock into the soil.
The bold arrow leaves the animals. - B. ✓ Dead remains and waste
- C. Absorbed by rootsAbsorbed by roots runs from the soil up into the plants.
The bold arrow leaves the animals and enters the soil.
Why: The bold arrow leaves the animals and enters the soil.
An animal’s dead body and its droppings return their phosphate to the soil.
So the bold arrow is dead remains and waste.
Suppose a herd of cattle grazes a hillside pasture above a pond. On rainy days, rain washes the cattle’s dung down the slope into the pond.
(a) Predict how the pond’s store of phosphate changes over the grazing season. (1 pt)
- Award 1 point for: the pond’s store of phosphate rises (increases).
(b) Explain how this change shows that the phosphorus cycle depends on the water cycle. (1 pt)
Frame The change shows this because …
On dry days no phosphate reaches the pond.
On rainy days rain carries the phosphate down the slope into the pond’s water.
So the phosphate moves into the pond when rain moves it, and stays on the hillside when no rain falls.
- Award 1 point for: rain (or flowing water), a process of the water cycle, is what carries the phosphate into the pond, so the phosphorus cycle’s arrow into the pond depends on the water cycle.
Glossary
- phosphate ion (PO₄³⁻)
- One phosphorus atom bonded to four oxygen atoms, carrying three negative charges. It is the form phosphorus takes in rock, in soil water and in living things: the phosphate group of a phospholipid’s head and of every nucleotide.
- weathering
- Rock slowly crumbling, under rain, frost and the acids in soil, and releasing what it holds; its phosphate dissolves into the soil water. On the cycle template it is the arrow from rock to the soil.
- phosphorus cycle
- Phosphorus’s reservoirs (rock, the soil and its water, plants and animals) and the processes that carry phosphate between them (weathering, absorbed by roots, eating, dead remains and waste). It has no reservoir in the air: rock is its slow store.
APBIO-U08-L19 Who makes the food
Photos: NOAA Photo Library; University of Washington / NOAA Office of Ocean Exploration and Research. Both via Wikimedia Commons, CC BY 2.0 (resized).
Here are two communities. On the left, a kelp forest sways in sunlit water. On the right, two kilometers down, where no light reaches, tube worms crowd around a vent pouring out hot, sulfurous water. Both communities are full of animals that eat.
What is making the food in the dark?
Unit 8 · Ecology
1Who makes the organic molecules
Unit 3 judged each living thing by one test: does it capture light energy to make sugar from carbon dioxide and water?
Which of the following photosynthesize?
- A. Plants onlyAlgae and cyanobacteria capture light energy to make sugar too.
An alga holds chloroplasts; a cyanobacterium catches light on membranes folded inside the cell. - B. Animals and fungiNo animal and no fungus captures light energy to make sugar.
Each animal and each fungus gets its sugar by eating. - C. ✓ Plants, algae and cyanobacteria
Why: An organism photosynthesizes when it captures light energy to make sugar from carbon dioxide and water.
Plants, algae and cyanobacteria do this.
Animals and fungi do not.
Who makes the organic molecules everything else eats, and from what energy?
Kelp is a large seaweed. It makes its own sugar from carbon dioxide and water, using the energy of sunlight.
The fish in the kelp forest make no sugar of their own. They get their organic molecules by eating other living things.
At the vent, no light reaches and no kelp grows. Bacteria there make sugar from carbon dioxide, using the energy in a chemical dissolved in the vent water.
The animals crowded around the vent get their organic molecules from what those bacteria make.
Video: Watch: Who makes the organic molecules
The kelp forest in sunlight and the vent two kilometers down in the dark; in each, the organisms that make their own sugar from carbon dioxide, and the animals that eat instead.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L19a.mp4
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Here is the word equation for photosynthesis, as Unit 3 wrote it. Kelp is an alga, so kelp does this in sunlit water.
For example, kelp makes its own organic molecules. It makes sugar from carbon dioxide and water, using the energy of sunlight.
But a fish among the kelp makes no organic molecules of its own. It gets its organic molecules by eating other living things.
And grass in a meadow makes its own organic molecules. It makes sugar from carbon dioxide and water, using the energy of sunlight.
But a cow grazing the meadow makes no organic molecules of its own. It gets its organic molecules by eating the grass.
Now consider the bacteria at the vent, two kilometers down, where no light reaches. These bacteria make their own organic molecules too: they make sugar from carbon dioxide, using the energy in a chemical dissolved in the vent water.
But the mussels and shrimp crowded around the vent make no organic molecules of their own. They get their organic molecules by eating.
Here are the six cases in a table, each with its verdict: does it make its own organic molecules from carbon dioxide, or not?
So every organism gets its organic molecules in one of two ways. It makes them itself from carbon dioxide, or it eats other organisms or their remains.
An organism that makes its own organic molecules from carbon dioxide, using an energy source from its surroundings, is called an . Auto means self, and troph means feeding: an autotroph feeds itself.
An organism that gets its organic molecules by eating other organisms or their remains is called a . Hetero means other: a heterotroph feeds on others.
Ecologists have a second name for each group. Here is a table comparing the two groups: where the organic molecules come from, the ecologist’s name, and one example.
An autotroph is also called a producer, because it produces the organic molecules the whole community lives on.
A heterotroph is also called a consumer, because it consumes what the producers made.
The two pairs of names describe the same two groups. Ecologists mostly say producer and consumer.
The grass is the cow’s food. But being eaten does not make the grass a heterotroph: the grass made its own sugar, and it respires that sugar itself.
What you are expected to know Classify an organism as an autotroph (a producer) or a heterotroph (a consumer) from where its organic molecules come from: made from carbon dioxide, or eaten.
Kelp is an autotroph.
Where does the kelp get its organic molecules?
- A. From the dead remains that settle around itKelp takes in no organic molecules from its surroundings.
An autotroph makes its own organic molecules, from carbon dioxide. - B. ✓ From carbon dioxide, by making them itself
Why: Kelp is an autotroph.
An autotroph makes its own organic molecules from carbon dioxide.
So the kelp makes its organic molecules itself, from carbon dioxide.
An oak tree makes sugar from carbon dioxide and water, using the energy of sunlight.
Which is the oak tree?
- A. ✓ An autotroph
- B. A heterotrophThe oak makes its own organic molecules from carbon dioxide.
An organism that does that is an autotroph.
Why: The oak makes its own organic molecules from carbon dioxide, using the energy of sunlight.
So the oak is an autotroph.
A cyanobacterium in pond scum captures light energy and makes sugar from carbon dioxide and water.
Which is the cyanobacterium?
- A. ✓ An autotroph
- B. A heterotrophThe cyanobacterium makes its own sugar from carbon dioxide.
An organism that does that is an autotroph, chloroplast or no chloroplast.
Why: The cyanobacterium makes its own organic molecules from carbon dioxide, using the energy of light.
So the cyanobacterium is an autotroph.
A caterpillar eats oak leaves.
Which is the caterpillar?
- A. An autotrophThe caterpillar makes no organic molecules of its own.
It gets them by eating the leaves. - B. ✓ A heterotroph
Why: The caterpillar gets its organic molecules by eating another organism, the oak.
So the caterpillar is a heterotroph.
A fungus grows on a dead log and takes its organic molecules from the rotting log.
Which is the fungus?
- A. An autotrophThe fungus makes no organic molecules from carbon dioxide.
It takes them from the remains of a tree. - B. ✓ A heterotroph
Why: The fungus gets its organic molecules from the remains of another organism, the dead tree.
So the fungus is a heterotroph.
A bacterium in dark vent water makes sugar from carbon dioxide, using the energy in a chemical dissolved in the water.
Which is the bacterium?
- A. ✓ An autotroph
- B. A heterotrophThe bacterium makes its own sugar from carbon dioxide.
Where the energy comes from does not change that: an organism that makes its own organic molecules is an autotroph.
Why: The bacterium makes its own organic molecules from carbon dioxide, using an energy source from its surroundings.
So the bacterium is an autotroph, even in the dark.
A person eats bread, vegetables and meat.
Which is the person?
- A. An autotrophA person makes no organic molecules from carbon dioxide.
Every organic molecule in the meal came from another organism. - B. ✓ A heterotroph
Why: A person gets every organic molecule by eating other organisms.
So a person is a heterotroph.
A deer eats the leaves of a hazel bush. A student says: “Once the hazel is eaten, it counts as a heterotroph, because it is now part of the deer’s food.”
Is the student correct?
- A. Yes: once the hazel is eaten, it counts as a heterotroph, part of the deer’s foodThe hazel made its own organic molecules from carbon dioxide.
Being eaten changes nothing about how the hazel got them. - B. ✓ No: the hazel made its own organic molecules, so it is an autotroph, eaten or not
Why: A heterotroph is an organism that gets its organic molecules by eating.
The hazel made its own organic molecules from carbon dioxide, using the energy of sunlight.
So the hazel is an autotroph, eaten or not.
The deer is the heterotroph.
34Quick quiz: autotroph (producer), heterotroph (consumer) mixed practice
Moss on a damp rock makes sugar from carbon dioxide and water, using the energy of sunlight.
Which is the moss, in the ecologist’s words?
- A. ✓ A producer
- B. A consumerThe moss makes its own organic molecules from carbon dioxide.
An organism that does that is an autotroph: a producer.
Why: The moss makes its own organic molecules from carbon dioxide.
So the moss is an autotroph, which ecologists call a producer.
A snail eats the moss on the rock.
Which is the snail, in the ecologist’s words?
- A. A producerThe snail makes no organic molecules from carbon dioxide.
It gets them by eating the moss: a heterotroph, a consumer. - B. ✓ A consumer
Why: The snail gets its organic molecules by eating the moss.
So the snail is a heterotroph, which ecologists call a consumer.
An ecologist lists a castle moat’s autotrophs: the rushes at its edge, and the algae on its mud.
Which of the following is the ecologist’s other name for that list?
- A. ✓ The moat’s producers
- B. The moat’s consumersConsumer is the ecologist’s name for a heterotroph, which consumes what others made.
The rushes and the algae make their own organic molecules.
Why: The rushes and the algae make their own organic molecules: they are autotrophs.
An autotroph produces the organic molecules the community lives on.
So ecologists call the moat’s autotrophs its producers.
Every organism needs organic molecules.
What is an autotroph?
- A. ✓ An organism that makes its own organic molecules from carbon dioxide
- B. An organism that gets its organic molecules by eating other organisms or their remainsGetting organic molecules by eating other organisms or their remains is what a heterotroph does.
- C. An organism that lives only where sunlight reaches the water or the groundWhere an organism lives decides nothing here.
The vent bacteria live in the dark and are autotrophs.
Why: An autotroph makes its own organic molecules from carbon dioxide, using an energy source from its surroundings.
Three living things share a rocky shore.
Which of the following is a heterotroph?
- A. Sea lettuce, a seaweed on the rocks, making sugar from carbon dioxide and waterThe sea lettuce makes its own organic molecules from carbon dioxide.
An organism that does that is an autotroph. - B. ✓ A shore crab eating dead fish washed up on the sand
- C. The red seaweeds in the rock pools, making sugar from carbon dioxide and waterThe red seaweeds make their own organic molecules from carbon dioxide.
An organism that does that is an autotroph.
Why: The shore crab makes no organic molecules from carbon dioxide.
It gets its organic molecules by eating the remains of another organism, the dead fish.
An organism that gets its organic molecules by eating other organisms or their remains is a heterotroph.
Ecologists sort every organism in a community into two groups by where its organic molecules come from.
(a) State what an autotroph is and what a heterotroph is. (1 pt)
A heterotroph gets its organic molecules by eating other organisms or their remains.
- Award 1 point for: an autotroph makes its own organic molecules from carbon dioxide (using an energy source from its surroundings) AND a heterotroph gets its organic molecules by eating other organisms or their remains.
41Energy from light, or energy from a chemical
Unit 3 followed light energy into a leaf cell’s chloroplast.
Where do the light reactions capture light energy?
- A. ✓ In the thylakoid membranes of the grana
- B. In the stroma, the fluid that fills the chloroplastThe stroma is where the Calvin cycle uses the ATP and NADPH that the light reactions made.
- C. In the mitochondrion, beside the chloroplastThe mitochondrion is where respiration happens, in the light and in the dark.
Why: The light reactions capture light energy in the thylakoid membranes of the grana.
The ATP and NADPH they make then move into the stroma.
Both the kelp and the vent bacteria make their own sugar from carbon dioxide. Where does each get the energy to do it?
Kelp gets its energy from sunlight, like a plant. The vent bacteria get theirs from a chemical in the vent water, hydrogen sulfide.
Making sugar with the energy in a chemical needs no light. So it can happen two kilometers down, where no photosynthesis is possible.
Video: Watch: Energy from light, or energy from a chemical
The kelp in sunlight and the vent bacteria in the dark side by side, each making sugar from carbon dioxide; the energy source labeled over each: sunlight, and hydrogen sulfide dissolved in the vent water.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L19b.mp4
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Here are the two communities side by side, with the energy source of each labeled.
Kelp captures the energy in sunlight. With that energy, it makes sugar from carbon dioxide and water.
Kelp is an alga, and its cells hold chloroplasts. So kelp captures light energy where a leaf does: in the thylakoid membranes of its chloroplasts.
The vent bacteria capture the energy in hydrogen sulfide. With that energy, they make sugar from carbon dioxide.
Hydrogen sulfide (H₂S) is a small molecule that pours out of the vent, dissolved in the hot water. It is not an organic molecule: it holds no carbon.
Small molecules that hold no carbon chain, such as hydrogen sulfide, are inorganic molecules. The bacteria take the energy they need from these molecules.
Making organic molecules from carbon dioxide using the energy in sunlight is photosynthesis, the process Unit 3 taught. An autotroph that does this is a photosynthetic autotroph, and kelp is one.
Making organic molecules from carbon dioxide using the energy in small inorganic molecules, such as hydrogen sulfide, is called . Chemo means chemical, and synthesis means building: building with a chemical’s energy.
An autotroph that does this is a chemosynthetic autotroph. The vent bacteria are chemosynthetic autotrophs.
Chemosynthesis needs no light. So the vent bacteria make sugar two kilometers down, in the dark, where no photosynthetic autotroph could live.
Some kinds of chemosynthesis happen where there is no oxygen. So some chemosynthetic bacteria live in water that holds no oxygen at all.
The vent bacteria on this page need oxygen. They use oxygen dissolved in the sea water when they take energy from hydrogen sulfide.
The vent water is hot. But heat is not the bacteria’s energy source: their energy source is the hydrogen sulfide dissolved in that water.
What you are expected to know Distinguish a photosynthetic autotroph, which captures the energy in sunlight, from a chemosynthetic autotroph, which captures the energy in small inorganic molecules such as hydrogen sulfide; some kinds can do so where there is no oxygen.
The bacteria at a deep-sea vent make sugar from carbon dioxide.
Which is their energy source?
- A. Sunlight reaching down through the waterNo sunlight reaches two kilometers down.
The bacteria make sugar there in the dark. - B. The heat of the hot water pouring from the ventThe water is hot, but the bacteria take their energy from a chemical dissolved in it, hydrogen sulfide.
- C. ✓ Hydrogen sulfide dissolved in the vent water
Why: The vent bacteria capture the energy in hydrogen sulfide, a small inorganic molecule dissolved in the vent water.
With that energy they make sugar from carbon dioxide.
Suppose a bacterium floating in a sunlit pond makes sugar from carbon dioxide and water, using the energy in sunlight.
Which kind of autotroph is this bacterium?
- A. ✓ A photosynthetic autotroph
- B. A chemosynthetic autotrophA chemosynthetic autotroph takes its energy from a small inorganic molecule.
This bacterium takes its energy from sunlight.
Why: The bacterium makes its organic molecules from carbon dioxide using the energy in sunlight.
So it is a photosynthetic autotroph.
Suppose a bacterium deep in a dark cave makes sugar from carbon dioxide, using the energy in a small inorganic molecule dissolved in the cave water.
Which kind of autotroph is this bacterium?
- A. A photosynthetic autotrophA photosynthetic autotroph takes its energy from sunlight.
No light reaches this bacterium; its energy comes from a chemical. - B. ✓ A chemosynthetic autotroph
Why: The bacterium makes its organic molecules from carbon dioxide using the energy in a small inorganic molecule.
So it is a chemosynthetic autotroph.
A student says: “Chemosynthesis needs oxygen, so chemosynthetic bacteria die in water with no oxygen in it.”
Is the student correct?
- A. ✓ No: some kinds of chemosynthesis happen where there is no oxygen
- B. Yes: chemosynthetic bacteria die in water that holds no oxygenSome kinds of chemosynthesis happen where there is no oxygen.
Chemosynthetic bacteria of those kinds live in water that holds no oxygen at all.
Why: Chemosynthesis is making organic molecules from carbon dioxide using the energy in small inorganic molecules.
It needs no light, and some kinds of it happen where there is no oxygen.
So some chemosynthetic bacteria live in water with no oxygen in it, and ‘needs oxygen’ is wrong as a rule.
65Quick quiz: chemosynthesis mixed practice
Bacteria in the mud at the bottom of a lake make sugar from carbon dioxide, using the energy in ammonia (NH₃) dissolved in the mud water. Ammonia holds no carbon.
Is this chemosynthesis?
- A. ✓ Yes
- B. NoThe bacteria make organic molecules from carbon dioxide using the energy in a small inorganic molecule.
That is chemosynthesis, whatever the molecule.
Why: Chemosynthesis is making organic molecules from carbon dioxide using the energy in small inorganic molecules.
Ammonia is a small molecule that holds no carbon: an inorganic molecule.
So this is chemosynthesis.
A seaweed on a sunlit rock makes sugar from carbon dioxide and water, using the energy in sunlight.
Is this chemosynthesis?
- A. YesThe seaweed takes its energy from sunlight.
Making organic molecules with the energy in sunlight is photosynthesis. - B. ✓ No
Why: Chemosynthesis uses the energy in small inorganic molecules.
The seaweed uses the energy in sunlight.
So this is photosynthesis, not chemosynthesis.
A shrimp at a deep-sea vent eats the bacteria that coat the rocks.
Is the shrimp carrying out chemosynthesis?
- A. YesThe shrimp makes no organic molecules from carbon dioxide.
It eats the bacteria that did. - B. ✓ No
Why: Chemosynthesis is making organic molecules from carbon dioxide using the energy in a small inorganic molecule.
The shrimp makes no organic molecules from carbon dioxide; it eats.
So the shrimp is a heterotroph, and the bacteria it eats are the chemosynthetic autotrophs.
Some autotrophs make their organic molecules with no light at all.
What is chemosynthesis?
- A. Making organic molecules from carbon dioxide using the energy in sunlightUsing the energy in sunlight to make organic molecules is photosynthesis.
- B. Getting organic molecules ready-made by eating other organisms or their remainsGetting organic molecules by eating is what a heterotroph does; it makes nothing from carbon dioxide.
- C. ✓ Making organic molecules from carbon dioxide using the energy in small inorganic molecules
Why: Chemosynthesis is making organic molecules from carbon dioxide using the energy in small inorganic molecules, such as hydrogen sulfide.
Photosynthesis and chemosynthesis both make organic molecules from carbon dioxide.
(a) State what chemosynthesis is. (1 pt)
- Award 1 point for: chemosynthesis makes organic molecules from carbon dioxide using the energy in small inorganic molecules (such as hydrogen sulfide). Accept with or without: the contrast with photosynthesis, which uses the energy in light.
71How fast the producers make food
Unit 3 gave the speed of a reaction a number.
Which of the following is a rate?
- A. ✓ 3.0 mL of oxygen per minute
- B. 12 mL of oxygen12 mL of oxygen is an amount.
A rate says how much in each unit of time.
Why: A rate is an amount per unit of time.
3.0 mL of oxygen per minute is an amount of oxygen in each minute: a rate.
How much food do a community’s producers make, and how would an ecologist put a number on it?
A field of grass in June and the same field in December hold very different amounts of grass. So the amount standing in the field on one day says little about how fast the grass grows.
An ecologist measures how much energy the producers turned into organic matter on a set area, in a set time: a rate.
Video: Watch: How fast the producers make food
One square meter of meadow through one year; the energy its grass turned into organic matter growing as one bar, with its unit, kilojoules per square meter a year.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L19c.mp4
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Now consider a meadow. Its grass is the meadow’s producer, and through the year it makes sugar from carbon dioxide and water.
Suppose an ecologist measures how fast the meadow’s producers make organic matter. She marks out a set area, one square meter, and follows it for a set time, one year.
She measures the energy the producers turned into organic matter on that square meter in that year. Here it is as one bar.
The result is 10 000 kilojoules per square meter a year: an amount of energy, per area, per time.
A quantity per unit of time is a rate. So this quantity is a rate, just like a rate of reaction.
The rate at which an ecosystem’s producers turn energy into organic matter, per area per time, is called . Primary means first: the producers are the first organisms to hold the energy.
The producers respire some of the sugar they made. That energy leaves the producers as heat.
The whole amount the producers made, before respiration, is called gross primary productivity.
What is left after the producers’ respiration is called . Net means what is left after taking something away.
Both figures are rates, in kilojoules per square meter a year. Only the net figure is there for the animals that eat the grass.
Productivity is a rate. The grass standing in the meadow on one day is an amount, and an amount is not a productivity.
Two meadows can hold the same amount of grass today and differ in productivity. One grew that grass in a month; the other took a year.
What you are expected to know State what primary productivity measures: the rate at which an ecosystem’s producers turn energy into organic matter, per area per time.
An ecologist reports a lake’s primary productivity.
Which kind of quantity is she reporting?
- A. ✓ A rate
- B. An amountAn amount would be the energy in the producers on one day, standing in the lake.
Productivity is how fast the producers made organic matter.
Why: Primary productivity is the rate at which the producers turn energy into organic matter, per area per time.
So she is reporting a rate.
A student says: “A meadow’s primary productivity is how much grass is standing in the field.”
Is the student correct?
- A. Yes: the grass standing in the field on one day is the meadow’s primary productivityThe grass standing in the field is an amount.
Primary productivity is a rate: the energy the producers turned into organic matter per area per time. - B. ✓ No: two meadows with the same grass today can differ in productivity, one grown in a month, one in a year
Why: Primary productivity is the rate at which the producers turn energy into organic matter, per area per time.
The grass standing in the field on one day is an amount, not a rate.
So the standing grass is not the meadow’s productivity.
An ecologist studies an upland meadow and a lowland meadow. The table below gives the one figure she reports for each meadow.
Which figure is a primary productivity?
- A. ✓ The upland meadow’s figure
- B. The lowland meadow’s figureThe lowland meadow’s figure is the energy in the grass standing on one square meter today: an amount.
A productivity is a rate, per area per time.
Why: Primary productivity is energy turned into organic matter, per area, per time.
The upland meadow’s figure is per square meter, in one year: a rate.
The lowland meadow’s figure is the energy standing on one day: an amount.
Here are the two communities again: the kelp forest in the light, and the tube worms crowded around the vent in the dark.
The kelp captures the energy in sunlight. The vent bacteria capture the energy in hydrogen sulfide.
So both are autotrophs, the producers. In the kelp forest the food is made by photosynthesis; at the vent it is made by chemosynthesis, in the dark.
Each community’s primary productivity is the rate at which its producers turn energy into organic matter, per area per time.
97Quick quiz: primary productivity, net primary productivity mixed practice
A forest’s trees turned 6 500 kJ of energy into organic matter per square meter in one year.
Is this quantity a rate or an amount?
- A. ✓ A rate
- B. An amountThe quantity is per square meter, in one year: energy per area per time.
A quantity per time is a rate.
Why: The quantity is energy turned into organic matter per square meter, in one year.
A quantity per unit of time is a rate.
A lagoon’s producers made 18 500 kJ of organic matter per square meter in one year and respired 11 300 kJ of it, leaving 7 200 kJ in new growth.
Which of the three values is the lagoon’s gross primary productivity?
- A. 7 200 kJ per square meter a year7 200 kJ per square meter a year is what is left after the producers’ respiration: the net primary productivity.
- B. 11 300 kJ per square meter a year11 300 kJ per square meter a year is what the producers respired: the part taken away.
- C. ✓ 18 500 kJ per square meter a year
Why: Gross means the whole amount, before respiration.
The lagoon’s producers made 18 500 kJ per square meter in the year in all.
So the gross primary productivity is 18 500 kJ per square meter a year.
A pond’s algae hold 90 kJ of energy on one afternoon in July.
Is this quantity a rate or an amount?
- A. A rateThe quantity is the energy in the algae at one moment.
Nothing is per unit of time, so it is an amount. - B. ✓ An amount
Why: The quantity is the energy the algae hold at one moment, with no time in it.
A quantity with no per-time in it is an amount, not a rate.
Ecologists put a number on how fast a community’s producers make food.
What is primary productivity?
- A. The amount of organic matter standing in an ecosystem on one dayOrganic matter standing on one day is an amount, not a rate.
- B. ✓ The rate at which the producers turn energy into organic matter, per area per time
- C. The number of producers living on one square meter of the ecosystemA count of producers is a number of organisms.
Productivity is energy turned into organic matter, per area per time.
Why: Primary productivity is the rate at which an ecosystem’s producers turn energy into organic matter, per area per time.
The producers of an ecosystem respire some of the organic matter they make.
What is net primary productivity?
- A. The rate at which the producers make organic matter, before their respirationThe whole rate before respiration is gross primary productivity.
- B. The rate at which the producers respire the organic matter they madeThe producers’ respiration is what is taken away, not what is left.
- C. ✓ The rate at which the producers make organic matter, after their respiration is taken away
Why: Net means what is left after something is taken away.
Net primary productivity is the rate at which the producers make organic matter after their own respiration is taken away.
An ecologist reports the primary productivity of a salt marsh.
(a) State what primary productivity measures, and give a unit it could be reported in. (1 pt)
- Award 1 point for: the rate at which the producers turn energy into organic matter per area per time, with a rate unit such as kJ per square meter a year (accept kJ/m²/yr or an equivalent per-area-per-time unit).
104Mixed practice mixed practice
Bacteria in the water draining from an old mine make their organic molecules from carbon dioxide, using the energy in inorganic iron compounds dissolved in the water.
Which are these bacteria?
- A. ✓ Chemosynthetic autotrophs
- B. Photosynthetic autotrophsA photosynthetic autotroph takes its energy from sunlight.
These bacteria take theirs from dissolved iron compounds. - C. HeterotrophsA heterotroph gets its organic molecules by eating.
These bacteria make theirs from carbon dioxide.
Why: The bacteria make their own organic molecules from carbon dioxide, so they are autotrophs.
Their energy comes from an inorganic chemical, not from light.
So they are chemosynthetic autotrophs.
A vulture eats a dead deer.
Which is the vulture?
- A. An autotrophThe vulture makes no organic molecules from carbon dioxide.
It gets them from the remains of another organism. - B. ✓ A heterotroph
Why: The vulture gets its organic molecules by eating the remains of another organism.
So the vulture is a heterotroph.
Suppose bacteria coat the rocks of a hot spring and make their organic molecules from carbon dioxide. No light reaches the rocks. A student says: “With no light, the bacteria must use the spring’s heat as their energy source.”
Is the student correct?
- A. ✓ No: the bacteria take their energy from a small inorganic molecule dissolved in the water
- B. Yes: with no light, the water’s heat is the only energy source leftHeat is not an energy source for making sugar.
An autotroph in the dark takes its energy from a small inorganic molecule dissolved in the water.
Why: An autotroph needs an energy source from its surroundings to make sugar from carbon dioxide.
Where no light reaches, that source is a small inorganic molecule dissolved in the water: chemosynthesis.
The water’s heat is not an energy source for making sugar.
An ecologist reports a prairie’s primary productivity as 4 800 kJ of energy turned into organic matter per square meter in one year. A student says: “The ecologist’s number is a rate: it is energy per area in a set time.”
Is the student correct?
- A. ✓ Yes: the number is energy per square meter in one year, a quantity per time
- B. No: 4 800 kJ is an amount of energy, so the prairie’s number is an amount tooThe 4 800 kJ is counted per square meter, in one year.
A quantity per unit of time is a rate, whatever amount of energy it names.
Why: The number is energy turned into organic matter per square meter in one year.
A quantity per unit of time is a rate.
So the prairie’s number is a rate, not an amount.
A student says: “A pond’s producers and its autotrophs are one and the same group of organisms.”
Is the student correct?
- A. No: the producers are a different group from the autotrophsProducer is the ecologist’s name for an autotroph.
The two words name one and the same group. - B. ✓ Yes: the pond’s producers and its autotrophs are one group
Why: An autotroph makes its own organic molecules from carbon dioxide.
Ecologists call the same organism a producer, because it produces the organic molecules the pond lives on.
So the pond’s producers and its autotrophs are one group.
A park pond and a quarry pond hold the same mass of algae on the same July day. The park pond’s algae grew that mass in two weeks. The quarry pond’s algae took four months.
Which pond has the higher primary productivity?
- A. ✓ The park pond
- B. The quarry pondThe quarry pond’s algae took four months to make the same organic matter.
The same amount in a longer time is a lower rate.
Why: Primary productivity is a rate: organic matter made per area per time.
The park pond’s algae made the same organic matter in two weeks; the quarry pond’s took four months.
The same amount in less time is a higher rate.
So the park pond has the higher primary productivity.
Suppose a lake deep inside a cave, where no light reaches. Bacteria coat its rocks and make their organic molecules from carbon dioxide, using the energy in hydrogen sulfide dissolved in the water. Small animals graze the bacteria, and blind fish eat the small animals.
(a) Identify the producers in this cave community. (1 pt)
- Award 1 point for: the bacteria (on the rocks).
(b) Explain how this community demonstrates that producers can make food in complete darkness. (2 pt)
Frame This community demonstrates it because …
A producer needs an energy source from its surroundings to make organic molecules from carbon dioxide.
These bacteria take that energy from hydrogen sulfide dissolved in the water, by chemosynthesis.
The grazing animals and the blind fish then get their organic molecules by eating what the bacteria made.
So the whole community is fed by producers that use a chemical’s energy, not sunlight.
- Award 1 point for: the bacteria are producers because they make their own organic molecules from carbon dioxide, using the energy in hydrogen sulfide (chemosynthesis), with no light.
- Award 1 point for: the animals of the community get their organic molecules by eating what the bacteria made, so the community is fed without sunlight.
Glossary
- autotroph (producer)
- An organism that makes its own organic molecules from carbon dioxide, using an energy source from its surroundings: sunlight, or the energy in a small inorganic molecule such as hydrogen sulfide. Ecologists call an autotroph a producer. Auto means self; troph means feeding.
- heterotroph (consumer)
- An organism that gets its organic molecules by eating other organisms or their remains. Ecologists call a heterotroph a consumer. Hetero means other.
- chemosynthesis
- Making organic molecules from carbon dioxide using the energy in small inorganic molecules, such as hydrogen sulfide. It needs no light, and some kinds of it happen where there is no oxygen; the bacteria at a deep-sea vent do it.
- primary productivity
- The rate at which an ecosystem's producers turn energy into organic matter, per area per time, such as kilojoules per square meter a year. A rate, not an amount standing in the field.
- net primary productivity
- The rate at which an ecosystem's producers turn energy into organic matter after taking away their own respiration, per area per time. The whole figure before respiration is the gross primary productivity.
APBIO-U08-L19B Net primary productivity
Suppose a meadow’s grass makes 10 000 kJ of sugar per square meter a year by photosynthesis. By the year’s end, only 6 000 kJ per square meter of that is stored in new grass.
Where did the other 4 000 kJ go? And which figure should an ecologist call the meadow’s productivity?
Unit 8 · Ecology
1Net primary productivity: what was made minus what was respired
An ecologist measures a hay field’s primary productivity.
Which of the following is the field’s primary productivity?
- A. The amount of grass standing in the field at the year’s endThe grass standing in the field is an amount, measured once.
Productivity is a rate: how fast the producers make organic matter, per area per time. - B. ✓ The rate at which the producers turn energy into organic matter
- C. The number of producers growing in the field at the year’s endA count of producers does not tell you how fast they make organic matter.
Productivity is a rate, per area per time.
Why: Productivity is a rate.
It is how fast the field’s producers turn energy into organic matter.
It is measured per area per time, in kilojoules per square meter a year.
In Unit 3, a plant sealed in a jar stood in the light, and its oxygen sensor showed a net change.
Which of the following is the net change in the jar’s oxygen?
- A. The oxygen the leaves made by photosynthesis in the lightThe oxygen the leaves made is one of the two amounts.
The net change is what is left after the oxygen respiration used is subtracted from it. - B. The oxygen respiration used, in the light and in the darkThe oxygen respiration used is one of the two amounts.
The net change is what is left after it is subtracted from the oxygen the leaves made. - C. ✓ The oxygen the leaves made minus the oxygen respiration used
Why: Net means what remains after the subtraction.
The leaves made oxygen, and respiration used some of it.
So the net change is the oxygen the leaves made minus the oxygen respiration used.
An ecosystem’s producers make sugar all year and respire some of it to stay alive.
Which of the following is the ecosystem’s gross primary productivity?
- A. ✓ The rate at which the producers make organic matter, before their respiration is taken away
- B. The rate at which the producers make organic matter, after their respiration is taken awayThe rate after the producers’ respiration is taken away is the net primary productivity.
- C. The rate at which the producers respire the organic matter they madeThe producers’ respiration is the part taken away from the gross, not the gross itself.
Why: Gross means the whole amount, before anything is taken away.
Gross primary productivity is the rate at which the producers make organic matter before their own respiration is taken away.
How much of what the producers make is left for anyone else?
The grass respired the missing 4 000 kJ to stay alive. That energy left the meadow as heat.
What is left is the net primary productivity: what the producers made minus what they respired. It is the same subtraction as Unit 3’s net change.
net primary productivity: the energy left in new plant material each year, in kJ per square meter a year
what the producers made: the energy in the sugar the producers made by photosynthesis each year, in kJ per square meter a year
what they respired: the energy in the sugar the producers respired to stay alive each year, in kJ per square meter a year
The net is the figure that matters, because only the net is there for the grasshoppers.
Write down the values, write down the equation, substitute, and state the answer in kilojoules per square meter a year. That is the whole routine.
Video: Watch: Net primary productivity: what was made minus what was respired
The meadow’s year drawn as one bar, 10 000 kJ per square meter; the bar splits, and the respired part, 4 000 kJ, slides away and leaves as heat; the 6 000 kJ that remains is the net primary productivity, the part a grasshopper can eat; the equation written on its own line and the working appearing line by line.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L19Ba.mp4
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Here is the meadow again. Suppose its grass makes 10 000 kJ of sugar per square meter a year by photosynthesis.
The grass is the meadow’s producer. The 10 000 kJ is what the producers made in the year.
The grass’s cells respire all year, in the light and in the dark, to stay alive.
Suppose the grass respires 4 000 kJ of that sugar over the year. The energy in the respired sugar leaves the meadow as heat.
The drawing shows the year as one bar. The shaded part is the 4 000 kJ the grass respired; the plain part is the 6 000 kJ stored in new grass.
Now take the respired part away. The bar that is left is the energy stored in new leaves, stems and roots.
That leftover is the net primary productivity. You find it by subtracting what the producers respired from what they made, as the equation below says.
net primary productivity: the energy left in new plant material each year, in kJ per square meter a year
what the producers made: the energy in the sugar the producers made by photosynthesis each year, in kJ per square meter a year
what they respired: the energy in the sugar the producers respired to stay alive each year, in kJ per square meter a year
Both values are in kilojoules per square meter a year. So the net primary productivity comes out in kilojoules per square meter a year too, as the line below shows.
kilojoules per square meter a year minus kilojoules per square meter a year leaves kilojoules per square meter a year: the net carries the same unit as the two values it comes from
The meadow’s grass makes 10 000 kJ per square meter a year and respires 4 000 kJ per square meter a year of it. Calculate the meadow’s net primary productivity.
This is Unit 3’s subtraction again. In the jar, the net change was the oxygen the leaves made minus the oxygen respiration used.
In the meadow, the net primary productivity is the energy the producers made minus the energy they respired.
The whole 10 000 kJ, the figure before respiration, is the gross primary productivity.
The 4 000 kJ the grass respired left as heat. No grasshopper can eat heat.
So the grasshoppers live on the net, the 6 000 kJ stored in new grass. The net is the meadow’s productivity that matters to everyone else.
Now consider a kelp forest. Suppose its kelp makes 22 000 kJ of sugar per square meter a year and respires 13 000 kJ per square meter a year of it.
The kelp makes 22 000 kJ per square meter a year and respires 13 000 kJ per square meter a year of it. Calculate the kelp forest’s net primary productivity.
The kelp forest made more than the meadow, and it also respired more. Its net primary productivity is 9 000 kJ per square meter a year, the energy left for the animals that eat kelp.
What you are expected to know Calculate an ecosystem’s net primary productivity from what its producers made and what they respired, in kilojoules per square meter a year.
An ecosystem’s producers made a known amount of sugar in a year and respired a known amount of it, both in kilojoules per square meter a year.
Which of the following gives the ecosystem’s net primary productivity?
- A. ✓ What the producers made minus what they respired
- B. What the producers respired minus what they madeWhat they respired minus what they made gives the right size with the wrong sign.
Producers respire less than they make, so the net would come out negative. - C. What the producers made plus what they respiredAdding the two values counts the respired sugar twice.
The net is what is left after the respired sugar is taken away.
Why: The net primary productivity is what is left in new plant material.
What is left is what the producers made with what they respired taken away.
So net primary productivity is what the producers made minus what they respired.
Suppose a pond’s producers, its algae and pondweed, make 3 600 kJ of sugar per square meter a year and respire 1 500 kJ per square meter a year of it.
Substitute the values into the equation and calculate the pond’s net primary productivity.
Part 1. Write down the values in the question. What did the pond’s producers make?
Answer: 3600 kJ per square meter a year (tolerance ±0)
Part 2. What did the pond’s producers respire?
Answer: 1500 kJ per square meter a year (tolerance ±0)
Answer: 2100 kJ per square meter a year (tolerance ±0)
Suppose a salt marsh’s grasses make sugar by photosynthesis all summer and respire some of it to stay alive.
Which of the following is the energy left for the marsh’s consumers?
- A. What the grasses respiredThe energy in the respired sugar left the marsh as heat.
No consumer can eat heat. - B. ✓ What the grasses made minus what they respired
- C. What the grasses madeSome of what the grasses made was respired, and that energy left as heat.
Only the part not respired is stored in new grass.
Why: The grasses respired some of the sugar they made.
The energy in that sugar left the marsh as heat.
So the consumers can eat only what the grasses made minus what they respired, the net primary productivity.
A student says: “The energy the grass respired is still in the meadow, so the grasshoppers can eat it later.”
Is the student correct?
- A. YesThe grass’s cells respired that sugar, and its energy left the meadow as heat.
Heat is not food. - B. ✓ No
Why: The grass’s cells respired the sugar to stay alive.
The energy in that sugar left the meadow as heat.
Heat is not stored in the grass, so no grasshopper can eat it later.
Suppose a survey of a lake finds that its producers make 2 900 kJ of sugar per square meter a year and respire 1 700 kJ per square meter a year of it.
Calculate the lake’s net primary productivity.
Answer: 1200 kJ per square meter a year (tolerance ±0)
Here is the meadow again, with its 10 000 kJ made and its 6 000 kJ stored.
The grass respired the other 4 000 kJ, and that energy left as heat.
The meadow’s net primary productivity is 6 000 kJ per square meter a year, the figure the grasshoppers live on.
37Numeric practice: net primary productivity mixed practice
Suppose a prairie’s grasses make 8 400 kJ of sugar per square meter a year and respire 3 900 kJ per square meter a year of it.
Calculate the prairie’s net primary productivity.
Answer: 4500 kJ per square meter a year (tolerance ±0)
Suppose a coral reef’s producers make 15 300 kJ of sugar per square meter a year and respire 9 200 kJ per square meter a year of it.
Calculate the reef’s net primary productivity.
Answer: 6100 kJ per square meter a year (tolerance ±0)
Suppose a desert’s sparse shrubs respire 600 kJ of sugar per square meter a year, out of the 1 400 kJ per square meter a year they make.
Calculate the desert’s net primary productivity.
Answer: 800 kJ per square meter a year (tolerance ±0)
Suppose a tropical forest’s trees make 36 400 kJ of sugar per square meter a year and respire 24 900 kJ per square meter a year of it.
Calculate the forest’s net primary productivity.
Answer: 11500 kJ per square meter a year (tolerance ±0)
Suppose the tiny floating producers of a patch of open ocean make 2 500 kJ of sugar per square meter a year and respire 1 850 kJ per square meter a year of it.
Calculate the patch’s net primary productivity.
Answer: 650 kJ per square meter a year (tolerance ±0)
APBIO-U08-L20 Who eats the food
Photo: Sarah Stierch, Wikimedia Commons, CC BY 4.0 (cropped and resized).
A dead deer lies beside a forest road. Vultures tear at it.
Beetles and fly larvae follow. Weeks later, fungi cover the last of the bones, and the soil around them is dark. Every one of these ate the deer. Are they all the same kind of eater?
Unit 8 · Ecology
1Five kinds of eater
Unit 8 sorted every organism by where its organic molecules come from.
Which of the following is a heterotroph?
- A. A living thing that builds its own organic molecules from carbon dioxideA living thing that builds its own organic molecules from carbon dioxide is an autotroph.
Ecologists call it a producer. - B. ✓ A living thing that eats other organisms, or their remains, to get its organic molecules
Why: A heterotroph gets its organic molecules by eating other organisms or their remains.
Ecologists call it a consumer.
Video: Watch: Five kinds of eater
The dead deer over the weeks: the vultures tear at it on the first day, the beetles and fly larvae arrive in the first week, and the fungi cover the bones weeks later; each eater is named as it arrives, and the vultures and the fungi are set side by side.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L20a.mp4
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How are the eaters at the dead deer sorted?
The vultures, the beetles, the fly larvae and the fungi all get their organic molecules by eating the deer. So every one of them is a heterotroph: a consumer.
Ecologists sort the heterotrophs by what they eat. There are five kinds.
Here is the dead deer over the weeks: the vultures on the first day, the beetles and fly larvae in the first week, the fungi weeks later.
Start with the deer while it was alive. The deer ate grass and leaves, which are producers.
A heterotroph that eats producers is called a .
Now consider a hawk. The hawk eats mice, which are animals.
A heterotroph that eats animals is called a .
Now consider a bear. The bear eats berries, and it eats fish: producers and animals.
A heterotroph that eats both producers and animals is called an .
Now consider the dead deer again. The vultures did not kill the deer; they eat a dead animal they found.
A heterotroph that eats dead animals it did not kill is called a .
A carnivore kills the animals it eats. A scavenger eats animals that were already dead when it found them.
The beetles and the fly larvae eat the dead deer too, and they did not kill it. So they are scavengers as well.
Weeks later, the fungi cover the bones. The fungi, and bacteria in the soil, break the last of the deer down into simple molecules: carbon dioxide, water and minerals.
Those simple molecules go into the soil and the air. The grass around the bones can take them up again.
A heterotroph that breaks dead matter and waste down into simple molecules is called a .
The fungi and microbes that eat dead material in the carbon cycle are the decomposers. Decomposers respire, like every other heterotroph.
The table below compares the five kinds of eater: what each eats, and one example.
The vultures and the fungi both ate the dead deer. Are they the same kind of eater?
The vultures are scavengers. They eat the carcass: the meat and skin of the dead deer.
The fungi are decomposers. They break down what is left into simple molecules that the producers can take up again.
The table below compares a scavenger with a decomposer: what each eats, and what is left when it has finished.
So a scavenger and a decomposer are not the same kind of eater. The scavenger eats the carcass; the decomposer eats what is left of it.
Here are five more eaters, each judged by one question: which kind of eater is it?
For example, a rabbit eats grass. The rabbit is a herbivore, because it eats producers.
But a fox eats rabbits and mice. The fox is a carnivore, because it eats animals.
And a person eats bread and fish. The person is an omnivore, because the person eats both producers and animals.
But a crow eats a dead squirrel it finds on a road. The crow is a scavenger, because it eats a dead animal it did not kill.
And bacteria in the soil break fallen leaves down into simple molecules. The bacteria are decomposers, because they break dead matter down into simple molecules.
The table below shows the five cases with their verdicts: which kind of eater each one is.
What you are expected to know Classify a heterotroph by what it eats: a herbivore eats producers, a carnivore eats animals, an omnivore eats both, a scavenger eats dead animals it did not kill, and a decomposer breaks dead matter and waste down into simple molecules.
A cow eats grass.
Which kind of eater is the cow?
- A. CarnivoreA carnivore eats animals.
Grass is a producer. - B. ✓ Herbivore
- C. OmnivoreAn omnivore eats both producers and animals.
The cow eats only grass, a producer.
Why: Grass is a producer.
A heterotroph that eats producers is a herbivore.
An owl eats mice.
Which kind of eater is the owl?
- A. ✓ Carnivore
- B. HerbivoreA herbivore eats producers.
Mice are animals. - C. OmnivoreAn omnivore eats both producers and animals.
The owl eats only animals.
Why: Mice are animals.
A heterotroph that eats animals is a carnivore.
A chicken eats grain, and it eats insects.
Which kind of eater is the chicken?
- A. CarnivoreA carnivore eats only animals.
The chicken also eats grain, a producer. - B. HerbivoreA herbivore eats only producers.
The chicken also eats insects, which are animals. - C. ✓ Omnivore
Why: Grain is a producer, and insects are animals.
A heterotroph that eats both producers and animals is an omnivore.
A bald eagle eats a dead salmon it finds on a riverbank.
Which kind of eater is the eagle here?
- A. CarnivoreA carnivore eats animals it kills.
The eagle did not kill this salmon. - B. DecomposerA decomposer breaks dead matter down into simple molecules.
The eagle eats the carcass itself. - C. ✓ Scavenger
Why: The salmon is dead, and the eagle did not kill it.
A heterotroph that eats dead animals it did not kill is a scavenger.
Bacteria in a pond’s mud break dead leaves down into simple molecules.
Which kind of eater are the bacteria?
- A. ✓ Decomposer
- B. HerbivoreA herbivore eats living producers.
These leaves are dead, and the bacteria break them down to simple molecules. - C. ScavengerA scavenger eats dead animals.
Leaves are not animals, and the bacteria break them down to simple molecules.
Why: The leaves are dead matter.
The bacteria break them down into simple molecules.
A heterotroph that does that is a decomposer.
A hyena hunts and kills other animals. It also eats dead animals it finds.
Is the hyena a scavenger as well as a carnivore?
- A. ✓ Yes
- B. NoThe hyena eats dead animals it did not kill.
That is what a scavenger does, whatever else it eats.
Why: The hyena kills and eats animals, so it is a carnivore.
The hyena also eats dead animals it did not kill, so it is a scavenger.
One animal can be both.
A student says: “A crow eating a dead mouse is a scavenger, and the bacteria breaking the last of the mouse down are decomposers: two different kinds of eater.”
Is the student correct?
- A. No: both eat a dead animal, so they are one kind of eaterThe crow eats the carcass.
The bacteria break what is left down into simple molecules: a different job. - B. ✓ Yes: a scavenger and a decomposer are two different kinds of eater
Why: The crow eats the carcass, a dead animal it did not kill: a scavenger.
The bacteria break what is left down into simple molecules: decomposers.
So the two are two kinds of eater.
43Quick quiz: herbivore, carnivore, omnivore, scavenger, decomposer mixed practice
A raccoon eats crayfish, and it eats berries.
Which kind of eater is the raccoon?
- A. HerbivoreA herbivore eats only producers.
The raccoon also eats crayfish, which are animals. - B. CarnivoreA carnivore eats only animals.
The raccoon also eats berries, from a producer. - C. ✓ Omnivore
Why: Berries come from a producer, and crayfish are animals.
A heterotroph that eats both producers and animals is an omnivore.
Mold spreads over a fallen apple and breaks it down into simple molecules.
Which kind of eater is the mold?
- A. CarnivoreA carnivore eats animals.
The apple is dead plant matter. - B. ScavengerA scavenger eats dead animals it did not kill.
An apple is not an animal, and the mold breaks it down to simple molecules. - C. ✓ Decomposer
Why: The fallen apple is dead matter.
The mold breaks it down into simple molecules.
A heterotroph that does that is a decomposer.
A fungus breaks a dead log down into simple molecules.
Is the fungus a scavenger?
- A. YesA scavenger eats dead animals it did not kill.
The fungus breaks dead matter down into simple molecules: a decomposer. - B. ✓ No
Why: The log is dead matter, and the fungus breaks it down into simple molecules.
A heterotroph that does that is a decomposer, so the fungus is a decomposer.
No: it is not a scavenger.
A snail eats pondweed.
Is the snail a herbivore?
- A. ✓ Yes
- B. NoPondweed is a producer.
A heterotroph that eats producers is a herbivore.
Why: Pondweed is a producer.
The snail eats producers, so the snail is a herbivore.
Heterotrophs are sorted by what they eat.
What is a herbivore?
- A. ✓ A heterotroph that eats producers
- B. A heterotroph that eats animalsA heterotroph that eats animals is a carnivore.
- C. A heterotroph that eats both producers and animalsA heterotroph that eats both producers and animals is an omnivore.
Why: A herbivore is a heterotroph that eats producers.
Two kinds of heterotroph feed on dead animals.
What is a scavenger?
- A. A heterotroph that breaks dead matter down into simple moleculesA heterotroph that breaks dead matter down into simple molecules is a decomposer.
- B. ✓ A heterotroph that eats dead animals it did not kill
- C. A heterotroph that hunts, kills and eats other animalsA heterotroph that kills and eats animals is a carnivore.
Why: A scavenger is a heterotroph that eats dead animals it did not kill.
Heterotrophs are sorted by what they eat.
What is a carnivore?
- A. ✓ A heterotroph that eats animals
- B. A heterotroph that eats both producers and animalsA heterotroph that eats both producers and animals is an omnivore.
- C. A heterotroph that eats producersA heterotroph that eats producers is a herbivore.
Why: A carnivore is a heterotroph that eats animals.
A scavenger and a decomposer both feed on a dead animal.
(a) State what a scavenger eats. (1 pt)
- Award 1 point for: dead animals it did not kill (a carcass it found).
(b) State what a decomposer does with dead matter. (1 pt)
- Award 1 point for: breaks it down into simple molecules (that producers can take up again). Accept with or without: waste.
52Two things from one meal
Unit 3 followed the energy released when a cell breaks its food down.
In Unit 3’s words, what does the cell do with that energy?
- A. ✓ Some of it makes ATP from ADP and Pi; the rest leaves as heat
- B. All of it leaves the cell as heat, and none of it makes ATPSome of the energy released makes ATP.
Only the rest leaves as heat. - C. All of it is stored in the cell’s fat for later useA cell does not store the released energy in fat.
Some makes ATP, and the rest leaves as heat.
Why: The cell breaks its food down and energy is released.
Some of that energy makes ATP from ADP and Pi.
The rest leaves as heat.
Unit 1 followed a stored sugar, such as starch, being used by a cell.
When does the cell get usable energy from the store?
- A. When hydrolysis frees the glucose units from the storeHydrolysis frees the glucose units and gives the cell no usable energy.
The energy comes out when the glucose reacts with oxygen. - B. ✓ When the cell reacts the glucose with oxygen, making carbon dioxide and water
Why: Hydrolysis frees the glucose units from the store.
The cell gets usable energy when it reacts the glucose with oxygen, making carbon dioxide and water.
Video: Watch: Two things from one meal
A fox eats a rabbit; some of the meal’s molecules are respired, ATP appears and heat leaves, and the rest are built into the fox’s muscle; then one carbon atom in that muscle is traced back to the rabbit, to the grass, and to carbon dioxide in the air.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L20b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L20b.mp4
What does a heterotroph get from a meal?
A heterotroph gets two things from one meal: energy, and the matter its body is built from.
Now consider a fox that has just eaten a rabbit. The rabbit’s meat is made of carbon compounds: protein, fat and a little carbohydrate.
The fox digests the meat into small molecules and takes them into its cells.
The fox’s cells respire some of those molecules, by the word equation of cellular respiration in Unit 3’s words.
The word equation for cellular respiration, in Unit 3’s words; the reaction releases energy
Some of the energy released makes ATP. The rest leaves the fox as heat.
So the fox gets energy from the meal: the ATP its cells use for everything they do.
The fox’s cells do not respire every molecule from the meal. They build the rest into the fox’s own muscle, fat and bone.
So the fox gets matter from the meal too: the carbon and the other atoms its body is made of.
The table below compares the two things the fox gets from one meal: what its cells do with the meal’s molecules, and what the fox ends up with.
Now follow one carbon atom that is in the fox’s muscle. Where was that carbon atom before the fox ate?
The carbon atom was in the rabbit’s muscle. The rabbit built that muscle from the grass it ate.
Before that, the carbon atom was in the grass. The grass built its sugar from carbon dioxide, by photosynthesis.
Here is the carbon atom’s path. Each arrow shows where the atom went next.
The rabbit rebuilt the grass’s compounds into rabbit. The fox rebuilt the rabbit’s compounds into fox.
Neither animal built a carbon compound from carbon dioxide. Only an autotroph does that.
So every carbon compound the fox eats was made, in the end, by an autotroph.
What you are expected to know Explain how a heterotroph gets energy and matter from one meal: its cells respire some of the meal’s carbon compounds to make ATP and build the rest into its own tissue.
What you are expected to know Explain why every carbon compound a heterotroph eats was made, in the end, by an autotroph: only an autotroph builds carbon compounds from carbon dioxide.
An owl eats a mouse. Some of the mouse’s carbon is now in the owl’s muscle.
Where was that carbon before it was in the mouse?
- A. In the air the mouse breathed inA mouse takes oxygen in from the air, not carbon.
Its carbon came in with its food. - B. ✓ In the seeds the mouse ate
- C. In the ATP the mouse madeATP is made from ADP and Pi.
The mouse built its muscle from the carbon compounds in its food.
Why: The mouse built its muscle from the carbon compounds in the seeds it ate.
The seeds’ compounds were built by the plant, from carbon dioxide.
So the carbon was in the seeds before it was in the mouse.
A heron catches and eats a perch.
(a) Explain how the heron gets both energy and matter from this one meal. (2 pt)
Frame The heron’s cells …
Some of the energy released makes ATP, so the heron gets energy from the meal.
The heron’s cells build the rest of the perch’s compounds into the heron’s own muscle and fat.
So the heron gets matter from the meal too.
- Award 1 point for: the heron’s cells respire some of the perch’s carbon compounds and the energy released makes ATP.
- Award 1 point for: the heron’s cells build the rest of the compounds into the heron’s own tissue (muscle, fat).
Follow a hawk’s meal back.
The hawk ate a snake.
The snake ate a frog.
The frog ate a grasshopper.
The grasshopper ate grass.
Which organism built the carbon compounds now in the hawk’s muscle from carbon dioxide?
- A. The snakeThe snake rebuilt the frog’s compounds into snake.
It built none from carbon dioxide. - B. The frogThe frog rebuilt the grasshopper’s compounds into frog.
It built none from carbon dioxide. - C. The grasshopperThe grasshopper rebuilt the grass’s compounds into grasshopper.
It built none from carbon dioxide. - D. ✓ The grass
Why: Each animal rebuilt the compounds in its food into its own body.
Only an autotroph builds carbon compounds from carbon dioxide.
The grass is the autotroph.
So the grass built the compounds, in the end.
A fungus grows on a fallen log and breaks the log down.
Which of the following does the fungus get from the log?
- A. Energy onlyThe fungus builds some of the log’s carbon compounds into its own body.
That is matter. - B. Matter onlyThe fungus’s cells respire some of the log’s compounds and make ATP.
That is energy. - C. ✓ Both energy and matter
Why: The fungus’s cells respire some of the log’s carbon compounds.
Some of the energy released makes ATP, so the fungus gets energy.
The fungus builds the rest into its own body, so it gets matter too.
A student says: “A lion gets energy from the zebra it eats, and its cells build the zebra’s carbon compounds into the lion’s own muscle.”
Is the student correct?
- A. No: the lion’s cells make the carbon atoms for its muscle themselvesNo cell makes an atom.
The lion’s cells build the zebra’s carbon compounds into lion muscle. - B. ✓ Yes: the lion gets both energy and matter from the zebra
Why: The zebra’s meat is carbon compounds.
The lion’s cells respire some of those compounds for energy.
They build the rest into the lion’s own muscle.
So the lion gets energy and matter from the zebra.
Here is the dead deer again: the vultures on the first day, the beetles and fly larvae in the first week, the fungi weeks later.
The vultures, the beetles and the fly larvae are scavengers. The fungi are decomposers.
Every one of them respires some of the deer’s carbon compounds to make ATP, and builds the rest into itself.
And every one of those compounds was built, in the end, by the grass and leaves the deer ate.
84Mixed practice mixed practice
A sheep eats grass and clover.
Which kind of eater is the sheep?
- A. CarnivoreA carnivore eats animals.
Grass and clover are producers. - B. ✓ Herbivore
- C. OmnivoreAn omnivore eats both producers and animals.
The sheep eats only producers.
Why: Grass and clover are producers.
A heterotroph that eats producers is a herbivore.
Bacteria in a cow’s dung break the dung down into simple molecules.
Which kind of eater are the bacteria?
- A. HerbivoreA herbivore eats living producers.
Dung is waste, and the bacteria break it down to simple molecules. - B. ScavengerA scavenger eats dead animals.
Dung is waste, not an animal, and the bacteria break it down to simple molecules. - C. ✓ Decomposer
Why: The dung is waste.
The bacteria break it down into simple molecules.
A heterotroph that does that is a decomposer.
A student says: “The carbon compounds in a pike’s meal were made by the perch it ate.”
Is the student correct?
- A. Yes: the perch made the carbon compounds in the pike’s mealThe perch only rebuilt compounds from its own food.
An autotroph built them from carbon dioxide, in the end. - B. ✓ No: an autotroph built them from carbon dioxide, in the end
Why: The perch built its body from the carbon compounds in its food.
Its food came from other eaters, and in the end from an autotroph.
Only an autotroph builds carbon compounds from carbon dioxide.
So the compounds were made, in the end, by an autotroph.
A pig eats acorns, and it eats worms.
Which kind of eater is the pig?
- A. CarnivoreA carnivore eats only animals.
The pig also eats acorns, which come from a producer. - B. HerbivoreA herbivore eats only producers.
The pig also eats worms, which are animals. - C. ✓ Omnivore
Why: Acorns come from a producer, and worms are animals.
A heterotroph that eats both producers and animals is an omnivore.
A scavenger eats a meal. Its cells respire some of the meal’s carbon compounds.
What happens to the rest of the meal’s carbon compounds?
- A. The rest leaves the scavenger’s body as heatHeat is energy, not carbon compounds.
The compounds that are not respired are built into the scavenger’s tissue. - B. ✓ The scavenger’s cells build the rest into its own tissue
- C. The rest becomes ATP inside the scavenger’s cellsATP is made from ADP and Pi, using energy from the respired compounds.
The compounds that are not respired are built into tissue.
Why: The scavenger’s cells respire some of the meal’s compounds for energy.
They build the rest into the scavenger’s own muscle, fat and bone.
So the rest of the compounds become the scavenger’s tissue.
Follow a heron’s meal back.
The heron ate a perch.
The perch ate a snail.
The snail ate pondweed.
Which organism built the carbon compounds now in the heron’s body from carbon dioxide?
- A. The heronThe heron rebuilt the perch’s compounds into heron.
It built none from carbon dioxide. - B. The perchThe perch rebuilt the snail’s compounds into perch.
It built none from carbon dioxide. - C. The snailThe snail rebuilt the pondweed’s compounds into snail.
It built none from carbon dioxide. - D. ✓ The pondweed
Why: Each animal rebuilt the compounds in its food into its own body.
Only an autotroph builds carbon compounds from carbon dioxide.
The pondweed is the autotroph.
So the pondweed built the compounds, in the end.
A dead fish lies on a lake shore. A gull eats much of it over two days. Weeks later, bacteria and fungi have broken the rest down, and only bones remain.
(a) State which kind of eater the gull is. (1 pt)
- Award 1 point for: scavenger.
(b) Explain how the dead fish demonstrates that a scavenger and a decomposer are two different kinds of eater. (2 pt)
Frame The dead fish demonstrates this because …
The bacteria and fungi break what is left down into simple molecules.
So the gull is a scavenger and the bacteria and fungi are decomposers.
The two eat different things from the same dead fish, so they are two kinds of eater.
- Award 1 point for: the gull eats the carcass (a dead animal it did not kill), so it is a scavenger.
- Award 1 point for: the bacteria and fungi break what is left down into simple molecules, so they are decomposers; the two eat different things.
(c) Explain why every carbon compound in the fish’s body was first built by an autotroph. (2 pt)
Frame Every carbon compound in the fish’s body was first built by an autotroph because …
Each animal it ate had rebuilt the compounds in its own food.
No animal builds a carbon compound from carbon dioxide.
Only an autotroph does, so the compounds were first built by an autotroph.
- Award 1 point for: the fish, and every animal it ate, only rebuilt the carbon compounds in its food.
- Award 1 point for: only an autotroph builds carbon compounds from carbon dioxide, so the compounds were first built by an autotroph.
Glossary
- herbivore
- A heterotroph that eats producers. A deer eating grass and leaves is a herbivore.
- carnivore
- A heterotroph that eats animals. A hawk eating mice is a carnivore.
- omnivore
- A heterotroph that eats both producers and animals. A bear eating berries and fish is an omnivore.
- scavenger
- A heterotroph that eats dead animals it did not kill. A vulture at a dead deer is a scavenger: it eats the carcass.
- decomposer
- A heterotroph that breaks dead matter and waste down into simple molecules the producers can take up again: the fungi and bacteria that eat dead material. Decomposers respire, like every other heterotroph.
APBIO-U08-L21 Levels, chains and the direction of the arrow
Photos: Sam Kieschnick, Wikimedia Commons, CC BY 4.0; Larry Rana / USDA, Courtney Celley / USFWS and Jacob W. Frank / NPS, Wikimedia Commons, public domain (all cropped and resized).
Here is a meadow: grass, grasshoppers, meadowlarks and a hawk. The grasshoppers eat the grass. The meadowlarks eat the grasshoppers. The hawk eats the meadowlarks.
Someone has drawn the four in a line with arrows between them, and the arrows point from the hawk to the grass. The drawing is upside down. Which way should the arrows point, and what is each step called?
Unit 8 · Ecology
1Each step of eating is a trophic level
Unit 8 sorted every organism into two groups by where its organic molecules come from.
What does a producer do?
- A. ✓ Makes its own organic molecules from carbon dioxide, using energy from its surroundings
- B. Gets its organic molecules by eating other living things or their remainsGetting organic molecules by eating other living things or their remains is what a consumer does.
Why: A producer makes its own organic molecules from carbon dioxide, using an energy source from its surroundings.
Grass is a producer: it makes its sugar by photosynthesis.
Unit 8 sorted the consumers by what they eat.
Which of the following is a decomposer?
- A. A vulture tearing meat from a deer it found deadA vulture eats dead animals it did not kill: a scavenger.
- B. ✓ A soil fungus breaking dead leaves down into simple molecules
- C. A deer grazing on grass and young leavesA deer eats producers: a herbivore.
Why: A decomposer breaks dead matter and waste down into simple molecules.
The soil fungus breaks the dead leaves down into simple molecules.
So the soil fungus is the decomposer.
How do you place an organism on a chain of eaters, and what does the arrow between two organisms mean?
Each step of eating, counted from the grass, has its own name. The number of steps sets the name, never the animal’s size.
Soil fungi and bacteria feed on the dead matter of every step. So they are drawn beside the chain, not in it.
The arrow points the way the energy goes, from the eaten to the eater. That holds for every chain.
Video: Watch: Each step of eating is a trophic level
The meadow chain standing up from the grass; the name of each step appearing beside its organism as the steps are counted from the grass.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L21a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L21a.mp4
Here is the meadow. The grass makes its own sugar by photosynthesis, so the grass is the producer.
A grasshopper eats the grass. That is one step of eating away from the producer.
A meadowlark eats the grasshopper. That is two steps of eating away from the producer.
The hawk eats the meadowlark. That is three steps of eating away from the producer.
Each of these steps of eating, counted from the producer, is called a . Trophic means feeding.
The producer is the first trophic level. Every eater above it is a consumer, named by its number of steps from the producer.
A consumer one step from the producer is called the . Primary means first.
A consumer two steps from the producer is called the secondary consumer. Secondary means second.
A consumer three steps from the producer is called the tertiary consumer. Tertiary means third.
A consumer four steps from the producer is called the quaternary consumer. Quaternary means fourth.
For example, the grasshopper eats the producer: one step from the producer. So the grasshopper is the primary consumer.
The meadowlark eats the grasshopper: two steps from the producer. So the meadowlark is the secondary consumer.
The hawk eats the meadowlark: three steps from the producer. So the hawk is the tertiary consumer.
Suppose an eagle ate the hawk: four steps from the producer. So the eagle would be the quaternary consumer.
Here is the meadow chain drawn standing up, the producer at the bottom and each eater above the one it eats. The name of each trophic level is written beside its box.
The table below compares the trophic levels: the level’s name, how many steps it is from the producer, and who is at that level in the meadow.
Here are two more examples. Size does not set the level.
Now consider a cow. Suppose the cow eats only grass.
The cow eats the producer: one step from the producer. So the cow is a primary consumer, though it is huge.
Now consider a small spider. Suppose the spider eats a fly, and the fly drank nectar from a flower.
The spider eats an eater of the producer: two steps from the producer. So the spider is a secondary consumer, though it is tiny.
The table below compares the cow and the spider: the size of each, its steps from the producer, and its trophic level.
The number of steps from the producer sets the trophic level. The animal’s size never does.
Soil fungi and bacteria feed on the dead matter of every trophic level: dead grass, dead grasshoppers, dead meadowlarks and a dead hawk.
So the decomposers do not stand at one step from the producer. We draw them beside the chain, not in it.
What you are expected to know Assign each organism in a described feeding sequence to its trophic level.
What you are expected to know Name the levels: the producer, then the primary, secondary, tertiary and quaternary consumers, by their steps from the producer.
What you are expected to know Place the decomposers beside the chain: they feed on the dead matter of every level.
In a chain of eaters, each organism has a trophic level.
Which of the following sets an organism’s trophic level?
- A. ✓ The number of eating steps between it and the producer
- B. The size of its body, from the smallest to the largestA huge grass-eater is one step from the producer: a primary consumer.
Size does not count the steps. - C. How many different kinds of food it eatsThe kinds of food do not count the steps.
A trophic level counts the eating steps from the producer.
Why: A trophic level is one step of eating, counted from the producer.
So the number of eating steps between an organism and the producer sets its trophic level.
Suppose a pond. Water fleas eat algae, minnows eat the water fleas, and a pike eats the minnows. Below, the chain stands with the algae at the bottom and each eater above the one it eats.
Which trophic level are the minnows?
- A. Primary consumerThe primary consumer is one step from the producer: the water fleas, which eat the algae.
- B. ✓ Secondary consumer
- C. Tertiary consumerThe tertiary consumer is three steps from the producer.
The minnows are two steps from the algae.
Why: The algae are the producer.
The water fleas eat the algae: one step.
The minnows eat the water fleas: two steps.
So the minnows are the secondary consumer.
Suppose a pond. Water fleas eat algae, minnows eat the water fleas, and a pike eats the minnows. Below, the chain stands with the algae at the bottom and each eater above the one it eats.
Which trophic level is the pike?
- A. Primary consumerThe primary consumer is one step from the producer: the water fleas, which eat the algae.
- B. Secondary consumerThe secondary consumer is two steps from the producer: the minnows, which eat the water fleas.
- C. ✓ Tertiary consumer
Why: The algae are the producer.
The water fleas are one step from the algae, the minnows two steps.
The pike eats the minnows: three steps.
So the pike is the tertiary consumer.
Suppose a caterpillar eats oak leaves.
Which trophic level is the caterpillar?
- A. ProducerThe oak makes its own sugar by photosynthesis: the oak is the producer, and the caterpillar eats it.
- B. ✓ Primary consumer
- C. Secondary consumerA secondary consumer is two steps from the producer.
The caterpillar eats the producer itself: one step.
Why: The oak leaves are the producer.
The caterpillar eats the producer: one step.
So the caterpillar is a primary consumer.
Suppose a frog eats caterpillars, and the caterpillars eat oak leaves.
Which trophic level is the frog?
- A. ProducerA producer makes its own organic molecules.
The frog eats caterpillars. - B. Primary consumerA primary consumer eats the producer.
The frog eats the caterpillars, which eat the producer: two steps. - C. ✓ Secondary consumer
Why: The oak leaves are the producer.
The caterpillars eat the producer: one step.
The frog eats the caterpillars: two steps.
So the frog is a secondary consumer.
Suppose a snake eats frogs, the frogs eat caterpillars, and the caterpillars eat oak leaves.
Which trophic level is the snake?
- A. Primary consumerA primary consumer eats the producer.
The snake eats frogs, three steps from the oak leaves. - B. Secondary consumerA secondary consumer is two steps from the producer: the frogs.
The snake eats the frogs: three steps. - C. ✓ Tertiary consumer
Why: The oak leaves are the producer.
The caterpillars are one step from them, the frogs two steps.
The snake eats the frogs: three steps.
So the snake is a tertiary consumer.
Suppose a pond’s algae make their own sugar by photosynthesis.
Which trophic level are the algae?
- A. ✓ Producer
- B. Primary consumerA primary consumer eats the producer.
The algae eat nothing: they make their own sugar. - C. DecomposerA decomposer breaks dead matter down into simple molecules.
The algae make their own sugar from carbon dioxide.
Why: The algae make their own organic molecules by photosynthesis.
So the algae are the producer, the first trophic level.
Suppose soil fungi in a meadow feed on dead grass, dead grasshoppers, dead meadowlarks and a dead hawk.
Where do the soil fungi stand in the meadow’s chain of eaters?
- A. At the top, one step above the hawkThe fungi feed on the dead matter of every level, not on the living hawk alone.
So they are not one step above the hawk. - B. ✓ Beside the chain, not in it
- C. At the bottom, one step below the grassThe grass makes its own sugar, so nothing feeds the grass.
The fungi feed on the dead matter of every level.
Why: The fungi feed on the dead matter of every trophic level.
So they are not one step from the producer.
Decomposers are drawn beside the chain, not in it.
Suppose an elephant eats only grass and leaves. A student says: “The elephant is the biggest animal on the plain, so it must be a tertiary consumer.”
Is the student correct?
- A. Yes: the biggest animal on the plain is the tertiary consumerThe elephant eats producers, so it is one step from the producer.
Size does not set the level. - B. ✓ No: the elephant is one step from the producer, a primary consumer
Why: A trophic level counts the eating steps from the producer.
The elephant eats grass and leaves, the producers.
So the elephant is one step from the producer: a primary consumer, whatever its size.
Suppose a stream. Mayfly larvae eat algae, minnows eat the mayfly larvae, trout eat the minnows, and an otter eats the trout.
Which trophic level is the otter?
- A. Secondary consumerA secondary consumer is two steps from the producer: the minnows.
- B. Tertiary consumerA tertiary consumer is three steps from the producer: the trout.
The otter eats the trout: four steps. - C. ✓ Quaternary consumer
Why: The algae are the producer.
The mayfly larvae are one step from them, the minnows two, the trout three.
The otter eats the trout: four steps.
So the otter is a quaternary consumer.
47Quick quiz: trophic level, primary consumer, secondary consumer, tertiary consumer, quaternary consumer mixed practice
Suppose a heron eats perch, the perch eat snails, and the snails graze on algae.
Which trophic level is the heron?
- A. Primary consumerA primary consumer eats the producer: the snails, which graze on the algae.
- B. Secondary consumerA secondary consumer is two steps from the producer: the perch, which eat the snails.
- C. ✓ Tertiary consumer
Why: The algae are the producer.
The snails are one step from them, the perch two steps.
The heron eats the perch: three steps.
So the heron is a tertiary consumer.
Suppose a deer eats grass.
Which trophic level is the deer?
- A. ProducerThe grass makes its own sugar: the grass is the producer, and the deer eats it.
- B. ✓ Primary consumer
- C. Secondary consumerA secondary consumer is two steps from the producer.
The deer eats the producer itself: one step.
Why: The grass is the producer.
The deer eats the producer: one step.
So the deer is a primary consumer.
A chain of eaters begins at a producer.
What is a trophic level?
- A. ✓ One step of eating, counted from the producer
- B. One kind of food that an animal eatsA kind of food is what an animal eats, not how many steps it stands from the producer.
- C. One species living in an ecosystemA species is one kind of organism; several species can stand at the same step from the producer.
Why: A trophic level is one step of eating, counted from the producer.
A chain of eaters begins at a producer.
Which consumer is one step from the producer?
- A. ✓ Primary consumer
- B. Secondary consumerA secondary consumer is two steps from the producer.
- C. Tertiary consumerA tertiary consumer is three steps from the producer.
Why: Primary means first.
The primary consumer is the first eater, one step from the producer.
A chain of eaters begins at a producer.
Which consumer is three steps from the producer?
- A. Secondary consumerA secondary consumer is two steps from the producer.
- B. ✓ Tertiary consumer
- C. Quaternary consumerA quaternary consumer is four steps from the producer.
Why: Tertiary means third.
The tertiary consumer is the third eater, three steps from the producer.
A chain of eaters begins at a producer.
Which consumer is four steps from the producer?
- A. Secondary consumerA secondary consumer is two steps from the producer.
- B. Tertiary consumerA tertiary consumer is three steps from the producer.
- C. ✓ Quaternary consumer
Why: Quaternary means fourth.
The quaternary consumer is the fourth eater, four steps from the producer.
A chain of eaters begins at a producer and has several consumers.
(a) State what a trophic level is. (1 pt)
- Award 1 point for: a trophic level is one step of eating (feeding) counted from the producer, or an organism’s position in the chain by the number of steps from the producer.
55The arrow points the way the energy goes
In a meadow, a grasshopper eats grass and uses the sugar’s energy to hop.
What happens to that energy once the grasshopper has used it?
- A. ✓ The energy passes through the meadow once
- B. The energy goes round the meadow again and againMatter goes round.
The sunlight’s energy spreads out as heat and never comes back.
Why: The grasshopper used the sugar’s energy to hop.
At every use some energy spreads out as heat, and no organism takes it back.
So the energy passes through the meadow once.
Video: Watch: The arrow points the way the energy goes
A token of energy moving along the meadow chain as each arrow is drawn from the eaten to the eater: grass to grasshopper, grasshopper to meadowlark, meadowlark to hawk.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L21b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L21b.mp4
Here is the meadow chain again: grass, grasshopper, meadowlark, hawk.
The grasshopper eats the grass. The sugar’s energy, and the grass’s atoms, move from the grass into the grasshopper.
The meadowlark eats the grasshopper. The energy and the atoms move from the grasshopper into the meadowlark.
The hawk eats the meadowlark. The energy and the atoms move from the meadowlark into the hawk.
So the energy moves one way along the chain: from the eaten into the eater.
An arrow between two organisms shows that movement. The arrow points from the eaten to the eater, the way the energy goes.
The grass’s energy goes into the grasshopper. So the arrow from the grass points at the grasshopper.
The opening drawing pointed its arrows from the hawk to the grass. The hawk’s energy does not go into the grass, so that drawing is upside down.
A line of organisms like this, each eaten by the next and drawn with arrows from the eaten to the eater, is called a .
The decomposers feed on the dead matter of every level. So they are drawn beside the food chain, with a dashed arrow from every level to them.
The energy the grass captured passes along the food chain once and leaves as heat. So no arrow returns from the hawk to the grass.
To draw a food chain from a description, do three things:
- Write the producer first, at the bottom.
- Write each eater above the organism it eats.
- Draw every arrow from the eaten to the eater.
For example, suppose a pond. Water fleas eat algae, minnows eat the water fleas, and a pike eats the minnows.
The algae are the producer, so the algae go first. The water fleas go above the algae, the minnows above the water fleas, and the pike above the minnows.
Every arrow points from the eaten to the eater: from the algae to the water fleas, from the water fleas to the minnows, from the minnows to the pike.
What you are expected to know Draw a food chain of three or four named organisms with each arrow pointing from the eaten to the eater, the direction the energy and matter move.
Suppose a coral reef. Parrotfish eat the algae growing on the reef, and barracudas eat the parrotfish. The food chain is to be drawn with the arrows pointing the way the energy goes.
Which organism comes first in the chain, where the energy enters it?
- A. ✓ The algae
- B. The parrotfishThe parrotfish eat the algae.
The energy enters the chain in the algae, the producer, so the algae come first. - C. The barracudasThe barracudas are the last eaters.
The energy enters the chain in the algae, the producer, so the algae come first.
Why: The algae make their own sugar by photosynthesis: the producer.
The energy enters the chain there.
So the algae come first.
Suppose a forest. The table below says what each of four organisms eats.
Which of the following is the forest’s food chain, written with each arrow pointing the way the energy goes?
- A. oak leaves → small birds → caterpillars → hawkThe small birds eat the caterpillars, not the oak leaves.
So the caterpillars come before the small birds. - B. ✓ oak leaves → caterpillars → small birds → hawk
- C. oak leaves → caterpillars → small birds → hawk → soil fungiSoil fungi feed on the dead matter of every level.
Decomposers are drawn beside the chain, not at its end.
Why: The oak leaves are the producer, so they come first.
The caterpillars eat the oak leaves, the small birds eat the caterpillars, the hawk eats the small birds.
Each arrow points from the eaten to the eater.
So the chain reads oak leaves → caterpillars → small birds → hawk.
Suppose a field. The table below says what each of four organisms eats.
Which of the following is the field’s food chain, written with each arrow pointing the way the energy goes?
- A. seeds → snakes → mice → hawkThe snakes eat the mice, not the seeds.
So the mice come before the snakes. - B. seeds → mice → snakes → soil bacteria → hawkSoil bacteria feed on the dead matter of every level.
Decomposers are drawn beside the chain, not inside it. - C. ✓ seeds → mice → snakes → hawk
Why: The seeds are the producer’s food store, so they come first.
The mice eat the seeds, the snakes eat the mice, the hawk eats the snakes.
Each arrow points from the eaten to the eater.
So the chain reads seeds → mice → snakes → hawk.
Suppose a garden. Slugs eat lettuce, thrushes eat the slugs, and a cat eats the thrushes. A student drew the food chain below and pointed one arrow the wrong way.
Which arrow points the wrong way?
- A. The arrow between the lettuce and the slugsThat arrow points from the lettuce to the slugs.
The lettuce’s energy goes into the slugs, so the arrow points the right way. - B. ✓ The arrow between the slugs and the thrushes
- C. The arrow between the thrushes and the catThat arrow points from the thrushes to the cat.
The thrushes’ energy goes into the cat, so the arrow points the right way.
Why: The thrushes eat the slugs.
So the slugs’ energy goes into the thrushes, and the arrow should point from the slugs to the thrushes.
The drawn arrow points from the thrushes to the slugs, toward the food.
So the arrow between the slugs and the thrushes points the wrong way.
Here is the meadow chain again: grass, grasshopper, meadowlark, hawk.
The grass is the producer. The grasshopper is the primary consumer, the meadowlark the secondary consumer, and the hawk the tertiary consumer.
The arrows now point from the grass to the grasshopper, from the grasshopper to the meadowlark, and from the meadowlark to the hawk: the way the energy goes.
81Quick quiz: food chain mixed practice
Suppose a food chain drawn with each arrow pointing from the eaten to the eater: algae → snails → perch → heron.
How many trophic levels does this food chain have?
- A. 3The algae are a trophic level too: the producer.
Algae, snails, perch and heron are four levels. - B. ✓ 4
- C. 5Four organisms stand in the chain, and each is one trophic level.
Four organisms make four levels.
Why: Each organism in the chain stands at its own step from the producer.
The algae are the producer, the snails the first eater, the perch the second, the heron the third.
So the food chain has 4 trophic levels.
Suppose a food chain drawn with each arrow pointing from the eaten to the eater: seeds → mouse → owl.
Which organism is the secondary consumer?
- A. The seedsThe seeds are the producer’s food store, where the chain begins.
- B. The mouseThe mouse eats the seeds: one step from the producer, the primary consumer.
- C. ✓ The owl
Why: The arrow points from the eaten to the eater, so the mouse eats the seeds and the owl eats the mouse.
The mouse is one step from the producer; the owl is two.
So the owl is the secondary consumer.
Ecologists draw who eats whom.
What is a food chain?
- A. ✓ A line of organisms, each eaten by the next, with arrows from the eaten to the eater
- B. A list of every organism that lives in one place, whatever each one eatsA list of every organism in one place does not show who eats whom.
A food chain shows the eating steps. - C. A cycle of atoms moving between living things and their surroundingsA cycle of atoms between living things and their surroundings is a biogeochemical cycle.
A food chain shows who eats whom.
Why: A food chain is a line of organisms, each eaten by the next, drawn with arrows from the eaten to the eater, the way the energy goes.
Ecologists draw who eats whom.
(a) State what a food chain is. (1 pt)
- Award 1 point for: a line (sequence) of organisms in which each is eaten by the next, with arrows from the eaten to the eater (the direction the energy moves).
Glossary
- trophic level
- One step of eating, counted from the producer. The producer is the first trophic level; each consumer is named by its number of steps from the producer. Decomposers feed on the dead matter of every level and are drawn beside the chain, not in it.
- primary, secondary, tertiary and quaternary consumer
- The consumer levels, named by their steps from the producer: the primary consumer is one step from the producer, the secondary consumer two steps, the tertiary consumer three steps, the quaternary consumer four steps. The number of steps sets the level, never the animal's size.
- food chain
- A line of organisms, each eaten by the next, drawn with each arrow pointing from the eaten to the eater, the way the energy goes. The producer is drawn first; decomposers are drawn beside the chain with a dashed arrow from every level.
APBIO-U08-L22 Build the web
Suppose a survey team lists what lives in a lake and what each one eats. Algae. Snails and mayfly larvae, which eat the algae. A perch, which eats snails and mayfly larvae. A duck, which eats algae and snails. A heron, which eats perch. Decomposer bacteria. There is no picture.
How do you turn that list into one drawing? And once it is drawn, who goes hungry if the snails disappear?
Unit 8 · Ecology
1Build the web from the table
In a food chain, each organism sits at a trophic level.
Which of the following sets an organism’s trophic level?
- A. The size of its bodyA large animal can sit one step from the producer, and a small one three steps away.
The trophic level is the number of steps, never the size. - B. The name of its diet, such as herbivore or carnivoreHerbivore and carnivore name what an organism eats.
The trophic level is the number of steps between the producer and the organism. - C. ✓ The number of steps it sits from the producer
Why: The producer is the first level.
Each organism sits one step above the organism it eats.
So the number of steps from the producer sets the trophic level.
A soil fungus is a decomposer.
Which of the following describes what a decomposer does?
- A. ✓ It breaks down dead matter and waste into simple molecules
- B. It eats dead animals it did not kill, as a scavenger doesEating a dead animal it did not kill is what a scavenger does.
A decomposer breaks the last of the dead matter down into simple molecules. - C. It eats living producers, such as plants and algaeEating living producers is what a herbivore does.
A decomposer feeds on dead matter and waste.
Why: A decomposer feeds on dead matter and waste from every organism.
It breaks that matter down into simple molecules.
Those molecules are what the producers can take up again.
How do you turn a table of who eats what into one drawing, and then read it?
Put every organism in the lake on one picture, joined by every arrow of who eats whom, and you get many food chains sharing their organisms.
Put each organism at its trophic level: the algae at the bottom, the snails and the mayfly larvae above them, the perch one level above both, the heron on top.
An omnivore can sit at two levels at once, and that is allowed.
Then read the drawing by tracing every arrow that leaves a box, never one chain alone: if the snails go, every eater of snails loses a food source.
Video: Watch: Build the web from the table
The lake survey as a table; each row becomes a box placed at its trophic level, the algae at the bottom and the heron on top; then each food named in the table becomes one arrow, drawn the way the energy goes, one at a time, until the web is complete, with the decomposer bacteria to one side and a dashed line from every row into their box.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L22a.mp4
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The table below lists the lake survey: each organism, and what the survey found it eats.
Many food chains drawn together on one picture, sharing their organisms, is called a .
A food web is built from a table like this one in two moves: first, place each organism at its level; second, draw the arrows.
The first move is placing. The algae make their own organic molecules by photosynthesis, so the algae are the lake’s producer.
The producer box goes on the bottom row.
The snails eat algae, and the mayfly larvae eat algae. Each of them is one step from the producer, so each of them is a primary consumer, on the row above the algae.
The perch eats snails and mayfly larvae. The snails and the mayfly larvae are both primary consumers, so the perch is two steps from the producer: a secondary consumer, on the third row.
The duck eats algae and snails. By its algae meal the duck is one step from the producer; by its snail meal it is two steps.
So the duck is an omnivore that sits at two levels.
In these drawings, an organism that sits at two levels is drawn once, on the higher of its two rows. So the duck’s box goes on the third row, beside the perch.
The heron eats perch. The perch is a secondary consumer, so the heron is three steps from the producer: a tertiary consumer, on the top row.
The decomposer bacteria feed on the dead matter and waste of every organism. Their box goes to one side, beside the rows, and with it every box has its place.
The second move is the arrows. Each food named in the table is one arrow.
The snails eat algae, so one arrow goes from the algae box to the snail box.
The mayfly larvae eat algae, and the duck eats algae. So three arrows leave the algae box: to the snails, to the mayfly larvae and to the duck.
The perch eats snails and mayfly larvae, so an arrow goes from the snails to the perch and another from the mayfly larvae to the perch. The duck eats snails, so an arrow goes from the snails to the duck.
The heron eats perch, so the last solid arrow goes from the perch to the heron.
The decomposer bacteria feed on dead matter from every row. In these drawings, a dashed line from each row into the decomposers’ box marks that dead matter.
The web is complete.
Check it against the table: every food the table names has one arrow, and every arrow has its food named in the table.
Trace the food chains inside the web. Algae → snails → perch → heron is one food chain.
Algae → mayfly larvae → perch → heron is another food chain. It shares the algae, the perch and the heron with the first.
That sharing is what makes the drawing a food web.
What you are expected to know Build a food web from a table of what each organism eats: place each organism at its trophic level, then draw one arrow for every food the table names, the way the energy goes.
Now consider a desert the survey team visited next. The table below lists its survey; build the desert’s web in your head as you answer the questions that follow.
The desert survey lists shrubs, kangaroo rats that eat shrub seeds, beetles that eat shrub leaves, lizards that eat beetles, and kit foxes that eat kangaroo rats and lizards.
Which of the following organisms are the desert’s primary consumers?
- A. The kangaroo rats onlyThe beetles eat shrub leaves, one step from the producer.
So the beetles are primary consumers too. - B. The beetles and the lizardsThe lizards eat beetles, and the beetles eat shrub leaves.
So the lizards are two steps from the producer. - C. ✓ The kangaroo rats and the beetles
Why: The shrubs are the producer.
The kangaroo rats eat shrub seeds and the beetles eat shrub leaves.
Each of them is one step from the producer, so each of them is a primary consumer.
In the desert survey, lizards eat beetles, and the beetles eat shrub leaves.
Which of the following is the lizard’s trophic level?
- A. Primary consumerA primary consumer eats the producer.
The lizard eats beetles, which eat the shrubs. - B. ✓ Secondary consumer
- C. Tertiary consumerA tertiary consumer is three steps from the producer.
Shrubs to beetles to lizards is two steps.
Why: The shrubs are the producer.
The beetles eat shrubs: one step.
The lizards eat beetles: two steps, so the lizard is a secondary consumer.
In the desert survey, kit foxes eat kangaroo rats, which eat shrub seeds, and lizards, which eat beetles that eat shrub leaves.
At how many trophic levels does the kit fox sit?
- A. OneThe kit fox has two foods: kangaroo rats and lizards.
Each food is at a different level, and each gives the kit fox a level. - B. ✓ Two
- C. ThreeThe kit fox has two foods: kangaroo rats and lizards.
Shrubs to rats to fox is two steps; shrubs to beetles to lizards to fox is three.
Why: Shrubs to kangaroo rats to kit fox is two steps: a secondary consumer by that meal.
Shrubs to beetles to lizards to kit fox is three steps: a tertiary consumer by that meal.
The kit fox has two meals at two levels, so it sits at two levels.
In the desert survey, the beetles eat shrub leaves.
Which of the following organisms eat the beetles?
- A. ✓ The lizards
- B. The kit foxesThe table names kangaroo rats and lizards as the kit fox’s foods.
Only the lizard row names beetles. - C. The lizards and the kit foxesOnly the lizard row of the table names beetles.
So only one arrow leaves the beetle box.
Why: An arrow leaves the beetle box for every organism whose row names beetles.
Only the lizard row names beetles.
So the lizards are the only organisms that eat the beetles.
Here is the desert web, built from its table: shrubs at the bottom; kangaroo rats and beetles above; lizards above them; the kit foxes on top, drawn once at the higher of their two levels.
Now consider a marsh. The table below lists its survey.
Before you continue, take a sheet of paper and build the marsh’s web yourself: place each organism at its level, then add one arrow for every food the table names, the way the energy goes.
A marsh survey lists these organisms.
Cattails.
Grasshoppers, which eat cattail leaves.
Muskrats, which eat cattail roots.
Frogs, which eat grasshoppers.
A mink, which eats muskrats and frogs.
Bacteria in the mud, which feed on dead matter and waste.
In a food web, each arrow is drawn from the eaten organism to the organism that eats it.
(a) Identify the trophic level of the cattails, the grasshoppers, the muskrats, the frogs and the mink. (2 pt)
Frame The cattails are …
The grasshoppers and the muskrats are primary consumers, one step from the producer.
The frogs are secondary consumers, two steps from the producer.
The mink is a secondary consumer by its muskrat meal and a tertiary consumer by its frog meal, so the mink sits at two levels.
- Award 1 point for: cattails producer; grasshoppers and muskrats primary consumers; frogs secondary consumers.
- Award 1 point for: the mink at two levels, secondary consumer (muskrats) and tertiary consumer (frogs).
(b) State which organisms the arrows arriving at the mink’s box come from. (1 pt)
Frame The arrows arriving at the mink’s box come from …
- Award 1 point for: two arrows, one from the muskrats and one from the frogs.
Here is the marsh web, built from its table. Check your drawing against it.
A finished web passes four checks.
- Every box sits on the row of its level.
- Every food named in the table has one arrow.
- An organism with two levels is drawn once, on the higher row.
- The decomposers sit to one side, with a dashed line from every row.
A tide pool survey lists seaweed; limpets that eat seaweed; crabs that eat limpets; and gulls that eat crabs and limpets. A student drew the web below from that table.
Which of the following is wrong in the student’s drawing?
- A. The gull box sits on the wrong row for what the gulls eatThe gulls eat crabs, three steps from the seaweed.
So the top row is the right row for the gulls. - B. The limpet box sits on the wrong row for what the limpets eatThe limpets eat seaweed, one step from the producer.
So the primary consumer row is the right row for the limpets. - C. ✓ The arrow from the seaweed to the crabs has no row in the table
Why: Every arrow must have its food named in the table.
The crab row names limpets only.
So the arrow from the seaweed to the crabs has no row in the table, and it should not be there.
46Quick quiz: food web mixed practice
The drawing below shows grass, a rabbit and a fox.
Is this drawing a food web?
- A. YesThe drawing is one food chain: one path from the grass to the fox.
No second chain shares an organism with it. - B. ✓ No
Why: The drawing has one path: grass to rabbit to fox.
A food web needs more than one food chain.
So this drawing is a food chain, and it is not a food web.
The drawing below shows grass, a rabbit, a mouse and a fox.
Is this drawing a food web?
- A. ✓ Yes
- B. NoGrass to rabbit to fox and grass to mouse to fox are two food chains.
Both chains share the grass and the fox.
Why: Grass to rabbit to fox is one food chain.
Grass to mouse to fox is another.
Both chains share the grass and the fox, so this drawing is a food web.
The drawing below shows grass, a rabbit and a fox beside seaweed, a limpet and a gull.
Is this drawing a food web?
- A. YesGrass to rabbit to fox and seaweed to limpet to gull are drawn, but no organism is in both.
Chains that share nothing are separate chains. - B. ✓ No
Why: Grass to rabbit to fox is one food chain, and seaweed to limpet to gull is another.
No organism sits in both chains, and no arrow joins them.
So this drawing is two food chains, and it is not a food web.
An ecologist studies a lake’s feeding.
Which of the following is a food web?
- A. One food chain of four organisms drawn in a single lineOne food chain is one path.
A food web is many food chains sharing organisms. - B. ✓ Many food chains drawn together on one picture, sharing organisms
- C. A table listing each organism and what it eats, with no drawingA table lists each organism’s foods; it is the web’s starting point.
The food web is the drawing built from it.
Why: A food chain is one path from the producer up through its eaters.
A food web draws many such chains together.
The chains share organisms, and that sharing makes the web.
A lake holds algae, snails, mayfly larvae, a perch and a heron.
(a) State what a food web of the lake shows that one food chain from the lake leaves out. (1 pt)
Frame A food web shows …
One food chain shows only one path.
- Award 1 point for: a food web shows every organism and every feeding link (or: many chains sharing organisms), where a chain shows one path.
52Read the web
Once the web is drawn, how do you read it?
Read it one box at a time. The arrows arriving at a box name what that organism eats; the arrows leaving it name what eats it.
Its trophic level is its number of steps from the producer, counted along the arrows.
Video: Watch: Read the web
The lake web with its row names taken off; a pointer rests on the perch box, then follows the two arrows arriving at it back to the snails and the mayfly larvae, then the one arrow leaving it to the heron, then counts the steps from the algae up to the perch.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L22b.mp4
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Here is the lake web again, with the row names taken off.
Take the perch: two arrows arrive at its box, one from the snails and one from the mayfly larvae. So the perch eats snails and mayfly larvae.
One arrow leaves the perch’s box, to the heron. So the heron eats the perch.
Count the perch’s steps from the producer along the arrows: algae to snails is one step, snails to perch is two. So the perch is a secondary consumer.
Every box is read the same way: the arrows arriving name its foods, the arrows leaving name its eaters, and the steps from the producer give its level.
What you are expected to know Read a drawn food web: identify an organism’s trophic level, what it eats and what eats it.
The lake web is drawn below.
Which of the following is the heron’s trophic level?
- A. Primary consumerA primary consumer is one step from the producer.
The heron is three steps from the algae. - B. Secondary consumerA secondary consumer is two steps from the producer.
The heron is three steps from the algae. - C. ✓ Tertiary consumer
Why: Algae to snails is one step.
Snails to perch is two.
Perch to heron is three, so the heron is a tertiary consumer.
The lake web is drawn below.
Which of the following organisms eat the mayfly larvae?
- A. The heronNo arrow goes from the mayfly larvae to the heron.
The heron’s only arrow arrives from the perch. - B. ✓ The perch only
- C. The perch and the duckThe duck’s arrows arrive from the algae and from the snails.
No arrow goes from the mayfly larvae to the duck.
Why: One arrow leaves the mayfly larvae box.
It goes to the perch.
So the perch is the only organism that eats the mayfly larvae.
The desert web is drawn below.
Which of the following organisms does the kit fox eat?
- A. The kangaroo rats onlyTwo arrows arrive at the kit fox box: one from the kangaroo rats and one from the lizards.
- B. ✓ The kangaroo rats and the lizards
- C. The kangaroo rats, the lizards and the beetlesNo arrow goes from the beetles to the kit foxes.
The beetles’ only arrow goes to the lizards.
Why: Two arrows arrive at the kit fox box: one from the kangaroo rats and one from the lizards.
The arrows arriving at a box name what that organism eats.
So the kit fox eats kangaroo rats and lizards.
The desert web is drawn below.
Which of the following organisms eat the kangaroo rats?
- A. ✓ The kit foxes
- B. The lizardsThe lizards’ only arrow arrives from the beetles.
No arrow goes from the kangaroo rats to the lizards. - C. The beetlesThe beetles eat shrub leaves.
No arrow goes from the kangaroo rats to the beetles.
Why: One arrow leaves the kangaroo rat box.
It goes to the kit foxes.
So the kit foxes are the only organisms that eat the kangaroo rats.
The lake web is drawn below.
Which of the following organisms eat the algae?
- A. The snails and the mayfly larvaeThree arrows leave the algae box: to the snails, to the mayfly larvae and to the duck.
- B. ✓ The snails, the mayfly larvae and the duck
- C. The snails, the mayfly larvae, the duck and the perchNo arrow goes from the algae to the perch.
The perch’s arrows arrive from the snails and the mayfly larvae.
Why: Three arrows leave the algae box: to the snails, to the mayfly larvae and to the duck.
The arrows leaving a box name what eats that organism.
So the snails, the mayfly larvae and the duck eat the algae.
The desert web is drawn below.
Which of the following is the beetle’s trophic level?
- A. ✓ Primary consumer
- B. Secondary consumerA secondary consumer is two steps from the producer.
The beetles are one step from the shrubs. - C. Tertiary consumerA tertiary consumer is three steps from the producer.
The beetles are one step from the shrubs.
Why: The beetles’ one arrow arrives from the shrubs.
The shrubs are the producer.
One step from the producer: the beetle is a primary consumer.
69Take one organism out
What happens to the rest of the web when one organism disappears?
Every arrow that left its box goes with it. So every eater at the head of one of those arrows loses a food source.
To find them all, trace every arrow that left the box, never the first one alone.
Video: Watch: Take one organism out
The lake web; the mayfly larvae box fades and the arrows that touch it fade with it; a pointer follows the faded arrow to the perch, then shows the perch’s other arrow, from the snails, still solid.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L22c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L22c.mp4
Suppose a disease kills every mayfly larva in the lake. Here is the web with the mayfly larvae box faded, and the arrows that touched it faded with it.
One arrow left the mayfly larvae box, to the perch. So the perch loses a food source.
The perch still has the snails. So the perch does not go hungry, because the web gave it a second path.
An eater with two arrows arriving has two paths. An eater with one arrow arriving has one path, and when that food disappears it has nothing left to eat.
What you are expected to know Predict which organisms lose a food source when one organism disappears from a food web, by tracing every arrow that left its box.
Suppose instead that a disease kills every snail in the lake. The web below shows the snail box faded.
Which of the following organisms lose a food source?
- A. The perch onlyTwo arrows left the snail box: one to the perch and one to the duck.
So the duck loses a food source too. - B. The duck onlyTwo arrows left the snail box: one to the perch and one to the duck.
So the perch loses a food source too. - C. ✓ The perch and the duck
Why: Two arrows left the snail box: one to the perch and one to the duck.
Both arrows go with the snails.
So the perch and the duck each lose a food source.
Suppose a pond survey lists pondweed; tadpoles that eat pondweed; water beetles that eat tadpoles; and newts that eat tadpoles and water beetles. A disease kills every tadpole in the pond.
(a) Explain why the water beetles are left with no food while the newts still have food. (2 pt)
Frame The water beetles are left with no food because …
One arrow arrived at the water beetle box, from the tadpoles, and that arrow goes with the tadpoles.
Two arrows arrived at the newt box, from the tadpoles and from the water beetles.
So when the tadpoles die, the newts still have the water beetles.
- Award 1 point for: the tadpoles were the water beetles’ only food (one arrow arriving), so the water beetles lose all their food.
- Award 1 point for: the newts had two foods (two arrows arriving), so the newts still have the water beetles.
A student looks at the lake web below and says: “If the algae die off, only the snails are affected, because the snails are the animals that eat algae.”
Is the student correct?
- A. Yes: the snails are the only organisms in the lake that eat the algaeThree arrows leave the algae box: to the snails, to the mayfly larvae and to the duck.
Every one of those three eaters loses a food source. - B. ✓ No: the mayfly larvae and the duck eat algae too, so they lose a food source as well
Why: Three arrows leave the algae box: to the snails, to the mayfly larvae and to the duck.
All three arrows go with the algae.
So the snails, the mayfly larvae and the duck each lose a food source; the student traced one arrow of three.
Suppose instead that the perch disappears from the lake. The web below shows the perch box faded.
Which of the following organisms loses all of its food?
- A. ✓ The heron
- B. The duckThe duck’s arrows arrive from the algae and the snails.
The perch was never one of the duck’s foods. - C. The snailsThe snails eat algae.
The perch ate the snails; the snails lose an eater, never a food.
Why: One arrow left the perch box, to the heron.
Only one arrow arrived at the heron box, and it came from the perch.
So when the perch goes, the heron has nothing left to eat.
Here is the lake table again, drawn as a web: algae at the bottom, snails and mayfly larvae above them, perch and duck one level above, the heron on top, decomposer bacteria to one side.
If the snails go, the perch still has the mayfly larvae and the duck still has the algae.
If the perch goes, the heron, which ate only perch, has nothing left to eat.
86Mixed practice mixed practice
A river web is drawn below.
Which of the following is the kingfisher’s trophic level?
- A. Primary consumerA primary consumer is one step from the producer.
The kingfisher is three steps from the algae. - B. Secondary consumerA secondary consumer is two steps from the producer.
The kingfisher is three steps from the algae. - C. ✓ Tertiary consumer
Why: Algae to caddisfly larvae is one step.
Caddisfly larvae to minnows is two.
Minnows to kingfisher is three, so the kingfisher is a tertiary consumer.
Suppose the caddisfly larvae in the river below die out.
Which of the following organisms lose a food source?
- A. The minnows onlyTwo arrows leave the caddisfly larvae box: one to the minnows and one to the bass.
So the bass loses a food source too. - B. ✓ The minnows and the bass
- C. The bass and the kingfisherNo arrow goes from the caddisfly larvae to the kingfisher.
The kingfisher’s one arrow arrives from the minnows.
Why: Two arrows leave the caddisfly larvae box: one to the minnows and one to the bass.
Both arrows go with the caddisfly larvae.
So the minnows and the bass each lose a food source.
A river survey lists algae; caddisfly larvae that eat algae; minnows that eat caddisfly larvae; a bass that eats minnows and caddisfly larvae; and a kingfisher that eats minnows. A student drew the web below from that survey.
Which box sits on the wrong row for what it eats?
- A. ✓ The minnow box
- B. The bass boxThe bass sits one row above the minnows, its higher-level food.
So its row is right. - C. The kingfisher boxThe kingfisher sits one row above the minnows it eats.
So its row is right.
Why: The minnows eat caddisfly larvae, which eat algae.
That is two steps from the producer: a secondary consumer.
The student drew the minnows on the primary consumer row, one row beneath their true row.
A river web is drawn below.
Which of the following organisms eat the minnows?
- A. The bass onlyTwo arrows leave the minnow box: one to the bass and one to the kingfisher.
- B. The kingfisher onlyTwo arrows leave the minnow box: one to the bass and one to the kingfisher.
- C. ✓ The bass and the kingfisher
Why: Two arrows leave the minnow box: one to the bass and one to the kingfisher.
The arrows leaving a box name what eats that organism.
So the bass and the kingfisher both eat the minnows.
Suppose a hedge survey lists hawthorn; aphids that eat hawthorn sap; ladybugs that eat aphids; and a wren that eats aphids and ladybugs.
At how many trophic levels does the wren sit?
- A. OneThe wren has two foods: aphids and ladybugs.
Each food is at a different level, and each gives the wren a level. - B. ✓ Two
- C. ThreeThe wren has two foods: aphids and ladybugs.
Hawthorn to aphids to wren is two steps; hawthorn to aphids to ladybugs to wren is three.
Why: Hawthorn to aphids to wren is two steps: a secondary consumer.
Hawthorn to aphids to ladybugs to wren is three steps: a tertiary consumer.
Two meals at two levels, so the wren sits at both.
Suppose the minnows in the river below die out.
Which of the following organisms loses all of its food?
- A. ✓ The kingfisher
- B. The bassTwo arrows arrive at the bass box: one from the minnows and one from the caddisfly larvae.
The bass still has the caddisfly larvae. - C. The caddisfly larvaeThe caddisfly larvae eat algae.
The minnows ate the caddisfly larvae; the larvae lose an eater, never a food.
Why: One arrow arrives at the kingfisher box, from the minnows.
That arrow goes with the minnows.
So the kingfisher has nothing left to eat.
Suppose an invasive crayfish arrives in the river drawn below. The crayfish eats caddisfly larvae, and nothing in the river eats the crayfish. In a food web, each arrow is drawn from the eaten organism to the organism that eats it.
(a) An ecologist adds the crayfish to the web. Describe where the ecologist places the crayfish’s box and the arrow the ecologist draws for it. (1 pt)
Frame The ecologist places the crayfish …
- Award 1 point for: the crayfish one level above the caddisfly larvae (the minnows’ row) with one arrow from the caddisfly larvae to the crayfish.
(b) The crayfish eat most of the caddisfly larvae. Explain why the food web predicts that the minnows go hungry before the bass does. (2 pt)
Frame The minnows go hungry first because …
Two arrows arrive at the bass box, from the caddisfly larvae and from the minnows.
So when the caddisfly larvae become scarce, the minnows have no other path, while the bass still has the minnows.
- Award 1 point for: the caddisfly larvae are the minnows’ only food (one arrow arriving), so the minnows have no other path.
- Award 1 point for: the bass has two foods (two arrows arriving), so the bass still has the minnows.
Glossary
- food web
- Many food chains drawn together on one picture, sharing their organisms: every organism in a community at its trophic level, with one arrow for every food it eats, drawn from the eaten to the eater, and the decomposers to one side with a dashed line from every row.
APBIO-U08-L23 Ten percent
Suppose a meadow. Its grass stores 10 000 kJ of energy. Its grasshoppers, which eat the grass, store about 1 000 kJ. Its meadowlarks, which eat the grasshoppers, store about 100 kJ.
One hawk hunts the whole meadow. Why does one bird need a whole field?
Unit 8 · Ecology
1About 10 % moves up a level
A meadow’s food chain is grass, then grasshoppers, then meadowlarks, then a hawk.
Which of the following is the primary consumer?
- A. The grassThe grass makes its own food, so it is the producer.
The primary consumer is the eater one step above the producer. - B. ✓ The grasshopper
- C. The meadowlarkThe meadowlark eats the grasshopper, which eats the grass.
It is two steps above the producer: the secondary consumer.
Why: The grass is the producer.
The grasshopper eats the grass, one step above the producer.
So the grasshopper is the primary consumer.
Sunlight’s energy enters a meadow through its grass and passes from organism to organism.
Which of the following happens to that energy in the end?
- A. It cycles back into the grass to be used againMatter cycles; energy does not.
Once the energy has left as heat, no grass can take it back in. - B. ✓ It leaves the meadow as heat and never returns
- C. It is stored in the soil for everThe soil stores some dead matter for a while.
The energy in that matter is respired by decomposers and leaves as heat.
Why: Energy passes once through an ecosystem.
Every organism respires some of it, and that energy leaves as heat.
The heat never comes back in.
How much of the energy stored at one trophic level is stored in the next? And where does the rest go?
Video: Watch: About 10 % moves up a level
The meadow’s levels stack into slices from the bottom up. The kJ value appears on each slice as it forms: 10 000, then 1 000, then 100. Each new slice is 10 % as wide as the one below. The hawk’s hairline slice arrives last, at about 10 kJ.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L23a.mp4
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Here is the meadow’s food chain again: grass, grasshoppers, meadowlarks, a hawk.
Suppose the meadow’s grass stores 10 000 kJ of energy in its leaves, stems and roots.
The grasshoppers eat that grass all summer. By the end of the summer, the grasshoppers store about 1 000 kJ.
The meadowlarks eat the grasshoppers all summer. By the end of the summer, the meadowlarks store about 100 kJ.
Each level stores about 10 % of the energy stored in the level below it.
Ecologists draw the levels as slices stacked one on top of another, with the producers’ slice at the bottom. Each slice is drawn 10 % as wide as the slice below it.
The energy stored at a level is written on its slice, or beside the slice where the slice is too thin to hold it.
A drawing of the trophic levels as slices like this, each slice’s width showing the energy stored at that level, is called an .
When only about 10 % of the energy stored at one trophic level is stored in the next, we call the pattern . It is the working rule for energy between trophic levels.
Now consider the hawk. It eats meadowlarks, so it sits one level above them, at the fourth level.
By the 10 % rule, the hawk’s level stores about 10 kJ. The table below lists all four levels and the energy each stores.
Here is the whole energy pyramid, with the hawk’s slice at the top. The hawk’s slice is so thin that the drawing shows it as a hairline.
The hawk’s level stores the least energy of the four. Of the 10 000 kJ stored in the whole meadow’s grass, only about 10 kJ ends up stored at the hawk’s level.
That is why one hawk needs a whole field: the grass of the whole meadow stands beneath the small amount of energy the hawk lives on.
One simplification we made here. The 10 % is a rounded working figure.
Measured transfers between trophic levels vary from about 5 % to about 20 %. We use 10 % because it makes the pattern easy to see and easy to calculate with.
What you are expected to know State the working rule for energy between trophic levels: only about 10 % of the energy stored at one level is stored in the next, drawn as an energy pyramid.
Suppose a lake’s algae store some energy, and the water fleas that eat the algae store some of it in turn.
By the 10 % rule, about what share of the algae’s stored energy is stored in the water fleas?
- A. ✓ About 10 %
- B. About 50 %Half the energy does not move up.
Only about 10 % of the energy stored at one level is stored in the next. - C. About 90 %About 90 % is the share that is missing from the next level.
Only about 10 % is stored there.
Why: The 10 % rule says only about 10 % of the energy stored at one trophic level is stored in the next.
The water fleas are the next level above the algae.
So they store about 10 % of the algae’s energy.
The energy pyramid below is drawn for a pond: algae, then the snails that eat them, then the carp that eat the snails.
Which level stores the most energy?
- A. ✓ The algae
- B. The snailsThe snails’ slice is narrower than the algae’s slice.
A narrower slice shows less energy stored. - C. The carpThe carp’s slice is the narrowest of the three.
The narrowest slice shows the least energy stored.
Why: A slice’s width shows the energy stored at that level.
The algae’s slice is the widest, at 6 200 kJ.
So the algae store the most energy.
The energy pyramid below is drawn for a pond: algae, then the snails that eat them, then the carp that eat the snails.
How much energy do the carp store?
- A. 6 200 kJ6 200 kJ is written on the widest slice, the algae’s.
- B. 620 kJ620 kJ is written beside the middle slice, the snails’.
- C. ✓ 62 kJ
Why: The carp are the top level, so their slice is the top slice.
The value written beside the top slice is 62 kJ.
So the carp store 62 kJ.
The energy pyramid below is drawn for an oak wood: oak leaves, then caterpillars, then small birds. The caterpillars’ slice is drawn 10 % as wide as the oak leaves’ slice.
Which of the following does the width of a slice show?
- A. ✓ The energy stored at that level
- B. The number of organisms at that levelA level with many small organisms can store less energy than a level with fewer large ones.
The slice shows energy, not a count. - C. The size of one organism at that levelOak leaves are small and the birds are larger, yet the leaves’ slice is the widest.
The slice shows energy stored, not body size.
Why: An energy pyramid draws each level’s slice to the energy stored at that level.
The caterpillars store about 10 % of the leaves’ energy.
So their slice is 10 % as wide.
A student says: “The 10 % rule is a law of nature. Exactly 10 % of the energy always passes from one level to the next.”
Is the student correct?
- A. Yes: every measured transfer is exactly 10 %Measured transfers between levels vary from about 5 % to about 20 %.
10 % is the rounded figure we work with. - B. ✓ No: the 10 % is a rounded working figure
Why: The 10 % is a rounded working figure.
Measured transfers between trophic levels vary from about 5 % to about 20 %.
So 10 % is a useful rule, and it is a simplification.
28Quick quiz: energy pyramid, the 10 % rule mixed practice
An energy pyramid of a lake has slices for the algae, the insects that eat the algae, and the fish that eat the insects.
Which slice is the base?
- A. ✓ The algae’s slice
- B. The fish’s sliceThe fish are the top level, so their slice is the top slice.
The producers’ slice is the base.
Why: The producers’ slice is drawn at the base of an energy pyramid.
The algae are the lake’s producers.
So the algae’s slice is the base.
In an energy pyramid, the third slice is 10 % as wide as the second slice.
Does the pyramid follow the 10 % rule?
- A. ✓ Yes
- B. NoThe 10 % rule says each level stores about 10 % of the level below.
A slice 10 % as wide as the one below shows that.
Why: The 10 % rule says each level stores about 10 % of the energy stored in the level below.
The third slice is 10 % as wide as the second.
So the pyramid follows the rule.
In an energy pyramid, the second slice is half as wide as the base.
Does the pyramid follow the 10 % rule?
- A. YesHalf as wide shows 50 % of the energy moving up.
The 10 % rule says about 10 % moves up. - B. ✓ No
Why: The 10 % rule says each level stores about 10 % of the energy stored in the level below.
A slice half as wide shows 50 %.
So the pyramid does not follow the rule.
Energy passes from one trophic level to the next.
What does the 10 % rule say?
- A. About 10 % of the organisms at one level are eaten by the nextThe 10 % rule is about energy stored, not about how many organisms are eaten.
- B. About 10 % of the sunlight reaching the producers is stored in themThe 10 % rule is about the step from one trophic level to the next, not about sunlight.
- C. ✓ About 10 % of the energy stored at one level is stored in the next
Why: The 10 % rule says only about 10 % of the energy stored at one trophic level is stored in the next.
An ecologist draws the trophic levels of a meadow.
What is an energy pyramid?
- A. ✓ Stacked slices, each slice’s width showing the energy stored at its level
- B. Stacked slices, each slice’s width showing the number of organisms at its levelThe slices show energy stored, not how many organisms there are.
- C. Boxes joined by arrows pointing from the eaten to the eaterBoxes joined by arrows from the eaten to the eater is a food chain or a food web.
Why: An energy pyramid draws the trophic levels as slices stacked with the producers at the bottom.
Each slice’s width shows the energy stored at that level.
Energy passes from one trophic level to the next.
(a) State the 10 % rule. (1 pt)
- Award 1 point for: about 10 % of the energy stored at one trophic level is stored in the next (accept 'passes to' or 'reaches' the next level).
35Where the other 90 % went
A cell breaks down glucose to make ATP.
By the second law of thermodynamics, what happens to some of the energy at every transfer?
- A. ✓ It spreads out as heat that can do no more work
- B. It is destroyed, so the total energy fallsEnergy is never created or destroyed.
Some of it spreads out as heat that can no longer do work. - C. It stays stored in the ATP for the cell to use laterThe cell uses its ATP within seconds.
At every transfer, some energy spreads out as heat that can no longer do work.
Why: The second law of thermodynamics says that at every energy transfer, some energy spreads out as heat.
Heat that has spread out can no longer do work.
So no transfer is fully efficient.
A mouse, an endotherm, stays warm through a cold night.
Where does the heat that keeps the mouse warm come from?
- A. From the sunlight the mouse soaked up by dayA lizard warms on a sunlit rock; a mouse makes its own heat.
The heat comes from respiration of its food. - B. From the nest material around the mouseNest material slows the loss of heat.
The heat itself comes from respiration of the mouse’s food. - C. ✓ From respiration of the food the mouse ate
Why: An endotherm makes its own heat.
Its cells respire the food it ate, and that respiration gives off heat.
So the heat comes from respiration of the food.
Why is so little of the grass’s energy stored in the grasshoppers?
Video: Watch: Where the other 90 % went
The meadow’s energy pyramid is on screen. Three labeled arrows leave each slice in turn as the fates are named: left as heat, never eaten, to decomposers. The heat arrow is drawn widest, because respiration takes the biggest share.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L23b.mp4
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Suppose a cow eats 100 kJ of grass in a morning.
The cow’s cells respire most of the sugar in that grass to make ATP. The cow uses the ATP to keep warm and to move.
Every time the cow’s cells respire, some energy spreads out as heat. That heat passes into the air around the cow and leaves the meadow.
At every transfer most of the energy is respired and leaves the ecosystem as heat, so only about a tenth is stored in the next trophic level.
Respiration takes the biggest share of the missing energy. Three smaller fates take the rest.
Some of the grass is never eaten. Its energy stays in the grass until the grass dies, and then decomposers take it.
On the pyramid, the never-eaten arrow carries the energy in living grass that no grasshopper eats, whatever happens to that grass later. The to-decomposers arrow carries the energy in the dead and the waste.
Some of the grass the cow eats passes through the cow undigested. Its energy leaves in the cow’s dung, and decomposers take it.
When the cow dies, decomposers take the energy still stored in its body.
The table below lists the four fates and where each one’s energy ends up.
Here is the meadow’s energy pyramid again, with three labeled arrows leaving each slice: left as heat, never eaten, to decomposers.
Undigested food, dead bodies and other waste all go to decomposers, so one arrow carries all three.
The decomposers respire what they take, so that energy leaves as heat too. None of the missing 90 % comes back to the grass as food.
Nothing destroyed the missing energy. The first law of thermodynamics says that energy is never created or destroyed.
The missing 90 % is heat in the surroundings, or energy that decomposers took and respired.
What you are expected to know Explain why so little energy passes up a level: most of what an organism eats is respired and leaves as heat, some is never eaten, some passes through undigested, and the dead and the waste go to decomposers.
Suppose a sheep eats grass all day.
Which of the following happens to most of the energy in the grass the sheep eats?
- A. ✓ The sheep’s cells respire it, and it leaves as heat
- B. The sheep stores it as new wool and fleshThe sheep stores only a small share as new tissue.
Its cells respire most of the sugar to keep it warm and moving. - C. It passes out in the sheep’s dungSome grass passes through undigested, but that is a small share.
The sheep’s cells respire most of the sugar.
Why: The sheep’s cells respire most of the sugar in the grass to make ATP.
Every time they respire, energy spreads out as heat.
So most of the grass’s energy leaves the sheep as heat.
Suppose a field’s rabbits eat 4 800 kJ of grass in a month. By the end of the month, the rabbits have stored only about 480 kJ of that energy as new tissue.
(a) Explain why the rabbits stored so little of the energy in the grass they ate. (2 pt)
Frame The rabbits stored so little because …
Every time the cells respired, some energy spread out as heat.
That heat left the rabbits and the field.
Some of the grass also passed through the rabbits undigested, and its energy left in their droppings.
So only a small share of the grass’s energy was left to build new tissue.
- Award 1 point for: the rabbits respired most of the energy in the grass, and that energy left as heat.
- Award 1 point for: some grass passed through undigested (its energy left in droppings) OR only what was left after respiration and waste could be built into new tissue.
In the energy pyramid below, three labeled arrows leave every slice. A grasshopper’s body gives off warmth as its cells respire.
Which arrow carries that warmth away from the primary consumers’ slice?
- A. ✓ Left as heat
- B. Never eatenNever eaten is energy in organisms that no eater took, still stored in their bodies.
Warmth given off is heat. - C. To decomposersTo decomposers is energy in dung and dead bodies.
Warmth given off is heat, and heat leaves the ecosystem.
Why: The grasshopper’s cells respire and give off heat.
Heat leaves the ecosystem.
So the warmth leaves by the arrow labeled left as heat.
In the energy pyramid below, three labeled arrows leave every slice. A grasshopper leaves droppings on the soil.
Which arrow carries the energy in the droppings away from the primary consumers’ slice?
- A. Left as heatDroppings are matter that still stores energy.
Decomposers take that energy; it has not yet left as heat. - B. Never eatenNever eaten is energy in organisms no eater took.
Droppings are waste, and decomposers take them. - C. ✓ To decomposers
Why: Droppings are waste that still stores energy.
Decomposers break down waste and take its energy.
So the droppings’ energy leaves by the arrow labeled to decomposers.
In the energy pyramid below, three labeled arrows leave every slice. Some of the meadow’s grass stands all summer, and no grasshopper eats it. At the end of summer it is still standing.
At the end of summer, which arrow carries that grass’s energy away from the producers’ slice?
- A. Left as heatThe living grass still stores its energy in its leaves, stems and roots.
That energy has not left as heat. - B. ✓ Never eaten
- C. To decomposersDecomposers take the grass’s energy only once the grass dies.
While the grass stands uneaten, its energy is in the never-eaten share.
Why: No grasshopper eats that grass, so none of its energy reaches the primary consumers.
The grass still stores the energy while it lives.
So that energy is in the arrow labeled never eaten.
A student says: “Of the energy stored in the grass, the 90 % missing from the grasshoppers has been destroyed.”
Is the student correct?
- A. ✓ No: the missing energy left as heat or went to decomposers
- B. Yes: the missing energy was destroyed when the grasshoppers used itEnergy is never created or destroyed.
The missing 90 % spread out as heat, or decomposers took it.
Why: The first law of thermodynamics says energy is never created or destroyed.
The grasshoppers respired most of the missing energy, and it left as heat.
The rest stayed in uneaten grass, in dung and in dead bodies, and decomposers took it.
Suppose a fish farm’s fish eat 14 300 kJ of pellets in a week and store 2 600 kJ of that energy as new flesh.
Where did most of the other 11 700 kJ go?
- A. It stayed in the fish’s dung on the tank floorSome food passes through undigested, but that is a small share.
The fish’s cells respire most of the food. - B. ✓ It left the fish as heat after their cells respired the food
- C. It was destroyed when the fish used itEnergy is never created or destroyed.
The fish’s cells respired most of the food, and that energy left as heat.
Why: The fish’s cells respire most of the food they eat to make ATP.
Every time they respire, energy spreads out as heat.
So most of the energy the fish did not store left as heat.
63Calculate the energy at the next level
How do you work out the energy at the level above from the energy at the level below?
Video: Watch: Calculate the energy at the next level
The calculation is worked on screen for a new starting figure. The values are written down, then the equation on its own line, then the substitution, then the answer with its unit, kJ per square meter a year. The second level is worked the same way from the first.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L23c.mp4
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The 10 % rule gives you the energy at the next level: you multiply the energy at this level by 0.10. Here is the equation on its own line, with its key beneath.
energy at the next level: the energy stored at the trophic level one step up, in kJ per square meter a year
energy at this level: the energy stored at the level you start from, in kJ per square meter a year
0.10: 10 % written as a decimal; it has no unit
Both energies are in kilojoules per square meter a year. The 0.10 has no unit, so the answer carries the same unit as the value you started with, as the line below shows.
kilojoules per square meter a year times 0.10 is still kilojoules per square meter a year: 0.10 has no unit, so the answer carries the unit you started with
Now consider a salt marsh. Suppose its producers store 25 000 kJ of energy per square meter a year in new growth.
The marsh’s producers store 25 000 kJ per square meter a year. Calculate the energy at the primary consumers’ level.
The secondary consumers eat the primary consumers. So apply the rule again, starting from the primary consumers’ level.
The marsh’s primary consumers store 2 500 kJ per square meter a year. Calculate the energy at the secondary consumers’ level.
Here is the marsh’s energy pyramid, with the three values on its slices.
Two levels up means the rule applied twice. Three levels up means the rule applied three times.
What you are expected to know Calculate the energy available at each higher trophic level from the producers’ energy using the 10 % rule, with units.
An ecosystem’s producers store a known amount of energy, in kilojoules per square meter a year.
Which of the following gives the energy at the primary consumers’ level?
- A. The producers’ energy × 10Multiplying by 10 gives a level storing ten times more than the producers.
The next level stores about 10 %, so the answer must be smaller. - B. The producers’ energy × 0.900.90 is the share that is missing from the next level.
The next level stores the other 10 %, so multiply by 0.10. - C. ✓ The producers’ energy × 0.10
Why: The 10 % rule says the next level stores about 10 % of this level’s energy.
10 % written as a decimal is 0.10.
So the primary consumers’ energy is the producers’ energy multiplied by 0.10.
Suppose a lake’s producers store 18 400 kJ of energy per square meter a year.
Calculate the energy at the tertiary consumers’ level.
Part 1. Calculate the energy at the primary consumers’ level.
Answer: 1840 kJ per square meter a year (tolerance ±0)
Part 2. Calculate the energy at the secondary consumers’ level.
Answer: 184 kJ per square meter a year (tolerance ±0)
Answer: 18.4 kJ per square meter a year (tolerance ±0)
Suppose a forest’s producers store 3 400 kJ of energy per square meter a year.
Calculate the energy at the secondary consumers’ level, two levels above the producers.
Answer: 34 kJ per square meter a year (tolerance ±0)
Suppose a coral reef’s producers store 7 900 kJ of energy per square meter a year.
Calculate the energy at the tertiary consumers’ level, three levels above the producers.
Answer: 7.9 kJ per square meter a year (tolerance ±0)
Here is the meadow again: 10 000 kJ stored in its grass, about 1 000 kJ in its grasshoppers, about 100 kJ in its meadowlarks, and about 10 kJ at the hawk’s level.
At each step, the other 90 % left as heat, stayed in grass no grasshopper ate, or went to decomposers.
About 10 kJ at the top is why one hawk needs the whole meadow.
82Numeric practice: the 10 % rule mixed practice
Suppose a prairie’s producers store 3 700 kJ of energy per square meter a year.
Calculate the energy at the primary consumers’ level.
Answer: 370 kJ per square meter a year (tolerance ±0)
Suppose an estuary’s producers store 12 600 kJ of energy per square meter a year.
Calculate the energy at the secondary consumers’ level, two levels above the producers.
Answer: 126 kJ per square meter a year (tolerance ±0)
Suppose a desert’s sparse shrubs store 740 kJ of energy per square meter a year.
Calculate the energy at the primary consumers’ level.
Answer: 74 kJ per square meter a year (tolerance ±0)
Suppose an eelgrass meadow’s producers store 21 300 kJ of energy per square meter a year.
Calculate the energy at the secondary consumers’ level, two levels above the producers.
Answer: 213 kJ per square meter a year (tolerance ±0)
Suppose a river’s producers store 5 800 kJ of energy per square meter a year.
Calculate the energy at the secondary consumers’ level, two levels above the producers.
Answer: 58 kJ per square meter a year (tolerance ±0)
88Mixed practice mixed practice
Suppose deer browse the leaves of a forest all year.
Which of the following takes the biggest share of the energy in the leaves the deer eat?
- A. ✓ Respiration, with the energy leaving as heat
- B. Growth, with the energy built into new deerOnly a small share of the leaves’ energy is built into new tissue.
- C. Decomposers, taking the energy in the deer’s dungDung carries a small share to decomposers.
The deer’s cells respire most of the energy.
Why: The deer’s cells respire most of the sugar in the leaves to make ATP.
Every time they respire, energy spreads out as heat.
So respiration takes the biggest share.
The energy pyramid below is drawn for a stream: algae, then the mayfly larvae that eat them, then the trout that eat the larvae.
How much energy do the trout store?
- A. 2 800 kJ2 800 kJ is written on the widest slice, the algae’s.
- B. 280 kJ280 kJ is written beside the middle slice, the mayfly larvae’s.
- C. ✓ 28 kJ
Why: The trout are the top level, so their slice is the top slice.
The value written beside the top slice is 28 kJ.
So the trout store 28 kJ.
Suppose a mangrove swamp’s producers store 9 300 kJ of energy per square meter a year.
Calculate the energy at the primary consumers’ level.
Answer: 930 kJ per square meter a year (tolerance ±0)
A student looks at an energy pyramid and says: “The top slice is narrow because the top predators are small animals.”
Is the student correct?
- A. Yes: a slice’s width shows the size of the animals at that levelA hawk is larger than a grasshopper, yet the hawk’s slice is narrower.
The width shows energy stored, not body size. - B. ✓ No: a slice’s width shows the energy stored at that level
Why: A slice’s width shows the energy stored at that level.
Each level stores only about 10 % of the level below.
So the top level stores the least energy, whatever the size of its animals.
Suppose some of a meadow’s grass dies in the autumn, and no cow has eaten it.
Who takes the energy stored in that dead grass?
- A. The cows, when they eat the next year’s grassNext year’s grass stores new energy from next year’s sunlight.
The dead grass’s energy goes to decomposers. - B. ✓ The decomposers in the soil
- C. The sun, which takes back the energy it gaveEnergy never returns to the sun.
Decomposers take the energy in dead grass and respire it, and it leaves as heat.
Why: The dead grass still stores energy in its leaves and stems.
Decomposers break down dead matter and take its energy.
So the decomposers in the soil take it.
The energy pyramid below is drawn for a lake: algae, then the insects that eat the algae, then a heron that eats the insects.
Which slice is drawn narrowest?
- A. The algae’s sliceThe algae are the producers, and their slice is the widest.
- B. The insects’ sliceThe insects’ slice is narrower than the algae’s and wider than the heron’s.
- C. ✓ The heron’s slice
Why: Each level stores only about 10 % of the energy stored in the level below.
The heron is the top level, so it stores the least energy.
So the heron’s slice is the narrowest.
Suppose a pond’s producers store 8 600 kJ of energy in a summer. The pond’s small fish, two trophic levels above the producers, store about 86 kJ. A heron eats the small fish.
(a) Calculate the energy the heron’s level stores, by the 10 % rule. (1 pt)
Answer: 8.6 kJ (tolerance ±0)
- Award 1 point for: 8.6 kJ.
(b) Explain why the small fish store so much less energy than the producers. (2 pt)
Frame The small fish store so much less because …
Some of the food is also never eaten or passes through undigested, and decomposers take it.
So only about 10 % of the energy stored at one level is stored in the next.
The small fish are two levels above the producers, so the loss happens twice.
- Award 1 point for: at each transfer most of the energy is respired and leaves as heat (accept: some is never eaten, passes through undigested or goes to decomposers as an addition).
- Award 1 point for: only about 10 % is stored in the next level, and the fish are two levels up, so the loss happens twice.
Glossary
- energy pyramid
- A drawing of an ecosystem's trophic levels as slices stacked with the producers at the bottom. Each slice's width shows the energy stored at that level, so each slice is about 10 % as wide as the one below it.
- the 10 % rule
- The working rule for energy between trophic levels: only about 10 % of the energy stored at one level is stored in the next. It is a rounded figure; measured transfers vary from about 5 % to about 20 %.
APBIO-U08-L24 Measuring the loss, and why the chain stops
Photo: Ebyabe, Wikimedia Commons, CC BY-SA 2.5 (resized).
Here is Silver Springs, Florida: clear water wells up from the ground and flows away as a river. Water plants grow on its bed, and insects and snails eat the plants. Ecologists measured the energy each level stored over a year. The plants stored about 31 900 kJ per square meter a year. The insects and snails stored about 4 620 kJ per square meter a year.
The 10 % working rule says about 10 % of one level’s energy is stored in the next. Is that what Silver Springs shows?
Unit 8 · Ecology
1Measure the transfer as a percentage
Sea urchins eat the kelp of a kelp forest.
By the 10 % rule, about what share of the energy stored in the kelp is stored in the sea urchins?
- A. ✓ About 10 %
- B. About 50 %Half of the kelp’s energy does not move up.
Only about 10 % of the energy stored at one level is stored in the next. - C. About 90 %About 90 % is the share that is missing from the sea urchins’ level.
Only about 10 % is stored there.
Why: The 10 % rule says only about 10 % of the energy stored at one trophic level is stored in the next.
The sea urchins are the level above the kelp.
So they store about 10 % of the kelp’s energy.
The 10 % rule is a rounded working figure.
Which of the following is the range of transfers ecologists have measured between trophic levels?
- A. Exactly 10 % every timeA measured transfer is rarely exactly 10 %.
Measured transfers vary from about 5 % to about 20 %. - B. ✓ About 5 % to about 20 %
- C. About 50 % to about 90 %Half or more of a level’s energy never moves up.
Measured transfers vary from about 5 % to about 20 %.
Why: The 10 % is a rounded working figure.
Measured transfers between trophic levels vary from about 5 % to about 20 %.
How do you measure a real transfer between two trophic levels? And is it the 10 % of the working rule?
You divide the energy stored at the level above by the energy stored at the level below. Then you multiply by 100, and the answer is the transfer as a percentage.
transfer (%): the share of the lower level’s stored energy that the level above stored, as a percentage
energy at the level above: the energy stored at the higher of the two trophic levels, in kJ per square meter a year
energy at the level below: the energy stored at the lower of the two trophic levels, in kJ per square meter a year
× 100: turns the share into a percentage
Video: Watch: Measure the transfer as a percentage
The two Silver Springs slices with their measured values; the equation written on its own line with its key; the values substituted line by line to 14.5 %; the result set beside the 10 % working rule and inside the 5 % to 20 % range.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L24a.mp4
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Here is Silver Springs again. It is a spring in Florida: clear water wells up from the ground and flows away as a river.
Water plants grow on the spring’s bed. Insects and snails eat the plants, and fish eat the insects and snails.
Ecologists measured the energy each level stored in new growth over one year, for each square meter of the spring.
The plants stored about 31 900 kJ per square meter a year. The insects and snails stored about 4 620 kJ per square meter a year.
Suppose the insects and snails had stored half of what the plants stored. Then the transfer would be 50 %.
Suppose they had stored much less, 10 % of what the plants stored. Then the transfer would be 10 %, the working rule’s figure.
The transfer is the share of the lower level’s stored energy that the level above stored. The equation below gives that share as a percentage, with its key beneath.
transfer (%): the share of the lower level’s stored energy that the level above stored, as a percentage
energy at the level above: the energy stored at the higher of the two trophic levels, in kJ per square meter a year
energy at the level below: the energy stored at the lower of the two trophic levels, in kJ per square meter a year
× 100: turns the share into a percentage
Both energies are in kilojoules per square meter a year. When you divide one by the other, the two units cancel, as the line below shows.
the same unit above and below the line cancels, so the share has no unit; multiplying by 100 writes that share as a percentage
So the share has no unit. Multiplying by 100 writes the share as a percentage.
At Silver Springs the plants stored about 31 900 kJ per square meter a year, and the insects and snails that ate them stored about 4 620 kJ per square meter a year. Calculate the transfer between the two levels, to one decimal place.
The transfer at Silver Springs came out at 14.5 %. So the insects and snails stored a little more than the working rule’s tenth.
14.5 % sits inside the range of measured transfers, about 5 % to about 20 %. So the working rule held roughly, as it usually does.
Now consider a lake. Suppose its algae store 11 600 kJ per square meter a year, and the water fleas that eat the algae store 812 kJ per square meter a year.
The lake’s algae store 11 600 kJ per square meter a year, and its water fleas store 812 kJ per square meter a year. Calculate the transfer between the two levels, to one decimal place.
The lake’s transfer is below the working rule, and the spring’s transfer is above it. The two transfers both sit inside the measured range.
The table below sets the two measured transfers beside the 10 % working rule.
What you are expected to know Calculate the percentage of energy transferred between two trophic levels from measured values, and compare it with the 10 % working rule.
An ecologist measures the energy stored at two trophic levels of a pond, one directly above the other, both in kJ per square meter a year.
Which of the following gives the transfer between the two levels as a percentage?
- A. ✓ The upper level’s energy ÷ the lower level’s energy × 100
- B. The lower level’s energy ÷ the upper level’s energy × 100Dividing the lower level’s energy by the upper level’s gives a number above 100.
The transfer is the share of the lower level’s energy that the upper level stored. - C. The lower level’s energy − the upper level’s energySubtracting gives the energy that is missing, in kJ per square meter a year.
The transfer is a share, so it is a division, times 100.
Why: The transfer is the share of the lower level’s energy that the upper level stored.
A share is the part divided by the whole.
So the transfer is the upper level’s energy divided by the lower level’s energy, times 100.
Suppose a fen’s sedges store 9 640 kJ per square meter a year, and the insects that eat the sedges store 1 254 kJ per square meter a year.
Substitute the values into the equation and calculate the transfer between the two levels, to one decimal place. Answers within 0.05 % are accepted.
Part 1. Write down the values in the question. What is the energy stored by the sedges, the producers?
Answer: 9640 kJ per square meter a year (tolerance ±0)
Part 2. What is the energy stored by the insects, the primary consumers?
Answer: 1254 kJ per square meter a year (tolerance ±0)
Answer: 13 % (tolerance ±0.05)
A survey of a heath measures a transfer of 12.3 % between its heather and the insects that eat it.
Is the heath’s transfer above or below the 10 % working rule?
- A. ✓ Above the working rule
- B. Below the working ruleA transfer below the working rule is less than 10 %.
12.3 % is more than 10 %.
Why: The working rule is 10 %.
12.3 % is more than 10 %.
So the heath’s transfer is above the working rule.
Suppose a forest’s trees store 6 250 kJ per square meter a year in new leaves and wood, and the caterpillars that eat the leaves store 425 kJ per square meter a year.
Calculate the transfer between the two levels, to one decimal place. Answers within 0.05 % are accepted.
Answer: 6.8 % (tolerance ±0.05)
The transfer between a pond’s algae and the snails that eat them is measured at 15 %. A student says: “So the snails lost 15 % of the algae’s energy.”
Is the student correct?
- A. Yes: the transfer is the share of the energy that was lostThe transfer is the share of the algae’s energy that the snails stored.
The missing share is the rest: 85 %. - B. ✓ No: the snails stored 15 % of the algae’s energy, and 85 % is missing
Why: The transfer is the energy at the level above divided by the energy at the level below, times 100.
So 15 % is the share of the algae’s energy that the snails stored.
The other 85 % is the share that is missing from the snails’ level.
29Why the chain stops
A stream’s producers store a known amount of energy, in kilojoules per square meter a year.
By the 10 % rule, which of the following gives the energy stored at the primary consumers’ level?
- A. The producers’ energy × 10Multiplying by 10 makes the next level store ten times more.
The next level stores about 10 % of this level’s energy. As a decimal, that share is 0.10. - B. The producers’ energy × 0.900.90 is the share that is missing from the next level.
The next level stores about 10 % of this level’s energy. As a decimal, that share is 0.10. - C. ✓ The producers’ energy × 0.10
Why: The 10 % rule says the next level stores about 10 % of this level’s energy.
10 % written as a decimal is 0.10.
So the primary consumers’ energy is the producers’ energy multiplied by 0.10.
Why does a food chain rarely have more than four or five levels?
Video: Watch: Why the chain stops
The river’s pyramid built from 100 000 kJ at the bottom; each new slice a tenth as wide as the one below; the fourth, fifth and sixth slices thinning to hairlines; the words too little to feed a population written beside the empty top.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L24b.mp4
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Now consider a river. Suppose its producers store 100 000 kJ per square meter a year.
By the 10 % rule, the primary consumers store about 10 000 kJ per square meter a year. The table below applies the rule level by level, up to a sixth level.
By the fifth level, only about 10 kJ per square meter a year is left. By a sixth level, only about 1 kJ per square meter a year would be left.
Here is the river’s energy pyramid, drawn up to the sixth level. The top three slices are too thin to see, so the drawing shows them as hairlines.
A population needs energy every year to keep its members alive, to grow and to raise young.
About 10 kJ per square meter a year is too little to keep a population of predators alive and breeding. About 1 kJ per square meter a year is far too little.
So no population can live at the sixth level of this river, and the fifth level is already close to the limit.
That is why a food chain rarely has more than four or five levels: after each 90 % loss, there is too little energy left to feed another level.
An ecosystem whose producers store more energy starts with more at the bottom. Its chain can reach about one level more before too little energy is left.
The energy pyramid stops where the energy does.
What you are expected to know Explain why a food chain rarely has more than four or five levels: after each 90 % loss there is too little energy left to feed another level.
Suppose a reservoir’s producers store 52 400 kJ per square meter a year. By the 10 % rule, the fifth level of its food chain stores about 5 kJ per square meter a year.
Could a population at a sixth level be sustained in this reservoir?
- A. YesA sixth level would store about 0.5 kJ per square meter a year.
That is far too little to keep a population alive and breeding. - B. ✓ No
Why: The fifth level stores about 5 kJ per square meter a year.
By the 10 % rule, a sixth level would store about 0.5 kJ per square meter a year.
That is too little energy to keep a population alive and breeding.
A survey of a lake finds a food chain of four levels: algae, water fleas, minnows and pike. Nothing in the lake hunts the pike.
(a) Explain why the lake’s chain is unlikely to gain a fifth level, even though nothing hunts the pike. (2 pt)
Frame The chain is unlikely to gain a fifth level because …
The pike are four levels up, so their level stores a small share of the algae’s energy.
A fifth level would store about 10 % of that.
A population needs energy every year to stay alive, grow and raise young.
So too little energy is left to feed a population above the pike.
- Award 1 point for: each level stores only about 10 % of the level below, so very little energy is left by the pike’s level.
- Award 1 point for: a fifth level would have too little energy to feed a population (accept: to keep its members alive and breeding).
A student looks at a marsh whose food chain ends with herons and says: “The chain stops at the herons because nothing in the marsh is big enough to eat a heron.”
Is the student correct?
- A. Yes: a chain ends when the top animal has no predatorA larger animal could eat herons.
The chain stops because too little energy is left at the herons’ level to feed a population of them. - B. ✓ No: too little energy is left at the herons’ level to feed another population
Why: Each level stores only about 10 % of the level below.
By the herons’ level very little energy is left.
So too little energy is left to feed a population of anything that eats herons.
Suppose one estuary’s producers store five times more energy per square meter a year than a nearby bay’s producers.
By the 10 % rule, how many more trophic levels can the estuary’s food chain support than the bay’s?
- A. ✓ At most about one more level
- B. About five more levelsFive times more energy at the bottom does not add five levels.
Each level keeps only about 10 %, so five times more is used up within one extra level. - C. The same number of levels, whatever the producers storeMore energy at the bottom means more energy left at every level above.
So the estuary’s chain can reach about one level more.
Why: Each level stores about 10 % of the level below.
Five times more energy at the bottom leaves five times more at every level.
Five times more is less than one tenfold step between levels.
So the estuary’s chain can reach at most about one level more.
Here is Silver Springs again, with about 31 900 kJ per square meter a year stored in its plants and about 4 620 kJ per square meter a year in the insects and snails that ate them.
The transfer came out at 14.5 %, above the 10 % working rule and inside the measured range.
Each level above keeps only a small share of the one below. So a fifth level above the plants would have too little energy left to feed a population.
51Mixed practice mixed practice
The table below gives the measured transfer between the producers and the primary consumers in four ecosystems.
Which ecosystem’s transfer sits above the 10 % working rule?
- A. The pondThe pond’s transfer is 8.2 %.
8.2 % is less than 10 %. - B. ✓ The seagrass bed
- C. The forestThe forest’s transfer is 5.9 %.
5.9 % is less than 10 %. - D. The coral reefThe coral reef’s transfer is 9.1 %.
9.1 % is less than 10 %.
Why: The working rule is 10 %.
Only the seagrass bed’s transfer, 16.4 %, is more than 10 %.
So the seagrass bed’s transfer sits above the working rule.
A pond’s food chain is algae, water fleas, minnows and perch. A student expects a fifth level, an eagle that eats perch. The survey finds only the four levels.
Which of the following explains why the chain ends at the perch?
- A. The perch respire all the energy in their food, so none is stored for an eagleThe perch store about 10 % of the minnows’ energy, so some energy is stored at their level.
The trouble is how little that is. - B. The perch have no predator, so the chain cannot continueA chain does not need a predator to be present to continue.
It stops where too little energy is left to feed another population. - C. ✓ Too little energy is stored at the perch’s level to feed a population of eagles
Why: Each level stores only about 10 % of the level below.
The perch are four levels up, so their level stores very little energy.
So too little energy is left to feed a population of eagles.
A survey measures a transfer of 7 % between a river’s algae and the snails that eat them.
Does this transfer sit inside the range of measured transfers, about 5 % to about 20 %?
- A. ✓ Yes
- B. NoThe range runs from about 5 % to about 20 %.
7 % is more than 5 % and less than 20 %.
Why: The measured range runs from about 5 % to about 20 %.
7 % is more than 5 % and less than 20 %.
So this transfer sits inside the range.
The transfer between a meadow’s grass and its rabbits is measured at 8 %.
Which share of the grass’s stored energy is missing from the rabbits’ level?
- A. 8 %8 % is the share the rabbits stored.
The missing share is the rest of the 100 %. - B. ✓ 92 %
- C. 100 %The rabbits stored 8 % of the grass’s energy, so not all of it is missing.
Only the rest of the grass’s energy is missing.
Why: The transfer is the share of the grass’s energy that the rabbits stored: 8 %.
The rest of the grass’s energy is missing from the rabbits’ level.
So the missing share is 92 %.
The table below gives the measured transfer at two steps of a canal’s food chain: pondweed, then the water snails that eat it, then the roach that eat the snails.
At which step did the level above store the larger share of the level below’s energy?
- A. ✓ Pondweed to water snails
- B. Water snails to roachThe transfer from water snails to roach is 9 %.
9 % is a smaller share than 12 %. - C. The two shares are the sameThe two transfers are 12 % and 9 %.
12 % is the larger share.
Why: A transfer is the share of the level below’s energy that the level above stored.
The pondweed-to-water-snails transfer is 12 %, and the water-snails-to-roach transfer is 9 %.
So the water snails stored the larger share.
Suppose a tundra’s producers store only 20 % of the energy a rainforest’s producers store, per square meter a year.
By the 10 % rule, which food chain is likely to have fewer trophic levels?
- A. The rainforest’sThe rainforest starts with more energy at the bottom, so more is left at every level above.
Its chain reaches too little energy later, not sooner. - B. Neither: the two chains have the same number of levelsThe number of levels depends on how much energy is left to feed each level.
The tundra starts with less, so it reaches too little sooner. - C. ✓ The tundra’s
Why: Each level stores about 10 % of the level below.
The tundra starts with 20 % of the rainforest’s energy, so every level above has 20 % as much.
So the tundra’s chain reaches too little energy sooner, and it has fewer levels.
Ecologists survey a brook. Its algae store 15 720 kJ per square meter a year, and the stonefly larvae that eat the algae store 1 886 kJ per square meter a year. The brook’s food chain has four levels: algae, stonefly larvae, trout and otters.
(a) Calculate the transfer between the algae and the stonefly larvae, to one decimal place. Answers within 0.05 % are accepted. (1 pt)
Answer: 12 % (tolerance ±0.05)
- Award 1 point for: 12.0 % (accept 11.9 % to 12.1 %).
(b) Explain why a fifth level, above the otters, is unlikely to be sustained in this brook. (2 pt)
Frame A fifth level is unlikely because …
The otters are four levels up, so their level stores only a very small share of the algae’s energy.
A fifth level would store about 10 % of that.
A population needs energy every year to stay alive, grow and raise young.
So too little energy would be left to feed a population at a fifth level.
- Award 1 point for: each level stores only about 10 % of the level below, so very little energy is left by the otters’ level.
- Award 1 point for: a fifth level would have too little energy left to feed a population.
APBIO-U08-L25 Less sun, fewer hawks, and why maize beats beef
Suppose a year of volcanic haze begins. The light reaching a lake’s algae falls by a third. Nothing else changes.
What happens to the pike at the top of the lake’s food chain?
Unit 8 · Ecology
1Less light at the bottom, fewer pike at the top
Two trophic levels sit one above the other in an energy pyramid.
By the 10 % rule, about how much of the energy stored in the lower level is stored in the level above it?
- A. ✓ About 10 %
- B. About 50 %Half the energy would make each slice half the width of the one below.
The 10 % rule makes each slice a tenth as wide. - C. About 90 %About 90 % of the energy is respired and leaves as heat, or is never eaten, or goes to decomposers.
Only about 10 % is stored in the level above.
Why: Most of the energy stored at one level is respired and leaves as heat.
Some is never eaten, and some goes to decomposers.
So only about 10 % of the energy is stored in the next trophic level up.
Energy pyramids rarely have more than four or five trophic levels.
Why does a food chain rarely have more than four or five trophic levels?
- A. The consumers at the top of the chain have no predators to eat themA top consumer with no predator could still be eaten by a new, larger predator.
The chain stops because too little energy is left for one. - B. ✓ After each 90 % loss, too little energy is left for another level
- C. The organisms at each level grow too large for anything to eat themSize does not set the trophic level.
The chain stops because too little energy is left to feed another level.
Why: Each trophic level stores only about 10 % of the level below.
After four or five such steps, very little energy is left.
Too little energy is left to feed a population at another level, so the chain stops.
What happens to the trophic levels above when less energy enters an ecosystem at the bottom?
The change passes up the energy pyramid level by level. Each trophic level stores only about 10 % of the level below it, so when the bottom level shrinks, every level above it shrinks too.
The top level had the least energy to begin with. So the top level is the first that may disappear.
Video: Watch: Less light at the bottom, fewer pike at the top
The lake’s energy pyramid in a normal year, the algae at the base and the pike at the top. The sun dims behind haze, the base slice narrows, and every slice above it narrows in turn until the pike’s slice is barely there.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L25a.mp4
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Suppose a lake’s food chain is algae, zooplankton, small fish and pike.
Tiny animals drift in the water and eat the algae. These tiny drifting animals are called zooplankton.
Small fish eat the zooplankton. Pike eat the small fish.
By the end of a normal summer, the algae in the lake hold 9 600 kJ of energy.
The pyramid below shows the lake’s four trophic levels. The energy stored in each level is written on or beside its slice.
The top two slices are so thin that the drawing shows them as hairlines. Their true widths would be too small to see.
Now suppose a year of volcanic haze begins. The light reaching the lake’s algae falls by a third, and nothing else changes.
With less light, the algae make less sugar over the summer. So by the end of the summer the algae hold a third less energy: 6 400 kJ instead of 9 600 kJ.
The zooplankton eat the algae. The zooplankton store only about 10 % of what the algae hold, so the zooplankton’s level shrinks by a third too.
The same happens at every level up. The small fish store about 10 % of the zooplankton’s energy, and the pike store about 10 % of the small fish’s energy.
The table below compares the four trophic levels in the normal year and in the hazy year.
Here are the two pyramids drawn to one scale. The whole pyramid shrinks from the bottom up.
Every level lost the same third. But the pike’s level had the least energy to begin with: 9.6 kJ in the normal year, and 6.4 kJ in the hazy year.
A pike population needs enough energy in its level to feed every pike. If 6.4 kJ is too little for that, the pike disappear from the lake.
Fewer producers, or smaller producers, do the same as less light. Either way, less energy enters the pyramid at the bottom, so every trophic level above it shrinks.
More light, or more producers, does the opposite. More energy enters the pyramid at the bottom, so every trophic level above it grows.
What you are expected to know Predict how a change in the energy entering an ecosystem, from less sunlight or from fewer or smaller producers, changes the size of the trophic levels above it.
Suppose a river’s food chain is algae, insect larvae, sticklebacks and kingfishers. A new dam upstream makes the water muddy, so less light reaches the algae. Nothing else changes.
What happens to the number of kingfishers the river can feed?
- A. ✓ Fewer kingfishers
- B. The same number of kingfishersThe kingfishers’ level stores about 10 % of the sticklebacks’ energy.
The sticklebacks store less, so the kingfishers store less. - C. More kingfishersMore kingfishers would need more energy at the top.
Less light at the bottom means less energy at every level.
Why: Less light reaches the algae, so the algae store less energy.
Each level above stores only about 10 % of the level below.
So the insect larvae, the sticklebacks and the kingfishers all store less.
The river can feed fewer kingfishers.
Suppose a shallow sea bay’s food chain is seaweed, sea snails, crabs and gulls. For one year a river carries silt into the bay, and the cloudy water cuts the light reaching the seaweed.
(a) Predict how the number of gulls the bay can feed changes by the end of the year. (1 pt)
Frame By the end of the year the bay can feed …
- Award 1 point for: fewer gulls.
(b) Justify your prediction using the 10 % rule. (2 pt)
Frame Less light reaches the seaweed, so …
The sea snails store only about 10 % of the seaweed’s energy, so they store less too.
The crabs store about 10 % of the sea snails’ energy, and the gulls about 10 % of the crabs’ energy.
So the loss passes up level by level to the gulls, whose level had the least energy to begin with.
Therefore the bay can feed fewer gulls.
- Award 1 point for: less light means the seaweed stores less energy, and each level stores only about 10 % of the level below it.
- Award 1 point for: the loss passes up every level to the gulls, whose level had the least energy to begin with, so fewer gulls can be fed.
Suppose a haze cuts the light reaching a lake’s algae for a year. A student says: “The haze changes only the algae, because only the algae use light.”
Is the student correct?
- A. Yes: the animals in the lake eat each other, not light, so the haze leaves them unchangedThe animals do not use light, but they eat what the light grew.
Less energy in the algae means less energy for every level above. - B. ✓ No: less energy in the algae means less energy at every trophic level above them
Why: The zooplankton eat the algae.
With less energy in the algae, the zooplankton store less.
Each level above stores about 10 % of the level below.
So the loss passes up to the small fish and to the pike.
Suppose a reed bed’s food chain is reeds, reed-eating insects, frogs and herons. One spring, farmers cut half of the reed bed’s reeds. The light is unchanged.
What happens to the number of herons the reed bed can feed?
- A. More heronsMore herons would need more energy at the top of the pyramid.
Fewer reeds put less energy in at the bottom. - B. The same number of heronsThe light is unchanged, but half the reeds are gone.
Fewer reeds store less energy, and the loss passes up every level. - C. ✓ Fewer herons
Why: Half the reeds are gone, so the producers store less energy.
Each level above stores only about 10 % of the level below.
So the insects, the frogs and the herons all store less.
The reed bed can feed fewer herons.
29Why maize feeds more people as maize than as beef
You know the energy stored in one trophic level.
To find the energy stored in the next trophic level up, what do you multiply the energy at this level by?
- A. ✓ 0.10
- B. 0.90Multiplying by 0.90 gives the energy that never reaches the level above.
The level above stores only about 10 %. - C. 10Multiplying by 10 makes the level above ten times larger.
Each level above stores less, not more.
Why: Each trophic level stores about 10 % of the level below.
10 % written as a decimal is 0.10.
So you multiply the energy at this level by 0.10.
Why does a field of maize feed more people eaten as maize than fed to cattle first?
Eaten directly, the maize is one trophic level below the people. Fed to cattle first, the maize is two trophic levels below the people, and each level keeps only about 10 % of the energy.
So every extra trophic level between the producers and the people leaves only 10 % of the energy for the people.
Video: Watch: Why maize feeds more people as maize than as beef
One field of maize with a slice for the people above it, 840 kJ. Beside it the same field with a slice for the cattle, 840 kJ, and a hairline slice for the people above that, 84 kJ.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L25b.mp4
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Suppose a field of maize holds 8 400 kJ of energy in its grain.
Now consider two ways to eat that field.
In one way, people eat the maize. In the other, cattle eat the maize, and then people eat the beef.
Either way, the people are a trophic level. Here is the equation from the 10 % rule again, on its own line with its key.
energy at the next level: the energy stored in the trophic level one step up, in kJ
energy at this level: the energy stored in the trophic level you start from, in kJ
0.10: 10 % written as a decimal; it has no unit
People eat the field’s 8 400 kJ of maize directly. Calculate the energy stored in the people’s level.
Eaten directly, the people’s level stores 840 kJ.
Cattle eat the field’s 8 400 kJ of maize, and people eat the beef. Calculate the energy stored in the people’s level.
Fed to cattle first, the people’s level stores 84 kJ.
The table below compares maize eaten directly with maize fed to cattle first, trophic level by trophic level.
Here are the two pyramids drawn to one scale, with their bases level.
Eaten directly, the people’s slice is 840 kJ. Fed to cattle first, the people’s slice is 84 kJ.
So the same field stores ten times more energy in people when the people eat the maize themselves. Ten times more energy feeds about ten times more people.
What you are expected to know Explain, using the 10 % rule, why the same producers feed more people eaten directly than fed to cattle first.
Suppose a field of wheat can be eaten in two ways. People can eat bread made from the wheat. Or pigs can eat the wheat, and people eat the pork.
Which way stores more energy in the people?
- A. Eating the porkThe pigs are an extra trophic level between the wheat and the people.
Each level stores only about 10 % of the level below. - B. ✓ Eating the bread
Why: Eaten as bread, the wheat is one level below the people.
Eaten as pork, the wheat is two levels below the people.
Each level keeps only about 10 % of the energy.
So the bread stores more energy in the people.
Suppose a village grows rice. In one year the villagers eat the rice. In the next year they feed the same amount of rice to chickens and eat the chickens.
(a) Explain, using the 10 % rule, why the rice feeds fewer villagers in the second year. (2 pt)
Frame In the first year the villagers store about …
In the second year the chickens store about 10 % of the rice’s energy.
The villagers then store about 10 % of the chickens’ energy.
So in the second year the villagers store only about 10 % of what they stored in the first year.
Therefore the same rice feeds fewer villagers in the second year.
- Award 1 point for: the chickens are an extra trophic level between the rice and the villagers, and each level stores only about 10 % of the level below.
- Award 1 point for: so the villagers store about 10 % of what they stored when they ate the rice directly, and the rice feeds fewer villagers.
Suppose a field of maize holds 15 500 kJ of energy in its grain.
Now suppose cattle eat all of that maize, and people eat the beef. Calculate the energy stored in the people’s level.
Part 1. Suppose people eat all of that maize directly. Calculate the energy stored in the people’s level.
Answer: 1550 kJ (tolerance ±0)
Answer: 155 kJ (tolerance ±0)
Suppose a catfish farm feeds grain to its catfish, and people eat the catfish.
By the 10 % rule, compared with people eating the grain themselves, about how much energy reaches the people through the catfish?
- A. About ten times as muchTen times as much would need the catfish to add energy.
The catfish respire most of the grain’s energy, and it leaves as heat. - B. About the sameThe catfish are an extra trophic level.
Each level stores only about 10 % of the level below, so the energy reaching the people falls. - C. ✓ About 10 % as much
Why: Eaten directly, the grain is one level below the people.
Through the catfish, the grain is two levels below the people.
The catfish store about 10 % of the grain’s energy, and the people about 10 % of the catfish’s.
So about 10 % as much energy reaches the people.
Here is the hazy lake again, with a third less light reaching its algae.
Fewer algae means fewer zooplankton, fewer small fish and fewer pike.
And here is the field of maize again, eaten two ways: 840 kJ in the people who eat the maize, 84 kJ in the people who eat the beef.
Each extra trophic level keeps only 10 % of the energy. That is why a field feeds ten times more people as maize than as beef.
56Mixed practice mixed practice
In Lake Ontario, snails eat green algae, a small fish called the slimy sculpin eats the snails, and Chinook salmon eat the sculpin. Suppose a string of cloudy summers cuts the light reaching the algae.
Which trophic level is the most likely to disappear from the lake?
- A. The green algaeThe algae lose energy first, but they had the most to begin with.
A third less of the largest level still leaves the most energy. - B. The snailsThe snails store about 10 % of the algae’s energy, so they lose energy too.
The salmon, two levels higher, had far less to begin with. - C. ✓ The Chinook salmon
Why: Every level stores less when less light reaches the algae.
The salmon’s level had the least energy to begin with.
So the salmon’s level is the first that may hold too little energy to feed a population.
Suppose a field of barley holds 7 300 kJ of energy in its grain. Cattle eat the barley, and people eat the beef.
By the 10 % rule, calculate the energy stored in the people’s level.
Answer: 73 kJ (tolerance ±0)
Suppose a ditch’s food chain is duckweed, tadpoles and dragonfly larvae. Over a decade of sunnier summers, more light reaches the duckweed. Nothing else changes.
What happens to the energy stored in the dragonfly larvae’s level?
- A. ✓ The dragonfly larvae’s level stores more energy
- B. The dragonfly larvae’s level stores the same energyThe dragonfly larvae store about 10 % of the tadpoles’ energy.
The tadpoles store more, so the dragonfly larvae store more. - C. The dragonfly larvae’s level stores less energyLess energy at the dragonfly larvae’s level would follow less light.
Here more light reaches the duckweed.
Why: More light reaches the duckweed, so the duckweed stores more energy.
The tadpoles store about 10 % of the duckweed’s energy, so the tadpoles store more.
The dragonfly larvae store about 10 % of the tadpoles’ energy.
So the dragonfly larvae’s level stores more energy.
A student says: “Feeding grain to cattle and eating the beef gives people more energy than eating the grain, because beef is richer food than grain.”
Is the student correct?
- A. Yes: the beef holds more energy than the grain the cattle ate heldThe cattle respire most of the grain’s energy, and it leaves as heat.
The beef holds only about 10 % of the grain’s energy. - B. ✓ No: the cattle are an extra level, so people get about 10 % of the grain’s energy
Why: The cattle are a trophic level between the grain and the people.
The cattle store only about 10 % of the grain’s energy.
The people store about 10 % of the cattle’s energy.
So the people get less energy from the beef than from the grain.
Suppose a hillside’s food chain is shrubs, shrub-eating insects, lizards and hawks. Goats graze the shrubs down to half their size. The light is unchanged.
What happens to the energy entering the hillside’s pyramid at the producer level?
- A. ✓ Less energy enters
- B. The same energy entersThe light is unchanged, but the shrubs are half their size.
Smaller producers capture less of that light. - C. More energy entersMore energy would need larger or more numerous producers.
The shrubs are half their size.
Why: The shrubs are the hillside’s producers.
Half-sized shrubs have half the leaf to capture light.
So the shrubs store less energy, and less energy enters the pyramid at the bottom.
Suppose a country wants its farmland to feed as many people as possible.
By the 10 % rule, which change makes the same fields feed more people?
- A. Feed more of the grain to cattle and eat the beefThe cattle are an extra trophic level.
Each extra level leaves only about 10 % of the energy for the people. - B. ✓ Eat more of the grain directly
- C. Feed the grain to chickens instead of cattleChickens keep more of the grain’s energy than cattle do, but they are still an extra trophic level.
Eating the grain directly feeds more people.
Why: Eaten directly, the grain is one level below the people.
Fed to any animal first, the grain is two levels below the people.
Each level keeps only about 10 % of the energy.
So eating more of the grain directly feeds more people.
Suppose a creek’s food chain is algae, water snails and sunfish. In one summer, mud from a building site clouds the water. The table below gives the energy stored in each trophic level by the end of a normal summer and by the end of the muddy summer.
(a) Make a claim about how the number of sunfish the creek can feed changed in the muddy summer. (1 pt)
Frame In the muddy summer the creek could feed …
- Award 1 point for: fewer sunfish.
(b) Support your claim with evidence from the table. (1 pt)
Frame The sunfish’s level stored …
So the sunfish’s level stored half as much energy to feed sunfish.
- Award 1 point for: the sunfish's level stored 41 kJ in the normal summer and 20.5 kJ in the muddy summer (or: half as much).
(c) Explain why less light reaching the algae changed the energy stored in the sunfish’s level. (2 pt)
Frame Less light reached the algae, so …
The water snails store only about 10 % of the algae’s energy, so they stored less.
The sunfish store only about 10 % of the water snails’ energy, so they stored less.
So the loss passed up the food chain, level by level, to the sunfish.
- Award 1 point for: less light means the algae store less energy.
- Award 1 point for: each level stores only about 10 % of the level below, so the loss passes up through the water snails to the sunfish.
APBIO-U08-L26 Explain the result by following the energy
Photo: Sam Fraser-Smith, Wikimedia Commons, CC BY 2.0 (resized).
In one class’s experiment, the students fed cabbage leaves to cabbage white butterfly larvae for a week. Before the week, they measured the mass of the leaves on a balance. After the week, they measured the mass the larvae had gained.
The larvae ate 80 g of leaf. The larvae gained 4 g. That is 5 % of the leaf they ate. Where did the other 76 g go?
Unit 8 · Ecology
1Where the other 76 g went
A grasshopper eats grass.
Where does most of the energy in the grass it digests end up?
- A. Stored in the grasshopper’s new tissueOnly about 10 % of the energy at one trophic level is stored in the next.
The grasshopper respires most of what it digests. - B. ✓ Given off as heat after the grasshopper respires it
- C. Passed on to the bird that eats the grasshopperThe bird gets only the energy stored in the grasshopper’s tissue.
Most of the grass’s energy has already left as heat.
Why: The grasshopper respires most of what it digests to make ATP.
The energy in the respired sugar leaves as heat.
So most of the grass’s energy ends up as heat.
A meadow’s grass makes sugar by photosynthesis and respires some of that sugar.
Which of the following is the meadow’s net primary productivity?
- A. What the producers made plus what they respiredThe respired sugar left the meadow as heat.
Adding it back includes energy that is no longer there. - B. What the producers respired minus what they madeThat order gives a negative number.
The net is what remains after the respired part is taken away from what was made. - C. ✓ What the producers made minus what they respired
Why: The grass respired part of what it made, and that energy left as heat.
What remains is the net: what the producers made minus what they respired.
A caterpillar’s body temperature follows the temperature of its surroundings.
Which is the caterpillar?
- A. An endothermAn endotherm keeps its body temperature with heat from its own metabolism.
The caterpillar’s temperature follows its surroundings. - B. ✓ An ectotherm
Why: An animal whose body temperature follows its surroundings is an ectotherm.
An enzyme’s solution is cooled to a temperature lower than the enzyme’s optimal temperature.
What happens to the rate of the enzyme’s reaction?
- A. ✓ It falls
- B. It stays the sameCooling makes the molecules move more slowly.
Enzyme and substrate collide less often. - C. It risesWarming raises the rate up to the optimal temperature.
Cooling below it does the opposite.
Why: Cooling a solution makes its molecules move more slowly.
Enzyme and substrate collide less often.
So the rate falls.
How do you explain a mass result by following the energy?
The larvae respired most of the leaf they digested, to make ATP for moving and for building their bodies. The energy in the respired sugar left as heat.
Some of the leaf passed through the larvae undigested.
Only what was left became new larva.
Larvae kept cold respire less. So a little more of the leaf they eat becomes larva.
To explain a result like this, you make a claim, give the evidence from the measurements, and write the reasoning that joins them.
Video: Watch: Where the other 76 g went
The 80 g of leaf as one bar, splitting three ways as each part is named: respired and left as heat, passed through undigested, built into new larva.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L26a.mp4
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Here is the class’s result again: 80 g of leaf eaten, 4 g of new larva, and 76 g to account for.
Every mass here is a dry mass. The class left out the water in the leaf and in the larvae, so the grams count only the leaf’s own matter.
The larvae did three things with the leaf they ate.
1. Some of the leaf passed straight through the larvae’s guts. The larvae never digested it, and it left them as droppings.
In this class’s experiment, 32 g of the 80 g passed through as droppings.
2. The larvae digested the rest of the leaf and absorbed it into their cells. Their cells respired most of it.
Respiration made ATP. The larvae used the ATP for moving, for feeding and for building new body.
The energy in the respired sugar left the larvae as heat.
The atoms of the respired sugar left too, as carbon dioxide and water. So respiration took mass out of the larvae as well as energy.
In this class’s experiment, the larvae respired 44 g of the 80 g.
3. The larvae built what was left into new body: new muscle, new gut, new skin. That is the 4 g the class measured.
Here is the whole 80 g as one bar, split three ways.
The table below compares the three parts: what happened to each, its mass, and where its energy went.
This is net primary productivity’s subtraction again, with one more part taken away. The equation below shows it in grams.
new larva: the mass the larvae built into their bodies, in g
leaf eaten: the mass of leaf the larvae ate, in g
leaf respired: the mass of leaf whose atoms left the larvae as carbon dioxide and water, in g
leaf passed through: the mass of leaf that left the larvae undigested, as droppings, in g
Now follow the energy to its end.
The energy in the 44 g the larvae respired left as heat. That heat spread into the room, and no living thing can eat it.
The energy in the 32 g of droppings is still chemical energy. Decomposers in the soil will respire it.
The energy in the 4 g of new larva is stored in the larvae. A bird that eats a larva gets only that part.
So the class’s result fits what energy flow predicts. Most of the leaf’s energy left the larvae, and only a small part was stored in the next trophic level.
In this class’s data, respiration took the biggest share. Other classes find that more of the leaf passes through as droppings, and that 10 % to 20 % of it becomes larva.
To explain a measured result, write three things.
1. The claim: what the result shows.
2. The evidence: the measurements that support the claim.
3. The reasoning: the biology that joins the evidence to the claim.
The table below shows the three, written for the class’s larvae.
Now imagine the same larvae kept in a cold room for the week.
A larva is an ectotherm. Its body temperature follows the room’s, so its cells are colder.
Below their optimal temperature, enzymes work more slowly. So colder cells respire more slowly.
So the cold larvae respire less of the leaf they eat. Less of the leaf’s energy leaves as heat.
A larger share of the leaf they eat is left to build into new larva. The 5 % rises a little.
Cold larvae also eat less and grow more slowly. So they gain fewer grams in the week, but a larger share of what they eat becomes larva.
The table below compares the warm room and the cold room.
What you are expected to know Explain a measured result about mass in terms of where the energy went: respired and left as heat, passed through undigested, or built into new tissue.
Suppose a class fed 50 g of grass to grasshoppers for a week. The grasshoppers respired most of the grass they digested.
Where did the energy of the grass they respired end up?
- A. Stored in the grasshoppers’ new bodyThe grasshoppers store only the part they build into new body.
The respired part is gone from them. - B. ✓ As heat, spread into the surroundings
- C. In the grasshoppers’ droppingsThe droppings are the grass that passed through undigested.
The respired grass left as carbon dioxide and water.
Why: The grasshoppers respired the grass to make ATP.
The energy in the respired sugar left them as heat.
That heat spread into the surroundings.
Suppose a class fed 75 g of lettuce leaves to garden snails for a week. At the end of the week the snails had gained 3 g.
(a) Make a claim about what happened to most of the lettuce the snails ate. (1 pt)
- Award 1 point for: most of the lettuce was respired (its energy left as heat) rather than built into snail.
(b) Support the claim with the evidence from the class’s measurements. (1 pt)
The snails gained only 3 g.
That is 4 % of the lettuce they ate, so most of the lettuce did not become snail.
- Award 1 point for: 3 g gained out of 75 g eaten (4 %), so most of the lettuce did not become snail.
(c) Explain the reasoning that joins the evidence to the claim. (2 pt)
Frame The snails digested the lettuce and …
Their cells respired most of it to make ATP.
The energy in the respired sugar left the snails as heat.
The atoms of the respired sugar left as carbon dioxide and water, so that mass left the snails.
Some lettuce passed through undigested as droppings.
Only the part built into new body stayed as snail mass.
- Award 1 point for: the snails respired most of what they digested, and the energy left as heat (the carbon dioxide and water carrying the mass away).
- Award 1 point for: some lettuce passed through undigested, and only the part built into new body was measured as mass gained.
Suppose a class fed 30 g of clover to caterpillars for a week, and the caterpillars gained 2 g. A student says: “The other 28 g of clover was destroyed inside the caterpillars.”
Is the student correct?
- A. ✓ No: the mass left the caterpillars as carbon dioxide, water and droppings
- B. Yes: respiration destroys the food that is respiredRespiration rearranges the atoms of sugar into carbon dioxide and water.
No atom is destroyed; the carbon dioxide and water leave the caterpillars.
Why: The caterpillars respired most of the clover they digested.
The atoms of that clover left as carbon dioxide and water.
The rest of the missing clover passed through as droppings.
So no clover was destroyed; its mass left the caterpillars.
Suppose a thrush eats garden snails that have been eating lettuce.
From which part of the lettuce the snails ate can the thrush get energy?
- A. The part the snails respiredThe energy of the respired lettuce left the snails as heat.
The thrush cannot eat heat. - B. The part that passed through the snails as droppingsThe droppings left the snails.
The thrush eats snails, and the energy in the droppings goes to decomposers. - C. ✓ The part the snails built into new body
Why: The thrush eats the snails’ bodies.
Only the lettuce the snails built into new body is in those bodies.
So the thrush gets energy from that part alone.
Suppose a class keeps two boxes of mealworm larvae for two weeks and feeds both boxes the same oat flakes. The larvae in both boxes digest all the oats they eat. Box A sits in a room at 25 °C. Box B sits in a room at 15 °C. Mealworm larvae are ectotherms. At the end of the two weeks the class divides the mass each box’s larvae gained by the mass of oats they ate. That share is called the box’s growth efficiency.
(a) Predict which box has the higher growth efficiency. (1 pt)
- Award 1 point for: box B (the 15 °C box).
(b) Explain the reasoning behind your prediction. (2 pt)
Frame A mealworm larva is an ectotherm, so …
At 15 °C its cells are colder.
Below their optimal temperature, enzymes work more slowly, so the colder cells respire more slowly.
The larvae in box B respire less of the oats they eat, so less of the oats’ energy leaves as heat.
A larger share of what they eat is left to build into new larva.
So box B’s growth efficiency is higher.
- Award 1 point for: the colder larvae’s cells respire more slowly (an ectotherm’s body temperature follows the room; enzymes work more slowly when cooler).
- Award 1 point for: less of the food’s energy leaves as heat, so a larger share of what is eaten is built into new larva (a higher growth efficiency).
Here are the cabbage leaves and the larvae again: 80 g of leaf eaten, 4 g of new larva.
The larvae respired 44 g of the leaf, and its energy left as heat. Another 32 g passed through undigested as droppings.
Only 4 g, 5 % of the leaf, became larva. Colder larvae would respire less and keep a little more.
APBIO-U08-P82 Practice questions: Topic 8.2
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one ecosystem’s energy figures one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it.
Video: Watch first: Energy flow through ecosystems, summed up
What energy is spent on; endotherm and ectotherm; the levels up to the biome; energy flows through while matter cycles; one template for four cycles; autotrophs and heterotrophs; trophic levels, arrows the way the energy goes, the web from the table; about 10 % up a level and where the rest went; less sun, fewer hawks.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T82-summary.mp4
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An albatross drinks sea water. Glands above its eyes pump the extra salt out, so the salt in its blood stays at one level.
Which of the four things an organism uses energy for is the albatross doing?
- A. GrowingGrowing adds new body.
Pumping salt out adds nothing. - B. ReproducingReproducing makes offspring.
Pumping salt out makes none. - C. Keeping its parts in orderKeeping its parts in order is rebuilding worn-out molecules.
Holding the blood’s salt at one level is keeping the inside steady. - D. ✓ Keeping its inside steady
Why: The albatross drinks salty water and its blood’s salt stays at one level.
Holding the body’s inside steady against the outside is homeostasis.
Homeostasis is one of the four uses of energy: keeping the inside steady.
A barn owl in a hard winter takes in 340 kJ per day from the voles it catches and uses 455 kJ per day.
Which of the following is the owl’s net energy?
- A. −795 kJ per day−795 adds the two values with a minus sign.
Net energy is energy in minus energy out. - B. ✓ −115 kJ per day
- C. +115 kJ per day+115 is energy out minus energy in.
Net energy is energy in minus energy out: 340 − 455, a negative number. - D. +795 kJ per day+795 adds the two values.
Net energy is energy in minus energy out.
Why: Energy in is 340 kJ per day and energy out is 455 kJ per day.
Net energy = 340 − 455 = −115 kJ per day.
The net is negative, so the owl loses mass.
A crocodile lies on a river bank in the morning sun with its mouth open. Its body warms from 24 °C to 32 °C in an hour.
Which of the following explains where the crocodile’s warmth came from?
- A. ✓ Heat passed into it from the sun and the warm bank
- B. Its muscles made the heat by shiveringShivering muscles make heat inside an endotherm such as a mouse.
A crocodile warms from outside. - C. Its cells respired extra food to make the heatAn ectotherm respires no food to stay warm.
Its body temperature follows its surroundings. - D. Its open mouth took in warm air, which heated its bloodBreathing warm air moves little heat.
The sun and the warm bank warmed the crocodile’s body.
Why: A crocodile is an ectotherm, so its body temperature follows its surroundings.
On the sunlit bank, heat passes from the sun and the bank into the crocodile.
So the crocodile warmed from outside, by lying where heat passes in.
Rotifers are tiny animals that eat single-celled algae. A researcher keeps rotifers in tanks holding different amounts of algae and records how each tank’s rotifers reproduce, in the table below.
Which of the following strategies does the table show?
- A. Timing mating so the young arrive when the algae doTiming means one way of breeding, placed in one season.
These rotifers breed two ways, and the share of each follows the food. - B. ✓ Switching from asexual to sexual reproduction as the food falls
- C. Pausing breeding while the algae are scarce, then breeding againPausing means no breeding while food is scarce.
In the tank with the least algae, 98 % of the rotifers are still breeding, by mating. - D. Storing energy from the algae and breeding on it laterStoring first means breeding on energy saved earlier.
The table records how the rotifers breed, not a store.
Why: With 10 000 algal cells per mL, 98 % of the rotifers reproduce without mating: asexual reproduction while food is plentiful.
With 100 cells per mL, 98 % reproduce by mating: sexual reproduction as food grows scarce.
So the rotifers switch how they reproduce as the energy on offer changes.
A campfire burns 2 kg of dry wood down to 0.1 kg of ash.
Which of the following happened to the other 1.9 kg of the wood’s matter?
- A. It was destroyed as the fire released the wood’s energyNo process makes or destroys matter.
The wood’s atoms went into the air as carbon dioxide and water vapor. - B. It left as heatHeat is energy, not matter.
The wood’s atoms left as gases, and its energy left as heat. - C. ✓ It went into the air, mostly as carbon dioxide and water vapor
- D. It sank into the ground under the fireThe fire burned in the air, not in the ground.
Combustion sends the wood’s carbon into the air as carbon dioxide.
Why: Combustion joins the wood’s carbon to oxygen from the air.
The carbon leaves as carbon dioxide, and the wood’s hydrogen leaves as water vapor.
No atom is made or destroyed, so the 1.9 kg that left the wood arrived in the air.
Meteorologists follow one parcel of coastal air from dawn to noon and measure the water in it, in the table below. No rain reached the ground.
Which of the following accounts for the change between 06:00 and 12:00?
- A. The atmosphere lost 0.9 million kg of water, because the vapor reading fell between the two timesVapor and cloud droplets are both water in the atmosphere.
The 0.9 million kg that left the vapor column appears in the cloud column. - B. The atmosphere gained 0.9 million kg of water, because the cloud grew between the two timesThe cloud grew by 0.9 million kg, and the vapor fell by 0.9 million kg.
The atmosphere’s total is the same. - C. The vapor evaporated out of the parcel into the sea beneath it, so the atmosphere lost waterEvaporation is liquid water turning into vapor.
Here vapor turned into liquid drops: condensation. - D. ✓ The atmosphere held 1.9 million kg of water at both times, because vapor condensed into cloud droplets
Why: Water vapor and cloud droplets both sit in the atmosphere reservoir.
The water that left the vapor column appears in the cloud column, so the parcel’s total is the same at both times.
No molecule left: vapor turned into liquid droplets, which is condensation.
For years a town’s washing powders carried phosphate, and the town’s waste water carried it into the town’s lake. A new law removes phosphate from every washing powder. No process takes phosphate out of the lake’s water any faster than before.
Over the following years, what happens to the lake’s store of phosphate?
- A. It risesThe arrow into the lake has slowed.
A slower arrow in lowers the reservoir it fed. - B. It stays the sameThe waste water now carries far less phosphate in, while the arrows out are unchanged.
Less enters than leaves. - C. ✓ It falls
- D. It falls to zero within the first yearThe lake still holds phosphate in its water, its plants and its mud, and rivers still bring a little.
The store falls, but not to nothing.
Why: The waste water was an arrow carrying phosphate into the lake’s water.
The law slows that arrow.
No arrow out of the lake got slower.
So less phosphate enters than leaves, and the store falls.
An orca hunts, kills and eats a seal.
Which kind of eater is the orca here?
- A. ✓ A carnivore
- B. A herbivoreA herbivore eats producers.
A seal is an animal. - C. A scavengerA scavenger eats dead animals it did not kill.
The orca killed this seal. - D. A decomposerA decomposer breaks dead matter down into simple molecules.
The orca eats the seal’s flesh.
Why: The orca kills the seal and eats it.
A heterotroph that eats animals it kills is a carnivore.
On a heather moor, red grouse eat the heather’s shoots, and hen harriers eat the grouse.
Which trophic level are the hen harriers?
- A. ProducerThe heather is the producer: it makes its own sugar by photosynthesis.
The harriers eat animals. - B. Primary consumerThe grouse are the primary consumers, one step from the heather.
The harriers eat the grouse. - C. ✓ Secondary consumer
- D. Tertiary consumerA tertiary consumer is three steps from the producer.
The harriers are two steps from the heather.
Why: The heather is the producer.
The grouse eat the heather: one step from the producer.
The hen harriers eat the grouse: two steps.
So the harriers are secondary consumers.
The food web of an alpine meadow is drawn below. Each arrow runs from the eaten organism to the organism that eats it. Suppose a disease kills every marmot in the meadow.
Which of the following organisms loses a food source but still has food?
- A. The ptarmiganNo arrow joins the marmots to the ptarmigan.
The ptarmigan eat alpine plants and grasshoppers, so they lose nothing. - B. ✓ The golden eagles
- C. The grasshoppersThe grasshoppers eat alpine plants.
They lose no food when the marmots die. - D. The alpine plantsThe alpine plants are producers.
They make their own sugar and eat nothing.
Why: Two arrows arrive at the golden eagles’ box: from the marmots and from the ptarmigan.
The marmots die, so the eagles lose that food source.
The ptarmigan remain, so the eagles still have food.
Ecologists survey a fjord. Its floating algae are eaten by small crustaceans, and the crustaceans are eaten by herring. The table gives what the algae turned into sugar in a year and what they respired, in kilojoules per square meter a year. A later survey measures the energy stored in the crustaceans’ level at 588 kJ per square meter a year.
(a) Calculate the net primary productivity of the algae. (1 pt)
Frame Net primary productivity = made − respired = … kJ per square meter a year
Hint Which of the two values in the table is taken away from the other?
Answer: 4200 kJ per square meter a year (tolerance ±0)
- Award 1 point for: 4 200 kJ per square meter a year.
Slip Reporting 6 800, the whole amount made, before the algae’s respiration is taken away.
(b) By the 10 % rule, calculate the energy the crustaceans’ level would store. (1 pt)
Frame The crustaceans’ level would store about … kJ per square meter a year
Hint The crustaceans are one level above the algae. What do you multiply your answer to part (a) by?
Answer: 420 kJ per square meter a year (tolerance ±0)
- Award 1 point for: 420 kJ per square meter a year.
Slip Multiplying by 0.90. The level above stores about 10 % of the level below; the 90 % is what does not pass up.
(c) By the 10 % rule, calculate the energy the herring’s level would store. (1 pt)
Frame The herring’s level would store about … kJ per square meter a year
Hint The herring are one level above the crustaceans. Start from your answer to part (b).
Answer: 42 kJ per square meter a year (tolerance ±0)
- Award 1 point for: 42 kJ per square meter a year.
Slip Multiplying the algae’s value by 0.10 once more, giving 420 again. Each level up takes another 10 %.
(d) Using the measured value of 588 kJ per square meter a year for the crustaceans’ level, calculate the transfer between the algae and the crustaceans as a percentage, to one decimal place. Answers within 0.05 % are accepted. (1 pt)
Frame Transfer = … ÷ … × 100 = … %
Hint The transfer is the level above divided by the level below, as a percentage. Which of your values is the level below?
Answer: 14.0 % (tolerance ±0.05)
- Award 1 point for: 14.0 % (accept 13.95 % to 14.05 %).
Slip Dividing by 6 800, the whole amount made. The crustaceans can eat only what the algae stored after respiring.
(e) Determine whether the fjord’s measured transfer sits inside the range ecologists have measured, about 5 % to about 20 %, and whether it sits above or below the 10 % working rule. (1 pt)
Frame The transfer of … % sits … the measured range and … the 10 % working rule, because …
Hint Compare your answer to part (d) with 5 %, 20 % and 10 %.
- Award 1 point for: inside the range (between 5 % and 20 %) AND above the 10 % rule, with the comparison stated.
Slip Calling 14.0 % a mistake in the survey because it is not 10 %. The 10 % rule is a rounded working figure; measured transfers run from about 5 % to about 20 %.
Suppose a long drought dries out a bulrush marsh. The bulrushes die and lie on the exposed mud, and soil microbes feed on them. Two years later heavy rains flood the marsh again, and the flooded mud holds little oxygen.
(a) Describe the form in which the bulrushes took up their nitrogen while they were alive. (1 pt)
- Award 1 point for: as ammonium or nitrate (dissolved, through the roots). Accept with or without: not as nitrogen gas.
Slip Saying the bulrushes took nitrogen gas from the air. No plant enzyme breaks the triple covalent bond in nitrogen gas.
(b) Explain how the nitrogen in the dead bulrushes’ protein becomes nitrate in the dry mud. (1 pt)
Soil bacteria then turn the ammonium into nitrite and then into nitrate: nitrification.
- Award 1 point for: ammonification (dead matter to ammonium) followed by nitrification (ammonium to nitrate), both done by soil microbes. Accept nitrification written with or without the nitrite step.
Slip Naming nitrogen fixation. Fixation turns nitrogen gas into ammonia; the bulrushes’ nitrogen was already in a compound.
(c) Predict how the store of nitrate in the mud changes in the months after the marsh floods again. (1 pt)
- Award 1 point for: the nitrate store falls (decreases).
Slip Predicting a rise because the water carries nitrate in. The flooded mud holds little oxygen, and that is where bacteria turn nitrate into nitrogen gas.
(d) Justify your prediction using the steps of the nitrogen cycle. (1 pt)
Denitrification is an arrow out of the mud’s nitrate store, into the air.
A faster arrow out lowers the reservoir it leaves.
So the mud’s store of nitrate falls.
- Award 1 point for: denitrification (nitrate to nitrogen gas, which leaves to the air) speeds up in the flooded mud, a faster arrow out of the nitrate store, so the store falls.
Slip Saying the nitrate washed away with the flood. The flooded mud holds little oxygen, and the nitrate leaves as nitrogen gas.
APBIO-U08-T82 End-of-topic test: Energy Flow Through Ecosystems
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Every energy value carries its unit, kilojoules per square meter a year, and every percentage is written with a space: 10 %.
A female pond turtle feeds on insects and pondweed all summer. Each day she takes in more energy than she uses, and by autumn she carries a thick layer of fat.
Which of the following happens to the surplus energy she takes in each day?
- A. ✓ She stores it as fat, and later builds it into her eggs
- B. It leaves her body as heat, with no use to herEnergy leaves as heat only when her cells respire it.
The surplus is the part her cells did not respire. - C. She uses it to keep her body warmer than the pond waterA turtle is an ectotherm.
Its body temperature follows the pond’s, so it respires no food to stay warm. - D. It passes out of her body undigestedFood that passes through undigested was never taken in.
The surplus is energy she took in.
Why: Energy in is greater than energy out, so the turtle has a surplus.
An organism stores a surplus as fat.
Reproduction is one of the four uses of energy, so the stored fat later goes into her eggs.
A weasel in January takes in 610 kJ per day from the mice it catches. It uses 690 kJ per day keeping warm and hunting.
Which of the following is the weasel’s net energy?
- A. −1 300 kJ per day−1 300 adds the two values with a minus sign.
Net energy is energy in minus energy out. - B. ✓ −80 kJ per day
- C. +80 kJ per day+80 is energy out minus energy in.
Net energy is energy in minus energy out: 610 − 690, a negative number. - D. +1 300 kJ per day+1 300 adds the two values.
Net energy is energy in minus energy out.
Why: Energy in is 610 kJ per day and energy out is 690 kJ per day.
Net energy = 610 − 690 = −80 kJ per day.
The net is negative, so the weasel loses mass.
A hamster and a tree frog of the same mass live in the same cool room. The table gives the food each one eats per gram of body each day.
Which of the following explains the difference in the table?
- A. The tree frog digests its food less completely than the hamster doesHow completely food is digested does not set how much must be eaten.
The hamster respires most of its food for heat. - B. The hamster is active at night, and moving about uses most of the food it eatsMoving about uses energy, but the hamster respires food for heat all day and night, still or moving.
The frog respires no food for warmth. - C. The tree frog makes its own sugar, so it needs little foodA frog is a heterotroph.
It gets its organic molecules by eating, never by photosynthesis. - D. ✓ The hamster is an endotherm, so it respires food to make the heat that keeps its body warm
Why: The hamster is an endotherm.
Its own respiration makes the heat that keeps it warm, so it respires food for heat all day.
The tree frog is an ectotherm: its body follows the room’s temperature, and it respires no food for warmth.
So the hamster eats far more per gram.
Sockeye salmon feed in the ocean for two or three years and store fat. Then they swim up a river, eating nothing on the way, and breed once at the end of the journey.
Which of the following strategies is the salmon using to match its breeding to the energy on offer?
- A. Switching from asexual to sexual reproduction as food fallsSwitching means asexual reproduction while food is plentiful, then sexual.
The salmon breeds one way, once. - B. Timing its mating so the young hatch when their food doesTiming means mating placed so the young arrive with the food.
The salmon’s point is the fat it stored first, over years at sea. - C. Pausing breeding through a poor season, then breeding againPausing means stopping and restarting breeding.
The salmon has not bred before; it stores, then breeds once. - D. ✓ Storing energy in one period and breeding on it later
Why: The salmon stores fat for two or three years in the ocean.
It then breeds on that store, eating nothing on the journey.
So it stores energy in one period and breeds on it later.
Ecologists survey two salt marshes and record the rows in the table below.
Which of the following groups of adults is one population?
- A. The saltmarsh sparrows and the seaside sparrows seen in North Marsh on May 12A population is every member of one species living in one place at one time.
Two kinds of sparrow are two species. - B. The saltmarsh sparrows and the river otters seen in North Marsh on May 12The sparrows and the otters are two species.
Together they are part of North Marsh’s community, not one population. - C. ✓ The saltmarsh sparrows seen in North Marsh on May 12
- D. The saltmarsh sparrows seen in North Marsh on May 12 and on August 12August’s count is the same species in the same place at another time.
A population is counted at one time; some of August’s birds hatched after May.
Why: A population is every member of one species living in one place at one time.
The 18 saltmarsh sparrows seen in North Marsh on May 12 are one species, one place, one time.
So they are one population.
A sealed glass sphere holds water, algae, small snails and bacteria. Nothing but light and heat can cross the glass. Over one year the ecologists who keep it measured what entered and left the sphere, in the table below, and its living things stayed alive.
Which of the following explains why the sphere needs 1 500 000 kJ of light every year but no new carbon atoms or nitrogen atoms?
- A. ✓ Heat leaves at every transfer and cannot come back, while the same atoms go round between the water, the algae and the snails
- B. The algae use up light faster than they use up carbon atoms, so light must be replaced far more often than carbonLight is not a stock the algae use up.
Its energy leaves the sphere as heat after each transfer, so the same amount must enter again. - C. Photosynthesis makes brand-new carbon atoms inside the sphere, but nothing inside the sphere can make new lightNo process makes a carbon atom.
Photosynthesis moves carbon atoms from the water into sugar; the atoms were already there. - D. The snails breathe out carbon atoms they make themselves, so the sphere always has a fresh supply of themThe snails breathe out carbon atoms they took in as food.
Those atoms are the same ones, going round.
Why: Every time an organism uses energy, some spreads out as heat.
That heat leaves through the glass and can do no more work, so fresh light must enter.
No process makes or destroys an atom.
The snails breathe carbon dioxide out and the algae take it in.
Ecologists add a small amount of traceable phosphate to a fenced-off pond and, three weeks later, measure how much of the tracer sits in each part of the pond. The table gives the results.
Which of the following is the largest abiotic reservoir the tracer reached?
- A. The algaeThe algae are alive.
They are a biotic reservoir, whatever they hold. - B. ✓ The mud at the bottom
- C. The pond waterThe pond water is abiotic, but it holds 1.2 mg.
The mud holds 5.4 mg, more than four times as much. - D. The snailsThe snails are alive.
They are a biotic reservoir, whatever they hold.
Why: An abiotic reservoir is a store of the cycle’s matter that is not alive.
The mud and the water are the pond’s abiotic reservoirs; the algae, snails and fish are alive.
The mud holds 5.4 mg of tracer, the water 1.2 mg.
So the mud is the largest.
At dawn, drops of water form on a spider’s web from the water vapor in the air. By noon the web is dry.
In order, which two processes did the water in a drop pass through?
- A. Precipitation, then evaporationPrecipitation is water falling from a cloud.
The drops formed on the web from vapor in the air beside it. - B. Evaporation, then condensationThe drops formed first and left later.
Vapor turning into drops is condensation, and it came first. - C. Condensation, then transpirationTranspiration is water leaving a plant’s leaves as vapor.
The web is not a plant; the drops left it as vapor by evaporation. - D. ✓ Condensation, then evaporation
Why: Water vapor in the cool dawn air turned into liquid drops on the web: condensation.
By noon the sun had warmed the drops, and the liquid turned back into vapor: evaporation.
So condensation came first, then evaporation.
A city closes its coal-burning power station and gets the same amount of electricity from a solar farm. Nothing else about the city changes.
Which of the following happens to the carbon dioxide in the air above the city over the following years, and why?
- A. It falls, because the solar panels take carbon dioxide out of the airSolar panels take in light, not carbon dioxide.
Only photosynthesis takes carbon out of the air. - B. It stays the same, because the city uses the same amount of electricityThe electricity is the same, but the carbon is not.
Burning coal put carbon into the air; the solar farm burns nothing. - C. ✓ It falls, because less fossil-fuel carbon leaves as carbon dioxide by combustion
- D. It rises, because the coal left in the ground decomposesCoal in the ground is buried where fungi and microbes cannot feed on it.
Unburned coal puts no carbon into the air.
Why: Combustion is the arrow from the fossil fuel to the air.
The power station burned coal, so combustion put carbon into the air.
The solar farm burns nothing, so that arrow slows.
Photosynthesis takes carbon out as before, so the air’s carbon dioxide falls.
A nitrogen atom in the protein of a dead bean leaf on the soil ends up, months later, in a molecule of nitrogen gas in the air.
In order, which steps of the nitrogen cycle carried the atom from the leaf to the air?
- A. Nitrogen fixation, assimilation, denitrificationNitrogen fixation turns nitrogen gas into ammonia; it starts in the air, not in a dead leaf.
Assimilation builds nitrogen into a plant, not into the air. - B. ✓ Ammonification, nitrification, denitrification
- C. Nitrification, ammonification, denitrificationNitrification turns ammonium into nitrate, so it needs ammonium first.
Ammonification makes that ammonium from the dead leaf. - D. Ammonification, assimilation, nitrogen fixationAssimilation is a plant taking nitrogen up.
Nitrogen fixation takes nitrogen out of the air, the opposite of this path.
Why: Soil microbes turn the dead leaf’s protein into ammonium: ammonification.
Soil bacteria turn the ammonium into nitrite and then nitrate: nitrification.
Bacteria turn the nitrate into nitrogen gas, which returns to the air: denitrification.
A grower raises lettuce with its roots in a tank of water and bubbles air through the tank. The air is about 78 % nitrogen gas. The lettuce turns yellow until the grower adds nitrate to the water.
Which of the following explains why the lettuce stayed short of nitrogen while the air bubbled through the tank?
- A. ✓ No plant enzyme breaks the triple bond in nitrogen gas, so roots take up nitrogen only as ammonium or nitrate
- B. Nitrogen gas dissolves in water so poorly that none of it ever reached the lettuce’s roots in the tankSome nitrogen gas does dissolve in the water.
The roots cannot use it, because no plant enzyme breaks its triple covalent bond. - C. The roots fix nitrogen gas into ammonia only when soil bacteria are present in the tank to help themNo plant fixes nitrogen gas.
Bacteria fix it, and a root takes up the ammonium or nitrate that results. - D. Lettuce needs almost no nitrogen, so the yellowing came from too little light reaching the leavesEvery plant needs nitrogen for its proteins and nucleic acids.
Adding nitrate greened the lettuce, so nitrogen was what it lacked.
Why: The two atoms in nitrogen gas share three pairs of electrons: a triple covalent bond.
No plant enzyme breaks that bond, so the lettuce cannot use the gas.
A root takes up nitrogen only as ammonium or nitrate dissolved around it.
The tank held neither until the grower added nitrate.
For centuries a river’s floods have spread sediment, rich in phosphate, over the soil of its delta. A dam is built upstream, and the sediment now settles behind the dam instead. The delta’s farmers harvest their crops as before.
Over the following decades, what happens to the delta soil’s store of phosphate?
- A. It rises, because the water held behind the dam weathers more rock and releases more phosphateThe dam traps the sediment behind it.
Whatever weathers upstream now settles in the reservoir, not on the delta. - B. It stays the same, because the phosphorus cycle has no reservoir in the air to lose phosphate toHaving no store in the air does not hold a soil’s store steady.
The store falls when less enters than leaves. - C. ✓ It falls, because the floods that carried phosphate into the soil have stopped while the harvest still carries it out
- D. It falls, because the phosphate in the delta’s soil slowly turns into a gas and leaves for the airPhosphate is never a gas.
The store falls because the flowing water that carried phosphate in no longer reaches the delta.
Why: The river’s floods were the arrow carrying phosphate into the delta’s soil: water moving phosphate, the water cycle serving the phosphorus cycle.
The dam slows that arrow almost to nothing.
The harvest still carries phosphate out of the soil.
So less enters than leaves, and the store falls.
Researchers grow bacteria from water deep underground in four flasks. Every flask holds carbon dioxide and the minerals the bacteria need, but no sugar. The table gives what each flask received and how much sugar the bacteria made from carbon dioxide.
Which of the following are the bacteria?
- A. Photosynthetic autotrophsA photosynthetic autotroph captures the energy in light.
Flask 1 was lit and the bacteria made no sugar. - B. ✓ Chemosynthetic autotrophs
- C. Heterotrophs that eat the hydrogen sulfideA heterotroph gets its organic molecules by eating other organisms.
These bacteria made their own from carbon dioxide. - D. Decomposers of the flasks’ dead matterA decomposer breaks dead matter down into simple molecules.
The flasks held no dead matter, and the bacteria built sugar from carbon dioxide.
Why: The bacteria made sugar from carbon dioxide only where hydrogen sulfide was added, in the dark or in the light.
So their energy source is a small inorganic molecule, hydrogen sulfide, not light.
An autotroph that captures energy this way is a chemosynthetic autotroph.
Researchers keep two animals of the same mass, P and Q, for two hours in chambers at three air temperatures. The chambers stop the animals moving to a warmer or cooler spot. The table gives each animal’s body temperature at the end of the two hours.
Which kind of animal is P, and how would it hold its body temperature in the wild?
- A. An endotherm; its own respiration makes the heat that holds its body at one temperatureAn endotherm holds its body temperature whatever the air does, as Q does.
P’s body temperature was the chamber’s every time. - B. ✓ An ectotherm; it would move between sun and shade
- C. An ectotherm; its muscles would make heat when it movesAn ectotherm’s muscles make no heat to keep its body warm.
P’s body follows the air, so it holds its temperature by where it sits. - D. An endotherm; it would sweat to keep its body cooler than the airP’s body was at the air’s temperature in every chamber, never cooler.
An animal that sweats to stay cool is an endotherm.
Why: P’s body temperature was 15 °C, 24 °C and 33 °C: the chamber’s temperature each time.
An animal whose body temperature follows its surroundings is an ectotherm.
The chamber stopped P moving, and in the wild an ectotherm holds its temperature by what it does: moving between sun and shade.
On the deep sea floor, hagfish gather at the body of a whale that died and sank. They eat its flesh over many weeks.
Which kind of eater is a hagfish here?
- A. A carnivoreA carnivore kills the animals it eats.
The whale was already dead when the hagfish found it. - B. An omnivoreAn omnivore eats both producers and animals.
The hagfish here eats only the whale’s flesh. - C. A decomposerA decomposer breaks dead matter down into simple molecules.
The hagfish eats the flesh itself. - D. ✓ A scavenger
Why: The whale is dead, and the hagfish did not kill it.
A heterotroph that eats dead animals it did not kill is a scavenger.
A wolf kills and eats a moose. The moose’s meat is made of carbon compounds: protein, fat and a little carbohydrate.
Which of the following describes what happens to the moose’s carbon compounds inside the wolf?
- A. The wolf’s cells respire all of them, and their carbon leaves the wolf as carbon dioxideRespiring every compound would leave nothing to build the wolf from.
A heterotroph gets matter from its meal as well as energy. - B. The wolf’s cells build all of them into the wolf’s own muscle and fatBuilding every compound into tissue would give the wolf no energy.
Its cells respire some of the meal to make ATP. - C. ✓ The wolf’s cells respire some of them to make ATP and build the rest into the wolf’s own tissue
- D. The wolf’s cells break them down and rebuild the carbon compounds from carbon dioxideNo animal builds a carbon compound from carbon dioxide.
Only an autotroph does; the wolf rebuilds compounds its food already carried.
Why: The wolf’s cells respire some of the moose’s carbon compounds, and energy is released.
Some of that energy makes ATP: the wolf’s energy from the meal.
The cells build the rest of the compounds into the wolf’s own muscle and fat: the wolf’s matter from the meal.
A survey of a salt lake lists what each organism eats, in the table below. A student builds the lake’s food web from the table, drawing each arrow from the eaten organism to the organism that eats it.
Which of the following arrows is in the student’s web?
- A. ✓ From the brine shrimp to the flamingos
- B. From the flamingos to the brine shrimpThe flamingos eat the brine shrimp, so the energy goes from the shrimp to the flamingos.
The arrow points the way the energy goes. - C. From the algae to the avocetsThe avocets eat brine shrimp and brine fly larvae, not algae.
No arrow joins the algae to the avocets. - D. From the brine fly larvae to the flamingosThe flamingos eat brine shrimp only.
The brine fly larvae are eaten by the avocets.
Why: Each arrow runs from the eaten organism to its eater.
The table says the flamingos eat brine shrimp.
So the arrow runs from the brine shrimp to the flamingos.
A class feeds 45 g of mulberry leaves, measured as dry mass, to silkworm larvae for a week. At the end of the week the larvae have gained 5 g, and the class has collected 16 g of droppings, both as dry mass.
Which of the following accounts for the other 24 g of leaf?
- A. The larvae’s cells destroyed it as they used its energy for moving and growingNo process destroys matter.
The 24 g left the larvae as carbon dioxide and water when their cells respired it. - B. ✓ The larvae respired it, so its carbon left as carbon dioxide and water and its energy left as heat
- C. It is still inside the larvae’s guts, undigested, waiting to pass out as droppingsA week is far longer than a larva takes to digest a meal.
What the larvae could not digest left as the 16 g of droppings. - D. It became larva, and the class weighed the larvae wrongly at the end of the weekThe larvae gained 5 g, and that is the leaf built into larva.
Most of the leaf was respired, not built into body.
Why: The larvae digested the leaves and absorbed them into their cells.
Their cells respired most of that to make ATP.
The respired sugar left as carbon dioxide and water, and its energy left as heat.
Only the 5 g built into new body stayed as larva.
An estuary’s food chain is eelgrass, then the snails that eat it, then the crabs that eat the snails, then the herons that eat the crabs. One year a flood carries extra nutrients into the estuary, and the eelgrass stores twice as much energy as usual. The table gives the two years.
By the 10 % rule, what happens to the number of herons the estuary can feed in the year of extra nutrients?
- A. It stays the same, because the herons eat crabs, not eelgrassThe herons eat crabs, and the crabs eat what the eelgrass fed.
More energy in the eelgrass means more at every level above. - B. It falls, because the extra eelgrass shades the estuary’s waterThe stem says the eelgrass stores twice as much energy.
The change passes up the chain as more energy, not less. - C. It rises by the whole extra amount, because every level above stores all the energy added at the baseEach level stores only about 10 % of the level below.
The herons gain a share of the extra energy, not all of it. - D. ✓ It rises, because about 10 % of the extra energy passes up at each level
Why: The eelgrass stores twice as much energy in the year of extra nutrients.
The snails store about 10 % of that, so they store about twice as much too; so do the crabs, and then the herons.
So the estuary can feed more herons.
Ecologists survey a tidal mudflat. Algae growing on the mud are eaten by mud snails, the mud snails are eaten by wading birds, and the wading birds are hunted by peregrine falcons. The table gives what the algae turned into sugar in a year and what they respired. The pyramid gives the energy stored in the mud snails’ level and in the wading birds’ level, both in kilojoules per square meter a year.
(a) Calculate the net primary productivity of the algae, in kilojoules per square meter a year. (1 pt)
Answer: 8100 kJ per square meter a year (tolerance ±0)
- Award 1 point for: 8 100 kJ per square meter a year (13 500 − 5 400).
Slip Reporting 13 500: that is the whole amount made, before the algae’s respiration is taken away.
(b) Calculate the transfer of energy between the algae and the mud snails as a percentage of the algae’s stored energy, to one decimal place. (1 pt)
Answer: 11.0 % (tolerance ±0.05)
- Award 1 point for: 11.0 % (accept 10.95 % to 11.05 %), from 891 ÷ 8 100 × 100. Accept a transfer computed correctly from the student’s own part (a) value.
Slip Dividing by 13 500, the whole amount made. The consumers can eat only what the algae stored after respiring.
(c) A student says: “Exactly 10 % of the energy passes up at every step of this mudflat.” Use the data to support or refute the student’s claim. (1 pt)
The transfer between the algae and the mud snails is 11.0 %.
The transfer between the mud snails and the wading birds is 71 ÷ 891 × 100, about 8 %.
Neither step is exactly 10 %, though both sit inside the range ecologists measure, about 5 % to about 20 %.
So 10 % is a working rule, not an exact law of this mudflat.
- Award 1 point for: the judgement (the claim is not supported) AND the ground (at least one measured step is not 10 %: 11.0 % between the algae and the snails, or about 8 % between the snails and the birds). Accept with or without: both values sit inside the measured range of about 5 % to 20 %.
Slip Agreeing because the values are close to 10 %. The claim says exactly, and the data give 11.0 % and about 8 %.
(d) By the 10 % rule, explain why the mudflat is unlikely to sustain a population at a fifth trophic level, above the peregrine falcons. (1 pt)
A fifth level would store about 10 % of that, under 1 kJ per square meter a year.
A population needs energy every year to stay alive, grow and raise young.
So too little energy would be left to feed a population above the falcons.
- Award 1 point for: each level stores only about 10 % of the level below, so a fifth level would store under about 1 kJ per square meter a year (about 10 % of the falcons’ 7), too little to feed a population.
Slip Explaining the missing level by the lack of a hunter. The chain ends where too little energy is left, not where a hunter is missing.
The food web below was drawn from a survey of an Arctic sea. Ice algae grow on the underside of the sea ice. In the web, each arrow is drawn from the eaten organism to the organism that eats it.
(a) Identify the trophic level, or levels, at which the seabirds sit in this web. (1 pt)
- Award 1 point for: secondary consumer (copepods) AND tertiary consumer (Arctic cod), the seabirds at two levels.
Slip Giving one level only. An organism with two foods at different levels sits at both.
(b) Suppose fishing boats remove almost all the Arctic cod. Explain why the ringed seals lose their food while the seabirds still have food. (1 pt)
Two arrows arrive at the seabirds’ box, from the copepods and from the Arctic cod.
When the cod are removed, the seals have no other path, while the seabirds still have the copepods.
- Award 1 point for: the Arctic cod are the seals’ only food (one arrow arriving), while the seabirds have two foods (two arrows arriving), so the seabirds still have the copepods.
Slip Tracing the loss to one consumer only. Every path the removed organism fed must be followed.
(c) The cod recover. A company plans to feed a coastal town from this sea and must choose between harvesting Arctic cod and harvesting ringed seals. Predict which harvest feeds more people from the same sea. (1 pt)
- Award 1 point for: the Arctic cod.
Slip Choosing the seals because each seal is a larger animal. The size of one animal does not set how much energy its level stores.
(d) Justify your prediction using the 10 % rule. (1 pt)
The ringed seals are one level above the Arctic cod, so the seals’ level stores about 10 % of the cod’s.
People eating cod are one level above the cod; people eating seals are two levels above the cod.
So people eating seals store about 10 % of what people eating cod would store, and the cod harvest feeds more people.
- Award 1 point for: the seals are an extra trophic level between the cod and the people, and each level stores only about 10 % of the level below, so eating cod directly leaves about ten times as much energy for the people.
Slip Saying seals are richer food. The 10 % rule counts the levels crossed, not the richness of the meal.
APBIO-U08-L27 Counting frogs
Two ponds lie a mile apart. Each pond holds frogs of the same species.
No frog has ever crossed the mile of dry ground between the ponds. Is that one population of frogs, or two?
Unit 8 · Ecology
1One population, or two
In the first week of the course you sorted an oak forest into levels.
What did the first week call all the oaks of one kind living in that forest?
- A. ✓ A population
- B. A communityA community is all the populations living together in one place: the oaks with the deer and the beetles.
- C. An ecosystemAn ecosystem is the community together with its non-living surroundings: the soil, the rain and the sunlight.
Why: All the oaks of one kind in one forest are all the organisms of one species living in one place.
All the organisms of one species living in one place are a population.
Video: Watch: One population, or two
The two ponds a mile apart, each holding frogs of one species; the line that no frog has ever crossed is drawn between them; the frogs of each pond compete and breed only with each other, so the two ponds hold two populations.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L27a.mp4
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When are individuals one population, and what fits them to their place?
The first week of the course called all the organisms of one species living in one place a population.
One more test applies. The individuals of a population must be able to mix: to compete for the same food and, in a species with two parents, to breed with each other.
So the two ponds hold two populations, because the frogs of one pond never mix with the frogs of the other.
Here are the two ponds, a mile apart. Each pond holds frogs of the same species.
The frogs of pond A live together in pond A. They compete for the same insects, and they breed with each other.
The frogs of pond B do the same in pond B.
Between the ponds lies a mile of dry ground. No frog has ever crossed it.
So a frog of pond A never mixes with a frog of pond B. The frogs of the two ponds never compete, and they never breed together.
The frogs of the two ponds pass the first week’s test in one way: they are one species.
But two ponds a mile apart are not one place. And the frogs of the two ponds cannot compete or breed with each other.
The table below compares the frogs of pond A with the frogs of both ponds taken together, against the three tests.
So the frogs of pond A are one population. The frogs of pond B are a second population.
The word is the first week’s word, with one test added: the individuals must be able to mix.
Here are four more groups, each judged by one question: are they one population?
Consider all the deer of one species in one forest. They are one population, because they are one species, they live in one place, and they mix.
Now consider those deer together with deer of the same species in a forest 200 miles away, which they never reach. They are not one population, because the deer of the two forests never mix.
Now consider the deer of one forest that a road divides, where the deer cross the road every night. They are still one population, because they are one species, they live in one place, and they mix.
Now consider the deer and the wild boar of one forest. They are not one population, because they are two species.
The table below shows the four cases with their verdicts: one population, or not.
What you are expected to know Decide whether a described group is one population or two: the individuals must be one species, live in one place, and be able to mix, competing for the same food and breeding with each other.
Two ponds lie a mile apart. Each pond holds frogs of one species, and no frog has ever crossed between the ponds.
Which of the following tests for one population do the frogs of the two ponds fail?
- A. Being all one species, the same kind of frogThe frogs of both ponds are one species.
They fail the other test: they cannot compete or breed with each other. - B. ✓ Being able to mix, competing and breeding
Why: A mile of dry ground lies between the ponds, and no frog crosses it.
So the frogs of pond A cannot compete or breed with the frogs of pond B.
That is the test they fail.
Suppose all the mice of one species live in one barn.
Are the mice one population?
- A. ✓ Yes
- B. NoThe mice are one species in one barn, and they compete and breed with each other.
They are one population.
Why: The mice are one species.
They live in one barn, where they compete for food and breed with each other.
So the mice are one population.
Suppose pigeons of one species live on both banks of a river and fly across it every day.
Are the pigeons of the two banks one population?
- A. ✓ Yes
- B. NoThe pigeons cross the river every day.
So the pigeons of the two banks compete and breed with each other: one population.
Why: The pigeons fly across the river every day.
So the pigeons of the two banks compete for the same food and breed with each other.
They are one species that mixes in one place: one population.
Suppose sparrows of one species live in two towns 300 miles apart, and no sparrow flies between the towns.
Are the sparrows of the two towns one population?
- A. YesThe sparrows of one town never reach the other town.
So the two groups never compete or breed with each other. - B. ✓ No
Why: No sparrow flies between the towns.
So the sparrows of one town never compete or breed with the sparrows of the other.
The sparrows of the two towns never mix: two populations.
Suppose carp and perch live in one lake.
Are the carp and the perch together one population?
- A. YesCarp and perch are two species.
A population is one species. - B. ✓ No
Why: Carp and perch are two species.
A population is one species.
So the carp and the perch of the lake are two populations.
Suppose beetles of one species live on two oak trees 10 m apart and fly between the trees.
Are the beetles of the two trees one population?
- A. ✓ Yes
- B. NoThe beetles fly between the trees.
So the beetles of the two trees compete and breed with each other: one population.
Why: The beetles fly between the two trees.
So the beetles of the two trees compete for the same food and breed with each other.
They are one species that mixes in one place: one population.
Suppose snails of one species live on two islands 50 miles apart, and no snail crosses the sea between the islands.
Are the snails of the two islands one population?
- A. YesThe snails never cross the sea.
So the snails of one island never compete or breed with the snails of the other. - B. ✓ No
Why: No snail crosses the sea between the islands.
So the snails of one island never compete or breed with the snails of the other.
The snails of the two islands never mix: two populations.
A student says: “Frogs of one species are one population wherever they live, because a population is one species.”
Is the student correct?
- A. Yes: a population is all the frogs of one species, wherever in the world they liveBeing one species is one test.
The individuals must also live in one place and be able to compete and breed with each other. - B. ✓ No: frogs of one species in ponds far apart never mix, so they are separate populations
Why: A population is one species in one place.
Its individuals must be able to compete and breed with each other.
Frogs of one species in ponds far apart never mix.
So frogs of one species living far apart are separate populations.
33Built to get energy and matter here
Unit 7 followed a heritable variation through a population.
In Unit 7’s words, what is an adaptation?
- A. ✓ A heritable variation that raises fitness in its environment
- B. A change an individual makes during its own lifeA change made during one life is not heritable.
An adaptation is a heritable variation. - C. Any variation that raises fitness in every environmentA variation raises fitness in one environment and may lower it in another.
An adaptation raises fitness in its own environment.
Why: An adaptation is a heritable variation that raises fitness in its environment.
Unit 8 opened with a field mouse that eats seeds every day.
Why must the mouse keep eating?
- A. The mouse stores every seed’s energy as fat and never uses itThe mouse uses its energy for staying warm and moving.
At every step, some of that energy leaves the mouse as heat. - B. ✓ The energy the mouse uses leaves it as heat, and the mouse cannot get that heat back
Why: The mouse uses energy for staying warm and moving.
At every step, some of that energy leaves the mouse as heat.
The mouse cannot get the heat back, so it must keep eating.
Video: Watch: Built to get energy and matter here
A frog’s sticky tongue catches an insect at the pond; a cactus in a desert, its leaves reduced to spines and its thick stem storing the rain; a hummingbird’s long bill reaching the nectar at the base of a tubular flower; each feature is named as an adaptation for getting energy or matter in that place.
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What fits a frog to its pond, or a cactus to its desert?
Every organism needs a continuous input of energy and of matter. It must get both from the place where it lives.
Each frog carries adaptations: heritable variations that raise its fitness in the pond, in Unit 7’s word.
Many of an organism’s adaptations are for getting energy and matter in its particular place.
A frog’s sticky tongue catches the insects around its pond. The insects are the frog’s food: its energy and its matter.
Now consider a cactus in a desert.
In a desert, less than 30 cm of rain falls in a year. No one can say when it will fall.
Water is matter. Every cell of the cactus needs it, and the desert offers little of it.
A cactus’s leaves are reduced to spines. A spine has almost no surface.
So the cactus loses very little water through its leaves.
A cactus’s stem is thick. It stores water.
When rain does fall, the stem fills with water. The cactus then lives on that stored water through the dry months.
So the spines and the water-storing stem are adaptations for using matter. They let the cactus keep the little water its desert gives it.
The spines and the thick stem are heritable. A cactus grows them because of the alleles it inherited, not because it learned to.
Now consider a hummingbird at a flower.
Some flowers are long tubes. Their nectar sits at the base of the tube.
Nectar is a sugar solution. Sugar is the hummingbird’s energy.
A hummingbird’s bill is long and thin. It reaches down the tube to the nectar at the base.
So the long, thin bill is an adaptation for getting energy. It reaches food that a short bill cannot reach.
The table below compares the three adaptations: the organism, its adaptation, what the adaptation gets or keeps, and whether that is energy or matter.
Each adaptation helps in its own place. The water-storing stem raises fitness in a desert, where rain is rare.
In a place where rain falls most days, the same stem raises fitness much less.
Here are the two ponds again, a mile apart, with no frog crossing between them.
They hold two populations, because the frogs of one pond never mix with the frogs of the other.
Each frog’s sticky tongue is an adaptation for catching the insects around its pond: the food that gives the frog its energy and its matter.
What you are expected to know Explain how a named adaptation helps an organism get or use energy or matter in its particular environment.
A cactus’s leaves are reduced to spines.
What does that do for the cactus in a desert?
- A. It catches more light with its leavesA spine has almost no surface.
A smaller surface catches less light, not more. - B. It takes in more water through its leavesA leaf does not take water in; the roots do.
A spine has almost no surface, so less water leaves through it. - C. ✓ It loses less water through its leaves
Why: A spine has almost no surface.
Water leaves a plant through the surface of its leaves.
So the cactus loses less water through spines than it would through broad leaves.
Suppose a plant grows on a dark forest floor, where little light reaches. Its leaves are very large and broad.
(a) Explain how the large, broad leaves help the plant get energy in that place. (2 pt)
Frame The large, broad leaves help because …
Little light reaches the forest floor.
A large, broad leaf catches more of that light than a small leaf would.
So the plant gets more energy in a place with little light.
- Award 1 point for: the plant gets its energy from light (by photosynthesis), and little light reaches the forest floor.
- Award 1 point for: a large, broad leaf catches more of the light that reaches it, so the plant gets more energy there.
A student says: “A hummingbird’s long bill is an adaptation because each bird learns to grow a longer bill by reaching into flowers.”
Is the student correct?
- A. Yes: a bird that reaches into flowers grows a longer bill, and passes the longer bill onReaching into flowers changes nothing a bird passes on.
A hummingbird inherits its bill length. - B. ✓ No: a hummingbird inherits its bill length; reaching into flowers changes nothing it passes on
Why: A hummingbird inherits its bill length from its parents.
Reaching into flowers changes nothing the bird passes on.
An adaptation is a heritable variation that raises fitness.
So the long bill is an adaptation because it is inherited and reaches nectar, not because the bird learned it.
Suppose a tree grows where the soil is dry near the surface but wet 10 m down. One root grows straight down to the wet soil.
Which of the following does the long root help the tree get?
- A. LightLight reaches the tree’s leaves above the ground.
A root grows in the dark soil and catches no light. - B. Sugar from the soilA plant makes its own sugar by photosynthesis.
Its roots take in water and minerals, not sugar. - C. ✓ Water
Why: The soil near the surface is dry, and the wet soil lies 10 m down.
The long root reaches the wet soil.
So the long root helps the tree get water.
67Mixed practice mixed practice
Suppose wild sheep of one species live on both sides of a wide river, and no sheep ever swims across it.
Are the sheep of the two sides one population?
- A. YesNo sheep crosses the river.
So the sheep of one side never compete or breed with the sheep of the other. - B. ✓ No
Why: No sheep crosses the river.
So the sheep of one side never compete or breed with the sheep of the other.
The two groups never mix: two populations.
Suppose a water lily grows in a pond. Its leaves are broad and float on the surface, in full sun.
Which of the following does the broad, floating leaf help the plant get more of?
- A. WaterThe plant has water all around it in the pond.
The broad leaf floats at the surface, in the light. - B. Minerals from the mudMinerals come in through the roots in the mud.
The broad leaf floats at the surface, in the light. - C. ✓ Light
Why: The plant gets its energy from light.
A broad leaf at the sunlit surface catches more light than a small leaf would.
So the broad, floating leaf helps the plant get more light, and so more energy.
Suppose one meadow holds field mice and voles.
Which of the following describes the mice and the voles of the meadow?
- A. One populationA population is one species.
Mice and voles are two species. - B. ✓ Two populations
- C. One population made of two speciesA population never holds two species.
Two species in one place are two populations.
Why: A population is one species in one place.
The mice are one species, and the voles are another.
So the mice and the voles of the meadow are two populations.
A student says: “Frogs of one species in one pond are one population, so they compete with each other for the same insects.”
Is the student correct?
- A. No: members of one population leave each other’s insects aloneThe frogs of one pond eat the same insects.
When insects are few, each frog gets fewer. - B. ✓ Yes: members of one population compete for the same food
Why: The frogs of one pond eat the same insects.
When insects are few, each frog gets fewer.
Members of one population compete for the same food.
So frogs of one species in one pond compete with each other.
Suppose a seabird drinks seawater. A gland above each eye removes the extra salt from its blood.
Which of the following does the gland let the bird get from seawater?
- A. ✓ Water
- B. EnergySeawater carries no food; the bird’s energy comes from the fish it eats.
The gland removes salt so the bird can use the water. - C. OxygenThe bird gets oxygen from the air it breathes.
The gland removes salt so the bird can use the water.
Why: Seawater is water with a lot of salt in it.
The gland removes the extra salt from the bird’s blood.
So the bird can drink seawater and keep the water: the gland lets it get water.
Suppose starlings of one species roost in two woods 2 miles apart and fly between the woods every day, feeding and breeding in both.
Are the starlings of the two woods one population?
- A. ✓ Yes
- B. NoThe starlings fly between the woods every day.
So the starlings of the two woods compete and breed with each other: one population.
Why: The starlings fly between the woods every day.
So the starlings of the two woods compete for the same food and breed with each other.
They are one species that mixes: one population.
Suppose ground squirrels of one species live in two mountain meadows. A bare rock ridge 900 m high separates the meadows, and in ten years of study no squirrel has crossed it.
Each squirrel eats seeds all summer and stores fat under its skin. It lives on that fat through the winter, when no seeds can be found.
(a) Determine whether the squirrels of the two meadows are one population or two. (1 pt)
- Award 1 point for: two populations.
(b) Justify your answer to part (a). (1 pt)
So the squirrels of one meadow cannot compete or breed with the squirrels of the other.
A population’s individuals must be able to mix, so the two meadows cannot hold one population.
- Award 1 point for: the squirrels of the two meadows cannot compete or breed with each other (no squirrel crosses the ridge), and a population’s individuals must be able to mix.
(c) Explain how the squirrels’ fat store demonstrates an adaptation for getting or using energy in that place. (2 pt)
Frame The fat store is an adaptation for using energy because …
In summer the squirrel eats more seeds than it uses and stores the surplus as fat.
In winter the squirrel breaks the fat down for energy, so it gets energy when the meadow offers none.
So the fat store lets the squirrel use summer’s energy in winter.
- Award 1 point for: the meadow offers no seeds in winter, so the squirrel can take in no energy then. Accept with or without: it still needs energy all winter (a continuous input).
- Award 1 point for: the squirrel stores summer’s surplus as fat and breaks it down in winter, so it uses energy taken in when food was there; accept with or without the word heritable.
APBIO-U08-L28 More births, more deaths
Photo: National Park Service, Wikimedia Commons, public domain (resized).
Suppose a seal colony breeds on a beach. This season, more pups were born than last season, and about the same number of seals died.
Next season, suppose the same number of pups are born, but a hard winter kills more seals. Which way does the colony’s size move each time?
Unit 8 · Ecology
1More births, faster growth
Hundreds of seals of one species lie on a beach, with gulls and crabs among them.
Which of the following is a population?
- A. One sealOne seal is one organism.
A population is all the organisms of one species in one place. - B. ✓ The seals of that one species on the beach
- C. The seals, the gulls and the crabs on the beach togetherSeveral species living together in one place make a community.
A population is one species only.
Why: A population is all the organisms of one species living in one place.
The seals of one species on the beach fit that exactly.
Suppose a female seal takes in more energy than she uses all summer, and stores the surplus as fat. A second female takes in only as much energy as she uses.
Compared with the second female, which of the following describes the number of offspring the first female is likely to produce over the years?
- A. FewerFewer offspring follow when energy out beats energy in.
The first female has stored a surplus. - B. The same numberThe second female has no surplus.
The first has stored energy to spare for young. - C. ✓ More
Why: The first female stored a surplus of energy as fat.
Stored energy is energy to spare for young.
So she is likely to produce more offspring than the second female.
What makes a population’s size go up or go down?
Only two things change a population’s size: how many join the population, and how many leave it. On this beach, seals join by being born and leave by dying.
Video: Watch: More births, faster growth
The colony is drawn as a box on the beach. Pups enter the box along the births arrow, and seals that die leave it along the deaths arrow. The births arrow thickens while the deaths arrow stays the same. The colony’s size climbs faster.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L28a.mp4
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Suppose a seal colony breeds on a beach, like the colony in the photograph below.
Each season, pups are born into the colony. Each season, some seals in the colony die.
Ecologists count how many pups are born per season. The number of individuals born into a population per unit of time is called the , written B.
Ecologists also count how many seals die per season. The number of individuals in a population that die per unit of time is called the , written D.
For the seal colony, B is the number of pups born per season, and D is the number of seals that die per season.
The drawing below shows the colony as a box. Pups that are born join the colony along the births arrow, and seals that die leave it along the deaths arrow.
When more seals join the colony than leave it, the colony grows.
Now consider two seasons of the same colony. Last season, some pups were born and some seals died, and the colony grew.
This season, more pups are born than last season. The same number of seals die as last season.
So more seals join the colony than last season, and the same number leave it. The colony grows faster than last season.
In the drawing below, a thicker arrow means more seals per season. The births arrow is thicker this season, and the deaths arrow is the same.
Raise the birth rate, and hold the death rate still: the population grows faster.
Now suppose a population is shrinking: more members die each year than are born. If more are born and the same number die, the population shrinks more slowly.
What you are expected to know Predict how a population’s growth changes when the birth rate rises and the death rate stays the same: it grows faster, or shrinks more slowly.
Suppose a good acorn year in an oak forest. More mice are born each month than last year, and the same number of mice die.
Compared with last year, how does the mouse population grow?
- A. ✓ Faster
- B. More slowlyMore mice are born and the same number die.
More mice join than before, so the growth speeds up. - C. At the same speedThe birth rate rose.
More mice join each month than before, so the growth is not the same.
Why: More mice are born each month, and the same number die.
So more mice join the population each month than last year.
The population grows faster.
Suppose a pond holds a population of newts. This year fewer eggs hatch than last year, and the same number of newts die.
Compared with last year, how does the newt population grow?
- A. FasterFewer newts join the pond this year, and the same number leave.
The growth slows. - B. ✓ More slowly
- C. At the same speedThe birth rate fell.
Fewer newts join each year than before, so the growth is not the same.
Why: Fewer eggs hatch, so the birth rate fell.
The same number of newts die.
Fewer newts join the population than last year, so it grows more slowly.
The table below gives the births and the deaths in a population of voles in two years.
In which year did the vole population grow faster?
- A. Year 1Deaths were 30 in both years.
Year 2 had more births than Year 1. - B. ✓ Year 2
Why: Deaths were 30 per year in both years.
Year 2 had 90 births, and Year 1 had 70.
More voles joined in Year 2, so the population grew faster in Year 2.
Suppose gulls nest in a colony on a cliff. This year the same number of chicks hatch as last year, and the same number of gulls die.
Compared with last year, how does the gull colony grow?
- A. FasterThe birth rate is unchanged.
The same number of gulls join as last year, so the growth does not speed up. - B. More slowlyThe death rate is unchanged.
The same number leave as last year, so the growth does not slow. - C. ✓ At the same speed
Why: The same number of chicks hatch, and the same number of gulls die.
The same number join and leave the colony as last year.
So the colony grows at the same speed.
Suppose a deer population is shrinking: each year more deer die than are born. This year more fawns are born than last year, and the same number of deer die.
Compared with last year, how does the deer population change?
- A. Shrinks fasterMore fawns join the population, and the same number of deer leave.
The gap between deaths and births narrows, so the shrinking slows. - B. ✓ Shrinks more slowly
- C. Shrinks at the same speedThe birth rate rose.
More deer join than last year, so the loss each year is smaller.
Why: More fawns are born, and the same number of deer die.
Deaths still outnumber births, so the population still shrinks.
The gap is smaller, so it shrinks more slowly.
26Quick quiz: birth rate (B), death rate (D) mixed practice
In one season, 35 seals of a colony die.
Which of the following is that count?
- A. ✓ The colony’s death rate for the season
- B. The colony’s birth rate for the seasonThe birth rate counts pups born, not seals that die.
- C. The colony’s sizeThe size is the number of seals in the colony.
35 is how many died in the season.
Why: The death rate is the number of individuals that die per unit of time.
35 seals died in one season.
So 35 per season is the colony’s death rate.
Ecologists write a population’s birth rate and death rate as two letters.
Which letter is written for the birth rate?
- A. ✓ B
- B. DD is written for the death rate.
Why: The birth rate is written B.
The death rate is written D.
In one season, 62 pups are born into a colony and 40 seals die.
Which number is the colony’s birth rate, B?
- A. 40 per season40 is the number of seals that died: the death rate, D.
- B. ✓ 62 per season
Why: B is the number born per unit of time.
62 pups were born in the season.
So B is 62 per season.
A good spring raises a population’s birth rate.
Which of the following has risen?
- A. ✓ The number born per unit of time
- B. The number that die per unit of timeDeaths per unit of time are the death rate, D.
Why: The birth rate is the number born into a population per unit of time.
So when the birth rate rises, more are born per unit of time.
Ecologists follow a population of seals from season to season.
What is the population’s birth rate?
- A. The number of seals that die per seasonDeaths per season are the death rate.
- B. The number of seals in the populationThe number of seals in the population is its size, not a rate.
- C. ✓ The number of pups born into the population per season
Why: The birth rate is the number of individuals born into a population per unit of time.
For the seals, that is the pups born per season.
Ecologists count a population of seals every season.
(a) State what a population’s death rate is. (1 pt)
- Award 1 point for: the number of individuals that die per unit of time (accept 'per season', 'per year' or 'in a set time').
33More deaths, slower growth
Now the deaths change instead. What happens to the colony when more seals die, and the births stay as they were?
With more deaths and the same births, the colony grows more slowly. Once deaths outnumber births, it shrinks.
Video: Watch: More deaths, slower growth
The colony box is on screen again. The deaths arrow thickens while the births arrow stays the same, and the colony’s size climbs more slowly. The deaths arrow thickens further, past the births arrow, and the size falls. Then the two arrows are drawn the same width, and the size stays steady while pups are still born and seals still die.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L28b.mp4
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Here is the seal colony again, drawn as a box. Pups that are born join it along the births arrow, and seals that die leave it along the deaths arrow.
Now consider a hard winter. This season, the same number of pups are born as last season, but more seals die.
So the same number of seals join the colony as last season, and more leave it. The colony grows more slowly than last season.
In the drawing below, a thicker arrow means more seals per season. The deaths arrow is thicker this season, and the births arrow is the same.
Raise the death rate, and hold the birth rate still: the population grows more slowly.
Suppose the winter is harder still. Now more seals die than pups are born.
More seals leave the colony than join it. So the colony shrinks.
Now suppose the deaths exactly match the births: as many seals die as pups are born.
The same number leave the colony as join it. So the colony’s size stays steady.
A steady size does not mean that nothing is happening. Pups are still being born, and seals are still dying, in equal numbers.
The table below compares the three cases. For each, it gives how the births compare with the deaths, what the colony’s size does, and whether pups are still being born.
What you are expected to know Predict how a population’s growth changes when the death rate rises and the birth rate stays the same: it grows more slowly, shrinks once deaths outnumber births, and stays steady when births equal deaths.
Suppose herons nest in a colony beside a lake. This year the same number of chicks hatch as last year, and more herons die.
Compared with last year, how does the heron colony grow?
- A. FasterThe same number of herons join as last year, and more leave.
The growth slows. - B. ✓ More slowly
- C. At the same speedThe death rate rose.
More herons leave each year than before, so the growth is not the same.
Why: The same number of chicks hatch, and more herons die.
So more herons leave the colony than last year, and the same number join.
The colony grows more slowly.
Suppose hares live on a moor. This year fewer hares die than last year, and the same number of young are born.
Compared with last year, how does the hare population grow?
- A. ✓ Faster
- B. More slowlyFewer hares leave the population, and the same number join.
The growth speeds up. - C. At the same speedThe death rate fell.
Fewer hares leave each year than before, so the growth is not the same.
Why: Fewer hares die, so the death rate fell.
The same number of young are born.
Fewer hares leave the population than last year, so it grows faster.
The table below gives one year’s births and deaths in a population of terns.
What does the tern population’s size do over that year?
- A. ✓ Grows
- B. Stays steadyThe births and the deaths are not equal.
84 terns join and 51 leave. - C. ShrinksMore terns are born than die.
More join than leave, so the size does not fall.
Why: 84 terns are born and 51 die.
More terns join the population than leave it.
So the population grows.
The table below gives one year’s births and deaths in a population of shrews.
What does the shrew population’s size do over that year?
- A. GrowsAs many shrews die as are born.
The same number leave as join, so the size does not rise. - B. ✓ Stays steady
- C. ShrinksAs many shrews are born as die.
The same number join as leave, so the size does not fall.
Why: 120 shrews are born and 120 die.
The same number join the population as leave it.
So the population’s size stays steady.
The table below gives one year’s births and deaths in a population of pike.
What does the pike population’s size do over that year?
- A. GrowsMore pike die than are born.
More leave than join, so the size does not rise. - B. Stays steadyThe births and the deaths are not equal.
36 pike join and 58 leave. - C. ✓ Shrinks
Why: 36 pike are born and 58 die.
More pike leave the population than join it.
So the population shrinks.
The table below gives one year’s births and deaths in a population of tortoises.
What does the tortoise population’s size do over that year?
- A. ✓ Grows
- B. Stays steadyThe births and the deaths are not equal.
15 tortoises join and 9 leave. - C. ShrinksMore tortoises are born than die.
More join than leave, so the size does not fall.
Why: 15 tortoises are born and 9 die.
More tortoises join the population than leave it.
So the population grows.
The table below gives one year’s births and deaths in a population of sparrows.
What does the sparrow population’s size do over that year?
- A. GrowsMore sparrows die than are born.
More leave than join, so the size does not rise. - B. Stays steadyThe births and the deaths are not equal.
210 sparrows join and 260 leave. - C. ✓ Shrinks
Why: 210 sparrows are born and 260 die.
More sparrows leave the population than join it.
So the population shrinks.
The table below gives one year’s births and deaths in a population of beavers.
What does the beaver population’s size do over that year?
- A. GrowsAs many beavers die as are born.
The same number leave as join, so the size does not rise. - B. ✓ Stays steady
- C. ShrinksAs many beavers are born as die.
The same number join as leave, so the size does not fall.
Why: 27 beavers are born and 27 die.
The same number join the population as leave it.
So the population’s size stays steady.
A student says: “The number of voles in a field has stayed the same for three years, so voles are still being born there, and the same number die.”
Is the student correct?
- A. No: a steady size means no births and no deathsVoles are still born each year, and the same number die.
A steady size means the births and the deaths are equal, not zero. - B. ✓ Yes: a steady size means the births and the deaths are equal, not zero
Why: A population’s size stays steady when births equal deaths.
Voles are still born each year.
The same number of voles die each year.
So the count stays the same while births continue.
Here is the seal colony again, on its beach.
This season, more pups were born than last season, and about the same number of seals died. So the colony grew faster than last season.
Next season, if the same number of pups are born and a hard winter kills more seals, the colony will grow more slowly.
If as many seals died as pups were born, the colony’s size would stay steady, with pups still being born and seals still dying.
Glossary
- birth rate (B)
- The number of individuals born into a population per unit of time, such as the pups born into a seal colony per season. Written B.
- death rate (D)
- The number of individuals in a population that die per unit of time, such as the seals of a colony that die per season. Written D.
APBIO-U08-L28B The equation: births minus deaths
Photo: National Park Service, Wikimedia Commons, public domain (resized).
Here is the seal colony again, on its beach. Suppose that this season, 48 pups were born into the colony and 19 seals died.
How many more seals are there than last season? And how does the formula sheet write that?
Unit 8 · Ecology
1Births minus deaths, as the formula sheet writes it
Ecologists count a seal colony every season.
Which of the following is the colony’s birth rate, B?
- A. The number of seals in the colonyThe number of seals in the colony is its size, not a rate.
- B. ✓ The number of pups born into the colony per season
- C. The number of seals in the colony that die per seasonThe seals that die per season are the death rate, D.
Why: The birth rate is the number of individuals born into a population per unit of time.
For the colony, that is the pups born per season.
In an experiment, 12 mL of oxygen formed in 4 minutes.
Which of the following is the rate at which the oxygen formed?
- A. 12 mL12 mL is the amount that formed, not an amount per unit of time.
- B. 4 minutes4 minutes is the time taken, not an amount per unit of time.
- C. ✓ 3 mL per minute
Why: A rate is an amount of change per unit of time.
12 mL formed in 4 minutes.
So the rate is 3 mL per minute.
How does the formula sheet write a population’s births minus its deaths?
The rate at which a population’s size changes is its birth rate minus its death rate.
The formula sheet prints that as one line. Here it is, with what each symbol means beneath it.
dN: the change in population size, in individuals (for the colony, seals)
dt: the change in time (for the colony, one season)
B: the birth rate, in individuals born per unit of time (pups born per season)
D: the death rate, in individuals that die per unit of time (seals that die per season)
dN/dt: the rate of change of population size, in individuals per unit of time (seals per season)
Video: Watch: Births minus deaths, as the formula sheet writes it
The colony box is on screen, with this season’s counts on its two arrows. The equation appears beneath it on its own line, births minus deaths. Each symbol lights up as it is named, and its meaning and unit appear under it. The line with the seal units shows that a rate minus a rate leaves a rate.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L28Ba.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L28Ba.mp4
Here is the seal colony again, on its beach.
Suppose that this season, 48 pups were born into the colony and 19 seals died.
The drawing below shows the colony as a box. Pups that were born join it along the births arrow, and seals that died leave it along the deaths arrow.
More seals joined the colony than left it. So the colony grew this season.
How much the colony grew is its births minus its deaths.
Ecologists count the seals in the colony. The number of individuals in a population is called the , written N.
The letter d in front of a symbol means a change in that quantity. So dN is the change in the population size N, and dt is the change in time.
For the colony, dN is the number of seals gained this season, and dt is one season.
dN divided by dt is the change in the population size per unit of time. The change in a population’s size per unit of time is called the , written dN/dt.
A rate is an amount of change per unit of time. So dN/dt is a rate, in seals per season, not a count of seals.
The table below compares N with dN/dt. For each, it gives what it is and its unit for the colony.
The formula sheet prints the equation on its own line, exactly as below. Beneath it is what each symbol means, with its unit.
dN: the change in population size, in individuals (for the colony, seals)
dt: the change in time (for the colony, one season)
B: the birth rate, in individuals born per unit of time (pups born per season)
D: the death rate, in individuals that die per unit of time (seals that die per season)
dN/dt: the rate of change of population size, in individuals per unit of time (seals per season)
B and D are both in seals per season. So dN/dt is in seals per season too, as the line below shows.
seals per season minus seals per season leaves seals per season: dN/dt carries the same unit as B and D
Now suppose some seals swim in from another beach and join the colony, and some swim away and leave it.
Seals arriving count with the births, B. Seals leaving count with the deaths, D.
So the same equation covers movement in and out of the colony.
What you are expected to know Read the population-growth equation in its formula-sheet form, dN/dt = B − D, and state what each symbol means: dN the change in population size, dt the change in time, B the birth rate, D the death rate.
In the population-growth equation, each symbol has one meaning.
Which symbol is written for the change in time?
- A. NN is the population’s size.
- B. dNdN has a d in front of N, so dN is a change in the population’s size.
- C. ✓ dt
Why: A d in front of a symbol means a change in that quantity.
t is time.
So dt is the change in time.
In the population-growth equation, each symbol has one meaning.
Which symbol is written for the change in the population’s size?
- A. NN on its own is the population’s size, with no d in front.
- B. ✓ dN
- C. dtdt has a d in front of t, so dt is a change in time.
Why: A d in front of a symbol means a change in that quantity.
N is the population size.
So dN is the change in the population size.
In the population-growth equation, each symbol has one meaning.
Which symbol is written for the population’s size?
- A. ✓ N
- B. dNdN has a d in front of N, so dN is a change in the population’s size.
- C. dtdt has a d in front of t, so dt is a change in time.
Why: N is the population size, the number of individuals in the population.
A d in front of a symbol means a change in that quantity, so N on its own is the size itself.
In the population-growth equation, each symbol has one meaning.
Which symbol is written for the death rate?
- A. BB is the birth rate: the number born into the population per unit of time.
- B. ✓ D
- C. NN is the population size: the number of individuals in the population.
Why: The death rate is the number of individuals that die per unit of time.
The formula sheet writes the death rate as D.
Suppose ecologists follow a population of otters in a river. They find that dN/dt for the otters is 12 otters per year.
Which of the following is that 12 otters per year?
- A. A count of the otters in the riverA count of otters would be a number of otters.
12 otters per year is a number of otters per unit of time. - B. ✓ A rate of change in the number of otters
Why: dN/dt is the change in the population’s size per unit of time.
A change per unit of time is a rate.
So 12 otters per year is a rate, not a count.
A student reads that a population of badgers on a hillside has a dN/dt of 8 badgers per year, and says: “So the hillside gains 8 badgers each year.”
Is the student correct?
- A. ✓ Yes: 8 badgers per year is how much the population’s size changes each year
- B. No: dN/dt is the number of badgers on the hillside, not a change in itdN/dt is a rate, in badgers per year, not a count of badgers.
The number of badgers on the hillside is N.
Why: dN/dt is the rate of change of population size.
A rate of change is a change per unit of time.
So 8 badgers per year means the hillside gains 8 badgers each year.
31Quick quiz: dN/dt (rate of change of population size), N (population size) mixed practice
At the start of a year, ecologists count 350 wolves in a population.
Which symbol is written for that count?
- A. ✓ N
- B. dN/dtdN/dt is a change per unit of time.
350 wolves is a count at one moment.
Why: N is the population size, the number of individuals in the population.
350 wolves is that number at the start of the year.
Over one year, a wolf population gains 7 wolves.
Which symbol is written for 7 wolves per year?
- A. NN is a count of wolves at one moment.
7 wolves per year is a change per unit of time. - B. ✓ dN/dt
Why: dN/dt is the change in the population size per unit of time.
7 wolves per year is a change per year.
So 7 wolves per year is dN/dt.
A seal colony is counted every season.
Which unit does the colony’s dN/dt carry?
- A. ✓ Seals per season
- B. SealsSeals is the unit of N, a count.
dN/dt is a change per season.
Why: B and D are both in seals per season.
dN/dt is B minus D, so dN/dt is in seals per season too.
In the population-growth equation, each symbol has one meaning.
Which of the following is dN/dt?
- A. The number of individuals in the populationThe number of individuals in the population is N.
- B. The number of individuals that die per unit of timeThe number that die per unit of time is the death rate, D.
- C. ✓ The change in the population’s size per unit of time
Why: dN is the change in the population size, and dt is the change in time.
dN divided by dt is the change in the population’s size per unit of time.
Ecologists count a population of wolves every year and follow its growth with the population-growth equation.
(a) State what N is in the population-growth equation. (1 pt)
- Award 1 point for: the population size (accept 'the number of individuals in the population' or 'the number of wolves').
37Calculate the rate of change
How many more seals does the colony have than last season? The two counts go into the equation.
Video: Watch: Calculate the rate of change
The calculation is worked on screen for the colony. The values are written down with their units, then the equation on its own line, then the substitution, then the answer with its unit, seals per season. A second population is worked the same way, and its answer comes out negative.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L28Bb.mp4
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Here is the seal colony again. This season, 48 pups were born into the colony and 19 seals died.
To calculate dN/dt, write down the values, write down the equation, then substitute the values in and calculate. The answer carries its unit, seals per season.
This season, 48 pups were born into the seal colony and 19 seals died. Calculate dN/dt for the colony.
So the colony gained 29 seals this season. That is how many more seals it has than last season.
dN/dt came out positive. When more are born than die, B is larger than D, and dN/dt is positive: the population grows.
Now consider a population of foxes in a forest. Suppose that in one year, 24 fox cubs are born and 37 foxes die.
More foxes died than were born. So D is larger than B, and dN/dt comes out negative.
In one year, 24 fox cubs are born into a fox population and 37 foxes die. Calculate dN/dt for the population.
A negative dN/dt means the population is shrinking. Keep the minus sign in the answer, because the sign says which way the size moves.
The table below sets a positive, a zero and a negative dN/dt beside which rate is larger and what the population’s size does.
What you are expected to know Calculate dN/dt from a birth rate and a death rate, with its unit, keeping the minus sign when deaths outnumber births.
A population’s birth rate, B, and death rate, D, are both known.
Which of the following gives dN/dt?
- A. ✓ B − D
- B. D − BD minus B gives the right size with the wrong sign.
A growing population would come out negative. - C. B + DAdding the deaths to the births counts the individuals that died as if they had joined the population.
Why: Births add individuals to the population, and deaths remove them.
So the change per unit of time is the births minus the deaths.
The formula sheet writes that as dN/dt = B − D.
Suppose a herd of wild goats lives on an island. In one year, 63 kids are born into the herd and 27 goats die.
Calculate dN/dt for the herd.
Part 1. State the herd’s birth rate, B, for the year.
Answer: 63 goats per year (tolerance ±0)
Part 2. State the herd’s death rate, D, for the year.
Answer: 27 goats per year (tolerance ±0)
Answer: 36 goats per year (tolerance ±0)
Suppose a population of marmots lives in a mountain valley. In one year, 34 marmots are born and 51 marmots die.
Calculate dN/dt for the population, with its sign.
Answer: -17 marmots per year (tolerance ±0)
Here is the seal colony again, on its beach, with 48 pups born and 19 seals dead this season.
The formula sheet writes the colony’s change as dN/dt = B − D.
So the colony’s dN/dt is 29 seals per season, as the line below shows. The colony has 29 more seals than last season.
the colony this season: 48 pups born, 19 seals died, so dN/dt is 29 seals per season
57Numeric practice: births minus deaths mixed practice
Suppose rabbits live on a heath. In one year, 210 rabbits are born and 145 rabbits die.
Calculate dN/dt for the rabbit population.
Answer: 65 rabbits per year (tolerance ±0)
Suppose cranes nest in a marsh. In one year, 38 chicks hatch and 47 cranes die.
Calculate dN/dt for the crane population, with its sign.
Answer: -9 cranes per year (tolerance ±0)
Suppose elk live in a national park. In one year, 72 calves are born into the population, 15 elk walk in from a neighboring valley and join the population, and 41 elk die. Count each elk that walks in as one joiner, like a birth.
Calculate dN/dt for the elk population.
Answer: 46 elk per year (tolerance ±0)
Suppose bats roost in a cave. In one year, 97 bats are born and 58 bats die.
Calculate dN/dt for the bat population.
Answer: 39 bats per year (tolerance ±0)
Suppose lizards live on a sand dune. In one year, 154 lizards hatch and 176 lizards die.
Calculate dN/dt for the lizard population, with its sign.
Answer: -22 lizards per year (tolerance ±0)
Glossary
- N (population size)
- The number of individuals in a population, such as the number of seals in a colony. Written N.
- dN/dt (rate of change of population size)
- The change in a population’s size per unit of time, such as the seals a colony gains per season. A rate, not a count. Written dN/dt, and equal to the birth rate minus the death rate, B − D.
APBIO-U08-L29 How many next year?
Photo: National Park Service, Wikimedia Commons, public domain (resized).
Here is the seal colony again, on its beach. Suppose it holds 500 seals now, and it gains 29 seals every season.
Three seasons from now, how many seals will the colony hold?
Unit 8 · Ecology
1From a rate to a size
In one year, 57 calves are born into a herd of bison and 24 bison die.
Which of the following gives the herd’s rate of change of population size, dN/dt, for that year?
- A. The births plus the deaths, B + DAdding the deaths counts the bison that left the herd as if they had joined it.
The rate of change is the births minus the deaths. - B. ✓ The births minus the deaths, B − D
- C. The deaths minus the births, D − BThe deaths minus the births counts the bison that died as a gain and the calves born as a loss.
The rate of change is the births minus the deaths.
Why: Calves born join the herd, and bison that die leave it.
The rate of change of population size is the births minus the deaths.
So dN/dt = B − D.
Ecologists find that a herd of caribou has a rate of change of population size of 15 caribou per year.
What does that rate mean?
- A. The herd has 15 caribouThe number of caribou in the herd is its population size, N.
A rate of change is individuals per unit of time, not a count. - B. The herd gains 15 caribou once, in totalA rate is per unit of time.
15 caribou per year is a gain of 15 caribou in each year, not once. - C. ✓ The herd gains 15 caribou in every year
Why: A rate of change of population size is individuals per unit of time.
15 caribou per year means the herd gains 15 caribou in each year that passes.
How do you get from a population’s rate of change to its size at a later time?
Multiply the rate of change by the time, then add the result to the size you started with.
Video: Watch: From a rate to a size
The colony is drawn as a box holding 500 seals. One season passes, and 29 seals are added to the box. Two more seasons pass, and the same number of seals is added each time. The calculation is then written out step by step, and a question mark lands over the words “if the rate stays the same”.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L29a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L29a.mp4
Here is the seal colony again, on its beach, like the colony in the photograph below.
Suppose the colony holds 500 seals now. So its population size, N, is 500 seals now.
Suppose that in each season, 29 more pups are born into the colony than seals die.
So the colony’s rate of change of population size, dN/dt, is 29 seals per season.
A rate of 29 seals per season means the colony gains 29 seals in every season.
Now consider three seasons. In each of the three seasons, the colony gains 29 seals.
The drawing below shows the colony as a box, season by season. Each arrow is one season, and the size after each season is written in its box.
The gain over the three seasons is the rate of change multiplied by the time.
Add that gain to the 500 seals the colony started with. The result is the colony’s size after three seasons.
For any population: multiply the rate of change by the time, then add the result to the present size. Here is that rule as an equation on its own line, with its key beneath.
later size: the population size, N, after the time has passed, in individuals
present size: the population size, N, now, in individuals
rate of change: dN/dt, the individuals the population gains per unit of time, taken as the same in every unit of time
time: how many units of time pass, such as seasons or years
The rate of change is in seals per season, and the time is in seasons. So the gain is in seals, as the first line below shows.
Seals added to seals is a size in seals, as the second line below shows.
seals per season × seasons = seals, so the gain is a number of seals.
seals + seals = seals, so the later size is a number of seals.
The colony holds 500 seals and gains 29 seals per season. Calculate its size after 3 seasons.
We made one simplification here. We assumed the colony gains 29 seals in every one of the three seasons.
A real colony’s gain often changes from one season to the next. So the answer is the size the colony would reach if the rate stayed at 29 seals per season.
A rate of change can be negative, when more die than are born. Then the gain is negative, and the population is smaller after the time.
The working is the same. Keep the minus sign on the rate, and the answer comes out smaller than the present size.
What you are expected to know Calculate a population’s size after a stated time from its present size and a constant rate of change.
An ecologist counts a herd today and measures how many it gains each year. She wants to know how many the herd will hold in several years.
Which of the following gives the number the herd will hold in several years?
- A. rate of change × timeThe rate of change multiplied by the time is the gain over that time.
The population also still has the individuals it started with. - B. ✓ present size + rate of change × time
- C. present size × rate of change × timeThe present size is added to the gain, not multiplied by it.
Multiplying gives a number far larger than the population ever reaches.
Why: The rate of change multiplied by the time is the gain over that time.
Add the gain to the present size.
The sum is the population’s size after that time.
Suppose coots live on a lake. The population holds 240 coots now, and its rate of change of population size is 17 coots per year.
Calculate the size of the coot population after 4 years.
Part 1. Calculate the gain in coots over the 4 years.
Answer: 68 coots (tolerance ±0)
Answer: 308 coots (tolerance ±0)
Suppose a puffin colony nests on an island. It holds 860 puffins now. Each year, 35 more puffins die than chicks hatch, so its rate of change of population size is −35 puffins per year.
Calculate the size of the puffin colony after 6 years.
Answer: 650 puffins (tolerance ±0)
Suppose pikas live on a mountain slope. The population holds 130 pikas now, and its rate of change of population size is 8 pikas per year.
Calculate the size of the pika population after 7 years.
Answer: 186 pikas (tolerance ±0)
Here is the seal colony again, on its beach: 500 seals now, gaining 29 seals every season.
After three seasons, the colony holds 587 seals, if the rate stays at 29 seals per season.
If the rate changes, the size after three seasons changes too.
32Numeric practice: size from a rate and a time mixed practice
Suppose spoonbills nest in a marsh. The population holds 410 spoonbills now, and its rate of change of population size is 23 spoonbills per year.
Calculate the size of the spoonbill population after 5 years.
Answer: 525 spoonbills (tolerance ±0)
Suppose ibex live on a ridge. The population holds 275 ibex now. Each year, 12 more ibex die than kids are born, so its rate of change of population size is −12 ibex per year.
Calculate the size of the ibex population after 4 years.
Answer: 227 ibex (tolerance ±0)
Suppose porcupines live in a forest. The population holds 96 porcupines now, and its rate of change of population size is 14 porcupines per year.
Calculate the size of the porcupine population after 6 years.
Answer: 180 porcupines (tolerance ±0)
Suppose egrets nest beside a lake. The colony holds 58 egrets now, and its rate of change of population size is 9 egrets per season.
Calculate the size of the egret colony after 4 seasons.
Answer: 94 egrets (tolerance ±0)
Suppose toads breed in a pond. The population holds 640 toads now, and its rate of change of population size is 31 toads per year.
Calculate the size of the toad population after 5 years.
Answer: 795 toads (tolerance ±0)
APBIO-U08-L30 Each one makes more
Suppose a flask of broth holds 1 000 bacteria, with food to spare. An hour later it holds 2 000. An hour after that, it holds 4 000.
The gain in the first hour was 1 000 bacteria, and the gain in the second hour was 2 000. Why did the gain grow, when nothing about the bacteria changed?
Unit 8 · Ecology
1Per head: how much each one adds
Suppose ecologists follow a population of brine shrimp in a tank. They find that its rate of change of population size, dN/dt, is 80 brine shrimp per day.
Which of the following is that dN/dt?
- A. The number of brine shrimp in the tankThe number of brine shrimp in the tank is the population size, N.
dN/dt is a rate: shrimp per unit of time. - B. ✓ The number of brine shrimp added to the tank in each day
- C. The number of days the population has been growingThe days are the time.
dN/dt is the change in the number of shrimp per unit of time.
Why: dN/dt is the rate of change of population size.
A rate is an amount of change per unit of time.
So 80 brine shrimp per day is the number of shrimp added to the tank in each day.
How much does each individual add to its population in an hour?
Each individual adds a certain number of new individuals per hour. Ecologists count that number per head: for each individual.
Video: Watch: Per head, how much each one adds
Two flasks hold the same number of bacteria, with food to spare. In one hour the left flask gains more bacteria than the right flask. So each bacterium in the left flask added more new bacteria than each bacterium in the right flask. The left flask’s species adds more per bacterium: its rate per head is higher.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L30a.mp4
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Here is the flask of broth again: 1 000 bacteria, then 2 000 an hour later, then 4 000 an hour after that.
The bacteria have food to spare. In one hour, each bacterium divides in two.
So each bacterium adds one new bacterium to the flask in that hour.
The number of new individuals that each individual adds per unit of time is called the . Per capita means per head: for each individual.
With food to spare, no predators and no disease, each individual adds the most new individuals it can.
That greatest per capita growth rate is called the , written rmax.
Now consider two flasks, each holding 400 bacteria of a different species, both with food to spare. In one hour the left flask gains 300 bacteria and the right flask gains 100 bacteria.
The same number of bacteria added different numbers of new bacteria. So each bacterium in the left flask added more, on average, than each bacterium in the right flask.
So the left flask’s species has the higher maximum per capita growth rate, rmax.
A bacterium can divide about once an hour. A human takes far longer to reproduce.
So a bacterium’s maximum per capita growth rate, rmax, is far higher than a human’s.
Now consider two populations of the same size, N, both with food to spare. Each individual in the population with the higher maximum per capita growth rate, rmax, adds more new individuals per hour.
The same number of individuals, each adding more, adds more in all. So the population with the higher maximum per capita growth rate, rmax, grows faster.
What you are expected to know Explain what rmax measures, the greatest number of new individuals each individual adds per unit of time, and predict that a population with a higher rmax grows faster from the same size.
Suppose a flask of bacteria has food to spare.
Which of the following does rmax measure?
- A. ✓ The greatest number of new bacteria each bacterium adds per hour
- B. The greatest number of new bacteria the whole flask adds per hourThe whole flask’s gain per hour is its rate of change of population size, dN/dt.
rmax is a rate per bacterium. - C. The number of bacteria in the flaskThe number of bacteria in the flask is its population size, N.
rmax is a rate per bacterium per hour.
Why: Per capita means per head.
rmax is the greatest number of new bacteria each bacterium adds per hour, with food to spare.
Suppose two ponds each hold 350 water fleas, with food to spare. The species in the north pond has a higher maximum per capita growth rate, rmax, than the species in the south pond.
Which pond’s population grows faster?
- A. ✓ The north pond
- B. The south pondEach water flea in the south pond adds fewer new water fleas per day.
The same number of water fleas, each adding fewer, adds fewer in all. - C. Neither: both grow at the same rateThe two populations are the same size, but each water flea in the north pond adds more new water fleas per day.
So the two gains differ.
Why: Both ponds hold 350 water fleas.
Each water flea in the north pond adds more new water fleas per day than each water flea in the south pond.
The same number of water fleas, each adding more, adds more in all.
So the north pond’s population grows faster.
Suppose two tanks each hold 620 yeast cells, with food to spare. Each cell in the left tank adds more new cells per hour than each cell in the right tank.
Which tank gains more yeast cells in the next hour?
- A. The right tankEach cell in the right tank adds fewer new cells per hour.
The same number of cells, each adding fewer, adds fewer in all. - B. ✓ The left tank
- C. Neither: both tanks gain the same numberThe two tanks hold the same number of cells, but each cell in the left tank adds more.
So the left tank gains more.
Why: Both tanks hold 620 yeast cells.
Each cell in the left tank adds more new cells per hour.
So the left tank gains more yeast cells in the next hour.
Suppose two islands each hold 90 pheasants, with food to spare. The pheasants on the west island belong to a species with a lower maximum per capita growth rate, rmax, than the species on the east island.
Which island’s population grows faster?
- A. The west islandEach pheasant on the west island adds fewer young per year.
The same number of pheasants, each adding fewer, adds fewer in all. - B. ✓ The east island
- C. Neither: both grow equally fastThe two populations are the same size, but each pheasant on the east island adds more young per year.
So the east island’s population grows faster.
Why: Both islands hold 90 pheasants.
Each pheasant on the east island adds more young per year than each pheasant on the west island.
So the east island’s population grows faster.
Suppose an upper meadow and a lower meadow each hold 120 moles of one species, with food to spare, so rmax is the same in both.
Which meadow gains more moles in the next year?
- A. The upper meadow, more than the lowerThe upper meadow’s moles each add the same number of young per year as the lower meadow’s.
The counts are equal, so the gains are equal. - B. The lower meadow, more than the upperThe lower meadow holds as many moles as the upper meadow, and each mole adds the same number of young.
So the gains are equal. - C. ✓ Neither: both meadows gain the same number
Why: Both meadows hold 120 moles.
Each mole adds the same number of young per year in both meadows.
The same number of moles, each adding the same number, adds the same in all.
So both meadows gain the same number of moles.
Suppose two aquaria each hold 80 snails, with food to spare. Each snail in the right aquarium adds more young per week than each snail in the left aquarium.
Which aquarium’s population gains fewer snails in the next week?
- A. ✓ The left aquarium
- B. The right aquariumEach snail in the right aquarium adds more young per week.
The same number of snails, each adding more, adds more in all. - C. Neither: both gain the same numberThe two aquaria hold the same number of snails, but each snail in the right aquarium adds more young.
So the gains differ.
Why: Both aquaria hold 80 snails.
Each snail in the left aquarium adds fewer young per week than each snail in the right aquarium.
So the left aquarium’s population gains fewer snails.
Suppose two bottles of pond water each hold 250 algal cells, with light and nutrients to spare. The species in the right bottle has the higher maximum per capita growth rate, rmax.
After one day, which bottle holds more algal cells?
- A. The left bottleEach algal cell in the left bottle adds fewer new cells per day.
The same number of cells, each adding fewer, adds fewer in all. - B. ✓ The right bottle
- C. Neither: both bottles hold the same numberThe two bottles start with the same number of cells, but each cell in the right bottle adds more per day.
So the right bottle gains more.
Why: Both bottles start with 250 algal cells.
Each cell in the right bottle adds more new cells per day.
So the right bottle gains more cells, and after one day it holds more.
27Quick quiz: rmax (maximum per capita growth rate), per capita mixed practice
Suppose the diatoms in a bottle of sea water add 480 new diatoms in a day, in all.
Which of the following is 480 diatoms per day?
- A. A per capita rate, for each diatomA per capita rate is per diatom.
480 diatoms per day is what all the diatoms together added. - B. ✓ The population’s rate of change, dN/dt
Why: 480 diatoms per day is the whole bottle’s gain per day.
So it is the population’s rate of change of population size, dN/dt.
Suppose each grasshopper in a field adds, on average, 30 young per year.
Which of the following is 30 young per year?
- A. ✓ A per capita rate, for each grasshopper
- B. The population’s rate of change, dN/dtThe whole population’s rate of change is what all the grasshoppers together add per year.
30 young per year is for each grasshopper.
Why: 30 young per year is what each grasshopper adds.
A rate for each individual is a per capita rate.
Suppose a population of yeast cells grows with food to spare.
Which of the following units does rmax carry?
- A. CellsCells is the unit of the population size, N.
rmax is a rate per cell per unit of time. - B. Cells per hourCells per hour is the unit of the whole population’s rate of change, dN/dt.
rmax is per cell. - C. ✓ New cells per cell per hour
Why: rmax is the greatest number of new cells each cell adds per hour.
So its unit is new cells per cell per hour.
Ecologists often give a population’s growth rate per capita.
Which of the following does per capita mean?
- A. ✓ For each individual
- B. For the whole populationThe whole population’s rate is its rate of change of population size, dN/dt.
Per capita is per head: for each individual. - C. For each hourPer hour is a unit of time.
Per capita means for each individual.
Why: Per capita means per head.
A per capita rate is the rate for each individual.
A flask of bacteria has food to spare, no predators and no disease.
(a) State what rmax measures for the bacteria in the flask. (1 pt)
- Award 1 point for: the greatest number of new individuals each individual adds per unit of time (the maximum per capita growth rate).
33The more there are, the more are added
Suppose a colony of storks holds 140 storks now, and this year it grew by 12 storks. An ecologist wants the colony’s size 5 years from now. She multiplies this year’s growth by the time and adds the result to the present size.
Which of the following did the ecologist assume?
- A. ✓ The colony’s yearly growth stays the same
- B. The colony’s yearly growth keeps risingMultiplying one yearly gain by the time treats every year’s gain as the same.
A growing gain would need a different gain for each year. - C. The colony’s yearly growth keeps fallingMultiplying one yearly gain by the time treats every year’s gain as the same.
A shrinking gain would need a different gain for each year.
Why: The calculation multiplies one rate of change by the time.
That is only right if the gain is the same in every year.
So the ecologist assumed the colony adds the same number of storks every year.
Why does a growing population add more individuals each hour, when nothing about its individuals has changed?
Each individual adds the same number of new individuals every hour. So the more individuals there are, the more new individuals they add each hour.
Video: Watch: The more there are, the more are added
The flask holds 1 000 bacteria, and each bacterium adds one new bacterium in the hour. The flask then holds 2 000, and each of those adds one more in the next hour. The gain grows every hour because more bacteria are adding. The curve of N against time bends upward into a J, and the bars for the gain in each hour grow with it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L30b.mp4
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Here is the flask again: 1 000 bacteria, then 2 000 an hour later, then 4 000 an hour after that.
Suppose the flask holds 1 000 bacteria at the start, with food to spare.
In the first hour, each of the 1 000 bacteria adds one new bacterium. So the flask gains 1 000 bacteria, and holds 2 000 at the end of the hour.
In the second hour, each of the 2 000 bacteria adds one new bacterium. So the flask gains 2 000 bacteria, and holds 4 000.
The table below sets out the flask hour by hour: the bacteria at the start of the hour, what each bacterium adds, the gain in the hour, and the bacteria at the end.
In every hour, each bacterium adds the same number of new bacteria: one. So the per capita growth rate stays the same from hour to hour.
The number of bacteria, N, does not stay the same. At the start of every hour, N is larger than it was an hour before.
More bacteria, each adding the same number, add more in all. So the flask’s gain grows every hour.
So a population does not always gain the same number every hour. In this flask, more bacteria are adding in every hour than in the hour before.
When a population’s gain grows from each hour, week or year to the next because more individuals are each adding the same number, it is called .
The graph below plots the flask’s population size, N, on the y-axis against time in hours on the x-axis.
The curve bends upward: it gets steeper every hour. A curve of this shape is called a , because it looks like the letter J.
Each individual adds the same number per unit of time, so a bigger population adds more.
So the curve bends upward.
Under the curve, each bar is the flask’s gain in one hour. Each bar is taller than the one before, because N was larger at the start of that hour.
The rate per bacterium stays the same: one new bacterium per bacterium per hour. The population’s rate does not stay the same, because N keeps rising.
What you are expected to know Explain why a population whose individuals each reproduce at the same per capita rate adds more individuals in each hour, week or year as it grows, so its size curve bends upward.
Suppose a jar of rotifers with food to spare gained 260 rotifers in the last day. Each rotifer adds the same number of young per day as before.
Which of the following is the jar’s gain in the next day?
- A. Fewer than 260 rotifersEach rotifer adds as many young as before, and there are more rotifers than before.
The gain cannot fall. - B. 260 rotifersA gain of 260 rotifers in every day would need the same number of rotifers adding in every day.
The jar now holds 260 more rotifers than a day ago. - C. ✓ More than 260 rotifers
Why: Each rotifer adds the same number of young per day.
The jar holds more rotifers than it did a day ago.
More rotifers, each adding the same number, add more in all.
So the gain in the next day is more than 260 rotifers.
Suppose a few duckweed plants float onto a pond with room and nutrients to spare. Every week, ecologists count the plants. The pond gains more duckweed plants each week than the week before.
(a) Explain why the pond’s gain grows each week. (2 pt)
Frame The gain grows each week because …
Each week the pond holds more plants than the week before.
More plants, each adding the same number, add more new plants in all.
So the pond gains more plants each week than the week before.
- Award 1 point for: each plant adds the same number of new plants every week (the per capita growth rate stays the same).
- Award 1 point for: the pond holds more plants each week, and more plants each adding the same number add more in all, so the gain grows.
A student watches a flask of bacteria with food to spare. The flask gains more bacteria every hour. The student says: “Each bacterium must be dividing faster and faster as the hours pass.”
Is the student correct?
- A. ✓ No: each bacterium adds the same number per hour, and more bacteria are adding
- B. Yes: the gain grows every hour, so each bacterium must be adding more each hourEach bacterium divides about once an hour, the same in every hour.
The gain grows because more bacteria are dividing.
Why: Each bacterium adds one new bacterium every hour, the same in every hour.
The flask holds more bacteria every hour.
More bacteria, each adding one, add more in all.
So the gain grows while the rate per bacterium stays the same.
Suppose a few starlings arrive on a large island with food to spare and no predators. Each starling adds the same number of young every year. Ecologists plot the starlings’ population size, N, on the y-axis against time in years on the x-axis.
Which of the following is the shape of the line?
- A. A straight line rising at the same steepness every yearA straight line rising at one steepness adds the same number of starlings every year.
More starlings each year, each adding the same number, add more every year. - B. ✓ A curve bending upward, steeper every year
- C. A line staying levelA level line is a population whose size does not change.
Each starling adds young every year, so N rises.
Why: Each starling adds the same number of young every year.
Each year there are more starlings than the year before.
So the gain grows every year, and the curve bends upward: a J-shaped curve.
Here is the flask again: 1 000 bacteria, then 2 000, then 4 000.
Each bacterium adds the same number of new bacteria every hour, so rmax stays the same.
N keeps rising, so the gain each hour rises with it, and the curve bends upward.
62Quick quiz: exponential growth, J-shaped curve mixed practice
Suppose a flask of bacteria grows exponentially, with food to spare.
Which of the following stays the same from one hour to the next?
- A. ✓ The number of new bacteria each bacterium adds in the hour
- B. The number of bacteria in the flaskThe number of bacteria in the flask, N, is larger every hour.
- C. The number of new bacteria the flask adds in the hourThe flask’s gain grows every hour, because more bacteria are adding.
Why: In exponential growth each bacterium adds the same number of new bacteria every hour.
N and the flask’s gain both grow.
Suppose a flask of bacteria grows exponentially, with food to spare.
Which of the following is the gain in the second hour, compared with the gain in the first hour?
- A. SmallerMore bacteria are adding in the second hour than in the first.
The gain grows. - B. The sameThe same gain in both hours would need the same number of bacteria adding in both.
There are more bacteria in the second hour. - C. ✓ Larger
Why: At the start of the second hour the flask holds more bacteria than at the start of the first.
Each bacterium adds the same number.
So the gain in the second hour is larger.
Suppose paramecia grow in a dish with food to spare. The graph below plots the population size, N, on the y-axis against time in days on the x-axis.
Which of the following does the curve show about the dish’s gain from one day to the next?
- A. The gain is smaller every dayA gain that shrinks would give a curve getting flatter.
This curve gets steeper every day. - B. The gain is the same every dayThe same gain every day would give a straight line.
This curve gets steeper every day. - C. ✓ The gain is larger every day
Why: The curve bends upward and gets steeper every day: a J-shaped curve.
A steeper curve is a larger rise in the day.
So the dish’s gain is larger every day.
A population grows with food to spare.
Which of the following is exponential growth?
- A. ✓ The gain grows every year, as more individuals each add the same number
- B. The gain is the same in every year, as the same individuals add the same numberA gain that is the same in every year comes from the same number of individuals adding each year.
In exponential growth the gain grows. - C. Each individual adds more new individuals in every year than the year beforeIn exponential growth each individual adds the same number every year.
The population adds more because there are more individuals.
Why: In exponential growth each individual adds the same number per year.
Every year there are more individuals.
So the population’s gain grows every year.
Suppose fruit flies breed in a jar with food to spare. The graph below plots the population size, N, on the y-axis against time in weeks on the x-axis.
(a) Describe what the shape of the curve shows about the jar’s gain from one week to the next. (1 pt)
- Award 1 point for: the curve gets steeper (bends upward, a J-shaped curve), so the gain per week grows.
68Mixed practice mixed practice
Suppose two fields hold locusts of one species, both with food to spare. The north field holds 200 locusts and the south field holds 800 locusts.
Which field’s population gains more locusts in the next week?
- A. The north fieldEach locust adds the same number of young per week in both fields, and the north field holds fewer locusts.
Fewer locusts add fewer young in all. - B. ✓ The south field
- C. Neither: both fields gain the same numberThe two fields hold different numbers of locusts.
Each locust adds the same number, so the fuller field gains more.
Why: The locusts are one species, so each locust adds the same number of young per week in both fields.
The south field holds more locusts.
More locusts, each adding the same number, add more in all.
Suppose ecologists study a population of water fleas with food to spare.
Which of the following is rmax?
- A. ✓ A rate per individual
- B. A rate for the whole populationThe whole population’s rate is its rate of change of population size, dN/dt.
rmax is per capita: per individual. - C. A count of individualsA count of individuals is the population size, N.
rmax is a rate.
Why: rmax is the maximum per capita growth rate.
Per capita means per individual.
So rmax is a rate per individual.
Suppose two islands hold finches of one species, both with food to spare. The larger population gains more finches per year. A student says: “The two populations have the same rmax; the larger one gains more because more finches are adding.”
Is the student correct?
- A. ✓ Yes: the finches are one species, so each finch adds the same number; more finches add more in all
- B. No: a bigger gain per year can only come from a higher maximum per capita growth rate, rmaxThe finches are one species, so each finch adds the same number of young per year on both islands.
More finches, each adding the same number, add more in all.
Why: rmax is per finch, and the finches are one species.
So each finch adds the same number of young per year on both islands.
The larger population gains more because more finches are adding.
Suppose two streams each hold 70 minnows, with food to spare. The minnows in the fast stream belong to a species with a lower maximum per capita growth rate, rmax, than the species in the slow stream.
Which stream’s population is larger after a year?
- A. The fast streamEach minnow in the fast stream adds fewer young per year.
The same number of minnows, each adding fewer, adds fewer in all. - B. ✓ The slow stream
- C. Neither: both are the same sizeThe two populations start the same size, but each minnow in the slow stream adds more young per year.
So the slow stream’s population gains more.
Why: Both streams start with 70 minnows.
Each minnow in the slow stream adds more young per year than each minnow in the fast stream.
So the slow stream’s population gains more, and is larger after a year.
Suppose ecologists draw a bar for a population’s gain in each year while it grows exponentially, with food to spare.
Which of the following describes the bars from one year to the next?
- A. Each bar is the same height as the one beforeBars of the same height are the same gain every year.
In exponential growth more individuals are adding every year, so the gain grows. - B. ✓ Each bar is taller than the one before
- C. Each bar is shorter than the one beforeShorter bars are a shrinking gain.
In exponential growth more individuals are adding every year, so the gain grows.
Why: In exponential growth each individual adds the same number every year.
Every year there are more individuals.
So the gain grows every year, and each bar is taller than the one before.
Suppose a dish of pond water with food to spare holds 170 ciliates, then 510 ciliates a day later, then 1 530 ciliates a day after that.
Which of the following happened to the number of new ciliates each ciliate added, from the first day to the second?
- A. ✓ It stayed the same
- B. It tripledThe dish’s gain tripled, because three times as many ciliates were adding.
Each ciliate added two new ciliates in both days. - C. It halvedEach ciliate added two new ciliates in both days.
The dish’s gain grew, not shrank.
Why: In the first day 170 ciliates added 340 new ciliates: two each.
In the second day 510 ciliates added 1 020 new ciliates: two each.
So the number each ciliate added stayed the same.
Suppose a few zebra mussels arrive in a lake with food to spare and no predators. Each mussel adds the same number of young every year.
(a) Predict the shape of the line on a plot of the mussels’ population size, N (y-axis), against time in years (x-axis). (1 pt)
- Award 1 point for: a curve bending upward and getting steeper (a J-shaped curve).
(b) Explain why the line has the shape you predicted. (2 pt)
Frame The line has that shape because …
Each year the lake holds more mussels than the year before.
More mussels, each adding the same number, add more young in all.
So the yearly gain grows, and the line gets steeper every year.
- Award 1 point for: each mussel adds the same number of young every year (the per capita rate stays the same) while the number of mussels grows.
- Award 1 point for: more mussels each adding the same number add more in all, so the yearly gain grows and the curve gets steeper.
Glossary
- per capita growth rate
- The number of new individuals each individual adds per unit of time. Per capita means per head: for each individual.
- rmax (maximum per capita growth rate)
- The greatest number of new individuals each individual adds per unit of time, when nothing limits the population: food to spare, no predators, no disease. Written rmax, as the formula sheet writes it.
- exponential growth
- Growth in which a population’s gain grows from each hour, week or year to the next, because each individual adds the same number and there are more individuals each time.
- J-shaped curve
- The shape of a graph of population size, N, against time during exponential growth: the curve bends upward and gets steeper, like the letter J.
APBIO-U08-L30B The equation for unchecked growth
Suppose 600 bacteria sit in fresh broth, with a maximum per capita growth rate, rmax, of 0.3 per hour. Nothing limits them: food to spare, no predator, no disease.
How many new bacteria appear in the next hour? And how does the formula sheet write it?
Unit 8 · Ecology
1The equation, as the formula sheet writes it
Suppose a population of gerbils on a large island has food to spare, no predators and no disease.
Which of the following is its maximum per capita growth rate, rmax?
- A. ✓ The greatest number of young each gerbil adds per year
- B. The number of young the whole population adds per yearThe whole population’s gain per year is its rate of change of population size, dN/dt.
rmax is a rate per gerbil. - C. The number of gerbils on the islandThe number of gerbils on the island is the population size, N.
rmax is a rate per gerbil per year.
Why: Per capita means per head.
rmax is the greatest number of young each gerbil adds per year, with nothing holding it back.
Ecologists give a population’s growth rate per capita.
Which of the following does per capita mean?
- A. For the whole populationThe whole population’s rate is its rate of change of population size, dN/dt.
Per capita is per head: for each individual. - B. ✓ For each individual
- C. For each yearPer year is a unit of time.
Per capita means for each individual.
Why: Per capita means per head.
A per capita rate is the rate for each individual.
Suppose ecologists count the midge larvae in a pond every day. They find that dN/dt for the larvae is 56 larvae per day.
Which of the following is that 56 larvae per day?
- A. The number of larvae in the pondThe number of larvae in the pond is the population size, N.
dN/dt is a rate: larvae per unit of time. - B. The number of days the larvae were countedThe days are the time.
dN/dt is the change in the number of larvae per unit of time. - C. ✓ The number of larvae the pond gains in each day
Why: dN/dt is the rate of change of population size.
A rate is an amount of change per unit of time.
So 56 larvae per day is the number of larvae the pond gains in each day.
How does the formula sheet write the growth of a population in which each individual adds the same number?
The rate of change of a population’s size is the rate per individual times the number of individuals.
The formula sheet prints that as one line. Here it is, with what each symbol means beneath it.
N: the population size, in individuals (for the flask, bacteria)
rmax: the maximum per capita growth rate of the population, in new individuals per individual per unit of time (per bacterium per hour)
dN: the change in population size, in individuals (bacteria)
dt: the change in time (for the flask, one hour)
dN/dt: the rate of change of population size, in individuals per unit of time (bacteria per hour)
Video: Watch: The equation, as the formula sheet writes it
The flask of 600 bacteria is on screen, with its condition written beside it: food to spare, no predator, no disease. The equation appears beneath the flask on its own line. Each symbol lights up as it is named, and its meaning and unit appear under it. The condition stays written beside the equation.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L30Ba.mp4
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Here is a second flask: 600 bacteria in fresh broth, with food to spare, no predator and no disease.
Suppose that in one hour, 30 % of the bacteria in the flask divide, each adding one new bacterium. So each bacterium adds, on average, 0.3 new bacteria per hour.
That 0.3 per hour is the flask’s maximum per capita growth rate, rmax: the most each bacterium adds, with nothing holding it back.
The flask’s gain in the hour is what each bacterium adds, times the number of bacteria.
The number of bacteria in the flask is its population size, N. The flask’s gain per hour is its rate of change of population size, dN/dt.
The formula sheet prints the equation on its own line, exactly as below. Beneath it is what each symbol means, with its unit.
N: the population size, in individuals (for the flask, bacteria)
rmax: the maximum per capita growth rate of the population, in new individuals per individual per unit of time (per bacterium per hour)
dN: the change in population size, in individuals (bacteria)
dt: the change in time (for the flask, one hour)
dN/dt: the rate of change of population size, in individuals per unit of time (bacteria per hour)
rmax N means rmax multiplied by N. The sheet writes the two symbols side by side, with no multiplication sign between them.
rmax is in new bacteria per bacterium per hour, and N is a number of bacteria. So rmax N is in bacteria per hour, as the line below shows.
rmax is per bacterium per hour and N is a number of bacteria, so rmax N is in bacteria per hour: the unit of dN/dt
Bacteria per hour is the unit of dN/dt. So rmax N is a rate, the bacteria added per hour, not the number of bacteria in the flask.
The table below compares the three symbols. For each, it gives what it is and its unit for the flask.
What you are expected to know Read the exponential-growth equation in its formula-sheet form, dN/dt = rmax N, and state what each symbol means: N the population size, rmax the maximum per capita growth rate, dN/dt the rate of change of population size.
In the exponential-growth equation dN/dt = rmax N, each symbol has one meaning.
Which symbol is written for the maximum per capita growth rate?
- A. NN is the population size: a number of individuals.
- B. ✓ rmax
- C. dN/dtdN/dt is the whole population’s rate of change, not a rate per individual.
Why: The maximum per capita growth rate is the most each individual adds per unit of time.
The formula sheet writes it as rmax.
In the exponential-growth equation dN/dt = rmax N, each symbol has one meaning.
Which symbol is written for the rate of change of population size?
- A. NN is the number of individuals in the population, a count.
- B. rmaxrmax is a rate per individual, not the whole population’s rate.
- C. ✓ dN/dt
Why: The rate of change of population size is the individuals the whole population adds per unit of time.
The formula sheet writes it as dN/dt.
In the exponential-growth equation dN/dt = rmax N, each symbol has one meaning.
Which symbol is written for the population size?
- A. ✓ N
- B. rmaxrmax is the maximum per capita growth rate: the most each individual adds per unit of time.
- C. dN/dtdN/dt is the rate of change of population size: the individuals added per unit of time.
Why: N is the population size, the number of individuals in the population.
The exponential-growth equation is dN/dt = rmax N.
Which of the following does rmax N mean?
- A. rmax added to NTwo symbols written side by side are multiplied, not added.
A sum is written with a plus sign. - B. ✓ rmax multiplied by N
- C. rmax divided by NTwo symbols written side by side are multiplied, not divided.
A division is written as a fraction, like dN over dt.
Why: The sheet writes rmax and N side by side, with no sign between them.
Two symbols side by side are multiplied.
So rmax N is rmax multiplied by N.
Suppose copepods breed in a tide pool with food to spare. For the copepods, rmax N is 26 copepods per day.
Which of the following is that 26 copepods per day?
- A. A count of the copepods in the poolA count of copepods would be a number of copepods.
26 copepods per day is a number of copepods per unit of time. - B. ✓ A rate: the copepods the pool gains each day
Why: rmax N equals dN/dt, the rate of change of population size.
A change per unit of time is a rate.
So 26 copepods per day is the copepods the pool gains each day.
A student reads that, for the gnats over a meadow, rmax N is 52 gnats per week, and says: “So there are 52 gnats over the meadow.”
Is the student correct?
- A. ✓ No: 52 gnats per week is how many gnats the population gains each week
- B. Yes: rmax N is the number of gnats in the populationrmax N is a rate, in gnats per week, not a count of gnats.
Why: rmax N equals dN/dt, the rate of change of population size.
52 gnats per week means the population gains 52 gnats each week.
The number of gnats is N, and the student was not given N.
26When the equation applies
Does the equation apply to every population, at every moment?
A population grows with food to spare.
Which of the following is exponential growth?
- A. ✓ The gain grows each year, as more individuals each add the same number
- B. The gain is the same in every year, as the same number of individuals addA gain that stays the same each year comes from the same number of individuals adding each year.
In exponential growth N grows, so the gain grows. - C. Each individual adds more new individuals in every year than the year beforeIn exponential growth each individual adds the same number every year.
The population adds more because there are more individuals.
Why: In exponential growth each individual adds the same number per year.
Every year there are more individuals.
So the gain grows every year.
Suppose a population grows exponentially. Ecologists plot its population size, N, on the y-axis against time in years on the x-axis.
Which of the following is the shape of the line?
- A. A straight line rising at one steepnessA straight line rising at one steepness is a gain that stays the same every year.
In exponential growth the gain grows. - B. ✓ A curve bending upward, steeper every year
- C. A line staying levelA level line is a population whose size does not change.
Why: In exponential growth the gain grows every year.
So the curve gets steeper every year: a J-shaped curve.
The equation applies only while nothing holds the population back: unlimited resources, no predators and no disease.
Look at the equation again. rmax is the maximum per capita growth rate: the most each individual can add per unit of time.
N: the population size, in individuals (for the flask, bacteria)
rmax: the maximum per capita growth rate of the population, in new individuals per individual per unit of time (per bacterium per hour)
dN: the change in population size, in individuals (bacteria)
dt: the change in time (for the flask, one hour)
dN/dt: the rate of change of population size, in individuals per unit of time (bacteria per hour)
Each individual adds its most only when nothing holds it back.
Anything that kills or starves individuals holds a population back. Three kinds are common:
- too little food, water or space
- predators
- disease
A population with unlimited resources, no predators and no disease has nothing holding it back. Its reproduction is without constraints.
So the equation applies only while reproduction is without constraints: unlimited resources, no predators, no disease.
A few bacteria in a fresh flask of broth have food to spare, no predator and no disease. So the equation applies to them.
Now consider the same flask a day later. Bacteria crowd the broth, and the food is nearly gone.
Each bacterium can no longer add its most. So the equation no longer applies.
The table below compares three populations: what each has, and whether the equation applies.
If any of these is present, something holds the population back. So the equation does not apply to that population.
While the equation applies, each individual keeps adding the same number per unit of time. N grows, so the gain grows too: that is exponential growth.
Plotted against time, N is then a J-shaped curve.
What you are expected to know State the condition for exponential growth: reproduction without constraints, with unlimited resources, no predators and no disease, as for a few bacteria in a fresh flask, not a full one.
The exponential-growth equation is dN/dt = rmax N.
When does the equation apply to a population?
- A. Once the population has too little foodA population with too little food has something holding it back.
Each individual can no longer add its most. - B. ✓ While nothing holds the population back
- C. At every stage of the population’s growthOnce food, space, predators or disease hold the population back, each individual adds less than its most.
The equation then no longer applies.
Why: rmax is the most each individual adds when nothing holds it back.
So the equation applies only while nothing holds the population back: unlimited resources, no predators, no disease.
Suppose a few mold spores land on a fresh loaf of bread. The loaf gives them food to spare, and nothing eats them.
Does the equation dN/dt = rmax N apply to this population?
- A. ✓ Yes
- B. NoThe mold has food to spare and nothing eats it.
Nothing holds it back.
Why: The mold has unlimited food, no predators and no disease.
Nothing holds it back, so the equation applies.
Suppose a few gerbils are released on a large island with seeds to spare, no predators and no disease.
Does the equation dN/dt = rmax N apply to this population?
- A. ✓ Yes
- B. NoThe gerbils have food to spare, no predators and no disease.
Nothing holds them back.
Why: The gerbils have unlimited food, no predators and no disease.
Nothing holds them back, so the equation applies.
Suppose a flask of bacteria has been growing for two days. The bacteria crowd the broth, and the food is nearly gone.
Does the equation dN/dt = rmax N apply to this population?
- A. YesThe food is nearly gone, so each bacterium can no longer add its most.
- B. ✓ No
Why: The bacteria are short of food.
Something holds them back, so the equation does not apply.
Suppose cattle graze a fenced field. They have eaten the grass down to bare earth.
Does the equation dN/dt = rmax N apply to this population?
- A. YesThe grass is gone, so the cattle are short of food.
- B. ✓ No
Why: The cattle have eaten their food down to bare earth.
Something holds them back, so the equation does not apply.
Suppose a few dandelion seeds land on a freshly cleared field, with light, water and space to spare, and nothing eats the plants.
Does the equation dN/dt = rmax N apply to this population?
- A. ✓ Yes
- B. NoThe dandelions have light, water and space to spare, and nothing eats them.
Nothing holds them back.
Why: The dandelions have unlimited resources, no predators and no disease.
Nothing holds them back, so the equation applies.
Suppose a flask of bacteria has food to spare. A technician adds an antibiotic to the broth, and it kills many of the bacteria every hour.
Does the equation dN/dt = rmax N apply to this population?
- A. YesThe antibiotic kills many bacteria every hour, so something holds the population back.
- B. ✓ No
Why: The antibiotic kills many of the bacteria every hour.
Something holds the population back, so the equation does not apply.
51Calculate the flask’s rate of change
How many new bacteria appear in the flask in the next hour? The two values go into the equation.
Video: Watch: Calculate the flask’s rate of change
The calculation is worked on screen for the flask. The values are written down with their units, then the equation on its own line, then the substitution, then the answer with its unit, bacteria per hour.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L30Bb.mp4
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Here is the flask again: 600 bacteria in fresh broth, with a maximum per capita growth rate, rmax, of 0.3 per hour.
To calculate dN/dt, write down the values, write down the equation, then substitute the values in and calculate. The answer carries its unit, bacteria per hour.
A flask holds 600 bacteria in fresh broth, with food to spare, no predator and no disease. Its maximum per capita growth rate, rmax, is 0.3 per hour. Calculate dN/dt for the flask.
So the flask adds 180 bacteria in the next hour, while nothing holds the bacteria back.
dN/dt came out in bacteria per hour. It is the number of bacteria the flask adds in an hour, not the number of bacteria in the flask.
rmax is always positive, and N is always positive. So dN/dt from this equation is always positive: while the equation applies, the population grows.
What you are expected to know Calculate dN/dt from rmax and N, with its unit: individuals per unit of time.
A population’s maximum per capita growth rate, rmax, and its population size, N, are both known, and nothing holds the population back.
Which of the following gives dN/dt?
- A. ✓ rmax × N
- B. rmax + NAdding a rate per individual to a count of individuals mixes two different kinds of quantity.
Each individual adds rmax, so N individuals add N times that. - C. N ÷ rmaxDividing N by rmax gives a smaller rate for a larger rmax.
More added per individual means a larger gain, not a smaller one.
Why: Each individual adds rmax new individuals per unit of time.
N individuals add N times that.
So dN/dt is rmax multiplied by N, which the sheet writes as rmax N.
Suppose hydra live in a pond with food to spare, no predators and no disease. The pond holds 460 hydra, and the hydra’s maximum per capita growth rate, rmax, is 0.15 per day.
Calculate dN/dt for the hydra.
Part 1. State the population size, N, for the hydra.
Answer: 460 hydra (tolerance ±0)
Part 2. State the maximum per capita growth rate, rmax, for the hydra.
Answer: 0.15 per day (tolerance ±0)
Answer: 69 hydra per day (tolerance ±0)
Suppose 270 gerbils live on a large island with seeds to spare, no predators and no disease. Their maximum per capita growth rate, rmax, is 0.4 per year.
Calculate dN/dt for the gerbils.
Answer: 108 gerbils per year (tolerance ±0)
Here are the 600 bacteria again, in fresh broth with food to spare, no predator and no disease.
Their maximum per capita growth rate, rmax, is 0.3 per hour. The formula sheet writes their rate of change as dN/dt = rmax N.
So dN/dt is 180 bacteria per hour, as the line below shows, while nothing holds the bacteria back.
the flask: rmax 0.3 per hour, N 600 bacteria, so dN/dt is 180 bacteria per hour while nothing holds the bacteria back
67Numeric practice: the rate of change from rmax and N mixed practice
Suppose mosquitoes breed in a rain barrel with food to spare and nothing that eats them. The barrel holds 1 250 mosquitoes, and their maximum per capita growth rate, rmax, is 0.6 per day.
Calculate dN/dt for the mosquitoes.
Answer: 750 mosquitoes per day (tolerance ±0)
Suppose water hyacinth plants float on a lake with room and nutrients to spare, and nothing eats them. The lake holds 132 plants, and their maximum per capita growth rate, rmax, is 0.25 per week.
Calculate dN/dt for the water hyacinth.
Answer: 33 plants per week (tolerance ±0)
Suppose 360 quail live on a farm with grain to spare, no predators and no disease. Their maximum per capita growth rate, rmax, is 0.35 per year.
Calculate dN/dt for the quail.
Answer: 126 quail per year (tolerance ±0)
Suppose Euglena cells grow in a jar of pond water with light and nutrients to spare, and nothing eats them. The jar holds 1 700 cells, and their maximum per capita growth rate, rmax, is 0.5 per hour.
Calculate dN/dt for the Euglena.
Answer: 850 cells per hour (tolerance ±0)
Suppose 95 earwigs live under the bark of a fallen tree, with food to spare and nothing that eats them. Their maximum per capita growth rate, rmax, is 0.8 per month.
Calculate dN/dt for the earwigs.
Answer: 76 earwigs per month (tolerance ±0)
73Mixed practice mixed practice
The exponential-growth equation is dN/dt = rmax N.
Which of the following populations does the equation describe?
- A. ✓ A few springtails in fresh compost, with food to spare and nothing eating them
- B. Reindeer on an island that have eaten the lichen down to bare rockThe reindeer have eaten their food down to bare rock, so they are short of food.
- C. Caterpillars on an oak tree that blue jays pick off every dayBlue jays eat the caterpillars every day, so predators hold the population back.
Why: The equation applies only while nothing holds a population back.
The springtails have food to spare and nothing eats them.
So the equation describes the springtails.
Suppose a population grows with nothing holding it back, so dN/dt = rmax N applies. Over a week, N doubles.
Which of the following happens to dN/dt over that week?
- A. dN/dt halvesrmax N grows when N grows.
A halved dN/dt would need N to halve. - B. dN/dt stays the samermax stays the same, but N has doubled.
rmax N is twice what it was. - C. ✓ dN/dt doubles
Why: dN/dt is rmax multiplied by N.
rmax stays the same while nothing holds the population back.
N doubles, so rmax N doubles.
So dN/dt doubles.
Suppose 110 mealworms live in a bin of flour with food to spare and nothing that eats them. Their maximum per capita growth rate, rmax, is 0.6 per month.
Which of the following is dN/dt for the mealworms?
- A. ✓ 66 mealworms per month
- B. 110.6 mealworms per monthAdding the rate per mealworm to the count of mealworms mixes two kinds of quantity.
Each mealworm adds 0.6 new mealworms per month, and there are 110 mealworms. - C. 183 mealworms per monthDividing the count by the rate per mealworm gives a rate larger than the population.
Each mealworm adds 0.6 new mealworms per month, and there are 110 mealworms.
Why: Each mealworm adds 0.6 new mealworms per month.
The bin holds 110 mealworms.
dN/dt is rmax multiplied by N: 66 mealworms per month.
The exponential-growth equation is dN/dt = rmax N.
Which of the following is dN/dt?
- A. The number of individuals in the populationThe number of individuals in the population is the population size, N.
- B. The most each individual adds per unit of timeThe most each individual adds per unit of time is rmax.
- C. ✓ The individuals the whole population adds per unit of time
Why: dN is the change in population size, and dt is the change in time.
dN divided by dt is the individuals the whole population adds per unit of time.
A flask of bacteria grows with food to spare, no predator and no disease, so dN/dt = rmax N applies.
Which of the following would make the equation stop applying?
- A. ✓ The food in the flask is nearly gone
- B. The number of bacteria risesN rising is what the equation describes: a larger N gives a larger dN/dt.
- C. Another hour passesTime passing changes nothing about what holds the bacteria back.
Why: The equation applies while nothing holds the bacteria back.
Too little food holds them back: each bacterium can no longer add its most.
So the equation stops applying.
A student looks at a tank of guppies that has been breeding for months. The guppies crowd the tank, and their food is gone by the end of each day. The student says: “The guppies are short of food each day, so dN/dt = rmax N stops applying to this tank.”
Is the student correct?
- A. ✓ Yes: the equation applies only while nothing holds the guppies back, and they are short of food
- B. No: the equation applies to a tank of guppies at every stage of its growthThe food is gone by the end of each day, so each guppy can no longer add its most.
The equation applies only while nothing holds the population back.
Why: rmax is the most each guppy adds when nothing holds it back.
The food is gone by the end of each day, so something holds the guppies back.
So the equation no longer gives the tank’s rate of change.
Suppose a few cockroaches move into a warm, empty warehouse where grain has spilled across the floor. Nothing in the warehouse eats them, and no disease reaches them. After some weeks the warehouse holds 105 cockroaches, and their maximum per capita growth rate, rmax, is 0.4 per week.
(a) Calculate dN/dt for the cockroaches. (1 pt)
Answer: 42 cockroaches per week (tolerance ±0)
- Award 1 point for: 42 cockroaches per week, with the unit (accept 42 per week).
(b) Explain why the equation dN/dt = rmax N applies to the cockroaches in the warehouse. (2 pt)
Frame The equation applies because …
The spilled grain gives them food to spare, nothing eats them, and no disease reaches them.
So each cockroach adds its most: the maximum per capita growth rate, rmax.
Therefore the population’s rate of change is rmax multiplied by N.
- Award 1 point for: the cockroaches have unlimited food (resources), no predators and no disease, so nothing constrains their reproduction.
- Award 1 point for: with nothing holding them back, each cockroach adds its maximum per capita growth rate, rmax, so the population adds rmax N cockroaches per week.
APBIO-U08-L31 The J and the straight line
Suppose an ecologist counts the whiteflies on the tomato plants in a greenhouse once a week, for six weeks. She plots the seven counts two ways. On the first graph the y-axis climbs in equal steps of 1 000 whiteflies, and the curve bends upward. On the second graph the y-axis reads 10, 100, 1 000, 10 000, and the same seven points lie on a straight line.
Same counts. Why two shapes?
Unit 8 · Ecology
1The J: exponential growth on an ordinary axis
A population grows with food to spare.
Which of the following is exponential growth?
- A. The gain is the same in every week, as the same number of individuals add the same numberA gain that is the same in every week comes from the same number of individuals adding each week.
In exponential growth the gain grows. - B. ✓ The gain grows every week, as more individuals each add the same number
- C. Each individual adds more new individuals in every week than the week beforeIn exponential growth each individual adds the same number every week.
The population adds more because there are more individuals.
Why: In exponential growth each individual adds the same number per week.
Every week there are more individuals.
So the population’s gain grows every week.
Ecologists plot a population’s size, N, on an ordinary y-axis against time on the x-axis while the population grows exponentially.
Which of the following is the shape of the plot?
- A. A line that stays levelA level line is a population whose size does not change.
In exponential growth N rises every week. - B. A straight line rising at one steepnessA straight line rising at one steepness adds the same number every week.
In exponential growth the gain grows every week. - C. ✓ A curve that bends upward and gets steeper
Why: In exponential growth the gain grows every week.
So each week the point climbs further than the week before.
The curve bends upward and gets steeper: a J-shaped curve.
How do you recognize exponential growth on a graph, or in a table of counts?
On an ordinary y-axis, the curve of exponential growth bends upward and gets steeper.
The count doubles in equal times. However large the count has grown, it takes the same time to double again.
A straight rising line is not exponential growth. It adds the same number in every equal time.
Video: Watch: The J on an ordinary axis
The ecologist’s seven whitefly counts are plotted one by one on an ordinary y-axis. The first counts sit close to the x-axis. Each week the point climbs further than the week before. The curve bends upward into a J.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L31a.mp4
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Here are the whitefly counts again, on an ordinary y-axis: population size, N, in whiteflies on the y-axis, and time in weeks on the x-axis.
Each week the ecologist counts about twice as many whiteflies as the week before: 55, then 110, then 220, then 440.
So the gain grows every week. In the first week the count rose by 55 whiteflies; in the sixth week it rose by 1 760 whiteflies.
Each week the point climbs further than the week before. So the curve bends upward and gets steeper: a J-shaped curve.
Now look at the time the count takes to double.
From 55 to 110 whiteflies took one week. From 880 to 1 760 whiteflies also took one week.
In exponential growth, the count doubles in equal times, however large it has grown.
Now consider two populations that both start at 40 and both reach 80 a year later: a flock of geese on a lake, and the thistles on a hillside.
The table below compares the two counts, year by year, over four years.
The geese add the same number every year: 40 geese. So their count rises by the same step each year.
The thistles double every year. So their gain grows each year: 40, then 80, then 160, then 320 thistles.
The graph below plots both counts on the same axes: population size, N, on the y-axis, and time in years on the x-axis.
The geese lie on a straight rising line. A straight rising line is not exponential growth: it adds the same number each year.
The thistles lie on a curve that bends upward. The thistles took one year to double from 40 to 80 thistles, and one year to double from 320 to 640 thistles: exponential growth.
What you are expected to know Identify exponential growth from a graph or a table of counts: the curve bends upward, and the population doubles in equal times.
Suppose a gardener counts the mealybugs on a houseplant every week. The graph below plots the counts, with population size, N, on the y-axis and time in weeks on the x-axis.
Which of the following does the graph show?
- A. ✓ Exponential growth
- B. Growth by the same number every weekGrowth by the same number every week is a straight rising line.
This curve gets steeper every week. - C. No growthA population with no growth is a level line.
This curve rises.
Why: The curve bends upward and gets steeper every week.
So the gain grows every week.
A curve that bends upward is exponential growth.
Suppose ecologists count two populations once a year: the slugs in a garden and the barnacles on a rock. The table below shows the counts.
Which population is growing exponentially?
- A. The slugs in the gardenThe slugs add 30 every year: 30, 60, 90, 120, 150.
Adding the same number each year is not exponential growth. - B. ✓ The barnacles on the rock
Why: The barnacles double every year: 30, 60, 120, 240, 480.
The count doubles in equal times, one year each time.
So the barnacles are growing exponentially.
Suppose a librarian counts the silverfish in a library store every week. The graph below plots the counts.
Is this population growing exponentially?
- A. ✓ Yes
- B. NoThe curve bends upward and gets steeper every week.
That is the shape of exponential growth.
Why: The curve bends upward and gets steeper every week.
Each week the count rises by more than the week before.
So the population is growing exponentially.
Suppose a baker counts the flour mites in a sack of flour every month. The table below shows the counts.
Is this population growing exponentially?
- A. ✓ Yes
- B. NoThe count doubles every month: 14, 28, 56, 112, 224.
Doubling in equal times is exponential growth.
Why: The count doubles every month: 14, 28, 56, 112, 224.
The count doubles in equal times.
So the population is growing exponentially.
Suppose a warden counts the trout in a reservoir every year. The graph below plots the counts.
Is this population growing exponentially?
- A. YesThe line rises at one steepness all the way.
The count adds the same number every year, so the growth is not exponential. - B. ✓ No
Why: The plotted points lie on a straight rising line.
A straight rising line adds the same number every year.
Exponential growth bends upward, so this population is not growing exponentially.
Suppose a council counts the magpies in a town every year. The table below shows the counts.
Is this population growing exponentially?
- A. YesThe count rises by 12 every year: 22, 34, 46, 58, 70.
Adding the same number every year is not exponential growth. - B. ✓ No
Why: The count rises by 12 magpies every year: 22, 34, 46, 58, 70.
The count adds the same number in every year, and does not double.
So the population is not growing exponentially.
Suppose ecologists count the muskrats in a marsh every two years. The table below shows the counts.
Is this population growing exponentially?
- A. ✓ Yes
- B. NoThe count doubles every two years: 16, 32, 64, 128, 256.
Doubling in equal times is exponential growth.
Why: The count doubles every two years: 16, 32, 64, 128, 256.
Two years is an equal time each time.
So the population is growing exponentially.
Suppose ecologists count the kestrels on a moor every year. The graph below plots the counts.
Is this population growing exponentially?
- A. YesThe line rises at one steepness all the way.
The count adds the same number every year, so the growth is not exponential. - B. ✓ No
Why: The plotted points lie on a straight rising line.
The count adds the same number every year.
So the population is not growing exponentially.
A student looks at the graph below, which plots the wrens in a park over ten years, and says: “The line rises, so the wrens are growing exponentially.”
Is the student correct?
- A. ✓ No: a straight rising line adds the same number every year; exponential growth bends upward
- B. Yes: a rising line on a graph of N against time is exponential growthThe line rises at one steepness, so the wrens add the same number every year.
Exponential growth bends upward and gets steeper.
Why: The plotted points lie on a straight rising line.
A straight rising line adds the same number every year.
In exponential growth the gain grows every year, so the curve bends upward.
So the wrens are not growing exponentially.
33Reading a log-scale y-axis
How do you read a count off a y-axis marked 10, 100, 1 000, 10 000?
Suppose an ecologist counts the crickets in a field once a week for eight weeks, and plots the counts week by week.
Which quantity goes on the x-axis?
- A. The number of cricketsThe number of crickets is the quantity the ecologist measured.
The measured quantity goes on the y-axis. - B. ✓ The time in weeks
Why: The ecologist set the times: once a week for eight weeks.
The set quantity goes on the x-axis.
So the time in weeks goes on the x-axis.
Suppose a biologist labels a y-axis 10, 100, 1 000 and 10 000 at equal spacing.
Which of the following does each equal step along the y-axis do to the count?
- A. Adds ten to the countAdding ten would make the mark above 10 read 20.
That mark reads 100: ten times the mark before it. - B. ✓ Multiplies the count by ten
- C. Doubles the countDoubling would make the mark above 10 read 20.
That mark reads 100: ten times the mark before it.
Why: The marks read 10, 100, 1 000, 10 000.
Each mark is ten times the mark before it.
So each equal step along the y-axis multiplies the count by ten: a logarithmic (log) scale.
Each equal step along a logarithmic (log) scale multiplies the count by ten. So the labeled gridlines read 10, 100, 1 000, 10 000.
Between two labeled gridlines the counts are not evenly spaced. So you read a point against the thinner dashed gridlines, never by eye between the labels.
Video: Watch: The same points on a log-scale y-axis
The ecologist’s y-axis is relabeled 10, 100, 1 000, 10 000, each label the same distance above the last. The same seven whitefly counts are plotted again. The points settle onto a straight line.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L31b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L31b.mp4
Here are the whitefly counts again, on a log-scale y-axis. The labeled gridlines read 10, 100, 1 000, 10 000, each the same distance above the last.
The seven points now lie on a straight line.
Between each labeled gridline and the next, the graph carries three thinner dashed lines, called minor gridlines. They sit at 2, 3 and 5 times the labeled gridline below them.
Above the 100 gridline, the minor gridlines read 200, 300 and 500. Above the 1 000 gridline, they read 2 000, 3 000 and 5 000.
Suppose a biologist counts the planarians in a tank every two days, and plots the counts on the same kind of y-axis.
To read the count at 4 days, find 4 on the x-axis and go up to the point. Then read straight across to the y-axis.
The point at 4 days sits on the 1 000 gridline. So the count at 4 days is 1 000 planarians, read at the 1 000 gridline.
The point at 0 days sits on the third minor gridline above the 10 gridline. The third minor gridline reads 5 times the labeled gridline below it.
So the count at 0 days is 50 planarians, read at the 50 gridline.
A point between two gridlines is an estimate. Say where it was read: about 400, between the 300 and 500 gridlines.
Now look at one step of the axis made larger, from the 100 gridline to the 1 000 gridline.
The three minor gridlines sit unevenly up the step:
- the 200 gridline sits 30 % of the way up;
- the 300 gridline sits 48 % of the way up;
- the 500 gridline sits 70 % of the way up.
Suppose a point sits exactly halfway up this step. On an ordinary axis, halfway between 100 and 1 000 would be 550.
On a log scale, the halfway point sits just above the 300 gridline. So the point reads about 300, not 550.
Read against the gridlines, never by eye between the labels.
What you are expected to know Read a count off a log-scale y-axis, on a labeled gridline or on a minor gridline, and say where it was read.
Suppose a grower counts the thrips on a bean crop every day. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
Between which two labeled gridlines does the point at 3 days sit?
- A. Between the gridlines labeled 10 and 100The point at 3 days sits well above the gridline labeled 100.
Go up from 3 on the x-axis to the point, then read across. - B. Between the gridlines labeled 100 and 1 000The point at 3 days sits above the gridline labeled 1 000.
Go up from 3 on the x-axis to the point, then read across. - C. ✓ Between the gridlines labeled 1 000 and 10 000
Why: Find 3 on the x-axis and go up to the point.
Read straight across to the y-axis.
The point sits above the gridline labeled 1 000 and below the gridline labeled 10 000.
The graph below plots a grower’s daily counts of the thrips on a bean crop on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
Counting up from the lower labeled gridline of its step, which minor gridline does the point at 3 days sit on?
- A. The first of the three minor gridlines in its stepThe first minor gridline is the lowest dashed line in the step.
The point sits one dashed line higher. - B. ✓ The second of the three minor gridlines in its step
- C. The third of the three minor gridlines in its stepThe third minor gridline is the highest dashed line in the step.
The point sits one dashed line lower.
Why: Above the gridline labeled 1 000 there are three dashed minor gridlines.
The point at 3 days sits on the middle one.
So the point sits on the second minor gridline above the gridline labeled 1 000.
The thrips on a bean crop again: the graph below plots the grower’s daily counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 3 days.
Answer: 3000 thrips (tolerance ±0)
Suppose a student counts the fairy shrimp in a rain pool every week. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 2 weeks.
Answer: 5000 fairy shrimp (tolerance ±0)
Suppose a farmer counts the tilapia in a farm pond every month. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 2 months.
Answer: 20000 tilapia (tolerance ±0)
The graph below plots one count of millipedes under a log on a log-scale y-axis. The point at 2 weeks sits exactly halfway between the gridlines labeled 100 and 1 000.
Which of the following is the count at 2 weeks?
- A. ✓ About 300, just above the minor gridline at 300
- B. 550, halfway between the two valuesBetween two labeled gridlines on a log scale, the counts are not evenly spaced.
The minor gridline at 300 sits just below the halfway point. - C. About 900, just below the gridline labeled 1 000The minor gridline at 500 sits 70 % of the way up the step, and 900 sits above it.
The point sits halfway up the step.
Why: Between the gridlines labeled 100 and 1 000 the counts are not evenly spaced.
The minor gridline at 300 sits 48 % of the way up the step, just below halfway.
So the halfway point reads about 300.
62Quick quiz: logarithmic (log) scale mixed practice
Suppose a student counts the tardigrades in a sample of wet moss every day. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
(a) State the count at 3 days, and name the gridline it was read at. (1 pt)
It was read at the 30 000 gridline, the second minor gridline above the 10 000 gridline.
- Award 1 point for: 30 000 tardigrades, read at the 30 000 gridline (the second minor gridline above the 10 000 gridline).
Suppose a beekeeper counts the wasps in a nest every week. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 2 weeks.
Answer: 200 wasps (tolerance ±0)
Suppose a student counts the patches of moss on a shed roof every year. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 2 years.
Answer: 30 moss patches (tolerance ±0)
Suppose ecologists count the cichlids in a lake every year. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
State the count at 2 years.
Answer: 100000 cichlids (tolerance ±0)
A biologist needs a y-axis for counts that climb from about 20 to about 20 000.
Which of the following y-axes is a logarithmic (log) scale?
- A. An axis marked 0, 6 000, 12 000, 18 000, 24 000 at equal spacingEach step adds 6 000 to the mark before it.
A log scale multiplies by ten at each step. - B. ✓ An axis marked 10, 100, 1 000, 10 000, 100 000 at equal spacing
- C. An axis marked 0, 20, 40, 60, 80 at equal spacingEach step adds 20 to the mark before it.
A log scale multiplies by ten at each step.
Why: The marks read 10, 100, 1 000, 10 000, 100 000.
Each mark is ten times the mark before it.
So each equal step multiplies the count by ten, and the axis is a log scale.
68Why the straight line
Here are the two whitefly graphs again: the same seven counts, on an ordinary y-axis at the left and on a log-scale y-axis at the right.
Suppose each whitefly in a greenhouse adds the same number of young every week.
Which of the following is that number, the new individuals each individual adds per unit of time?
- A. The population size, NThe population size, N, is the number of whiteflies in the greenhouse.
The number each whitefly adds is a rate per individual. - B. The rate of change of population size, dN/dtdN/dt is the whole population’s gain per week.
The number each whitefly adds is a rate per individual. - C. ✓ The per capita growth rate
Why: Per capita means per head: for each individual.
The number of new individuals each individual adds per unit of time is the per capita growth rate.
Why does exponential growth appear as a straight line on a log-scale y-axis?
On a log scale, equal distances along the y-axis are equal factors, not equal amounts. Doubling is the same distance at 55 as at 1 760.
A population growing exponentially multiplies by the same factor in each equal time. So each week its point climbs the same distance: a straight line.
Each whitefly adds the same number of young every week. So the per capita growth rate stays the same from week to week.
So the count multiplies by the same factor every week. Here it doubles: 55, 110, 220, 440 whiteflies.
On the ordinary y-axis, each equal distance along the axis is the same number of whiteflies added.
The gain grows every week: 55 whiteflies in the first week, 1 760 in the sixth. So each week the point climbs further than the week before, and the curve bends upward.
On the log-scale y-axis, each equal distance along the axis is the same factor.
From the 10 gridline to the 100 gridline is ten times. From the 100 gridline to the 1 000 gridline is ten times again, in the same distance.
So a count twice another sits the same distance above it, wherever the two counts are on the axis. From 55 to 110 is the same distance as from 880 to 1 760.
The graph below marks the rise between each week’s point and the next week’s point, on the log-scale y-axis.
Every rise is the same height, because every week the count doubles. Equal steps along the x-axis and equal rises along the y-axis: the points lie on a straight line.
The steepness of the line shows the factor. A population that triples every week climbs further each week than one that doubles, so its line is steeper.
What you are expected to know Explain why exponential growth appears as a straight line on a log-scale y-axis: the population multiplies by the same factor in each equal time, and each equal distance along the axis is the same factor.
Suppose the bark beetles in a forest triple every year. A student plots the yearly counts twice: once with an ordinary y-axis, once with a log-scale y-axis.
Which y-axis gives points that lie on a straight line?
- A. The ordinary y-axisOn an ordinary y-axis the gain grows every year, so each year the point climbs further.
The curve bends upward. - B. ✓ The log-scale y-axis
- C. Both y-axesOn the ordinary y-axis the curve bends upward.
Only the log-scale y-axis gives a straight line.
Why: The count triples every year: the same factor each year.
On a log-scale y-axis, the same factor is the same distance, wherever the count is.
So each year the point climbs the same distance, and the points lie on a straight line.
Suppose the spider mites on a bean crop multiply by the same factor every week: each week there are three times as many as the week before. A student plots the weekly counts on a log-scale y-axis and finds that the points lie on a straight line.
(a) Explain why the points lie on a straight line. (2 pt)
Frame The points lie on a straight line because …
On a log-scale y-axis, equal distances along the axis are equal factors.
So a count three times another sits the same distance above it, wherever the two counts are.
Each week the point climbs the same distance.
Equal steps along the x-axis and equal rises along the y-axis give a straight line.
- Award 1 point for: the count multiplies by the same factor (three) every week.
- Award 1 point for: on a log-scale y-axis equal distances are equal factors, so each week the point climbs the same distance, giving a straight line.
A student plots the weekly counts of the hornets in a nest on the graph below, whose y-axis is a log scale, and sees the points on a straight rising line. The student says: “A straight line means the nest gains the same number of hornets every week.”
Is the student correct?
- A. ✓ No: on a log-scale y-axis a straight line means the count multiplies by the same factor every week
- B. Yes: a straight line on any y-axis means the nest adds the same number of hornets every weekOn a log-scale y-axis equal distances are equal factors, not equal amounts.
A straight line means the count multiplies by the same factor each week.
Why: On a log-scale y-axis, each equal distance along the axis is the same factor.
A straight line climbs the same distance every week.
So the count multiplies by the same factor every week, and the gain grows.
The nest does not gain the same number each week.
Suppose a fungus kills half of the ash trees in a forest every year. Ecologists plot the number of living ash trees on a log-scale y-axis against time in years on the x-axis.
Which of the following is the shape of the plot?
- A. A curve that bends downward, getting steeperA curve that gets steeper would need the factor to change from year to year.
The count halves every year: the same factor each time. - B. A straight line sloping upA line sloping up is a count that rises.
The number of living trees falls every year. - C. ✓ A straight line sloping down
Why: The count halves every year: the same factor each year.
On a log-scale y-axis the same factor is the same distance, so each year the point drops the same distance.
The points lie on a straight line sloping down.
Here are the two whitefly graphs again, the same seven counts on each.
On the ordinary y-axis the curve bends upward into a J. On the log-scale y-axis the points lie on a straight line.
Halfway between the 100 and the 1 000 gridlines reads about 300, not 550.
92Mixed practice mixed practice
Suppose ecologists count the cane toads on an island every three years. The table below shows the counts.
If the growth continues in the same way, which count do you predict at year 9?
- A. 725 cane toads725 adds the first gain, 145, once more.
The count doubles every three years: 145, 290, 580. - B. 870 cane toads870 adds the last gain, 290, once more.
The count doubles every three years: 145, 290, 580. - C. ✓ 1 160 cane toads
Why: The count doubles every three years: 145, 290, 580.
Doubling in equal times is exponential growth, and the doubling continues.
So at year 9 the count is 1 160 cane toads.
Suppose a student counts the Chlamydomonas cells in a bottle of pond water every week. The graph below plots the counts on a log-scale y-axis. The point at 0 weeks sits on the gridline labeled 100.
By what factor did the count multiply over the three weeks?
- A. By 2Two is the number of labeled steps between the two points.
Each labeled step multiplies the count by ten. - B. ✓ By 100
- C. By 9 9009 900 is the number of cells added, not the factor.
Each labeled step multiplies the count by ten, and the points sit two labeled steps apart.
Why: From the gridline labeled 100 to the gridline labeled 10 000 is two labeled steps.
Each labeled step multiplies the count by ten.
Ten times ten is one hundred, so the count multiplied by 100.
Suppose two ponds hold water striders of one species, and both populations double every month. The east pond starts with more water striders than the west pond. A student plots both counts on one log-scale y-axis against time in months.
Which of the following describes the two plots?
- A. Two straight lines, the east pond’s line steeper than the west pond’sThe steepness of a line on a log-scale y-axis shows the factor.
Both populations double every month, so the two lines are equally steep. - B. ✓ Two straight lines equally steep, the east pond’s line above the west pond’s
- C. Two curves that bend upward, the east pond’s curve steeper than the west pond’sOn a log-scale y-axis, doubling every month is a straight line, not a curve.
Both populations double every month.
Why: Both populations multiply by the same factor every month: two.
On a log-scale y-axis the same factor is the same rise each month, so both plots are straight lines of the same steepness.
The east pond starts higher, so its line sits above the west pond’s.
The graph below plots one count of Volvox colonies in a jar on a log-scale y-axis. The point at 3 days sits exactly halfway between the gridlines labeled 1 000 and 10 000. A student says: “Halfway up this step is not 5 500. On this axis the count is about 3 000.”
Is the student correct?
- A. ✓ Yes: on a log scale the counts between two gridlines are not evenly spaced; halfway reads about 3 000
- B. No: halfway between the two gridlines is halfway between the two values, so the count is 5 500Between two labeled gridlines on a log scale, the counts are not evenly spaced.
The minor gridline at 3 000 sits just below the halfway point.
Why: Between the gridlines labeled 1 000 and 10 000 the counts are not evenly spaced.
The minor gridline at 3 000 sits 48 % of the way up the step, just below halfway.
So the halfway point reads about 3 000, not 5 500.
Suppose ecologists count the mink on a river in four of the five years. The table below shows the counts.
Is this population growing exponentially?
- A. ✓ Yes
- B. NoThe count doubles each year: 20, 40, 80.
In the two years from year 2 to year 4 it doubles twice: 80, 160, 320.
Why: From year 0 to year 2 the count doubles each year: 20, 40, 80.
From year 2 to year 4, two years, it doubles twice: 80 to 160 to 320.
The count doubles in equal times, so the growth is exponential.
On a log-scale y-axis, a biologist measures the distance from the gridline at 30 up to the gridline at 60.
Which other pair of gridlines is the same distance apart?
- A. The gridlines at 60 and at 90From 60 to 90 is a factor of one and a half, not two.
On a log scale equal distances are equal factors. - B. The gridlines at 3 000 and at 3 030From 3 000 to 3 030 adds 30, a factor barely above one.
On a log scale equal distances are equal factors, not equal amounts. - C. ✓ The gridlines at 3 000 and at 6 000
Why: From the gridline at 30 to the gridline at 60 the count doubles.
On a log-scale y-axis equal distances are equal factors, wherever the counts are.
From the gridline at 3 000 to the gridline at 6 000 the count also doubles, so the distance is the same.
Suppose an ecologist counts the sea urchins on a reef once a year for six years. The table below shows the seven counts.
(a) State the evidence in the table that the sea urchin population is growing exponentially. (1 pt)
- Award 1 point for: the count doubles in equal times (one year each time), with two doublings quoted from the table.
(b) The ecologist plots the seven counts on a log-scale y-axis against time in years. Predict the shape of the plotted points. (1 pt)
- Award 1 point for: a straight rising line.
(c) Explain why the points have that shape on a log-scale y-axis. (2 pt)
Frame The points have that shape because …
On a log-scale y-axis, equal distances along the axis are equal factors.
So each year the point climbs the same distance.
Equal steps along the x-axis and equal rises along the y-axis give a straight line.
- Award 1 point for: the count multiplies by the same factor (doubles) every year.
- Award 1 point for: on a log-scale y-axis equal distances are equal factors, so each year the point climbs the same distance, giving a straight line.
APBIO-U08-L32 Draw the growth graph
Suppose a technician counts the bacteria in a flask at the end of every hour, for eight hours. Her eight counts climb from 500 bacteria to 300 000 bacteria. Beside her table lies a sheet of graph paper.
How should she draw the graph, so that every one of the eight counts can be read?
Unit 8 · Ecology
1From the table to the graph
Suppose ecologists count 2 300 penguins in a colony.
Which symbol does the formula sheet use for the population size, the number of penguins?
- A. dN/dtdN/dt is the rate of change of population size: penguins per year.
The number of penguins is a count, not a rate. - B. ✓ N
- C. rmaxrmax is the maximum per capita growth rate: new penguins per penguin per year.
The number of penguins is a count.
Why: The number of individuals in a population is its population size.
The formula sheet writes the population size as N.
So the count of 2 300 penguins is N.
Suppose a student measures the mass of a sunflower seedling at the end of every day for ten days. She set the days on which to measure, and the mass was what she measured.
Which quantity goes on the x-axis of her graph?
- A. ✓ Time in days
- B. Mass in gramsThe mass was measured.
The quantity that was measured goes on the y-axis.
Why: The student set the days.
The quantity that was set goes on the x-axis.
So time in days goes on the x-axis.
A finished graph has a title on each of its two axes.
Which of the following does each axis title carry?
- A. The quantity onlyA value taken from an axis with no unit has no meaning: 5 could be 5 hours or 5 days.
- B. ✓ The quantity and its unit
Why: A reader takes every value from the axis.
A value has no meaning without its unit.
So each axis title carries the quantity and its unit.
Suppose a y-axis carries the marks 10, 100, 1 000 and 10 000 at equal spacing.
What is an axis marked this way called?
- A. An ordinary axisOn an ordinary axis each equal step adds the same number.
Here each equal step multiplies the count by ten. - B. ✓ A logarithmic (log) scale
Why: Each mark is ten times the mark before it, and the marks are equally spaced.
So each equal step along the y-axis multiplies the count by ten.
An axis marked this way is called a logarithmic (log) scale.
How do you build a graph of a population’s size against time from a table of counts, so that every count can be read?
The counts follow one flask through time. So the graph is points joined one to the next: a line graph.
Time in hours goes on the x-axis, and the population size, N, in bacteria goes on the y-axis. Each axis title carries the quantity and its unit.
The counts span three powers of ten, from the hundreds to the hundred thousands. So the y-axis is a log scale, on which every count can be read.
Plot the eight points, join each to the next, then check every point against the table.
Video: Watch: From the table to the graph
The eight counts sit in a table. Time in hours goes on the x-axis and the population size, N, goes on the y-axis. An ordinary y-axis presses five of the eight points against the x-axis, so the y-axis becomes a log scale. Each point is placed on its gridline, the points are joined, and each is checked against the table.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L32a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L32a.mp4
Here are the technician’s eight counts as a table: the time in hours, and the population size, N, in bacteria.
The smallest count is 500 bacteria, at 1 hour. The largest is 300 000 bacteria, at 8 hours.
Hours are not separate categories, the way three species are. Hours are amounts along a time scale, so each count is plotted as a point.
The counts follow one flask through time. So each point is joined to the next, and the graph is a line graph.
The technician set the times: she counted at 1 hour, at 2 hours, and so on to 8 hours. The quantity that was set goes on the x-axis, so time goes on the x-axis.
She measured the counts. The quantity that was measured goes on the y-axis, so the population size, N, goes on the y-axis.
Each axis title carries the quantity and its unit: “time (hours)” on the x-axis, and “population size, N (bacteria)” on the y-axis.
Now choose the y-axis scale so that the largest count fits. Try an ordinary axis first: 0 to 300 000 bacteria, with a gridline every 50 000.
On this axis, five of the eight points sit below the first gridline, pressed against the x-axis. Their counts cannot be read.
The counts span three powers of ten: from the hundreds, at 1 hour, to the hundred thousands, at 8 hours. When counts span several powers of ten, choose a log scale.
On a log scale each equal step along the y-axis multiplies the count by ten. The labeled gridlines read 100, 1 000, 10 000, 100 000 and 1 000 000 bacteria, at equal spacing.
The axis starts at the 100 gridline, below the smallest count. It ends at the 1 000 000 gridline, above the largest count.
Between each pair of labeled gridlines sit three dashed minor gridlines, at 2, 3 and 5 times the lower one. Between 100 and 1 000 they mark 200, 300 and 500.
Take the first count: 500 bacteria at 1 hour. The point sits on the third minor gridline above 100, the 500 gridline, directly above 1 hour.
At 3 hours the count is 5 000 bacteria. That count sits between the 1 000 and 10 000 gridlines, on the third minor gridline above 1 000: the 5 000 gridline.
At 8 hours the count is 300 000 bacteria. That count sits between the 100 000 and 1 000 000 gridlines, on the second minor gridline above 100 000: the 300 000 gridline.
The other five points go on the same way: each count on its own gridline, directly above its hour.
Join the points, each to the next, with straight lines.
Then check every point against the table. At 6 hours the point sits on the 50 000 gridline, and the table says 50 000 bacteria at 6 hours.
Get graph paper. Mark the two axes, choose the y-axis scale so that the largest count fits, plot the eight points, join them, then check your graph against the one below.
Check your graph against this list:
- 1 both axes titled with the quantity and its unit
- 2 time in hours on the x-axis
- 3 a log scale on the y-axis, its gridlines labeled 100 to 1 000 000 bacteria
- 4 eight points, each on its gridline directly above its hour
- 5 the points joined, each to the next
What you are expected to know Construct a graph of population size against time from a table: a line graph, time on the x-axis, population size on the y-axis, units on both, a scale that fits the counts, the points plotted and joined.
Now decide the scale for three more sets of counts: clover plants in a lawn, crows roosting in a park, and sawfly larvae on a stand of pines.
Suppose a class counts the clover plants in a lawn every week. The table below gives the smallest and the largest count.
Which y-axis scale fits the clover counts?
- A. ✓ An ordinary axis
- B. A log scaleFrom 120 plants to 480 plants is less than one power of ten.
On an ordinary axis from 0 to 500 plants, every count sits on its own gridline.
Why: The counts climb from 120 plants to 480 plants.
That span is within one power of ten.
An ordinary axis from 0 to 500 plants shows every count clear of the x-axis.
So an ordinary axis fits.
Suppose a birdwatcher counts the crows roosting in a park each winter evening. The smallest and the largest count are in the table below.
Which y-axis scale fits the crow counts?
- A. ✓ An ordinary axis
- B. A log scaleFrom 2 000 crows to 9 000 crows is less than one power of ten.
An ordinary axis to 10 000 crows shows every count clear of the x-axis.
Why: The counts climb from 2 000 crows to 9 000 crows.
That span is within one power of ten.
An ordinary axis from 0 to 10 000 crows shows every count clear of the x-axis.
So an ordinary axis fits.
Suppose a forester counts the sawfly larvae on a stand of pines every week. The smallest and the largest count are in the table below.
Which y-axis scale fits the sawfly larva counts?
- A. An ordinary axisFrom 30 larvae to 60 000 larvae spans more than three powers of ten.
On an ordinary axis to 60 000, the count of 30 sits pressed against the x-axis. - B. ✓ A log scale
Why: The counts climb from 30 larvae to 60 000 larvae.
That span is more than three powers of ten.
When counts span several powers of ten, a log scale shows every count.
So a log scale fits.
Now consider a student who counts the cells of a protist in a beaker of pond water once a day for six days, starting at 0 days. Her counts are in the table below.
A student counts the protist cells in one beaker of pond water once a day for six days.
Which kind of graph fits these counts?
- A. A bar graphA bar graph is for separate categories, such as three species.
Days are amounts along a time scale. - B. ✓ A line graph
Why: Days are amounts along a time scale, not separate categories.
The counts follow one beaker through time.
So the points are joined one to the next: a line graph.
A student graphs her daily counts of the protist cells in a beaker. On the y-axis she writes the title “population size, N”.
Which of the following is missing from her y-axis title?
- A. The quantityThe title names the quantity: the population size, N.
What it lacks is the unit the counts were made in. - B. Nothing: the title is completeA title with no unit leaves every value unreadable: 300 could be 300 cells or 300 of anything.
Each axis title carries the quantity and its unit. - C. ✓ The unit, cells
Why: Each axis title carries the quantity and its unit.
The title “population size, N” names the quantity but no unit.
The counts are in cells, so the title lacks the unit, cells.
A student counts the protist cells in a beaker once a day for six days, from day 0 to day 5. Her six counts climb from 20 cells to 5 000 cells. Three y-axes for these counts are drawn below, numbered 1, 2 and 3.
On which axis can every one of the six counts be read?
- A. Axis 1Axis 1 is an ordinary axis with its first gridline at 1 000 cells.
On it, 20, 100, 300 and 500 cells all sit below the first gridline. - B. Axis 2Axis 2 ends at 1 000 cells.
The counts of 2 000 and 5 000 cells would sit above its top. - C. ✓ Axis 3
Why: The counts climb from 20 cells to 5 000 cells, across three powers of ten.
Axis 3 is a log scale from 10 to 10 000 cells.
Every count from 20 to 5 000 sits on one of its gridlines, clear of the x-axis.
The counts go on the log-scale y-axis drawn below. Between each pair of labeled gridlines, the three dashed minor gridlines mark 2, 3 and 5 times the lower gridline. Three markers numbered 1, 2 and 3 sit above 2 days.
Which marker sits at 300 cells, the count at 2 days?
- A. ✓ Marker 1
- B. Marker 2Marker 2 sits between the 10 and 100 gridlines, on the minor gridline for 30 cells.
300 cells is a power of ten higher. - C. Marker 3Marker 3 sits between the 1 000 and 10 000 gridlines, on the minor gridline for 3 000 cells.
300 cells is a power of ten lower.
Why: 300 cells sits between the 100 gridline and the 1 000 gridline.
Within that step, the minor gridlines mark 200, 300 and 500.
Marker 1 sits on the second minor gridline above 100: the 300 gridline.
The student has placed all six protist counts on a log-scale grid and finished her graph. Three drawings of the result are numbered 1, 2 and 3 below.
Which drawing is the finished graph?
- A. Drawing 1In drawing 1 the six points stand apart.
A finished line graph joins the points, each to the next, with straight lines. - B. ✓ Drawing 2
- C. Drawing 3In drawing 3 the axis titles are “time” and “count”.
A finished graph names each quantity with its unit: time (days), population size, N (cells).
Why: A finished graph titles both axes with the quantity and its unit, and joins the points, each to the next.
Drawing 2 has both.
So drawing 2 is the finished graph.
Now consider a student who counts the earthworms in a compost bin every week for five weeks. Her smallest count is 40 worms and her largest is 90 worms. She says: “These counts fit on an ordinary y-axis from 0 to 100 worms, so no log scale is needed.”
Is the student correct?
- A. No: a graph of a population’s counts always needs a log-scale y-axis, whatever their spanA log scale is chosen when the counts span several powers of ten.
From 40 worms to 90 worms is within one power of ten. - B. ✓ Yes: 40 to 90 worms is within one power of ten, so an ordinary axis shows every count
Why: From 40 worms to 90 worms is within one power of ten.
On an ordinary axis from 0 to 100 worms, every count sits clear of the x-axis.
A log scale is for counts that span several powers of ten.
So the student is correct.
Now consider a farmer who counts the leafhoppers in a rice field at the end of every week for five weeks. The counts are in the table below.
A farmer counts the leafhoppers in a rice field at the end of every week for five weeks. Her five counts are in the table below, and three graphs of them are numbered 1, 2 and 3 beneath the table.
Which graph shows the five counts correctly, so that every count can be read?
- A. ✓ Graph 1
- B. Graph 2Graph 2 is an ordinary axis with its first gridline at 500 leafhoppers.
On it, 20, 50 and 200 leafhoppers all sit pressed against the x-axis. - C. Graph 3In graph 3 the 3-week point sits on the 2 000 gridline.
The 3-week count is 200 leafhoppers, a power of ten lower.
Why: The counts span three powers of ten, so the y-axis is a log scale.
Graph 1 is a log scale from 10 to 10 000.
Its five points sit at 20, 50, 200, 500 and 2 000, each on its own gridline.
So every count on graph 1 is readable.
A technician counts the bacteria in a flask at the end of every hour for eight hours; her table is below. A student looks at the eight counts, from the hundreds to the hundred thousands, and says: “An ordinary y-axis from 0 to the largest count, with a gridline every 50 000, would show all eight counts just as clearly as a log scale.”
Is the student correct?
- A. ✓ No: on that ordinary axis, five of the eight counts sit below the first gridline and cannot be read
- B. Yes: an ordinary axis that reaches the largest count fits all eight counts, so all eight can be readAn ordinary axis with a gridline every 50 000 has its first gridline at 50 000 bacteria.
Five of the counts sit below it, pressed against the x-axis.
Why: The counts span three powers of ten.
With a gridline every 50 000 bacteria, the first gridline sits at 50 000.
The five counts below 50 000 sit under that gridline, pressed against the x-axis.
So those five cannot be read, and the student is not correct.
Suppose a student counts the grain weevils in a silo at the end of every week for six weeks. Her six counts are in the table below. She plots them on an ordinary y-axis from 0 to 10 000 weevils, with a gridline every 2 000 weevils. Her graph is drawn beneath the table.
(a) Identify the feature of the student’s y-axis that stops the first four counts from being read. (1 pt)
- Award 1 point for: the ordinary (linear) y-axis, whose first gridline is at 2 000 weevils, places the four small counts below the first gridline / against the x-axis.
(b) Explain why a log scale on the y-axis would let every one of the six counts be read. (2 pt)
Frame A log scale would show every count because …
Its labeled gridlines read 10, 100, 1 000 and 10 000 weevils.
30 weevils sits between the 10 and 100 gridlines; 300 sits between 100 and 1 000; 100 and 1 000 sit on labeled gridlines.
Each small count sits at its own height above the x-axis.
So every count can be read against the gridlines.
- Award 1 point for: on a log scale each equal step multiplies the count by ten, so the gridlines read 10, 100, 1 000, 10 000 at equal spacing.
- Award 1 point for: the small counts (30, 100, 300, 1 000) then sit on or between their own gridlines, at their own heights above the x-axis, so each can be read.
Here is the technician’s graph of her eight counts again: one count at the end of every hour, from 500 bacteria at 1 hour to 300 000 bacteria at 8 hours.
The graph is a line graph, with time in hours on the x-axis and the population size, N, in bacteria on the y-axis. The counts span three powers of ten, so the y-axis is a log scale.
The eight points sit on their gridlines, joined each to the next. Each point has been checked against the table.
APBIO-U08-P83 Practice questions: Topic 8.3
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one population’s growth one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. The formula sheet gives the two growth equations, dN/dt = B − D and dN/dt = r<sub>max</sub> N.
Video: Watch first: Population ecology, summed up
A population sharpened: one species, one place, able to mix; births minus deaths as dN/dt = B − D; per head, r<sub>max</sub>; the more there are, the more are added, dN/dt = r<sub>max</sub> N; the J on an ordinary axis and the straight line on a log axis; drawing the graph.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T83-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T83-summary.mp4
Each of the following is a group of armadillos, or of armadillos and skunks.
Which of the following is one population?
- A. ✓ All the armadillos of one wood
- B. The armadillos and the skunks of one woodArmadillos and skunks are two species.
A population is the individuals of one species. - C. The armadillos of two islands 80 km apart, between which no armadillo swimsThe armadillos of the two islands never mix.
They are two populations of one species. - D. All the armadillos of the species, wherever they liveA population is the individuals of one species in one place.
The whole species lives in many places whose members never mix.
Why: A population is the individuals of one species, in one place, able to mix.
The armadillos of one wood are one species in one place.
They breed with each other and compete for the same food, so they mix.
So the armadillos of the wood are one population.
Nuthatches nest in a wood. Ecologists count how many hatch and how many die each year. This year more nuthatches hatched than died.
Which of the following would make the nuthatch population smaller next year than it is this year?
- A. More nuthatches hatch next year than this year, and the same number dieMore births with the same deaths make births minus deaths larger.
The population grows faster. - B. The same number hatch and the same number die next year as this yearThe same births and the same deaths leave births minus deaths as it was.
The population grows at the same speed. - C. The same number hatch next year as this year, more die, and the deaths are still fewer than the birthsMore deaths with the same births make births minus deaths smaller.
While births still outnumber deaths, the population grows more slowly. - D. ✓ Next year the deaths outnumber the births
Why: A population’s size changes by births minus deaths.
While births outnumber deaths, the population grows.
Once deaths outnumber births, births minus deaths is negative.
So the population shrinks, and next year it is smaller.
Ecologists write four numbers about a colony of guillemots: B is 310 guillemots per year, D is 190 guillemots per year, N is 2 400 guillemots, and dN/dt is 120 guillemots per year.
Which of the four numbers is a count at one moment?
- A. BB is the birth rate: the number born per year.
A number per year is a rate. - B. DD is the death rate: the number that die per year.
A number per year is a rate. - C. ✓ N
- D. dN/dtdN/dt is the rate of change of population size: the number gained per year.
A number per year is a rate.
Why: N is the population size: the number of guillemots in the colony at one moment.
B, D and dN/dt each carry “per year”.
A number per unit of time is a rate.
So N is the count, and the other three are rates.
Partridges live on a farm. In one year 130 chicks hatch and 74 partridges die.
Which of the following is dN/dt for the partridges?
- A. −56 partridges per year−56 is deaths minus births, 74 − 130.
dN/dt is births minus deaths, B − D. - B. ✓ 56 partridges per year
- C. 74 partridges per year74 is the death rate, D.
dN/dt is births minus deaths, B − D. - D. 204 partridges per year204 adds the births to the deaths.
dN/dt is births minus deaths, B − D.
Why: dN/dt = B − D.
B = 130 partridges per year and D = 74 partridges per year.
dN/dt = 130 − 74 = 56 partridges per year.
Tench live in a lake. The population holds 620 tench now, and its rate of change of population size is 45 tench per year.
Which of the following is the size of the population after 6 years, if the rate stays the same?
- A. 270 tench270 is the gain over the 6 years, 45 × 6.
The later size is the present size plus the gain. - B. 350 tench350 takes the gain of 270 from 620.
The rate is positive, so the population gains 270. - C. 665 tench665 adds one year’s gain to 620.
The rate acts in each of the 6 years. - D. ✓ 890 tench
Why: later size = present size + rate of change × time.
later size = 620 + 45 × 6 = 620 + 270.
later size = 890 tench.
Suppose 250 houseflies live in a barn with food to spare and nothing that eats them. Their maximum per capita growth rate, r<sub>max</sub>, is 0.2 per day.
Which of the following is dN/dt for the houseflies?
- A. 0.2 houseflies per day0.2 per day is r<sub>max</sub>, the number each housefly adds.
dN/dt is r<sub>max</sub> multiplied by N, the whole barn’s gain. - B. ✓ 50 houseflies per day
- C. 250.2 houseflies per day250.2 adds r<sub>max</sub> to N.
The two symbols written side by side are multiplied. - D. 1 250 houseflies per day1 250 divides N by r<sub>max</sub>.
The two symbols written side by side are multiplied.
Why: dN/dt = r<sub>max</sub> N.
r<sub>max</sub> = 0.2 per day and N = 250 houseflies.
dN/dt = 0.2 × 250 = 50 houseflies per day.
Krill in a large tank have food to spare. The tank held 400 krill a year ago and holds 500 krill now. Each krill adds the same number of young per year as before.
Which of the following is the tank’s gain over the next year?
- A. 25 krill25 krill would be a gain of 0.05 per krill.
Last year each krill added 0.25. - B. 100 krill100 krill was last year’s gain, when 400 krill were adding.
This year 500 krill are adding. - C. ✓ 125 krill
- D. 625 krill625 is the size after the year.
The gain is the size after the year minus the size now.
Why: Last year 400 krill added 100: 100 ÷ 400 = 0.25 per krill.
Each krill adds the same number as before.
This year 500 krill each add 0.25.
500 × 0.25 = 125 krill.
Ecologists count the sea anemones on four rocks once a year for three years. The table below shows the counts.
Which rock’s anemone population is growing exponentially?
- A. The north rock’sThe north rock’s count adds 40 every year: 40, 80, 120, 160.
Adding the same number each year is not exponential growth. - B. The east rock’sThe east rock’s count stays at 160.
A steady count is no growth. - C. ✓ The south rock’s
- D. The west rock’sThe west rock’s count halves every year: 320, 160, 80, 40.
The population shrinks.
Why: Exponential growth multiplies the count by the same factor in equal times.
The south rock’s count doubles every year: 40, 80, 160, 320.
So the south rock’s population is growing exponentially.
A student counts the leeches in a pond once a week. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
Which of the following is the count at 2 weeks?
- A. 15 leeches15 would sit halfway between the 10 and 20 gridlines.
The point at 2 weeks sits on the first minor gridline above 10. - B. ✓ 20 leeches
- C. 30 leeches30 is the second minor gridline above 10.
The point at 2 weeks sits on the first minor gridline above 10. - D. 200 leeches200 is the first minor gridline above 100.
The point at 2 weeks sits between the gridlines labeled 10 and 100.
Why: Find 2 weeks on the x-axis and go up to the point.
The point sits between the gridlines labeled 10 and 100.
It sits on the first minor gridline above 10.
The first minor gridline is 2 times the labeled gridline: 20 leeches.
A student counts the swans on a lake once a month for a year. Her counts run from 6 swans to 43 swans.
Which y-axis lets every one of her counts be read?
- A. ✓ An ordinary axis from 0 to 50 with a gridline every 10
- B. An ordinary axis from 10 to 50 with a gridline every 10The axis starts at 10, and the smallest count is 6 swans.
An ordinary axis starts at 0. - C. An ordinary axis from 0 to 500 with a gridline every 100On an axis to 500, every count sits below the first gridline.
The counts cannot be read there. - D. A log scale from 10 to 1 000The axis starts at 10, and the smallest count is 6 swans.
Counts within one power of ten fit an ordinary axis.
Why: The counts run from 6 to 43 swans, within one power of ten.
An ordinary axis from 0 to 50 puts every count between its gridlines, clear of the x-axis.
So every count can be read.
Razorbills nest on a small island off a rocky coast. The colony holds 260 razorbills now. In the past year 72 chicks hatched and 33 razorbills died. The birds have fish to spare, nothing on the island eats them, and no disease has reached them.
(a) Calculate dN/dt for the colony over the past year. (1 pt)
Frame dN/dt = B − D = … − … = … razorbills per year
Hint Which of the two counts is the birth rate, B, and which is the death rate, D?
Answer: 39 razorbills per year (tolerance ±0)
- Award 1 point for: 39 razorbills per year, with the unit (accept 39 per year).
Slip Adding the two counts, 105. The rate of change is births minus deaths.
(b) Calculate the size of the colony after 4 years, using the rate of change from part (a). (1 pt)
Frame later size = present size + rate of change × time = … + … × … = … razorbills
Hint How many razorbills does the colony gain in one year, and how many years pass?
Answer: 416 razorbills (tolerance ±0)
- Award 1 point for: 416 razorbills.
Slip Adding one year’s gain only, 299. The rate acts in each of the four years.
(c) Explain what the calculation in part (b) assumes about the colony’s rate of change. (1 pt)
Frame The calculation assumes that the rate of change …
Hint In your working for part (b), did the number you multiplied by the years change from one year to the next?
It multiplies one year’s rate by four years.
So it takes the colony to gain the same number of razorbills in each of the four years.
- Award 1 point for: the rate of change (the yearly gain) stays the same in every one of the four years.
Slip Saying the calculation assumes no deaths. Deaths are inside the rate; the assumption is that the rate does not change.
(d) The birds have fish to spare, no predators and no disease. Predict how the colony’s yearly gain changes over the four years, and explain your prediction. (1 pt)
Frame The yearly gain … because …
Hint How many razorbills are adding chicks in year 2, compared with year 1?
Each year the colony holds more razorbills than the year before.
More razorbills, each adding the same number, add more chicks in all.
- Award 1 point for: the gain grows, because each bird adds the same number while more birds are adding each year.
Slip Predicting the same gain every year because nothing has changed for the birds. Nothing changes per bird; the number of birds adding changes.
(e) An ecologist plots the colony’s yearly counts, which climb from 260 to about 450 razorbills. Determine which y-axis scale, ordinary or log, fits these counts. (1 pt)
Frame An … y-axis fits, because the counts …
Hint How many powers of ten do the counts cross between 260 and 900?
On an ordinary axis from 0 to 500 razorbills, every count from 260 to 450 sits clear of the x-axis and can be read.
- Award 1 point for: an ordinary axis, because the counts span less than one power of ten (from the hundreds to the hundreds), so every count can be read on it.
Slip Choosing a log scale because the counts grow exponentially. The scale is chosen by the span of the counts, not by the shape of the growth.
Gobies of one kind live in two rock pools on a shore. At every high tide the sea covers both pools, and the gobies swim between them. The two pools hold 310 gobies now. In the past year 140 gobies hatched in the pools and 95 died. The gobies have food to spare, no predator reaches the pools, and no disease has been seen.
(a) Determine whether the gobies of the two pools are one population or two. (1 pt)
They are one species, and at every high tide they swim between the pools.
So the gobies of the two pools mix: they compete for the same food and breed with each other.
- Award 1 point for: one population, because the gobies are one species and swim between the pools at high tide, so they mix (compete and breed together).
Slip Answering two populations because there are two pools. Two places hold one population when the individuals move between them and mix.
(b) Calculate dN/dt for the gobies over the past year. (1 pt)
Answer: 45 gobies per year (tolerance ±0)
- Award 1 point for: 45 gobies per year, with the unit (accept 45 per year).
Slip Adding the two counts, 235. The rate of change is births minus deaths.
(c) A student predicts the count after three more years by adding three years of this year’s rate of change to the count now. Evaluate the student’s prediction. (1 pt)
The student’s method assumes the gobies gain 45 every year.
The gobies have food to spare, so each goby keeps adding the same number of young.
Each year more gobies are adding, so the yearly gain grows above 45.
- Award 1 point for: the judgement (too small) AND the ground (the method assumes a constant yearly gain, but with food to spare each goby adds the same number and more gobies are adding each year, so the gain grows).
Slip Judging the prediction right because the rate was measured. The measured rate is this year’s; next year more gobies are adding.
(d) The student plots the gobies’ yearly counts on a log-scale y-axis against time in years. Predict the shape of the plotted points while the pools stay as described, and justify your prediction. (1 pt)
With food to spare, the count multiplies by the same factor every year.
On a log-scale y-axis, equal distances along the y-axis are equal factors.
So each year the point climbs the same distance, and equal rises in equal steps make a straight line.
- Award 1 point for: a straight rising line, because the count multiplies by the same factor each year and on a log-scale axis equal factors are equal distances, so each yearly rise is the same.
Slip Predicting a curve that bends upward. That is the shape on an ordinary y-axis; a log-scale y-axis straightens it.
APBIO-U08-T83 End-of-topic test: Population Ecology
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. The formula sheet gives the two growth equations, dN/dt = B − D and dN/dt = r<sub>max</sub> N.
Ecologists survey the shore of an island and a second island 60 km away, which no bird flies between. Their counts are in the table below.
Which of the following groups of birds is one population?
- A. The adult oystercatchers and the adult curlews on the islandOystercatchers and curlews are two species.
A population is the individuals of one species. - B. ✓ The adult oystercatchers and the oystercatcher chicks on the island
- C. The adult oystercatchers on the island and those on the second islandNo bird flies between the islands, so the two sets of oystercatchers never mix.
They are two populations of one species. - D. The oystercatcher chicks and the adult curlews on the islandChicks and curlews are two species.
A population is the individuals of one species in one place.
Why: A population is the individuals of one species, in one place, able to mix.
The adults and the chicks are one species on one island.
The chicks were born to those adults, so the two sets mix.
So the adults and the chicks together are one population.
A student counts 150 bumblebees of one species and 120 bumblebees of a second species in one meadow, and writes: “The meadow’s bumblebee population is 270.”
Which of the following is the error in the student’s statement?
- A. The count should also include the bumblebees of the meadows nearbyA population is one species in one place.
The bumblebees of other meadows are in other places. - B. Only the queens breed, so only the queens count toward a populationA population is every individual of the species in the place, queens and workers alike.
- C. The bees were counted on one day, and a population is the number born in a yearA population’s size is a count at one moment.
The number born in a year is the birth rate. - D. ✓ The two species cannot breed with each other, so they are two populations
Why: A population is the individuals of one species in one place, able to mix.
The 150 bees are one species and the 120 bees are another.
Bees of two species do not breed with each other.
So the meadow holds two bumblebee populations, of 150 and of 120.
Each of the following is a feature of an organism in its own place.
Which feature helps its organism get energy?
- A. ✓ A moth’s long tongue, which reaches the nectar deep in a flower
- B. A wetland plant’s wide roots, which take up water and minerals from the mudThe roots take in water and minerals.
Water and minerals are matter, not a source of energy. - C. A desert rodent’s kidneys, which make very little urineMaking little urine keeps water in the rodent’s body.
Water is matter, not a source of energy. - D. A salt-marsh shrub’s leaves, which push salt out onto their surfacesPushing salt out keeps the salt from building up in the leaves.
Salt is matter, not a source of energy.
Why: Nectar is sugar dissolved in water, so nectar is food.
A long tongue reaches nectar that a short tongue cannot.
So the long tongue gets the moth energy.
The other three features get their organisms matter: water and minerals, water kept in the body, and a way to lose salt.
Gannets nest in a colony on a cliff. The table below gives the chicks hatched and the gannets that died in two years.
Compared with year 1, how does the colony grow in year 2?
- A. It shrinksIn both years more chicks hatch than gannets die.
A colony with more births than deaths grows. - B. It grows more slowlyThe births rose by 40 and the deaths rose by 40.
Births minus deaths is 120 in both years. - C. ✓ It grows at the same speed
- D. It grows fasterThe births rose, but the deaths rose by the same number.
Births minus deaths is 120 in both years.
Why: In year 1 the births outnumber the deaths by 120 gannets.
In year 2 the births rose by 40 and the deaths rose by 40.
So in year 2 the births again outnumber the deaths by 120.
The colony grows at the same speed.
Wallabies live in a national park. At the start of a year the park holds 480 wallabies. In that year 96 joeys are born and 51 wallabies die.
Which of the following is dN/dt for the wallabies?
- A. −45 wallabies per year−45 is deaths minus births, 51 − 96.
dN/dt is births minus deaths, B − D. - B. ✓ 45 wallabies per year
- C. 147 wallabies per year147 adds the births to the deaths.
dN/dt is births minus deaths, B − D. - D. 525 wallabies525 is the size after the year, 480 + 45.
dN/dt is the change in size per year, not the size.
Why: dN/dt = B − D.
B = 96 wallabies per year and D = 51 wallabies per year.
dN/dt = 96 − 51 = 45 wallabies per year.
The 480 wallabies are N, the size, and take no part in the rate.
Opossums live in a wood. The population holds 340 opossums now, and its rate of change of population size is −18 opossums per year.
Which of the following is the size of the population after 5 years, if the rate stays the same?
- A. 90 opossums90 is the change over the 5 years, 18 × 5.
The later size is the present size plus that change. - B. ✓ 250 opossums
- C. 322 opossums322 takes one year’s change from 340.
The rate acts in each of the 5 years. - D. 430 opossums430 adds 90 to 340.
The rate is negative, so the change of 90 is a loss.
Why: later size = present size + rate of change × time.
later size = 340 + (−18) × 5 = 340 − 90.
later size = 250 opossums.
Two islands hold nutria of one species, both with food to spare. At the start of this year the larger island held 400 nutria, and over the year it gained 80 nutria. The smaller island held 100 nutria at the start of the year.
Which of the following is the smaller island’s gain this year?
- A. ✓ 20 nutria
- B. 40 nutria40 nutria would be a gain of 0.4 per nutria.
On the larger island each nutria added 80 ÷ 400 = 0.2. - C. 80 nutria80 nutria is the larger island’s gain.
The smaller island has 25% as many nutria adding. - D. 500 nutria500 divides the smaller island’s count by the gain per nutria, 100 ÷ 0.2.
The gain is the count multiplied by the gain per nutria.
Why: The nutria are one species with food to spare, so each nutria adds the same number.
On the larger island, 400 nutria added 80: 80 ÷ 400 = 0.2 per nutria.
On the smaller island, 100 nutria each add 0.2.
100 × 0.2 = 20 nutria.
A pond holds 800 mosquitofish and 200 killifish, both with food to spare. In one week the mosquitofish gain 160 fish and the killifish gain 80 fish.
Which species has the higher maximum per capita growth rate, r<sub>max</sub>?
- A. The mosquitofishThe mosquitofish gain more in all, 160 fish.
Per fish, 800 mosquitofish adding 160 is 0.2 each. - B. ✓ The killifish
- C. The two species have the same r<sub>max</sub>Per fish, the mosquitofish add 160 ÷ 800 = 0.2 each and the killifish add 80 ÷ 200 = 0.4 each.
The two rates differ. - D. Not enough information: the two gains are for different sizesA per capita rate is the gain divided by the number adding it.
The two gains and the two sizes are given.
Why: r<sub>max</sub> is the number each individual adds per week with food to spare.
Each mosquitofish adds 160 ÷ 800 = 0.2 fish per week.
Each killifish adds 80 ÷ 200 = 0.4 fish per week.
So the killifish have the higher maximum per capita growth rate.
A tank of zebrafish has been breeding for a year. The fish crowd the tank and their food is gone by the end of each day. A student moves ten of the zebrafish into a large tank of fresh water with food to spare.
Does the equation dN/dt = r<sub>max</sub> N apply to the fish in the new tank, and why?
- A. No: the equation applies to a population only once it has crowded its tankThe equation applies while nothing holds a population back.
A crowded tank with its food gone holds the fish back. - B. No: the fish came from a crowded tank, so they keep that tank’s rateA fish adds young at its most when it has food to spare.
Where the fish came from changes nothing about the new tank. - C. Yes: a small population always grows exponentially, whatever its foodA small population with its food gone is held back.
The equation rests on the food, not on the size. - D. ✓ Yes: the fish have food to spare, so nothing holds them back
Why: dN/dt = r<sub>max</sub> N applies while nothing holds the population back.
In the new tank the ten fish have food to spare, no predators and no disease.
So each fish adds its most, r<sub>max</sub>.
So the equation applies to the new tank.
Suppose a lake holds loach and bream, both with food to spare. The table below gives each population’s size and its maximum per capita growth rate.
Which population gains more fish in the next year?
- A. ✓ The loach
- B. The breamThe bream have the higher r<sub>max</sub>, but fewer fish are adding.
0.5 × 150 = 75 bream per year, against 0.3 × 290 = 87 loach per year. - C. The two populations gain the same numberdN/dt = r<sub>max</sub> N.
0.3 × 290 = 87 loach per year and 0.5 × 150 = 75 bream per year. - D. Not enough information: the two species have different r<sub>max</sub> valuesdN/dt = r<sub>max</sub> N gives each population’s gain from its own r<sub>max</sub> and its own N.
Why: dN/dt = r<sub>max</sub> N.
Loach: 0.3 × 290 = 87 loach per year.
Bream: 0.5 × 150 = 75 bream per year.
87 is more than 75, so the loach gain more fish.
Ecologists count the crossbills in a pine wood every two years. The table below shows the counts.
Which of the following describes the growth of the crossbill population?
- A. It doubles every 2 years7 to 21 is not a doubling.
Each count is three times the count two years before. - B. It triples every yearThe counts are two years apart.
Each count is three times the count two years before, not the year before. - C. ✓ It triples every 2 years
- D. It adds the same number every 2 yearsThe gains are 14, 42 and 126.
The gains grow; the counts multiply by three each time.
Why: Each count is three times the count two years before: 7, then 21, then 63, then 189.
So the population triples every 2 years.
Multiplying by the same factor in equal times is exponential growth.
A student counts the cells of one green alga in a bottle of water with light and nutrients to spare, once a day. The graph below plots the counts on a log-scale y-axis; its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
Which of the following is the count at 3 days?
- A. 200 000 cells200 000 is the first minor gridline above 100 000.
The point at 3 days sits on the third minor gridline above 100 000. - B. 300 000 cells300 000 is the second minor gridline above 100 000.
The point at 3 days sits on the third minor gridline above 100 000. - C. ✓ 500 000 cells
- D. 5 000 000 cells5 000 000 is above the gridline labeled 1 000 000.
The point at 3 days sits below that gridline.
Why: Find 3 days on the x-axis and go up to the point.
The point sits between the gridlines labeled 100 000 and 1 000 000.
It sits on the third minor gridline above 100 000.
The third minor gridline is 5 times the labeled gridline: 500 000 cells.
Nettles spread in two abandoned fields. Ecologists count the plants in each field every year and plot both counts on one log-scale y-axis, shown below. Both sets of points lie on straight lines.
Which population multiplies by the larger factor each year?
- A. ✓ The east field’s nettles
- B. The west field’s nettlesThe west field’s line starts higher, so that field holds more nettles.
Its line is less steep, so its count multiplies by the smaller factor each year. - C. The two fields’ nettles multiply by the same factorThe two lines rise at different steepnesses.
On a log-scale axis, a steeper line is a larger factor per year. - D. It cannot be told from a log-scale graphOn a log-scale y-axis, equal distances along the y-axis are equal factors.
The steeper line climbs further each year, so its factor is larger.
Why: On a log-scale y-axis, equal distances along the y-axis are equal factors.
Each year the east field’s point climbs further than the west field’s point.
So the east field’s count multiplies by the larger factor each year.
A student follows the fungus gnats in a greenhouse for four months, counting them once a month, and draws the graph below.
Which of the following is the one fault in the student’s graph?
- A. The y-axis should be an ordinary axisThe counts climb from 50 to 5 000, across two powers of ten.
On an ordinary axis to 5 000 the count of 50 sits pressed against the x-axis. - B. Time belongs on the y-axisThe student set the weeks and measured the counts.
The quantity that was set goes on the x-axis. - C. The points should not be joinedThe counts follow one greenhouse through time.
A line graph joins the points, one after another. - D. ✓ The x-axis title has no unit
Why: An axis title carries the quantity and its unit.
The y-axis title reads “population size, N (fungus gnats)”: quantity and unit.
The x-axis title reads “time” and gives no unit.
So the fault is the missing unit, months, on the x-axis.
Suppose a student counts the capybaras along a river once a year for six years. The counts climb from 12 capybaras to 1 900 capybaras.
Which row lists the right choices for showing these counts against time?
- A. A line graph; time on the x-axis; an ordinary y-axisFrom 12 to 1 900 spans more than two powers of ten.
On an ordinary y-axis the small counts sit pressed against the x-axis. - B. A bar graph; time on the x-axis; a log-scale y-axisThe counts follow one population through time.
Years are amounts along a time scale, so the graph is a line graph. - C. ✓ A line graph; time on the x-axis; a log-scale y-axis
- D. A line graph; population size on the x-axis; a log-scale y-axisThe student set the years and measured the counts.
The quantity that was set, time, goes on the x-axis.
Why: The counts follow one population through time, so the graph is a line graph.
The student set the years, so time goes on the x-axis.
From 12 to 1 900 spans more than two powers of ten.
So the y-axis is a log scale.
A student plots a count of 700 individuals on the log-scale y-axis drawn below. Its minor gridlines sit at 2, 3 and 5 times each labeled gridline.
Between which two gridlines does the point sit?
- A. Between the gridlines at 100 and 200100 and 200 are the labeled gridline and the first minor gridline above it.
700 is more than 500. - B. ✓ Between the gridlines at 500 and 1 000
- C. Between the gridlines at 1 000 and 2 000700 is less than 1 000.
The point sits below the gridline labeled 1 000. - D. On a gridline at 700The minor gridlines above 100 read 200, 300 and 500.
No gridline reads 700.
Why: 700 is more than 100 and less than 1 000.
Above 100 the minor gridlines read 200, 300 and 500.
700 is more than 500 and less than 1 000.
So the point sits between the 500 gridline and the 1 000 gridline.
A population grows with nothing holding it back, so dN/dt = r<sub>max</sub> N applies.
Which of the following multiplies dN/dt by four?
- A. Doubling N onlyDoubling N with r<sub>max</sub> fixed doubles the product r<sub>max</sub> N.
One factor doubled doubles dN/dt; it does not multiply it by four. - B. Doubling r<sub>max</sub> onlyDoubling r<sub>max</sub> with N fixed doubles the product r<sub>max</sub> N.
One factor doubled doubles dN/dt; it does not multiply it by four. - C. Doubling either N or r<sub>max</sub> on its ownDoubling one factor on its own doubles r<sub>max</sub> N.
Multiplying by four needs r<sub>max</sub> and N doubled together. - D. ✓ Doubling N and r<sub>max</sub> together
Why: dN/dt is r<sub>max</sub> multiplied by N.
Doubling N doubles the product.
Doubling r<sub>max</sub> as well doubles it again.
So doubling N and r<sub>max</sub> together multiplies dN/dt by four.
The table below gives one year’s births and deaths in four populations of moorland birds.
Which population shrinks over the year?
- A. The lapwingsThe lapwings’ 64 births equal their 64 deaths.
The lapwing population stays steady. - B. The snipe120 snipe are born and 95 die.
Births outnumber deaths, so the snipe population grows. - C. ✓ The plovers
- D. The grouse88 grouse are born and 70 die.
Births outnumber deaths, so the grouse population grows.
Why: A population shrinks once deaths outnumber births.
The lapwings’ births equal their deaths, and the snipe and the grouse have more births than deaths.
The plovers have 33 births and 51 deaths, so their deaths outnumber their births.
So the plover population shrinks.
Gorse plants spread over a cleared hillside. An ecologist counts the plants every year for four years. The ecologist’s table of counts, and her graph of them on a log-scale y-axis, are below. In year 4, 360 seedlings took root and 36 plants died. The plants have space, light and water to spare, and nothing on the hillside eats them.
(a) Calculate dN/dt for the gorse population in year 4. (1 pt)
Answer: 324 gorse plants per year (tolerance ±0)
- Award 1 point for: 324 gorse plants per year, with the unit (accept 324 per year).
Slip Adding the births to the deaths, 396. The rate of change is births minus deaths.
(b) Determine whether the gorse population is growing exponentially, using the counts in the table. (1 pt)
Each count is three times the count the year before: 6, 18, 54, 162, 486.
Multiplying by the same factor in equal times is exponential growth.
- Award 1 point for: yes, growing exponentially, because each count is three times the year before (the count multiplies by the same factor in equal times). Accept: the yearly gain grows every year (12, 36, 108, 324), with the counts quoted.
Slip Answering yes because the count rises. A count that rises by the same number every year also rises; exponential growth multiplies by the same factor.
(c) A student predicts the count in year 6 by adding two years of year 4’s rate of change to the year-4 count. Evaluate the student’s prediction. (1 pt)
The student’s method keeps the gain at 324 plants every year: 486 + 2 × 324 = 1 134 plants.
The plants have space to spare, so each plant keeps adding the same number of seedlings.
Each year there are more plants adding, so the yearly gain grows above 324.
The count in year 5 alone is 486 × 3 = 1 458, already more than 1 134.
- Award 1 point for: the judgement (too small) AND the ground (the method assumes a constant yearly gain, but each plant adds the same number and more plants are adding each year, so the gain grows). Accept with or without: year 5 is already 1 458.
Slip Judging the prediction right because 324 is the latest rate. A rate that is right for year 4 is too small for year 5, when more plants are adding.
(d) Explain the shape of the ecologist’s graph. (1 pt)
Each year the count multiplies by the same factor, three.
On a log-scale y-axis, equal distances along the y-axis are equal factors.
So each year the point climbs the same distance.
Equal steps along the x-axis and equal rises along the y-axis give a straight line.
- Award 1 point for: the points lie on a straight line because the count multiplies by the same factor every year, and on a log-scale y-axis equal distances are equal factors, so each year’s point climbs the same distance.
Slip Saying the plants add the same number every year. The yearly gain grows; it is the factor that stays the same.
Suppose 60 ptarmigan are released on a large island with plants to spare, where nothing eats them and no disease reaches them. Their maximum per capita growth rate, r<sub>max</sub>, is 0.35 per year.
(a) Explain why dN/dt = r<sub>max</sub> N describes the ptarmigan population in the years after the release. (1 pt)
So each ptarmigan adds its most, r<sub>max</sub>, every year.
So the population adds r<sub>max</sub> N ptarmigan each year, and the equation describes it.
- Award 1 point for: nothing holds the population back (food to spare, no predators, no disease), so each bird adds its most, r<sub>max</sub>, and the population adds r<sub>max</sub> N.
Slip Listing the island’s features and stopping. The point needs the link: nothing holds the birds back, so each adds its most, so the population adds r<sub>max</sub> N.
(b) Calculate dN/dt for the 60 ptarmigan. (1 pt)
Answer: 21 ptarmigan per year (tolerance ±0)
- Award 1 point for: 21 ptarmigan per year, with the unit (accept 21 per year).
Slip Adding 0.35 to 60. The two symbols written side by side are multiplied.
(c) Predict how the population’s yearly gain changes over the following years while the island stays as described. (1 pt)
- Award 1 point for: the yearly gain grows (rises year on year). Accept: dN/dt rises; more ptarmigan are added each year than the year before.
Slip Predicting the same gain every year because r<sub>max</sub> stays the same. r<sub>max</sub> is the gain per bird; the gain in all is r<sub>max</sub> times a growing N.
(d) Justify your prediction. (1 pt)
Each year the island holds more ptarmigan than the year before.
More birds, each adding the same number, add more young in all.
So the yearly gain, r<sub>max</sub> N, grows as N grows.
- Award 1 point for: each bird adds the same number of young per year (r<sub>max</sub> is unchanged) while the number of birds adding grows each year, so the population adds more in all (dN/dt = r<sub>max</sub> N with N rising).
Slip Saying each bird adds more young each year. Each bird adds the same number; more birds are adding.
APBIO-U08-L33 How crowded?
Suppose 40 deer live in a fenced reserve of 5 square kilometers (km²). Suppose 60 deer live in another reserve of 30 km².
The second herd is the bigger herd. Which herd is the more crowded?
Unit 8 · Ecology
1The count per square kilometer
Suppose wild ponies of one species graze a moor. Some ponies live on the moor’s east side and some on its west side, and the two groups mix: they graze the same grass and breed with each other.
Which of the following are the ponies of the moor?
- A. One communityA community is all the populations of every species living together in one place, not one species alone.
- B. ✓ One population
- C. Two populations, one on each side of the moorThe east ponies and the west ponies graze the same grass and breed with each other, so the two groups mix.
Why: The ponies are one species, and they live in one place, the moor.
The ponies mix: they compete for the same grass and breed with each other.
So the ponies of the moor are one population.
How crowded is a population?
The count alone does not say.
Divide the count by the area the population lives in, and you get how many deer live in each square kilometer.
The count per unit of area or volume is called the population density.
The first herd holds 8 deer in each km². The second holds only 2. So the smaller herd is the more crowded.
When a limit in this topic depends on the population, it depends on how crowded the population is, not on how large.
Here are the two reserves again, drawn to scale. Each dot marks one deer.
The small reserve covers 5 km² and holds 40 deer. The large reserve covers 30 km² and holds 60 deer.
The large reserve holds the bigger herd. But its deer are spread over six times as much ground.
Imagine the small reserve’s fence moved outward, so that the reserve covered twice as much ground with the same 40 deer inside.
Each deer would then have twice the room. So the herd would be less crowded.
Now imagine the fence left where it is, and 40 more deer released inside.
The same ground would then hold twice the count. So the herd would be more crowded.
So how crowded a herd is depends on two things: the count of deer, and the area of ground the deer live on.
One number is made from both: the count divided by the area. Here it is as an equation, with what each symbol means beneath.
N: the population size, the number of individuals
area: the ground the population lives in, in km²
population density: the number of individuals per km²
For a population in water, divide by the volume of water, in liters (L), instead of the area
To find how many deer live in each square kilometer of the small reserve, divide the count by the area.
The small reserve covers 5 km² and holds 40 deer. Calculate the deer’s population density.
So the small reserve holds 8 deer in each square kilometer.
Per km² means in each square kilometer. So 8 deer per km² is 8 deer in each square kilometer of the reserve.
Now work the large reserve the same way: 60 deer on 30 km².
The large reserve covers 30 km² and holds 60 deer. Calculate the deer’s population density.
So the large reserve holds 2 deer in each square kilometer.
A count of individuals per unit of area, such as 8 deer per km², is called the .
Divide a population’s count by the area the population lives in, and you have its population density.
Here is a table comparing the two herds: the count, the area and the population density.
Here are the two reserves again, with each herd’s population density written beneath.
The large reserve holds the bigger herd: 60 deer against 40 deer.
The small reserve holds the denser herd: 8 deer per km² against 2 deer per km².
A bigger population is not always a denser one. The count says how many; the population density says how crowded.
A population that lives in water has no area of ground to divide by. Divide by the volume of water instead, in liters (L).
Suppose 280 tadpoles live in a tank holding 8 L of pond water.
A tank holds 8 L of pond water and 280 tadpoles. Calculate the tadpoles’ population density.
So the tank holds 35 tadpoles in each liter of water: the tadpoles’ population density.
The unit of a population density is the count’s unit per the area’s or the volume’s unit: deer per km², tadpoles per L.
What you are expected to know Calculate a population’s density: the number of individuals per unit of area or volume.
What you are expected to know Compare two populations by their population densities, not by their counts.
Suppose 90 llamas graze a high plateau of 15 km².
Calculate the population density of the llamas.
Part 1. State the population size, N, of the llamas.
Answer: 90 llamas (tolerance ±0)
Part 2. State the area the llamas live on.
Answer: 15 km² (tolerance ±0)
Answer: 6 llamas per km² (tolerance ±0)
Suppose 84 wombats live in a reserve of 12 km².
Calculate the population density of the wombats.
Answer: 7 wombats per km² (tolerance ±0)
Suppose 450 amphipods, small swimming crustaceans, live in a tank holding 18 L of seawater.
Calculate the population density of the amphipods.
Answer: 25 amphipods per L (tolerance ±0)
The table below gives the count and the area for two hedgehog populations.
Which population has the higher population density?
- A. ✓ Fen Park
- B. Heath ParkHeath Park holds 4 hedgehogs per km²; Fen Park holds 9 hedgehogs per km².
Why: Fen Park: 36 hedgehogs on 4 km², so 9 hedgehogs per km².
Heath Park: 44 hedgehogs on 11 km², so 4 hedgehogs per km².
The higher population density is Fen Park’s.
The table below gives the count and the area for two gazelle populations.
Which population has the higher population density?
- A. Kara ReserveKara Reserve holds 4 gazelles per km²; Tem Reserve holds 5 gazelles per km².
- B. ✓ Tem Reserve
Why: Kara Reserve: 96 gazelles on 24 km², so 4 gazelles per km².
Tem Reserve: 75 gazelles on 15 km², so 5 gazelles per km².
The higher population density is Tem Reserve’s.
The table below gives the count and the area for two koala populations.
Which population has the higher population density?
- A. East ForestEast Forest holds 6 koalas per km²; West Forest holds 7 koalas per km².
- B. ✓ West Forest
Why: East Forest: 78 koalas on 13 km², so 6 koalas per km².
West Forest: 49 koalas on 7 km², so 7 koalas per km².
The higher population density is West Forest’s.
The table below gives the count and the volume of water for two tanks of water boatmen, small insects that swim in ponds.
Which population has the higher population density?
- A. ✓ Tank J
- B. Tank PTank P holds 30 water boatmen per L; Tank J holds 55 water boatmen per L.
Why: Tank J: 385 water boatmen in 7 L, so 55 water boatmen per L.
Tank P: 720 water boatmen in 24 L, so 30 water boatmen per L.
The higher population density is Tank J’s.
The table below gives the count and the area for two petrel colonies on two islands.
Which population has the higher population density?
- A. Skerry IslandSkerry Island holds 70 petrels per km²; Long Island holds 80 petrels per km².
- B. ✓ Long Island
Why: Skerry Island: 210 petrels on 3 km², so 70 petrels per km².
Long Island: 640 petrels on 8 km², so 80 petrels per km².
The higher population density is Long Island’s.
A student reads that the north fen holds 80 capybaras and the south fen holds 30 capybaras, and says: “The north fen’s capybaras are the more crowded, because there are more of them.”
Is the student correct?
- A. Yes: the north fen holds more capybaras, so its capybaras are the more crowdedThe count alone does not say how crowded a population is.
The area the population lives on matters too. - B. ✓ No: the count per km² decides, and the two fens’ areas are not given
Why: Population density is the count divided by the area.
The student was given the two counts but not the two areas.
So the student cannot say which fen’s capybaras are the more crowded.
The table below gives two wallaby populations. A student says: “The two islands hold the same number of wallabies, but Island M’s wallabies are the more crowded.”
Is the student correct?
- A. ✓ Yes: Island M’s 30 wallabies live on less ground, so they are the more crowded
- B. No: the two islands hold the same count of wallabies, so their wallabies are equally crowdedIsland M’s 30 wallabies live on 3 km²; Island P’s 30 wallabies live on 10 km².
The counts match, but the areas do not.
Why: Island M: 30 wallabies on 3 km², so 10 wallabies per km².
Island P: 30 wallabies on 10 km², so 3 wallabies per km².
The same count on less ground is the higher population density.
Here are the two reserves again: 40 deer on 5 km², and 60 deer on 30 km².
The large reserve holds the bigger herd. The small reserve holds the denser herd: 8 deer per km² against 2 deer per km².
So the smaller herd is four times as crowded as the bigger herd.
When a limit in this topic depends on the population, it depends on how crowded the population is, not on how large.
53Quick quiz: population density mixed practice
Suppose 72 camels graze a desert reserve of 24 km².
Calculate the population density of the camels.
Answer: 3 camels per km² (tolerance ±0)
Suppose 405 flamingos feed on a salt lake of 9 km².
Calculate the population density of the flamingos.
Answer: 45 flamingos per km² (tolerance ±0)
Suppose 1 950 fulmars nest on a cliff-top island of 6 km².
Calculate the population density of the fulmars.
Answer: 325 fulmars per km² (tolerance ±0)
Suppose a 12 L sample of stream water holds 144 mayfly nymphs.
Calculate the population density of the mayfly nymphs.
Answer: 12 mayfly nymphs per L (tolerance ±0)
Suppose 2 100 shags, a seabird, nest on the ledges of a sea cliff whose island covers 7 km².
Calculate the population density of the shags.
Answer: 300 shags per km² (tolerance ±0)
Suppose dormice live in two parks. The table below gives each park’s count of dormice and its area.
(a) Calculate the population density of the dormice in Bramble Park. (1 pt)
Answer: 11 dormice per km² (tolerance ±0)
- Award 1 point for: 11 dormice per km², with the unit (accept 11 per km²).
(b) Determine which park’s dormice have the higher population density. (1 pt)
- Award 1 point for: Bramble Park.
(c) Justify your answer to part (b). (1 pt)
Thorn Park holds 3 dormice per km².
The decision rests on the count per km², not on the count alone, which is 66 dormice in both parks.
- Award 1 point for: the decision rests on the population density (the count per km²: 11 against 3), not on the count, which is the same in both parks.
Which of the following is the population density?
- A. The number of individuals in the populationThe number of individuals in the population is its population size, N.
- B. The area or volume the population lives inThe area or volume is what the count is divided by, not the density itself.
- C. ✓ The number of individuals per unit of area or volume
Why: Population density is the count divided by the area or the volume the population lives in.
So it is the number of individuals per unit of area or volume.
Glossary
- population density
- The number of individuals of a population per unit of area or volume, such as 8 deer per square kilometer. Found by dividing the population size, N, by the area (or, for a population in water, the volume) the population lives in.
APBIO-U08-L33B The curve that levels off
Imagine yeast cells growing in a test tube of sugar solution, counted every two hours. For the first ten hours the count climbs faster and faster, just like the bacteria in the flask of broth did. Then the climb slows.
By hour sixteen the count sits near 10 million cells per milliliter (mL) and stays there. Nothing goes into the tube and nothing comes out. What stopped the climb?
Unit 8 · Ecology
1The level the count settles at
Suppose a flask of bacteria has food to spare, no predator and no disease. The count of bacteria doubles every hour.
On an ordinary y-axis, which shape does the curve of N against time have?
- A. A straight rising lineA straight rising line adds the same number every hour.
A count that doubles adds more each hour than the hour before. - B. ✓ A curve that bends upward and gets steeper
Why: The count doubles every hour.
Each hour adds as many bacteria as the whole count before it, so the gain grows every hour.
So the curve bends upward and gets steeper: a J-shaped curve.
A technician measures out 1 mL of solution from a tube of yeast and counts 4 million cells in it.
Which of the following is a count of 4 million cells per mL?
- A. ✓ The population density
- B. The population size, NThe population size, N, is the count of the whole population, not the count in each milliliter.
- C. The maximum per capita growth rate, rmaxrmax is the number of new cells each cell adds per hour when nothing limits it, not a count of cells.
Why: The count is the number of cells in each milliliter of solution.
A count per unit of volume is the population density.
Suppose 85 oystercatcher chicks hatch on a beach in one season, and 85 oystercatchers die on that beach in the same season.
What does the size of the oystercatcher population do over that season?
- A. GrowsThe population grows only when more are born than die.
Here the same number are born as die. - B. ShrinksThe population shrinks only when more die than are born.
Here the same number die as are born. - C. ✓ Stays the same
Why: 85 chicks join the population and 85 oystercatchers leave it.
Births equal deaths.
So the population’s size stays the same.
Why does a growing population stop growing?
The tube holds a fixed amount of sugar, and every new cell takes a share of it.
As the cells multiply, each cell’s share shrinks.
Cells short of sugar divide more slowly and die sooner.
Births fall and deaths rise until births equal deaths, and then the count stays steady.
The level the count settles at is the carrying capacity, written K, because it is the most the tube’s resources can carry.
A curve that climbs and then levels off is an S-shaped curve.
Growth that follows an S-shaped curve is logistic growth.
Video: Watch: The curve that levels off
The yeast counts are plotted one every two hours. For the first ten hours each point climbs further above the one before. Then the climb slows. From hour 18 the points sit on the 10 million gridline. A dashed line is drawn along that level and labeled K.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L33Ba.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L33Ba.mp4
Here are the counts as a table: the time in hours, and the number of yeast cells in each milliliter of the solution, in millions.
Each count is the number of cells in 1 mL of solution. So each count is the yeast’s population density.
The tube holds the same volume of solution all the way through. So the count per mL rises and falls with the population size, N.
Here are the same counts plotted: time in hours on the x-axis, millions of cells per mL on the y-axis, a gridline every 2 million cells per mL.
For the first ten hours the curve bends upward and gets steeper: the J-shaped curve of the flask.
After hour 10 the curve bends the other way. Each two hours adds fewer cells than the two hours before.
From hour 18 the curve is flat along the 10 million gridline. The count stays at 10 million cells per mL.
Suppose nothing had limited the yeast. Each cell would have kept adding the same number, and the count would have followed the dashed J-shaped curve.
The real count left the J-shaped curve at about hour 10. Then the real count leveled off.
The count settled at 10 million cells per mL, read from the 10 million gridline. The tube’s sugar can feed no more yeast cells than that.
The largest population that a place’s resources can support is called the , written K.
On a growth graph, K is drawn as a dashed horizontal line at the level the curve settles at, labeled K at the right.
In this tube, K is 10 million cells per mL.
The tube’s sugar set this K, for yeast. A tube with more sugar can feed more cells, so its K for yeast is larger.
The same tube gives a different species a different K. A different species needs a different share of sugar.
A curve that climbs, slows and then levels off is called an , because it looks like a stretched letter S.
Growth that follows an S-shaped curve, slowing as the count nears K, is called .
Here are three more populations. For each, ask: has the count leveled off?
For example, the sticklebacks’ curve has leveled off. So the sticklebacks have reached K, and K is 250 sticklebacks, read from the 250 gridline.
But the chaffinches’ curve has not leveled off. So the chaffinches have not reached K yet, and no K can be read.
And the tench’s curve has leveled off. So the tench have reached K, and K is 160 tench, read from the 160 gridline.
The three are drawn below, each with its verdict.
A curve that has leveled off gives K: the gridline the level part sits on. A curve still climbing gives no K yet.
What you are expected to know Identify the carrying capacity, K, on an S-shaped growth curve: the level the count settles at.
The graph below shows the count of limpets on one stretch of rocky shore, once a year for 12 years.
Identify the carrying capacity, K, of this shore for limpets.
Answer: 1400 limpets (tolerance ±0)
The graph below shows the count of house flies in a laboratory cage, once a day for 6 days.
Has the count leveled off?
- A. ✓ No
- B. YesThe curve bends upward and the last gain is the largest of all.
A curve that has leveled off is flat at the end.
Why: The curve bends upward all the way to day 6.
The last point sits well above the one before.
So the count is still climbing, and no K can be read yet.
The graph below shows the count of terrapins in a lake, once a year for 10 years.
Has the count leveled off?
- A. NoFrom year 8 the curve is flat: the last three points sit on one gridline.
- B. ✓ Yes
Why: From year 8 the curve is flat along the 150 gridline.
So the count has leveled off.
K is 150 terrapins, read from the 150 gridline.
The graph below shows the count of roach in a canal, once a year for 6 years.
Has the count leveled off?
- A. ✓ No
- B. YesThe climb has slowed, but the last point sits above the one before.
A curve that has leveled off is flat at the end.
Why: After year 3 each year adds fewer roach than the year before.
But the year-6 point sits above the year-5 point.
So the count is still climbing, and no K can be read yet.
A student reads that the yeast in the tube leveled off, and says: “K belongs to yeast, so every tube of yeast has the same K.”
Is the student correct?
- A. ✓ No: K is set by the tube’s sugar, so a tube with more sugar has a larger K for yeast
- B. Yes: K belongs to the yeast, so every tube of yeast levels off at the same countK is the largest population the tube’s resources can support.
A tube with more sugar can support more yeast cells.
Why: The tube’s sugar sets how many cells the tube can feed.
A tube with more sugar can feed more cells.
So a tube with more sugar has a larger K for yeast: K is set by the tube’s resources, not by the yeast alone.
43Quick quiz: carrying capacity, S-shaped curve, logistic growth mixed practice
Suppose an angler counts the bream in a pond once a year for 12 years. The graph below shows the counts.
(a) Identify the pond’s carrying capacity, K, for bream. (1 pt)
Answer: 1800 bream (tolerance ±0)
- Award 1 point for: 1 800 bream (accept 1 800).
(b) Describe how the shape of the curve changes from year 0 to year 12. (2 pt)
After year 6 the climb slows: each year adds fewer bream than the year before.
From year 10 the curve is flat along the 1 800 gridline.
- Award 1 point for: the curve first bends upward and gets steeper (accept: climbs faster and faster).
- Award 1 point for: the climb then slows and the curve levels off (accept: becomes flat; an S-shaped curve).
Which of the following is the carrying capacity, K?
- A. The fastest rate at which a population can growThe fastest rate at which a population can grow is rmax, a rate, not a level.
- B. The count of individuals in each square kilometer or liter of a placeThe count in each square kilometer or liter is the population density: how crowded, not how many the place can support.
- C. ✓ The largest population that a place’s resources can support
Why: A place’s resources can feed only so many individuals.
The count settles at that number.
So the carrying capacity, K, is the largest population the place’s resources can support.
Which of the following is an S-shaped curve?
- A. A curve of N against time that climbs faster and faster and never levels offA curve that climbs faster and faster and never levels off is the J-shaped curve of a population that nothing limits.
- B. ✓ A curve of N against time that climbs, slows and then levels off
Why: An S-shaped curve climbs, slows and then levels off at K.
Its shape is a stretched letter S.
Which of the following is logistic growth?
- A. Growth that adds the same number in every equal timeGrowth that adds the same number in every equal time is a straight rising line, not an S-shaped curve.
- B. Growth whose gain grows in every equal time, because more individuals are addingGrowth whose gain grows in every equal time is exponential growth, the J-shaped curve.
- C. ✓ Growth that slows as the count nears K and then stops at K
Why: Logistic growth follows an S-shaped curve.
The count climbs, then slows as it nears K, then stays at K.
48Why the climb slows
After a meal, a person’s blood glucose rises. The rise triggers the release of insulin, and the insulin brings the glucose back down toward its set point. As the glucose falls, the insulin release fades.
What is this kind of control called?
- A. ✓ Negative feedback
- B. Positive feedbackIn positive feedback the response makes the change bigger.
Here the response, insulin, reduces the change that triggered it.
Why: The rise in glucose triggered the insulin.
The insulin brought the glucose back down.
So the response reduced its own trigger: negative feedback.
Here is the tube of yeast again: sugar solution, counted every two hours, the curve that climbed and then leveled off at K, 10 million cells per mL.
Why did the climb slow, when nothing went into the tube and nothing came out?
Video: Watch: The sugar shared out
The tube’s sugar is drawn as one bar. The bar is shared among 4 cells, then 16, then 64, and each cell’s share narrows. Cells with a narrow share divide more slowly and die sooner. Births fall and deaths rise until they are equal, and the curve levels off at K.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L33Bb.mp4
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The tube held a fixed amount of sugar at the start. No more sugar ever went in.
Every yeast cell takes in sugar. A cell uses the sugar to grow and to divide.
Imagine the tube’s sugar as one bar, shared equally among the cells.
With 4 cells, the bar is split four ways. Each cell’s share is wide.
With 16 cells, the same bar is split sixteen ways. Each cell’s share is narrower.
With 64 cells, the same bar is split sixty-four ways. Each cell’s share is a sliver.
So as the cells multiply, each cell’s share of sugar shrinks.
A cell with plenty of sugar divides quickly. A cell short of sugar divides more slowly, and it dies sooner.
Early on, every cell has sugar to spare. Many cells divide and few die, so births far outnumber deaths and the count climbs fast.
As the count nears K, each cell’s share is small. Fewer cells divide and more die.
Births fall and deaths rise until births equal deaths. Then the count stays steady, at K.
At K the yeast have not stopped living. Cells still divide and cells still die, in equal numbers.
In Unit 4, a rise in blood glucose triggered insulin, and the insulin brought the glucose back down. The response reduced its own trigger: negative feedback.
In the tube, the change is the count rising. The rising count shrinks each cell’s share of sugar.
The smaller share slows the births and speeds the deaths. So the smaller share slows the very rise that caused it.
As a population nears its carrying capacity each individual’s share of resources falls, births fall and deaths rise until they balance: negative feedback on growth.
Here is a table comparing the two loops: the change, the response, and what the response does.
What you are expected to know Explain why a population’s growth slows as its count nears K: each individual’s share of resources shrinks, so births fall and deaths rise until they are equal.
Suppose a jar of pond water holds a growing population of ostracods, tiny crustaceans, and a fixed amount of food. The count is nearing the jar’s K.
As the count nears K, what happens to the number of young born per ostracod each day?
- A. RisesMore young per ostracod would need each ostracod to have more food.
Each ostracod’s share of food is shrinking. - B. ✓ Falls
- C. Stays the sameEach ostracod’s share of food is shrinking, so the births per ostracod cannot stay as they were.
Why: The count is rising, so each ostracod’s share of the fixed food shrinks.
An ostracod short of food breeds more slowly.
So fewer young are born per ostracod each day.
Suppose a jar of pond water holds a growing population of ostracods, tiny crustaceans, and a fixed amount of food. The count is nearing the jar’s K.
What happens to the number of ostracods that die each day as the count nears K?
- A. ✓ Rises
- B. FallsFewer deaths would need each ostracod to have more food.
Each ostracod’s share of food is shrinking. - C. Stays the sameEach ostracod’s share of food is shrinking, so the deaths cannot stay as they were.
Why: The count is rising, so each ostracod’s share of the fixed food shrinks.
An ostracod short of food dies sooner.
So more ostracods die each day.
Suppose stick insects live in a cage. Every week the keeper puts in the same amount of bramble leaves, their food, and takes out what is left uneaten. The count of stick insects climbed for six weeks and then leveled off.
(a) Explain why the count of stick insects stopped climbing. (3 pt)
As the stick insects multiplied, each stick insect’s share of the bramble shrank.
A stick insect short of food breeds more slowly and dies sooner, so births fell and deaths rose.
Births and deaths became equal.
So the count stopped climbing and stayed steady.
- Award 1 point for: each stick insect’s share of the food shrank as the count rose (accept: the food per insect fell).
- Award 1 point for: births fell and deaths rose (accept either half stated with its cause: less food per insect).
- Award 1 point for: births became equal to deaths, so the count stayed steady (accept: the cage had reached its carrying capacity, K, with births equal to deaths).
A student looks at the yeast counts at hours 18 and 20, both 10 million cells per mL, and says: “At K the population has stopped: no cell divides and no cell dies.”
Is the student correct?
- A. ✓ No: cells still divide and cells still die, in equal numbers
- B. Yes: a steady count means no cell divides and no cell diesA steady count means the same number of cells divide as die.
Cells are still dividing and still dying.
Why: At K each cell has a small share of sugar.
Some cells still divide, and some cells still die.
The number that divide equals the number that die, so the count stays steady.
Suppose donkeys graze an island whose grass grows back by the same amount each year. The herd has grown for many years, and its count is now near the island’s K.
Compared with an early year, when the herd was small, how large is each donkey’s share of grass now?
- A. LargerThe grass grows back by the same amount each year, and there are more donkeys to share it.
- B. ✓ Smaller
- C. The sameThe grass grows back by the same amount each year, but more donkeys now share it.
Why: The island grows the same amount of grass each year.
The herd is larger than in the early year.
So the same grass is shared among more donkeys, and each donkey’s share is smaller.
Here is the tube of yeast again: sugar solution, counted every two hours, the curve that climbed and then leveled off at 10 million cells per mL.
The count climbed while every cell had sugar to spare.
Each cell’s share shrank as the cells multiplied.
Births fell and deaths rose until births equaled deaths.
So the count settled near 10 million cells per mL: the tube’s K.
81Mixed practice: the curve that levels off mixed practice
The graph below shows the count of dace in a stream pool, once a month for 12 months.
Identify the carrying capacity, K, of this pool for dace.
Answer: 700 dace (tolerance ±0)
Two tanks of the same size hold the same species of gudgeon, a small fish. The keeper gives Tank J more food each day than Tank P. A student says: “Tank J’s carrying capacity for gudgeon is larger than Tank P’s.”
Is the student correct?
- A. ✓ Yes: Tank J’s gudgeon get more food each day, so Tank J can feed more gudgeon than Tank P
- B. No: the extra food makes each gudgeon grow larger, so Tank J levels off at the same count as Tank PMore food each day feeds more gudgeon, not only bigger gudgeon.
So Tank J’s K is larger than Tank P’s.
Why: K is the largest population a place’s resources can support.
The keeper gives Tank J’s gudgeon more food each day than Tank P’s.
So Tank J can support more gudgeon: its K is larger.
Suppose flour moths breed in a bin of flour that a baker tops up each week. As the count nears the bin’s K, more moths die each week than before.
Which of the following is the reason more moths die each week?
- A. ✓ A moth’s share of the flour has shrunk
- B. The bin’s flour is used up, so no moth can feedNear K the flour is not gone: many moths share each week’s flour, so each moth’s share is small.
- C. Near K the moths have stopped breeding, so the count is fallingNear K moths still breed; fewer young are born per moth and more moths die, until births equal deaths.
Why: The count has risen, so each moth’s share of each week’s flour has shrunk.
A moth short of food dies sooner.
So more moths die each week than when the bin held few moths.
The two graphs below show two populations counted once a week for 8 weeks.
Which graph shows logistic growth?
- A. Graph 1Graph 1 bends upward and gets steeper all the way to week 8: exponential growth, a J-shaped curve.
- B. ✓ Graph 2
Why: Logistic growth follows an S-shaped curve: it climbs, slows and levels off.
Graph 2 climbs, slows and is flat at week 8.
So graph 2 shows logistic growth.
In the tube of yeast, the rising count shrank each cell’s share of sugar, and the smaller share slowed the rise.
Which kind of feedback is this?
- A. Positive feedbackIn positive feedback the response makes the change bigger.
Here the smaller share slowed the rise that caused it. - B. ✓ Negative feedback
Why: The change was the count rising.
The rising count shrank each cell’s share of sugar.
The smaller share slowed the rise.
The response reduced its own trigger: negative feedback.
The graph below shows the count of damselfly larvae in a pond, once a week for 6 weeks.
Has the count leveled off?
- A. ✓ No
- B. YesThe climb has slowed, but the week-6 point sits above the week-5 point.
A curve that has leveled off is flat at the end.
Why: After week 3 each week adds fewer larvae than the week before.
But the week-6 point sits above the week-5 point.
So the count is still climbing, and no K can be read yet.
Three tubes of yeast are counted for one hour. In Tube J, 40 cells divide and 12 die. In Tube M, 25 cells divide and 25 die. In Tube R, 10 cells divide and 30 die.
Which tube’s population is at its K?
- A. Tube JIn Tube J more cells divide than die, so the count is still climbing.
- B. ✓ Tube M
- C. Tube RIn Tube R more cells die than divide, so the count is falling.
Why: At K the count stays steady.
A steady count means the number of cells that divide equals the number that die.
In Tube M 25 cells divide and 25 die.
So Tube M’s population is at its K.
Suppose a farmer stocks a new pond with catfish. The pond’s supply of food is fixed. The count of catfish climbs for two years and then stays the same for the next three years.
(a) Determine whether the catfish have reached the pond’s carrying capacity. (1 pt)
- Award 1 point for: the catfish have reached the pond’s carrying capacity (accept: have reached K).
(b) Justify your answer to part (a). (1 pt)
So the count has leveled off, at the pond’s carrying capacity.
- Award 1 point for: the count stayed the same for three years after climbing (accept: the count has leveled off).
(c) A student says that in year 4 no catfish are born. Explain why the student is wrong. (2 pt)
Catfish are still born each year.
The same number of catfish die each year as are born.
Births equal deaths, so the count stays the same although catfish are still born.
- Award 1 point for: catfish are still born at K (accept: births continue).
- Award 1 point for: the count stays the same because births equal deaths (accept: as many die as are born).
Glossary
- carrying capacity (K)
- The largest population that a place’s resources can support. On a growth graph it is the level an S-shaped curve settles at, drawn as a dashed horizontal line labeled K.
- S-shaped curve
- A curve of population size, N, against time that climbs, slows and then levels off at the carrying capacity, K. It looks like a stretched letter S.
- logistic growth
- Growth that follows an S-shaped curve: the population climbs, slows as its count nears the carrying capacity, K, and then stays at K.
APBIO-U08-L34 Limits that bite harder in a crowd
Photo: Jessica Reeder, Wikimedia Commons, CC BY-SA 2.0 (resized).
Suppose two rabbit pens stand side by side, the same size. One pen holds 10 rabbits and the other holds 40. The keeper puts one rabbit carrying a disease into each pen.
A month later, 1 of the 10 rabbits has died in the sparse pen, but 12 of the 40 have died in the crowded pen: 10 % against 30 %. Now suppose instead that a night of hard frost had hit both pens. The frost would have killed 1 of the 10 rabbits and 4 of the 40: 10 % and 10 %. Why did the disease harm the crowded pen more, while the frost would harm both pens the same?
Unit 8 · Ecology
1What holds the growth back
Suppose water fern floats on a tank of pond water. The tank holds a fixed supply of the minerals the water fern needs. The count of water-fern plants climbed for three weeks and then leveled off.
What set the level the count settled at?
- A. ✓ The tank’s supply of minerals
- B. The water fern’s own nature, the same in any tankK is set by a place’s resources for the species.
A tank with more minerals has a larger K for water fern. - C. The number of plants at the startThe count at the start sets where the curve begins, not where it levels off.
Why: Each new plant takes a share of the tank’s minerals.
When the minerals can feed no more plants, births equal deaths and the count stays steady.
So the tank’s supply of minerals set K.
What holds a population’s growth back?
Does crowding change how much harm a limit does?
Anything that holds a population’s growth below what it would otherwise be is a limiting factor.
Some limits harm a larger fraction of a population the more crowded it is: a disease, a shortage of food, a predator.
In a crowd each animal comes into contact with more others, gets a smaller share of the food, and is found more easily by a predator.
Other limits harm the same fraction whether the pen holds 10 rabbits or 40: a frost, a flood, a fire.
Every limit in this topic is one kind or the other.
Video: Watch: What holds the growth back
The tube of yeast is on screen, then the crowded pen, then a river with kingfishers. On each, the thing that held the growth back is named: the sugar, the disease, the earth banks.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L34a.mp4
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Here are the two pens again, seen from above: 10 rabbits in one and 40 in the other, each rabbit a dot.
In the tube of yeast, each cell’s share of sugar shrank as the cells multiplied. So the count of yeast stopped climbing.
In the crowded pen, the disease killed 12 rabbits in a month. So the pen’s count grew less than it would have without the disease.
Now suppose kingfishers nest along a river. Each pair digs its nest tunnel into a soft earth bank.
The river has only 12 such banks. So no more than 12 pairs ever nest, however many fish the river holds.
The sugar, the disease and the earth banks each hold a population’s growth below what it would otherwise be. Anything that does that is called a .
A limiting factor can be a resource that falls short: the sugar, the earth banks. Or it can be a hazard that kills: the disease.
Here is a table comparing the three cases: what held the growth back, and whether that thing is a resource or a hazard.
What you are expected to know Identify the limiting factor in a described case: the resource or hazard that holds the population’s growth below what it would otherwise be.
Suppose a keeper houses a flock of budgerigars in a large aviary with floor space to spare. Every day the keeper gives the flock the same amount of seed. The count of budgerigars climbed and then leveled off.
Which of the following is the limiting factor?
- A. ✓ The seed
- B. The floor spaceThe aviary has floor space to spare, so space holds no budgerigar back.
Why: The keeper gives the same amount of seed every day.
As the flock grew, each budgerigar’s share of the seed shrank.
So the seed holds the flock’s growth back: the seed is the limiting factor.
Suppose blue tits breed in a forest where insects to eat are plentiful. Only 20 holes in the forest are fit for a blue tit’s nest. Every year exactly 20 pairs nest.
Which of the following is the limiting factor?
- A. The insectsThe insects are plentiful, so food holds no blue tit back.
- B. ✓ The nest holes
Why: A pair of blue tits breeds only in a nest hole.
The forest has 20 holes, so only 20 pairs breed each year.
So the nest holes hold the population’s growth back: the nest holes are the limiting factor.
Suppose shield bugs live in a meadow with plants to feed on and ground to spare. Every summer a fungal disease spreads through the shield bugs and kills 30 % of them.
Which of the following is the limiting factor?
- A. The plantsThe meadow has plants to spare, so food holds no shield bug back.
- B. The groundThe meadow has ground to spare, so space holds no shield bug back.
- C. ✓ The fungal disease
Why: The shield bugs have plants and ground to spare.
The fungal disease kills 30 % of them every summer.
So the disease holds the population’s growth back: the fungal disease is the limiting factor.
Suppose a herd of impala grazes a reserve with grass to spare. In the dry season the reserve’s one waterhole shrinks to a puddle, and many impala die of thirst.
Which of the following is the limiting factor?
- A. ✓ The water
- B. The grassThe reserve has grass to spare, so food holds no impala back.
Why: The impala have grass to spare.
In the dry season the water falls short, and impala die of thirst.
So the water holds the herd’s growth back: the water is the limiting factor.
Suppose grayling, a river fish, live in a river with plenty of insects to eat and deep pools with room to spare. A pair of cormorants fishes the river every day, and each year the cormorants eat 30 % of the young grayling.
Which of the following is the limiting factor?
- A. The insectsThe river has insects to spare, so food holds no grayling back.
- B. ✓ The cormorants
- C. The poolsThe pools have room to spare, so space holds no grayling back.
Why: The grayling have food and room to spare.
The cormorants eat 30 % of the young grayling every year.
So the cormorants hold the population’s growth back: the cormorants are the limiting factor.
25Quick quiz: limiting factor mixed practice
Suppose swallows nest inside a barn. Flying insects to eat are plentiful all summer. A swallow pair builds its nest only on a ledge, and the barn has 8 ledges. Every year 8 pairs nest in the barn, and the count of nesting swallows has stayed at 8 pairs for ten years.
(a) Identify the limiting factor for the barn’s swallows. (1 pt)
- Award 1 point for: the ledges (accept: the nest sites; the number of ledges).
(b) Describe how this limiting factor holds the swallows’ growth back. (2 pt)
The barn has 8 ledges, so only 8 pairs can nest.
So the count of nesting pairs cannot climb above 8, however many insects there are.
- Award 1 point for: a pair needs a ledge to nest, and the barn has only 8 ledges.
- Award 1 point for: so no more than 8 pairs can nest, and the count grows no further (accept: the count stays at 8 pairs).
Which of the following is a limiting factor?
- A. A resource that falls short, but never a hazard that killsA hazard that kills, such as a disease, holds growth back too: a limiting factor can be a resource or a hazard.
- B. The level a population’s count settles at when a resource falls shortThe level the count settles at is the carrying capacity, K; a limiting factor is what holds the growth there.
- C. ✓ Anything that holds a population’s growth below what it would otherwise be
Why: A resource that falls short, or a hazard that kills, holds a population’s growth below what it would otherwise be.
Anything that does that is a limiting factor.
28Two kinds of limit
Suppose 45 partridges live on a moor of 21 km², and another 45 partridges live on a moor of 7 km².
Which population has the higher population density?
- A. The partridges on the 21 km² moorThe 21 km² moor spreads the same 45 partridges over three times the ground.
So it holds fewer partridges in each square kilometer. - B. ✓ The partridges on the 7 km² moor
Why: Population density is the count per unit of area.
Both moors hold 45 partridges.
The 7 km² moor is the smaller, so it packs the same count onto less ground.
So the partridges on the 7 km² moor have the higher population density.
Here are the two pens again: 10 rabbits in one, 40 in the other, and the pens are the same size. So the crowded pen has four times the population density of the sparse pen.
Four different limits are about to hit these two pens, one at a time. For each, compare the fraction of rabbits it harms in the sparse pen with the fraction in the crowded pen.
Video: Watch: Two kinds of limit
The two pens are on screen. The disease hits both: 1 of 10 rabbits and 12 of 40 are marked. The frost hits both: 1 of 10 and 4 of 40. Each fraction is written under its pen, and each limit is sorted into one of two columns.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L34b.mp4
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Here are the four limits, one at a time. For each, ask: did the factor harm a larger fraction of the crowded pen’s rabbits?
For example, the disease killed 1 of the 10 rabbits in the sparse pen and 12 of the 40 in the crowded pen: 10 % against 30 %.
So the disease harmed a larger fraction of the crowded pen’s rabbits.
But the frost killed 1 of the 10 rabbits in the sparse pen and 4 of the 40 in the crowded pen: 10 % against 10 %.
So the frost harmed the same fraction of both pens’ rabbits.
Now suppose the stream beside the pens had flooded both pens to the same depth.
And the flood would drown 2 of the 10 rabbits in the sparse pen and 8 of the 40 in the crowded pen: 20 % against 20 %.
So the flood harmed the same fraction of both pens’ rabbits, too.
Now suppose the keeper had given both pens the same small amount of hay each day. In the crowded pen each rabbit’s share is 25 % as large as in the sparse pen.
But the hay shortage left 0 of the 10 rabbits in the sparse pen hungry and 20 of the 40 in the crowded pen: 0 % against 50 %.
So the hay shortage harmed a larger fraction of the crowded pen’s rabbits.
The four rows are drawn below, each with its verdict. A filled dot is a rabbit the factor harmed.
A limiting factor that harms a larger fraction the denser the population is, like the disease and the hay shortage, is called a . Its harm depends on the density.
A limiting factor that harms the same fraction at any density, like the frost and the flood, is called a . Its harm does not depend on the density.
To sort a limiting factor, compare the fraction it harms at a low density with the fraction it harms at a high density.
A larger fraction when denser means density-dependent. The same fraction means density-independent.
The frost killed 4 rabbits in the crowded pen and only 1 in the sparse pen. But 4 of 40 and 1 of 10 are the same fraction: 10 %.
So the frost is density-independent, although it killed more rabbits where there were more rabbits.
What you are expected to know Classify a limiting factor as density-dependent or density-independent from the fraction it harms at two densities.
Two coops of the same size hold hens. A cough-like disease reaches both coops. The table below shows the hens in each coop and how many fell ill in a month.
Which kind of limiting factor is the disease?
- A. ✓ Density-dependent
- B. Density-independentThe disease harmed 15 % of the sparse coop and 40 % of the crowded coop.
The fraction rose with the density.
Why: Coop J: 3 of 20 hens fell ill, 15 %.
Coop P: 24 of 60 fell ill, 40 %.
The disease harmed a larger fraction of the denser coop, so it is density-dependent.
Two beds of the same size hold young bean plants. A night of frost reaches both beds. The table below shows the plants in each bed and how many the frost killed.
Which kind of limiting factor is the frost?
- A. Density-dependentThe frost killed 20 % of the sparse bed and 20 % of the crowded bed.
The fraction did not change with the density. - B. ✓ Density-independent
Why: Bed J: 10 of 50 plants killed, 20 %.
Bed P: 50 of 250 killed, 20 %.
The frost harmed the same fraction at both densities, so it is density-independent.
Two paddocks of the same size grow the same grass. One holds 4 horses and the other holds 16. The keeper gives neither paddock any extra feed. In the crowded paddock each horse gets 25 % of the grass a horse in the sparse paddock gets, and several horses in the crowded paddock go hungry.
Which kind of limiting factor is the shortage of grass?
- A. ✓ Density-dependent
- B. Density-independentIn the crowded paddock each horse’s share is smaller, so more of its horses go hungry.
The fraction harmed rose with the density.
Why: The same grass is shared among 4 horses in one paddock and 16 in the other.
In the crowded paddock each share is smaller, and a larger fraction of the horses go hungry.
The fraction harmed rises with the density, so the shortage of grass is density-dependent.
Two hedges of the same length hold nesting dunnocks, small brown birds. Sparrowhawks hunt along both hedges. The table below shows the dunnocks in each hedge and how many the sparrowhawks took over the winter.
Which kind of limiting factor are the sparrowhawks?
- A. ✓ Density-dependent
- B. Density-independentThe sparrowhawks took 20 % of the sparse hedge’s dunnocks and 40 % of the crowded hedge’s.
The fraction rose with the density.
Why: Hedge J: 3 of 15 dunnocks taken, 20 %.
Hedge P: 18 of 45 taken, 40 %.
The sparrowhawks harmed a larger fraction of the denser hedge, so they are density-dependent.
Sand martins dig their nest tunnels into the earth banks of a river. One bank has 20 nests dug into it, and another bank of the same size has 80. Both banks hold nests at the same heights. The river rises to the same height on both banks and drowns every nest the water reaches.
Which kind of limiting factor is the flood?
- A. Density-dependentThe water reaches the same height on both banks.
A nest the water reaches drowns whether its bank has 20 nests or 80. - B. ✓ Density-independent
Why: The flood reaches the same height on both banks.
Every nest the water reaches drowns, on the sparse bank and on the crowded bank alike.
So the flood harms the same fraction at both densities: it is density-independent.
Two ponds of the same size hold pond skaters, insects that live on the water’s surface. One night in late autumn a sudden cold snap chills both ponds. The table below shows the pond skaters in each pond and how many the cold killed.
Which kind of limiting factor is the cold snap?
- A. Density-dependentThe cold killed 20 % of the sparse pond’s skaters and 20 % of the crowded pond’s.
The fraction did not change with the density. - B. ✓ Density-independent
Why: Pond J: 6 of 30 pond skaters died, 20 %.
Pond P: 24 of 120 died, 20 %.
The cold snap harmed the same fraction at both densities, so it is density-independent.
Two tanks of the same size hold goldfish: 6 in one tank and 36 in the other. Every goldfish releases waste into its tank’s water, and the keeper changes neither tank’s water. After a month the crowded tank’s water holds six times as much waste in every liter.
Which kind of limiting factor is the waste build-up?
- A. ✓ Density-dependent
- B. Density-independentMore goldfish in the same water put more waste into every liter.
So a larger fraction of the crowded tank’s goldfish are poisoned.
Why: Every goldfish adds waste to the same volume of water.
The crowded tank’s water holds six times as much waste in every liter, so a larger fraction of its goldfish are poisoned.
The fraction harmed rises with the density: the waste build-up is density-dependent.
Two patches of heath of the same size hold slow-worms, legless lizards. One patch holds 14 slow-worms and the other holds 56. A heath fire burns across half of each patch and kills every slow-worm in the ground it burns: 7 of the 14 and 28 of the 56.
Which kind of limiting factor is the fire?
- A. Density-dependentThe fire killed 7 of 14 and 28 of 56: 50 % of each patch.
The fraction did not change with the density. - B. ✓ Density-independent
Why: The fire burns half of each patch, whatever lives there.
7 of 14 is 50 %, and 28 of 56 is 50 %.
The fire harmed the same fraction at both densities, so it is density-independent.
Here is a table comparing the two kinds of limiting factor: what happens to the fraction harmed as the density rises, and four cases of each.
61Quick quiz: density-dependent factor, density-independent factor mixed practice
Rooks nest in colonies at the tops of trees. Two patches of forest of the same size each hold a rook colony. A gale blows through both patches on the same night. The table below shows the nests in each colony and how many the gale blew down.
(a) Determine which kind of limiting factor the gale is. (1 pt)
- Award 1 point for: density-independent.
(b) Justify your answer to part (a). (1 pt)
So the fraction harmed did not change with the density.
- Award 1 point for: the gale blew down the same fraction, 20 %, of both colonies’ nests (accept: the fraction did not rise with the density; 3 of 15 and 12 of 60 are the same fraction).
Ecologists sort limiting factors into two kinds.
Which of the following is a density-dependent factor?
- A. A limiting factor that harms a larger count of individuals when the population is denserA larger count alone does not decide: the frost killed 4 of 40 and 1 of 10, the same fraction.
- B. ✓ A limiting factor that harms a larger fraction of a population when the population is denser
- C. A limiting factor that harms a population only when the population is crowdedA density-dependent factor harms a sparse population too, only a smaller fraction of it.
Why: The sort compares the fraction harmed at two densities.
A density-dependent factor harms a larger fraction of a population when the population is denser.
Ecologists sort limiting factors into two kinds.
Which of the following is a density-independent factor?
- A. ✓ A limiting factor that harms the same fraction of a population at any density
- B. A limiting factor that harms a smaller fraction of a population when the population is denserA frost or a flood harms the same fraction at a high density as at a low one, not a smaller fraction.
- C. A limiting factor that harms only sparse populationsA frost or a flood harms crowded populations too, and by the same fraction.
Why: The sort compares the fraction harmed at two densities.
A density-independent factor harms the same fraction of a population at any density.
65Why the crowded pen lost more
Suppose a keeper puts the same weight of pellets into a hutch of 5 chinchillas every evening. Later the keeper moves 15 more chinchillas into the hutch, and still puts in the same weight of pellets.
Compared with before, how large is each chinchilla’s share of the pellets now?
- A. LargerThe same pellets are now shared among more chinchillas.
- B. ✓ Smaller
- C. The sameThe same pellets are now shared among 20 chinchillas instead of 5.
Why: The keeper puts in the same weight of pellets.
20 chinchillas now share them instead of 5.
So each chinchilla’s share is smaller.
Here are the two pens again, 10 rabbits in one and 40 in the other, each rabbit a dot. A line joins two rabbits close enough to touch noses.
Why did the disease kill 30 % of the crowded pen’s rabbits and only 10 % of the sparse pen’s?
Video: Watch: Why the crowded pen lost more
The two pens are on screen. A line is drawn between every two rabbits close enough to touch: many lines in the crowded pen, few in the sparse pen. The disease passes along the lines from one sick rabbit. Then a frost settles over both pens, and 1 rabbit in 10 is marked in each.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L34c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L34c.mp4
This disease passes from one rabbit to another when the two touch.
In the crowded pen each rabbit comes into contact with many others every day: the drawing shows many lines.
So the disease passes from the sick rabbit to many others, and from them to many more.
In the sparse pen each rabbit comes into contact with few others: the drawing shows few lines.
So most rabbits in the sparse pen never touch a sick rabbit, and the disease reaches few of them.
The denser the pen, the larger the fraction of its rabbits the disease reaches.
Now suppose the keeper gives both pens the same small amount of hay each day. Imagine the hay as one bar, shared equally among the rabbits.
In the sparse pen the bar is split 10 ways. Each rabbit’s share is wide, and every rabbit eats its fill.
In the crowded pen the same bar is split 40 ways. Each rabbit’s share is 25 % as wide, and the weakest rabbits go hungry.
The denser the pen, the smaller each rabbit’s share, and the larger the fraction that goes hungry.
Now consider a heath where stoats hunt young rabbits. A stoat finds its prey by searching.
Where the rabbits are dense, a stoat finds a rabbit after a short search, eats, and hunts there again. Other stoats gather where the finding is easy.
Where the rabbits are sparse, a stoat searches for a long time between rabbits, and it catches few.
The denser the rabbits, the larger the fraction of them the stoats catch.
Now consider the frost. The cold settles over both pens, and it reaches every rabbit in each pen.
A rabbit’s chance of dying in the cold is the same whether 9 other rabbits share its pen or 39.
So the frost kills the same fraction in both pens, 10 %. Crowding changes nothing for a frost, a flood or a fire.
A density-dependent factor harms a larger fraction of a denser population because each individual comes into contact with more others, gets a smaller share of the food, or is found more easily by a predator.
What you are expected to know Explain why a density-dependent factor harms a larger fraction of a denser population: more contacts, a smaller share, or an easier find for a predator.
Suppose two hutches of the same size hold guinea pigs: 20 in one hutch and 80 in the other. A disease that passes from one guinea pig to another when they touch reaches both hutches.
Predict which hutch loses the larger fraction of its guinea pigs to the disease.
- A. ✓ The crowded hutch
- B. The sparse hutchIn the sparse hutch each guinea pig touches few others, so the disease reaches few of them.
- C. The two hutches lose equal fractionsA disease that passes by touch reaches a larger fraction where each guinea pig touches more others.
Why: In the crowded hutch each guinea pig touches many others every day.
So the disease passes to many of them, and from them to more.
So the crowded hutch loses the larger fraction.
Suppose two hutches of the same size hold guinea pigs: 20 in one hutch and 80 in the other. One winter night the air in both hutches falls far below freezing, and the cold kills some of the guinea pigs in each hutch.
Predict which hutch loses the larger fraction of its guinea pigs to the cold.
- A. The sparse hutchThe same cold fills both hutches.
A guinea pig in the sparse hutch is no more likely to die of the cold than one in the crowded hutch. - B. The crowded hutchThe crowded hutch is no colder than the sparse hutch.
So its guinea pigs are no more likely to die of the cold than the sparse hutch’s. - C. ✓ The two hutches lose equal fractions
Why: The same cold fills both hutches.
A guinea pig’s chance of dying of the cold is the same whether 19 or 79 others share its hutch.
So the two hutches lose equal fractions.
Suppose a gardener grows cucumber plants in two beds of the same size: 12 plants in one bed and 60 in the other. A leaf mold that passes from plant to plant where their leaves touch gets into both beds. By August the mold has reached 2 of the 12 plants in the sparse bed and 36 of the 60 plants in the crowded bed.
(a) Explain why the mold reached a larger fraction of the plants in the crowded bed. (3 pt)
In the crowded bed each plant’s leaves touch the leaves of many neighbors.
So the mold passes from each infected plant to many others, and from them to more.
In the sparse bed most plants touch no infected neighbor, so the mold reaches few of them.
So the mold reached a larger fraction of the crowded bed’s plants.
- Award 1 point for: the mold passes between plants whose leaves touch (accept: spreads by contact).
- Award 1 point for: in the crowded bed each plant touches more neighbors, so the mold passes to more plants (accept: more contacts per plant).
- Award 1 point for: in the sparse bed most plants touch no infected plant, so the mold reaches a smaller fraction (accept: fewer contacts, so fewer plants are reached).
A student reads that the frost killed 4 of the crowded pen’s 40 rabbits and 1 of the sparse pen’s 10, and says: “The frost killed more rabbits where the rabbits were crowded, so the frost is density-dependent.”
Is the student correct?
- A. Yes: the frost killed four times as many rabbits in the crowded pen, so it is density-dependentThe sort compares fractions, not counts.
4 of 40 and 1 of 10 are both 10 %. - B. ✓ No: 4 of 40 and 1 of 10 are both 10 %, so the frost is density-independent
Why: The crowded pen holds four times as many rabbits, so the frost killed four times as many there.
But 4 of 40 is 10 %, and 1 of 10 is 10 %.
The frost harmed the same fraction at both densities, so it is density-independent.
Suppose two tanks of the same size hold platies, small aquarium fish: 6 in one tank and 48 in the other. A skin parasite that passes from fish to fish when they brush past each other gets into both tanks. Each week a larger fraction of the crowded tank’s platies catch the parasite than of the sparse tank’s. Now suppose the keeper moves 24 of the crowded tank’s platies to a third tank.
Predict what happens to the fraction of the crowded tank’s platies that catch the parasite each week.
- A. RisesFewer platies in the same tank brush past each other less often, so the parasite passes to fewer of them.
- B. ✓ Falls
- C. Stays the sameThe tank now holds half as many platies, so each fish brushes past fewer others.
Why: The crowded tank now holds 24 platies instead of 48, so its density has halved.
Each platy brushes past fewer others each week.
So the parasite passes to a smaller fraction of them: the fraction falls.
Here are the two pens again: 10 rabbits in one and 40 in the other, the same size, and the keeper put one sick rabbit into each.
The disease killed 1 of the 10 and 12 of the 40: 10 % against 30 %.
The disease is density-dependent, because each rabbit in the crowded pen came into contact with more others every day.
A frost would have killed 1 of the 10 and 4 of the 40: 10 % and 10 %.
A frost is density-independent, because the cold reaches every rabbit alike, however many share its pen.
99Mixed practice: limits that bite harder in a crowd mixed practice
Suppose swifts nest in the gaps under the roofs of a village. Flying insects to eat are plentiful all summer. Only 30 roofs in the village have a gap a swift can nest in, and every summer 30 pairs nest. A student says: “The swifts’ limiting factor is the flying insects, because every swift has to eat.”
Is the student correct?
- A. ✓ No: the insects are plentiful; only 30 roofs have a gap to nest in, so the gaps cap the count at 30 pairs
- B. Yes: every swift has to eat, so the flying insects are what hold the swifts’ count backThe insects are plentiful all summer, so food holds no swift back.
Only 30 roofs have a gap a swift can nest in.
Why: A swift pair nests only in a gap under a roof.
The village has 30 such gaps, so only 30 pairs can nest, however many insects there are.
So the gaps hold the population’s growth back: the gaps are the limiting factor.
Lapwings nest on the ground in open fields. Two fields of the same size hold lapwing chicks. A hailstorm crosses both fields. The table below shows the chicks in each field and how many the hail killed.
Which kind of limiting factor is the hail?
- A. Density-dependentThe hail killed 20 % of the sparse field’s chicks and 20 % of the crowded field’s.
The fraction did not change with the density. - B. ✓ Density-independent
Why: Field J: 4 of 20 chicks killed, 20 %.
Field P: 16 of 80 killed, 20 %.
The hail harmed the same fraction at both densities, so it is density-independent.
A rust fungus attacks wheat. Its spores blow from plant to plant, and they land on more leaves where the plants stand close together. A farmer sows one field thinly and another field of the same size four times as thickly.
Which kind of limiting factor is the rust fungus?
- A. ✓ Density-dependent
- B. Density-independentIn the thickly sown field each plant’s leaves are closer to more neighbors, so the rust reaches a larger fraction of the plants.
Why: The rust’s spores blow from plant to plant.
In the thickly sown field each plant stands closer to more neighbors, so the spores reach a larger fraction of the plants.
The fraction harmed rises with the density: the rust fungus is density-dependent.
Two tubs of the same size hold eels, a farmed fish: 5 in one tub and 30 in the other. Nobody changes the water. A student says: “Every eel adds waste to the same water, so the crowded tub’s water fouls faster, and a larger fraction of its eels are poisoned. The waste build-up is density-dependent.”
Is the student correct?
- A. No: the water fouls in any tub, crowded or sparse, so the waste build-up is density-independentThe water fouls in both tubs, but six times faster in the crowded tub.
So a larger fraction of the crowded tub’s eels are poisoned. - B. ✓ Yes: the crowded tub’s water fouls faster, so a larger fraction of its eels are poisoned: density-dependent
Why: Every eel adds waste to the same volume of water.
The crowded tub holds six times as many eels, so its water fouls six times faster.
So a larger fraction of its eels are poisoned: the waste build-up is density-dependent.
Two ponds of the same size hold koi, an ornamental fish: 8 in one pond and 40 in the other. A skin infection that passes from koi to koi when they touch reaches both ponds. By the end of the month the infection has reached a larger fraction of the crowded pond’s koi.
Which of the following explains why the infection reached a larger fraction of the crowded pond’s koi?
- A. The infection is stronger in crowded waterThe same infection passes the same way in both ponds; what differs is how often koi touch.
- B. The crowded pond holds more koi, so more koi are there to catch itMore koi catching the infection would not by itself make the fraction larger; the fraction is what rose.
- C. ✓ Each koi in the crowded pond touches more other koi every day
Why: The infection passes from koi to koi when they touch.
In the crowded pond each koi touches more other koi every day.
So the infection passes to a larger fraction of the crowded pond’s koi.
A hobby, a small falcon, hunts dragonflies over two pools of the same size. The table below shows the dragonflies over each pool and how many the hobby caught in a week.
Which kind of limiting factor is the hobby?
- A. ✓ Density-dependent
- B. Density-independentThe hobby caught 10 % of the sparse pool’s dragonflies and 25 % of the crowded pool’s.
The fraction rose with the density.
Why: Pool J: 2 of 20 dragonflies caught, 10 %.
Pool P: 20 of 80 caught, 25 %.
The hobby harmed a larger fraction of the denser pool, so it is density-dependent.
Suppose a keeper gives a shed of 15 turkeys the same weight of feed every day, and every turkey eats its fill. The keeper then moves 15 more turkeys into the shed and keeps the daily feed the same.
Predict what happens to the fraction of the shed’s turkeys that go hungry.
- A. ✓ Rises
- B. FallsThe same feed is now shared among 30 turkeys instead of 15, so each share is smaller.
- C. Stays the sameEach turkey’s share is now half as large, so the fraction that go hungry cannot stay as it was.
Why: The keeper gives the same feed every day.
30 turkeys now share it instead of 15, so each turkey’s share is half as large.
So a larger fraction of the turkeys go hungry: the fraction rises.
Curlews nest on the ground on open moors. Two moors of the same size each hold a curlew population. In a cold wet spring, rain floods nests on both moors. The table below shows the nests on each moor and how many the rain flooded. Later that year a gut parasite that passes between curlews where they feed close together reaches both moors.
(a) Determine which kind of limiting factor the flooding is. (1 pt)
- Award 1 point for: density-independent.
(b) Justify your answer to part (a). (1 pt)
So the fraction harmed did not change with the density.
- Award 1 point for: the flooding harmed the same fraction, 20 %, on both moors (accept: the fraction did not rise with the density; 5 of 25 and 15 of 75 are the same fraction).
(c) Predict which moor loses the larger fraction of its curlews to the gut parasite, and justify your prediction. (2 pt)
The parasite passes between curlews where they feed close together.
On the crowded moor each curlew feeds close to more others, so the parasite passes to a larger fraction of them.
- Award 1 point for: the crowded moor (accept: Moor P; the moor with 75 nests).
- Award 1 point for: each curlew on the crowded moor feeds close to more others, so the parasite reaches a larger fraction (accept: more contacts per bird where the birds are denser).
Glossary
- limiting factor
- Anything that holds a population’s growth below what it would otherwise be: a resource that falls short, such as food or nest sites, or a hazard that kills, such as a disease or a frost.
- density-dependent factor
- A limiting factor that harms a larger fraction of a population the denser the population is: a disease, competition for food, a predator, waste build-up.
- density-independent factor
- A limiting factor that harms the same fraction of a population at any density: a frost, a flood, a drought, a fire.
APBIO-U08-L35 The braking term
Now consider a deer herd on a fenced reserve whose grass feeds 1 000 deer at most: K is 1 000 deer.
When the herd is 100 deer, the reserve has room for 900 more. When the herd is 500 deer, it has room for 500 more. When the herd is 1 000 deer, it has no room at all. The exponential equation, dN/dt = rmax N, keeps a population growing for ever. How do you write the room left into that equation, so that growth stops at K?
Unit 8 · Ecology
1Room left, as a fraction
Suppose a lake’s weed and nest sites can support no more than 540 moorhens. This spring the lake holds 90 moorhens, and the count is climbing.
Which of the following is the lake’s carrying capacity, K, for moorhens?
- A. 90 moorhens90 moorhens is the count now, N, not the most the lake can support.
- B. 450 moorhens450 moorhens is how many more the lake can support, not the most it can support.
- C. ✓ 540 moorhens
Why: The carrying capacity, K, is the largest population a place’s resources can support.
The lake’s weed and nest sites support no more than 540 moorhens.
So K is 540 moorhens.
Suppose ecologists count the pine martens in a forest every year, and find that their dN/dt is −7 pine martens per year.
What is the pine marten population’s size doing?
- A. GrowingA growing population has a positive dN/dt; −7 per year is negative.
- B. ✓ Shrinking
- C. Staying steadyA steady population has a dN/dt of 0; −7 per year is not 0.
Why: dN/dt is the rate of change of the population’s size.
A negative dN/dt means more pine martens die than are born.
So the pine marten population is shrinking.
How does the growth equation know when to stop?
The room left on the reserve is K − N: the most the reserve can feed, minus the herd it feeds now.
Divide the room left by K, and you have the room left as a fraction of the whole reserve: .
This fraction is a brake on the herd’s growth: the smaller the fraction, the more it slows the growth.
With 100 deer the fraction is 0.9: almost no brake.
With 500 deer the fraction is 0.5: half a brake.
With 1 000 deer the fraction is 0, and growth stops.
Above K the fraction is negative, and the herd shrinks.
Multiply the exponential equation by this fraction, and you have the formula sheet’s logistic equation, the old equation with one new term.
Video: Watch: Room left, as a fraction
The herd’s growth curve is on screen, with K drawn as a dashed line at 1 000 deer. At three herd sizes a bar is drawn from the x-axis up to K: the filled part is the herd, and the unfilled part is the room left. The room left is divided by K, and the fraction is written above each bar: 0.9, 0.5, 0.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L35a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L35a.mp4
Here is the herd on the growth-graph frame: time in years on the x-axis, population size N in deer on the y-axis. The herd climbed and then leveled off at K, the dashed line at 1 000 deer.
The reserve’s grass feeds 1 000 deer at most. So K is 1 000 deer, whatever size the herd is.
Suppose the herd stands at 100 deer. The grass feeds 1 000, so the reserve has room for 900 more deer, as the line below shows.
with 100 deer on a reserve that feeds 1 000, the room left is 900 deer
The room left is K − N: the most the reserve can feed, minus the herd it feeds now.
Now suppose the herd stands at 500 deer. The room left is 500 deer: the reserve is half full.
And when the herd stands at 1 000 deer, the room left is 0 deer: the reserve is full.
Room for 900 deer is nearly the whole of a reserve that feeds 1 000. It is a small part of a reserve that feeds 10 000.
So a count of the deer still to fit does not say how full a reserve is.
Divide the room left by K, the whole, and you have the room left as a fraction of the whole reserve, as the line below shows.
the room left, K − N, divided by the whole reserve, K: the room left as a fraction of the whole
This fraction slows the herd’s growth, the way a brake slows a car: the smaller the fraction, the harder it slows the growth.
Because it slows the growth, this fraction is called the .
With 100 deer on the reserve, most of the reserve is still free. The worked example below calculates the braking term.
The reserve’s grass feeds 1 000 deer at most, and the herd stands at 100 deer. Calculate the braking term for the herd.
With 500 deer on the reserve, the braking term is 0.5, as the line below shows.
500 deer on the reserve: the braking term is 0.5
With 1 000 deer on the reserve, the braking term is 0, as the line below shows.
1 000 deer on the reserve: the braking term is 0
Here is the herd’s graph again, with a bar at three herd sizes. Each bar reaches from the x-axis up to K.
The filled part of a bar is the herd, N. The unfilled part is the room left, K − N.
The unfilled part, as a fraction of the whole bar, is the braking term: 0.9, then 0.5, then 0.
At 100 deer the braking term is 0.9: the brake is barely on, and the herd grows almost as fast as it can.
At 500 deer the braking term is 0.5: the brake is half on.
At 1 000 deer the braking term is 0: the brake is fully on, and growth stops.
Now suppose a ranger releases 200 more deer onto the full reserve in year 42. The herd stands at 1 200 deer: 200 more than the grass can feed.
The grass cannot feed every deer, so more deer die than are born. dN/dt is negative: the herd shrinks.
The braking term is negative too, as the line below shows: with 1 200 deer on the reserve it is −0.2.
1 200 deer on the reserve: the braking term is −0.2, negative
A negative braking term means the herd is above K, and the herd shrinks.
Here is a table comparing the herd at four sizes: the room left, and the braking term.
The room left with 100 deer is 900 deer: a count of deer, with a unit.
The braking term is that room left divided by K: 0.9, a fraction of the whole reserve, with no unit.
Below K the braking term lies between 1 and 0. Above K it is negative.
What you are expected to know Evaluate the braking term for a given K and N: close to 1 when the population is small, 0 when the population is at K, negative above K.
Suppose a stretch of river has food and hiding places for no more than 560 chub, a river fish: K is 560 chub. The stretch holds 140 chub.
Calculate the braking term for the chub.
Part 1. Calculate the room left in the stretch of river, K − N.
Answer: 420 chub (tolerance ±0)
Answer: 0.75 (tolerance ±0)
Suppose a marsh’s plants feed no more than 320 wigeon, a duck, through the winter: K is 320 wigeon. In November the marsh holds 64 wigeon.
Calculate the braking term for the wigeon.
Answer: 0.8 (tolerance ±0)
Suppose a mountain pasture feeds no more than 75 chamois, a goat-like animal of mountain pastures, through the summer: K is 75 chamois. After a mild spring the pasture holds 105 chamois.
Calculate the braking term for the chamois. Keep the sign.
Answer: -0.4 (tolerance ±0)
Suppose an island’s grass feeds no more than 230 mouflon, a wild sheep: K is 230 mouflon. After several mild winters the herd stands at 290 mouflon.
Predict what the herd’s size does over the next year.
- A. RisesThe grass feeds 230 mouflon and 290 are on the island, so the grass cannot feed every mouflon.
- B. ✓ Falls
- C. Stays the sameA herd’s size stays steady at K, 230 mouflon; this herd is above K.
Why: The grass feeds 230 mouflon, and 290 mouflon are on the island.
The grass cannot feed every mouflon, so more mouflon die than are born.
dN/dt is negative: the herd’s size falls.
Suppose a mountain’s grass and moss feed no more than 380 wild yaks through the summer: K is 380 yaks. After a mild spring the mountain holds 475 yaks. Over the coming year the yak population will shrink.
(a) Explain why the yak population will shrink. Use the braking term in your answer. (3 pt)
So the braking term, the room left divided by K, is negative.
The grass and moss cannot feed every yak, so more yaks die than are born.
So dN/dt is negative, and the yak population shrinks.
- Award 1 point for: N is larger than K, so the room left, K − N, is negative (accept: K − N is −95 yaks).
- Award 1 point for: so the braking term, K − N divided by K, is negative (accept: −0.25).
- Award 1 point for: the grass and moss cannot feed every yak, so more yaks die than are born and dN/dt is negative (accept: deaths outnumber births, so the population shrinks).
Suppose a lake’s weed feeds no more than 90 gadwalls, a duck that grazes water weed: K is 90 gadwalls. The lake holds 27 gadwalls. A student says: “The braking term for the gadwalls is 63, because 63 more gadwalls can fit on the lake.”
Is the student correct?
- A. ✓ No: 63 gadwalls is how many more the lake can feed; the braking term is 63 out of 90, which is 0.7
- B. Yes: the lake has room for 63 more gadwalls, and that room, 63, is the braking termThe room left, K − N, is 63 gadwalls, a count with a unit.
The braking term is a fraction of the whole lake, with no unit.
Why: The room left on the lake, K − N, is 63 gadwalls: a count of gadwalls.
The braking term is the room left divided by K, the whole.
So the braking term is a fraction of the whole lake, with no unit, not a count of gadwalls.
50The logistic equation
Suppose harvest mice breed in a barn full of spilled grain, with no predators and no disease. Their maximum per capita growth rate is rmax, and their population size is N.
Which of the following gives the harvest mice’s dN/dt?
- A. rmax + NAdding a rate per individual to a count of individuals mixes two different kinds of quantity.
Each mouse adds rmax, so N mice add N times that. - B. ✓ rmax × N
- C. N ÷ rmaxDividing N by rmax gives a smaller rate for a larger rmax.
More added per mouse means a larger gain, not a smaller one.
Why: Each harvest mouse adds rmax new mice per unit of time.
N mice add N times that.
So dN/dt is rmax multiplied by N, which the formula sheet writes as rmax N.
Here is the reserve again: the grass feeds 1 000 deer at most, so K is 1 000 deer. The braking term is 0.9 at 100 deer, 0.5 at 500 deer and 0 at 1 000 deer.
The exponential equation, dN/dt = rmax N, has no K in it, so the growth it describes never stops. Multiply it by the braking term, and the growth stops at K.
Video: Watch: The logistic equation
The exponential equation, dN/dt = rmax N, is typeset on screen. The braking term is written beside it, and the two are multiplied. The formula sheet’s logistic equation appears on one line, and each symbol is named beneath it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L35b.mp4
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The formula sheet prints the logistic equation on one line, exactly as below. Beneath it is what each symbol means, with its unit.
dN/dt: the rate of change of population size, in individuals per unit of time (for the herd, deer per year)
rmax: the maximum per capita growth rate, in new individuals per individual per unit of time (per deer per year)
N: the population size, in individuals (deer)
K: the carrying capacity, the largest population the place’s resources can support, in individuals (deer)
the bracket, K − N over K: the braking term, the room left as a fraction of the whole; a fraction, no unit
rmax N is the exponential equation’s right-hand side: the rate of change with nothing holding the herd back.
The bracket holds the braking term: the room left as a fraction of the whole.
So the logistic equation is the exponential equation multiplied by the braking term. The braking term is the only new part.
Here is a table comparing the two growth equations: the right-hand side of each, and when each applies.
rmax is in new deer per deer per year, N is a number of deer, and the braking term has no unit. So dN/dt comes out in deer per year, as the line below shows.
rmax is per deer per year, N is a number of deer, and the braking term has no unit, so dN/dt is in deer per year
When the herd stands at K, the braking term is 0. Any number multiplied by 0 is 0.
So dN/dt is 0 deer per year, as the line below shows, whatever rmax is: the herd neither grows nor shrinks.
the herd at K, 1 000 deer: the braking term is 0, so dN/dt is 0 deer per year, whatever rmax is
When the herd is small, the braking term is close to 1.
Multiplying by a number close to 1 changes the rate very little.
So far below K the logistic equation gives almost the same rate as the exponential equation. The S-shaped curve starts out like the J-shaped curve.
What you are expected to know Read the logistic equation in its formula-sheet form and state what each symbol means: the exponential equation multiplied by the braking term.
The logistic equation is written for the deer herd on its reserve.
Which symbol stands for the most deer the reserve’s grass can feed?
- A. rmaxrmax is the most each deer adds per year, a rate, not a number of deer.
- B. NN is the number of deer in the herd now, whatever the grass can feed.
- C. ✓ K
Why: The most deer the reserve’s grass can feed is the reserve’s carrying capacity.
The carrying capacity is written K.
The logistic equation is written for the deer herd on its reserve.
Which symbol stands for the most each deer adds to the herd per year, with nothing holding the herd back?
- A. ✓ rmax
- B. NN is the number of deer in the herd, not a rate per deer.
- C. KK is the most deer the grass can feed, not a rate per deer.
Why: The most each individual adds per unit of time, with nothing holding the population back, is the maximum per capita growth rate.
The maximum per capita growth rate is written rmax.
The logistic equation is written for the deer herd on its reserve.
Which symbol stands for the number of deer in the herd now?
- A. rmaxrmax is a rate per deer, not a number of deer.
- B. ✓ N
- C. KK is the most deer the grass can feed, not the number in the herd now.
Why: The number of individuals in a population now is its population size.
The population size is written N.
The exponential equation is dN/dt = rmax N.
Which of the following does the logistic equation add to it?
- A. NN is already in the exponential equation: rmax N.
- B. dN/dtdN/dt is the left-hand side of both equations.
- C. ✓
Why: The exponential equation’s right-hand side is rmax N.
The logistic equation multiplies rmax N by the braking term.
So the braking term is the one new part.
For the deer herd, rmax is in new deer per deer per year, N is a number of deer, and the fraction in the bracket has no unit.
In which unit does the logistic equation give dN/dt?
- A. deerA number of deer is a population size, N, not a rate of change.
- B. per yearPer year alone has no deer in it; dN/dt counts the deer the herd adds each year.
- C. ✓ deer per year
Why: Per deer per year, multiplied by deer, gives deer per year.
The braking term has no unit, so it changes no unit.
So dN/dt is in deer per year.
A student reads the logistic equation for the deer herd when the herd has reached K and says: “At K the braking term is 0, so the whole right-hand side is 0. dN/dt is 0, and the herd’s size stays steady.”
Is the student correct?
- A. No: at K the herd is at its largest, so rmax N is at its largest and the herd grows fastestrmax N is at its largest at K, but it is multiplied by a braking term of 0.
Any number multiplied by 0 is 0. - B. ✓ Yes: at K the braking term is 0, so dN/dt is 0 and the herd’s size stays steady
Why: At K the room left, K − N, is 0, so the braking term is 0.
rmax N multiplied by 0 is 0, whatever rmax N is.
So dN/dt is 0 deer per year: the herd neither grows nor shrinks.
Suppose a forest’s food and nest holes can support no more than 2 400 jackdaws: K is 2 400 jackdaws. This year 96 jackdaws live in the forest.
Compared with the rate of change the exponential equation gives for these 96 jackdaws, the rate the logistic equation gives is which of the following?
- A. ✓ Almost the same
- B. About half as largeThe braking term is about half only when the forest is half full; 96 jackdaws fill a small part of a forest that supports 2 400.
- C. ZeroThe braking term is 0 only when the population is at K; 96 jackdaws are far below 2 400.
Why: 96 jackdaws fill a small part of a forest that supports 2 400.
So the room left is nearly the whole, and the braking term is close to 1.
Multiplying rmax N by a number close to 1 changes it very little, so the two rates are almost the same.
Here is the reserve that feeds 1 000 deer at most again: K is 1 000 deer, and the herd’s curve levels off at K.
With 100 deer the braking term is 0.9. With 500 deer it is 0.5.
With 1 000 deer the braking term is 0, and growth stops.
The logistic equation is the exponential equation multiplied by that fraction, as the line below shows.
the logistic equation for the reserve: the exponential equation, rmax N, multiplied by the braking term in the bracket
78Quick quiz: braking term mixed practice
Which of the following is the braking term?
- A. The room left, K − N, as a count of individualsThe room left is a count with a unit; the braking term divides that count by K.
- B. The population size, N, divided by the carrying capacity, KN divided by K is the part of the place already filled, not the room left.
- C. ✓ The room left, K − N, divided by the carrying capacity, K
Why: The braking term is the room left, K − N, divided by K.
So it is the room left as a fraction of the whole, with no unit.
Glossary
- braking term
- The fraction (K − N)/K in the logistic equation: the room left, K − N, divided by the carrying capacity, K. It is the room left as a fraction of the whole, with no unit: close to 1 when the population is small, 0 when the population is at K, negative above K.
APBIO-U08-L35B Calculate the rate, brake on
Now go back to the deer reserve whose grass feeds 1 000 deer at most: K is 1 000 deer.
Suppose the herd stands at 300 deer. With nothing holding the herd back, each deer adds, on average, 0.2 new deer a year. So the herd’s maximum per capita growth rate, rmax, is 0.2 per year. How many deer does the herd add this year?
Unit 8 · Ecology
1Calculate the rate
Suppose a forest’s nest holes can support no more than 400 nuthatches, a small bird: K is 400 nuthatches. This year the forest holds 100 nuthatches.
Which of the following is the braking term for the nuthatches?
- A. 0.250.25 is N divided by K: the part of the forest already filled, not the room left as a fraction of the whole.
- B. ✓ 0.75
- C. 300300 nuthatches is the room left, K − N: a count of birds, not a fraction of the whole.
Why: The room left is K − N: 300 nuthatches.
The braking term is the room left divided by K, the whole.
300 nuthatches is 75 % of K, 400 nuthatches, so the braking term is 0.75.
Suppose a student is calculating dN/dt for a herd of saiga, an antelope, on a steppe whose grass sets a carrying capacity. The student has already worked out rmax N and the braking term.
To get dN/dt, what does the student do with rmax N and the braking term?
- A. Adds the braking term to rmax NA braking term of 0 would then leave rmax N unchanged, and growth would not stop at K.
- B. Divides rmax N by the braking termDividing by a braking term close to 0 would make the rate huge near K, not small.
- C. ✓ Multiplies rmax N by the braking term
Why: The logistic equation is the exponential equation, rmax N, multiplied by the braking term.
So the student multiplies rmax N by the braking term to get dN/dt.
How fast does a braked herd grow?
Work the braking term first: with 300 deer on a reserve that feeds 1 000, the braking term is 0.7.
Then multiply the exponential part, rmax N, by the braking term.
For 300 deer with an rmax of 0.2 per year, dN/dt is 42 deer per year.
Work the same herd at 100, 500 and 900 deer, and the rates are 18, 50 and 18 deer per year.
The herd grows fastest halfway to K, at 500 deer.
Just before K the braking term is close to 0. So the herd grows slowly there, not fastest.
Video: Watch: Calculate the rate
The herd of 300 deer is ringed on its growth curve. The braking term is worked first, on its own line: 0.7. Then rmax N and the braking term are multiplied, line by line, and the answer is 42 deer per year.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L35Ba.mp4
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Here is the herd on the growth-graph frame again: time in years on the x-axis, population size N in deer on the y-axis. The ring marks the herd at 300 deer.
The reserve’s grass feeds 1 000 deer at most. So K is 1 000 deer.
Suppose that, with nothing holding the herd back, 2 fawns join the herd each year for every 10 deer in it.
So each deer adds, on average, 0.2 new deer a year: the herd’s maximum per capita growth rate, rmax, is 0.2 per year.
The formula sheet prints the logistic equation on one line, below. Every calculation on this page starts from that line.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Work the braking term first, on its own line: the room left as a fraction of the whole.
Then multiply rmax N by the braking term. The worked example below calculates the herd’s dN/dt.
The reserve’s grass feeds 1 000 deer at most, the herd stands at 300 deer, and rmax is 0.2 per year. Calculate dN/dt for the herd.
dN/dt is 42 deer per year: this year the herd adds 42 deer.
The braking term is 0.7. So the herd adds 70 % of the deer that the exponential equation, with no brake, would give it.
To calculate dN/dt for a braked population, take three steps.
1 Write down rmax, N and K.
2 Work the braking term.
3 Multiply rmax N by the braking term.
What you are expected to know Calculate dN/dt from rmax, N and K, with its unit, working the braking term on its own line before it multiplies rmax N.
Suppose a pine forest’s cones feed no more than 640 chipmunks: K is 640 chipmunks. The forest holds 160 chipmunks, and their rmax is 0.6 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate dN/dt for the chipmunks.
Part 1. Calculate the braking term for the chipmunks.
Answer: 0.75 (tolerance ±0)
Part 2. Calculate rmax N for the chipmunks.
Answer: 96 chipmunks per year (tolerance ±0)
Answer: 72 chipmunks per year (tolerance ±0)
Suppose a lake’s food and weed can support no more than 5 000 rudd, a fish: K is 5 000 rudd. The lake holds 2 000 rudd, and their rmax is 0.3 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate dN/dt for the rudd.
Answer: 360 rudd per year (tolerance ±0)
Suppose a greenhouse’s pepper plants can feed no more than 6 000 aphids: K is 6 000 aphids. The plants hold 2 400 aphids, and the aphids’ rmax is 0.4 per week.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate dN/dt for the aphids, in aphids per week.
Answer: 576 aphids per week (tolerance ±0)
27Quick quiz: the braking term, and rmax N mixed practice
Suppose a steppe’s grasses feed no more than 400 hamsters: K is 400 hamsters. The steppe holds 60 hamsters.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate the braking term for the hamsters.
Answer: 0.85 (tolerance ±0)
Suppose 700 skuas, large seabirds, nest on an island, and their rmax is 0.3 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate rmax N for the skuas.
Answer: 210 skuas per year (tolerance ±0)
Suppose a stream can support no more than 250 crayfish: K is 250 crayfish. The stream holds 50 crayfish.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate the braking term for the crayfish.
Answer: 0.8 (tolerance ±0)
Suppose 350 linnets nest on a gorse heath, and their rmax is 0.4 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate rmax N for the linnets.
Answer: 140 linnets per year (tolerance ±0)
Suppose an island’s plants feed no more than 300 iguanas: K is 300 iguanas. The island holds 285 iguanas.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate the braking term for the iguanas.
Answer: 0.05 (tolerance ±0)
33Where the herd grows fastest
Suppose a forest’s nuts and roots feed no more than 240 raccoons: K is 240 raccoons. This autumn the forest holds 228 raccoons.
Which of the following best describes the braking term for the raccoons?
- A. ✓ Close to 0
- B. Close to 0.5The braking term is 0.5 for a half-full forest; 228 of 240 raccoons is a nearly full forest.
- C. Close to 1The braking term is close to 1 for a nearly empty forest; 228 of 240 raccoons is a nearly full forest.
Why: The room left is 12 raccoons out of 240: a small part of the whole.
The braking term is the room left divided by K.
So the braking term is close to 0: the brake is nearly full on.
Here is the reserve again. Its grass feeds 1 000 deer at most: K is 1 000 deer, and rmax is 0.2 per year.
Now consider the herd at three sizes: 100 deer, 500 deer and 900 deer. At which size does the herd add the most deer in a year?
Video: Watch: Where the herd grows fastest
The herd’s dN/dt is worked at 100, 500 and 900 deer, one line each. The three rates are drawn as bars beneath the growth curve, on the same time axis. The tallest bar stands at 500 deer, halfway to K.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L35Bb.mp4
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At 100 deer the braking term is 0.9. The line below works the rate: 18 deer per year.
100 deer: the braking term is 0.9, and the herd’s dN/dt is 18 deer per year
At 500 deer the braking term is 0.5. The line below works the rate: 50 deer per year.
500 deer: the braking term is 0.5, and the herd’s dN/dt is 50 deer per year
At 900 deer the braking term is 0.1. The line below works the rate: 18 deer per year.
900 deer: the braking term is 0.1, and the herd’s dN/dt is 18 deer per year
Here is a table comparing the herd at the three sizes: the braking term, rmax N, and dN/dt.
rmax N is largest at 900 deer, because the most deer are there to breed.
But at 900 deer the braking term is 0.1: the brake is nearly full on.
So the herd at 900 deer adds as few deer a year as the herd at 100 deer.
Here is the herd’s curve, with the three rates as bars beneath it on the same time axis.
The curve is steepest at 500 deer, in year 11. That year the herd adds 50 deer: more than at any other size.
500 deer is half of K. So the herd grows fastest halfway to K.
Halfway to K the herd is already large, and many fawns are born each year.
And halfway to K the braking term is still 0.5: the brake is only half on.
Just before K the herd is at its largest.
But just before K the braking term is close to 0: the brake is nearly full on.
So just before K the herd adds few deer a year. The herd grows slowly there, not fastest.
What you are expected to know Determine, from dN/dt worked at several sizes, the size at which a logistic population grows fastest: halfway to K.
Suppose a forest reserve’s willow and other browse feed no more than 800 moose: K is 800 moose, and the moose’s rmax is 0.4 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate dN/dt for a herd of 600 moose.
Part 1. Calculate dN/dt for a herd of 200 moose.
Answer: 60 moose per year (tolerance ±0)
Part 2. Calculate dN/dt for a herd of 400 moose.
Answer: 80 moose per year (tolerance ±0)
Answer: 60 moose per year (tolerance ±0)
Suppose a forest’s acorns and beech nuts feed no more than 1 640 jays: K is 1 640 jays, and the jays’ rmax is 0.4 per year.
At which of the following flock sizes does the flock add the most jays in a year?
- A. 410 jaysAt 410 jays the brake is barely on, but few jays are there to breed.
- B. ✓ 820 jays
- C. 1 230 jaysAt 1 230 jays many jays are there to breed, but the braking term is 0.25.
Why: 820 jays is half of K, 1 640 jays.
A logistic population grows fastest halfway to K.
So the flock adds the most jays in a year at 820 jays.
Suppose a plain’s grass feeds no more than 1 800 kangaroos: K is 1 800 kangaroos. A student says: “The mob grows fastest just before it reaches K, because that is when there are the most kangaroos to breed.”
Is the student correct?
- A. ✓ No: just before K the plain has almost no room left for more kangaroos, so the mob adds few a year
- B. Yes: just before K the mob is at its largest, so it adds the most kangaroos a yearThe mob is at its largest just before K, but its rmax N is multiplied by a braking term close to 0.
Why: Just before K the room left is a small part of the whole.
So the braking term is close to 0.
dN/dt is rmax N multiplied by that braking term.
So the mob adds few kangaroos a year: it grows fastest halfway to K.
Suppose a reedbed’s reeds can support no more than 300 reed warblers: K is 300 warblers. This year 150 warblers nest in it. A student says: “At 150 warblers the braking term is 0.5, so the population grows at half the rate the exponential equation would give it.”
Is the student correct?
- A. ✓ Yes: at 150 warblers the braking term is 0.5, so the population grows at half the exponential equation’s rate
- B. No: below K the braking term changes nothing, so the population grows at the exponential equation’s full rateThe braking term is 1 for an empty reedbed; at 150 warblers of 300 it is 0.5, and it multiplies the rate.
Why: 150 warblers is half of K, 300 warblers.
So the room left is half of the whole, and the braking term is 0.5.
dN/dt is rmax N, the exponential equation’s rate, multiplied by 0.5.
So the population grows at half the exponential equation’s rate.
Here is the reserve that feeds 1 000 deer at most again, with the herd of 300 deer ringed on its curve.
The braking term for 300 deer is 0.7. So this year the herd adds 42 deer, as the line below shows.
the herd of 300 deer, braking term 0.7: this year the herd adds 42 deer
Here are the three rates again, as bars beneath the curve.
The herd grows fastest at 500 deer, halfway to K: 50 deer a year, the most of any size.
62Mixed practice: calculate the rate, brake on mixed practice
Suppose a plain’s grass feeds no more than 1 400 pronghorn, an antelope: K is 1 400 pronghorn. A student says: “Keep the herd small, because a small herd has a braking term close to 1 and so grows fastest.”
Is the student correct?
- A. Yes: a small herd’s braking term is close to 1, so the herd grows fastest while it is smallA braking term close to 1 lets the herd grow almost as fast as it can, but rmax N is small when the herd is small.
- B. ✓ No: a small herd has few pronghorn to breed, so its rmax N is small and it adds few pronghorn a year
Why: A small herd’s braking term is close to 1, but few pronghorn are there to breed, so rmax N is small.
At 700 pronghorn, half of K, many breed and the braking term is still 0.5.
So the herd adds most at 700 pronghorn, not while it is small.
Suppose an island’s burrows and fish can support no more than 1 500 shearwaters: K is 1 500 shearwaters. One colony on such an island holds 150 shearwaters; another colony, on an identical island, holds 600 shearwaters.
Which colony adds more shearwaters this year?
- A. The colony of 150 shearwatersThe braking term is larger for 150 shearwaters, but only 150 shearwaters are there to breed.
- B. ✓ The colony of 600 shearwaters
- C. Both colonies add the same numberThe two colonies differ in size and in braking term, so their rates differ.
Why: For 150 shearwaters the braking term is 0.9, but few shearwaters breed.
For 600 shearwaters the braking term is 0.6, and far more shearwaters breed.
600 is close to half of K, 750, where a colony grows fastest.
So the colony of 600 adds more shearwaters.
Suppose the logistic equation gives dN/dt for a flock of ibises on a marsh as 45 ibises per year.
Which of the following does this value mean?
- A. Each ibis adds 45 ibises a yearThe rate per ibis is rmax, a small fraction; 45 ibises per year is the whole flock’s gain.
- B. The flock holds 45 ibisesA count of ibises is N, the population size; dN/dt is a rate, in ibises per year.
- C. ✓ The flock gains 45 ibises each year
Why: dN/dt is the rate of change of population size.
Its unit is ibises per year: the ibises added to the flock in a year.
So the flock gains 45 ibises each year.
Suppose a student knows that a herd of springbok, an antelope, stands at 360 springbok and that its rmax is 0.3 per year.
Which of the following does the student still need, to calculate the herd’s dN/dt from the logistic equation?
- A. The number of springbok born last yearThe equation counts births through rmax N; last year’s births are not in it.
- B. The herd’s size a year agoThe equation uses the herd’s size now, N; last year’s size is not in it.
- C. ✓ The carrying capacity, K
Why: The logistic equation multiplies rmax N by the braking term.
The braking term is , and needs K.
So the student still needs the carrying capacity, K.
Suppose a plain’s grass and roots feed no more than 600 warthogs: K is 600 warthogs. A student compares a herd of 100 warthogs with a herd of 200 warthogs on that plain and says: “The herd of 200 warthogs adds less than twice as many warthogs a year as the herd of 100, because its braking term is smaller.”
Is the student correct?
- A. No: the herd of 200 warthogs is twice as large, so it adds twice as many warthogs a yearrmax N doubles from 100 to 200 warthogs, but the braking term falls, so dN/dt less than doubles.
- B. ✓ Yes: the herd of 200 warthogs has a smaller braking term, so it adds less than twice as many
Why: From 100 to 200 warthogs, rmax N doubles.
But the room left shrinks, so the braking term falls.
dN/dt is rmax N multiplied by the braking term.
So dN/dt less than doubles: the student is correct.
Suppose the tundra of a valley feeds no more than 960 musk oxen: K is 960 musk oxen.
At which of the following herd sizes is the braking term closest to 1?
- A. ✓ 120 musk oxen
- B. 480 musk oxenAt 480 musk oxen, half of K, the braking term is 0.5.
- C. 960 musk oxenAt 960 musk oxen the herd is at K, and the braking term is 0.
Why: The braking term is the room left as a fraction of the whole.
The smaller the herd, the more room is left.
So the braking term is closest to 1 at 120 musk oxen.
Suppose a stretch of desert scrub feeds no more than 600 meerkats: K is 600 meerkats. The scrub holds 240 meerkats, and their rmax is 0.25 per year.
the formula sheet’s logistic equation: rmax is the maximum per capita growth rate, N the population size, K the carrying capacity; the bracket is the braking term
Calculate dN/dt for the meerkats.
Answer: 36 meerkats per year (tolerance ±0)
Suppose a plain’s grass feeds no more than 1 200 wildebeest: that is the herd’s K. The herd’s rmax is 0.3 per year. Ten years ago the herd stood at 600 wildebeest; this year it stands at 1 080. The reserve’s manager expects the herd to add more wildebeest this year than it did ten years ago, because the herd is larger now. In fact the herd will add fewer.
(a) Explain why the herd will add fewer wildebeest this year than it did ten years ago. Use the braking term in your answer. (3 pt)
So the braking term then was 0.5.
This year the herd stands at 1 080 wildebeest, so the room left is 120 wildebeest.
So the braking term now is 0.1.
dN/dt is rmax N multiplied by the braking term.
rmax N is larger this year, but it is multiplied by 0.1 instead of 0.5.
So dN/dt is smaller, and the herd adds fewer wildebeest.
- Award 1 point for: ten years ago, at 600 wildebeest (half of K), the braking term was 0.5.
- Award 1 point for: this year, at 1 080 wildebeest, the braking term is 0.1 (accept: close to 0; the brake is nearly full on).
- Award 1 point for: dN/dt is rmax N multiplied by the braking term, so the braking term near 0 makes dN/dt smaller this year even though rmax N is larger (accept: the herd grows fastest halfway to K, not near K).
APBIO-U08-L36 Past K, and a moving K
Photo: National Park Service, Wikimedia Commons, public domain (resized).
Suppose harbor seals are counted on one stretch of coast at the end of every summer, for twelve years. In the first years the count climbs. Then the count climbs past the level the coast’s food can carry, and falls back below it. Then it climbs again.
Now consider a deer reserve whose grass feeds no more than 1 000 deer: K is 1 000. A drought comes, and the grass left feeds no more than 600 deer. What happens to a herd of 900 deer?
Unit 8 · Ecology
1Past K, and a K that moves
Suppose a reservoir’s food feeds no more than 1 150 ruffe, a small fish: K is 1 150. After a good breeding year the reservoir holds 1 250 ruffe.
Which sign does the braking term, the room left divided by K, have for this population?
- A. PositiveThe braking term is positive while N is below K.
Here N, 1 250, is above K, 1 150. - B. ZeroThe braking term is zero when N equals K.
Here N, 1 250, is above K, 1 150. - C. ✓ Negative
Why: N is 1 250 ruffe and K is 1 150 ruffe, so N is above K.
The room left, K − N, is negative.
So the braking term, the room left divided by K, is negative.
Suppose a hedge’s berries feed a flock of fieldfares, a thrush, through the winter. Each winter the flock’s count climbs and then levels off at the same count.
What sets the count the flock levels off at?
- A. The number of fieldfares that arrive in autumnThe berries feed only so many fieldfares.
The number that arrive does not change how many the hedge can feed. - B. The fieldfares themselves, the same in any hedgeA hedge with more berries feeds more fieldfares.
So the level differs from hedge to hedge. - C. ✓ The hedge’s supply of berries
Why: The hedge’s berries feed only so many fieldfares.
The count levels off at that number.
So the hedge’s supply of berries sets K for the fieldfares.
Suppose 53 shelduck ducklings hatch on an estuary in one season, and 91 shelducks die on that estuary in the same season.
What does the size of the shelduck population do over that season?
- A. ✓ Shrinks
- B. GrowsThe population grows when more are born than die.
Here 91 shelducks die and 53 are born. - C. Stays the sameThe size stays the same when births equal deaths.
Here 91 shelducks die and 53 are born.
Why: 53 ducklings join the population and 91 shelducks leave it.
Deaths outnumber births.
So the population shrinks.
What happens when a population climbs past K, or when K itself moves?
Above K, the room left, K − N, is negative, so the braking term is negative.
Deaths outnumber births, and the count falls back toward K.
The count can fall below K, and then climb again.
So a real population wanders around K rather than sitting on it.
K is set by the place’s resources, so when a drought cuts the grass, K falls.
A herd above the new K falls toward it.
When more food arrives, K rises, and the population climbs toward the new K.
The logistic model draws K as one fixed line, though in the field K moves with the seasons.
Video: Watch: Past K, and a moving K
The deer’s count is drawn climbing toward a dashed K at 1 000. At year 22 the dashed line drops to 600. The count falls until it reaches the new K. Then the seals’ count is drawn: it climbs past its K, falls below, and climbs again.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L36a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L36a.mp4
Here are the seal counts as a table: the year, and the number of seals counted on the coast at the end of that summer.
Here are the same counts plotted: time in years on the x-axis, population size N in seals on the y-axis, a gridline every 400 seals.
The count climbs for four years. Then it wanders: down below 3 200 seals, up past 3 200, down again.
The count keeps returning to the 3 200 gridline. So 3 200 seals is the coast’s K, and K is drawn as the dashed line at 3 200.
Between year 3 and year 4 the count climbs past K: it overshoots. In year 4 it stands at 3 500 seals, above K.
Why does the count fall back?
Above K, N is larger than K, so the room left, K − N, is negative.
So the braking term is negative.
In the seals’ own terms: the coast’s fish feed no more than 3 200 seals, and 3 500 seals share those fish.
So each seal’s share of fish is too small.
Seals short of food raise fewer pups, and more seals die.
So deaths outnumber births, and the count falls.
The count does not stop at K. In year 7 it stands at 2 950 seals, below K.
Below K each seal’s share of fish is large enough again.
So births outnumber deaths, and the count climbs.
A real population wanders around K rather than sitting on it. Each swing carries the count past K, and the braking term brings it back.
Above K each individual’s share of resources is too small, so deaths outnumber births and the count falls back toward K.
Now consider the deer reserve: its grass feeds no more than 1 000 deer, so K is 1 000. In year 22 the herd stands at 900 deer, still climbing toward K.
A drought comes. The grass left feeds no more than 600 deer.
K is set by the reserve’s grass. So K falls from 1 000 to 600.
The herd, 900 deer, is now above the new K, 600.
Here is the braking term after the drought, on its own line: K is 600 and N is 900.
the braking term after the drought: K is 600 deer and N is 900 deer, so the term is negative
The braking term is negative, so deaths outnumber births. The herd falls toward 600.
Here is the herd’s growth graph, with K drawn as a dashed line.
K steps down from 1 000 to 600 at year 22. The count falls, steeply at first and then more slowly, until it runs along the 600 gridline.
Now suppose instead that in year 22 a neighboring valley is fenced into the reserve. The grass of the reserve and the new valley together feeds no more than 1 600 deer.
K rises from 1 000 to 1 600.
A herd of 900 deer is now below K. So the room left is positive, and births outnumber deaths.
The herd climbs until it runs along the 1 600 gridline: the new K.
We made one simplification here: the logistic model draws K as one fixed line. In the field K moves with the seasons, because more grass grows in summer than in winter.
What you are expected to know Predict which way a population’s count moves when the count is above K or below K, whether or not K has just moved.
Suppose a plain’s grass feeds no more than 370 hartebeest, an antelope: K is 370. After a mild winter the herd stands at 420 hartebeest.
Predict what the herd’s count does over the following years.
- A. RisesA count rises while it is below K.
420 hartebeest is above K, 370. - B. ✓ Falls
- C. Stays the sameA count stays the same when it equals K.
420 hartebeest is above K, 370.
Why: The herd, 420, is above K, 370.
So the room left, K − N, is negative, and the braking term is negative.
Deaths outnumber births, and the count falls.
Suppose a bay’s mussels feed no more than 660 eider ducks: K is 660. The flock stands at 650 eiders. A storm buries half the mussel beds in sand, and the mussels left feed no more than 310 eiders.
Predict what the eiders’ count does over the following years.
- A. RisesA count rises while it is below K.
The new K is 310, and 650 eiders is above it. - B. ✓ Falls
- C. Stays the sameA count stays the same when it equals K.
The new K is 310, and 650 eiders is above it.
Why: The mussels set K, so K falls from 660 to 310 eiders.
The flock, 650, is above the new K.
So the braking term is negative.
Deaths outnumber births, and the count falls.
Suppose a river pool’s food feeds no more than 470 barbel, a river fish: K is 470. The pool holds 470 barbel. From this year the angling club adds fish food to the pool every week, and the food now feeds no more than 940 barbel.
Predict what the barbel’s count does over the following years.
- A. ✓ Rises
- B. FallsA count falls while it is above K.
The new K is 940, and 470 barbel is below it. - C. Stays the sameA count stays the same when it equals K.
The new K is 940, and 470 barbel is below it.
Why: The added food raises K from 470 to 940 barbel.
The count, 470, is now below K.
So the room left is positive.
Births outnumber deaths, and the count rises.
Suppose kittiwakes, a cliff-nesting gull, feed on the sand eels in the bay at the foot of their cliff. The bay’s sand eels feed no more than 5 200 kittiwakes: K is 5 200. After two good breeding seasons the count reaches 5 900 kittiwakes. Over the next two years the count falls back to about 5 200.
(a) Explain why the count fell back after it climbed past K. (3 pt)
5 900 kittiwakes shared those sand eels, so each kittiwake’s share was too small.
Kittiwakes short of food raised fewer chicks, and more kittiwakes died.
So deaths outnumbered births, and the count fell.
The count fell until it was back near 5 200, where births equal deaths.
- Award 1 point for: above K each kittiwake’s share of food was too small (accept: the sand eels could not feed 5 900 kittiwakes).
- Award 1 point for: deaths outnumbered births (accept: fewer chicks were raised and more kittiwakes died, so the count fell).
- Award 1 point for: the count fell until it was back near K, where births equal deaths (accept: the braking term is negative above K and brings the count back toward K).
A student reads that a drought on a deer reserve has left grass that feeds fewer deer than before, and says: “K belongs to the deer, so a drought leaves K where it was.”
Is the student correct?
- A. Yes: K belongs to the deer themselves, so the drought leaves K where it was whatever the grass doesK is the largest herd the reserve’s resources can support.
The drought cut those resources. - B. ✓ No: the reserve’s grass sets K for the deer, so a drought that cuts the grass lowers K
Why: K is the largest herd the reserve’s grass can feed.
The drought cut the grass, so the grass feeds fewer deer.
So K fell: K is set by the reserve’s resources, not by the deer alone.
A student reads that in one year the seals counted on a stretch of coast stood above the coast’s K, and says: “The count is above K, so over the next year more seals will die than pups are born, and the count will fall.”
Is the student correct?
- A. No: above K the count keeps climbing, because more seals on the coast means more pups are bornAbove K each seal’s share of fish is too small.
So fewer pups are raised and more seals die, and the count falls. - B. ✓ Yes: more seals share the coast’s fish than the fish can feed, so each seal’s share is too small
Why: The count is above K, so more seals share the coast’s fish than the fish can feed.
So each seal’s share of fish is too small.
Fewer pups are raised and more seals die: deaths outnumber births, and the count falls.
Suppose a moor’s insects feed no more than 830 meadow pipits, a small ground-nesting bird: K is 830. A wet spring drowns many of the insects, and the insects left feed no more than 550 pipits. The pipits stand at 375.
Predict what the pipits’ count does over the following years.
- A. ✓ Rises
- B. FallsA count falls while it is above K.
375 pipits is below the new K, 550. - C. Stays the sameA count stays the same when it equals K.
375 pipits is below the new K, 550.
Why: K fell from 830 to 550 pipits.
But the count, 375, is still below the new K.
So the room left is positive.
Births outnumber deaths, and the count rises.
Suppose a reedbed’s seeds and insects feed no more than 650 reed buntings, a small bird: K is 650. The flock has stood at 650 buntings for years.
In year 8 a fire burns much of the reedbed. The reeds left feed no more than 390 buntings.
Get graph paper. Mark the two axes: time in years from 0 to 16 years on the x-axis, and population size N in reed buntings on the y-axis.
Choose the y-axis scale so that 650 fits. Draw the dashed K line and the flock’s count from year 0 to year 16.
Then check your graph against the one below.
Here is the finished graph: K stepping down from 650 to 390 at year 8, and the count falling until it runs along the 390 gridline.
Check four features on your graph:
1 the count runs along the 650 gridline before year 8;
2 the dashed K line steps down from 650 to 390 at year 8;
3 the count falls after year 8, steeply at first and then more slowly;
4 the count runs along the 390 gridline once it has fallen.
Suppose a lake’s plants feed no more than 200 pochard, a diving duck: K is 200. The flock has stood at 200 for years. From year 6 the lake’s owner adds grain every day, and the food now feeds no more than 400 pochard. The three graphs below, labeled J, M and R, show three possible counts for the pochard from year 0 to year 12.
Which graph shows the pochard’s count?
- A. Graph JThe added grain raised K to 400, so the count does not stay at 200.
- B. Graph MA flock grows by births over years, so the count does not jump to 400 in one year.
- C. ✓ Graph R
Why: From year 6 the food feeds 400 pochard, so K rises to 400.
The flock, 200, is below the new K, so births outnumber deaths and the count climbs.
The climb takes years, and it slows as the count nears 400.
Graph R shows that climb.
Here are the harbor seals again: counted on one stretch of coast for twelve years.
Their count climbed past 3 200, fell back below it and climbed again, wandering around the coast’s K.
Here is the deer reserve again, after the drought: the grass that fed 1 000 deer feeds 600, so K has fallen to 600.
A herd of 900 deer is above the new K.
So deaths outnumber births, and the herd falls toward 600.
APBIO-U08-L36B Percent change
Suppose raptors, birds of prey, are counted on a reserve one year and again the next year: 240 raptors in the first year, and 180 in the second. The reserve lost 60 birds.
Is that a big loss or a small one? It depends what you compare the 60 birds with. How do you state a change so that reserves of any size can be compared?
Unit 8 · Ecology
1A change as a percent of the starting count
Suppose 8 % of the cells in a skin sample are in mitosis. A day later, 10 % of the cells in a second sample from the same skin are in mitosis.
Which of the following calculates the percent change from the first sample to the second?
- A. The change, divided by the second sample’s 10 %, times 100The change is measured against where the percent started: the first sample’s 8 %.
- B. The second sample’s 10 %, divided by the first sample’s 8 %, times 100The second percent divided by the first is the final value as a share of the start, not the change.
- C. ✓ The change, divided by the first sample’s 8 %, times 100
Why: The percent change measures the change against the starting value.
The starting value is the first sample’s 8 %.
So the change is divided by the first sample’s value, not the second’s.
Suppose a lake’s count of grebes is lower this year than it was last year.
Which sign does the percent change from last year to this year carry?
- A. ✓ Negative
- B. PositiveA positive percent change means the count rose.
This count fell. - C. No signA percent change carries a sign, and the sign says which way the count went.
Why: The count fell from last year to this year.
A fall is a negative change.
So the percent change is negative.
How do you state a change so that populations of different sizes can be compared?
Divide the change by the starting count, then multiply by 100: that is the percent change.
The sign says which way the count went: a minus sign for a fall, a plus sign for a rise.
The starting count, never the final one, is what you divide by.
Unit 4 used the same routine on two percents; here the two values are counts of animals.
Suppose raptors, birds of prey, are counted on a reserve one year and again the next year: 240 raptors in the first year, and 180 in the second.
The reserve lost 60 birds.
Is that a big loss or a small one?
Now consider a second reserve, where the count fell from 400 raptors to 340.
That reserve lost 60 birds too.
Here is a table comparing the two reserves: the starting count, the final count and the change.
The two losses are the same size: 60 birds each.
But 60 birds is a larger share of 240 birds than of 400 birds.
So the first reserve lost the larger share of its raptors.
To compare the two losses in one number each, state each loss as a percent of the reserve’s starting count.
Here is the equation, the one Unit 4 used for two percents, with what each word means beneath.
initial: the starting count, the count in the first year
final: the count in the later year
final − initial: the change in the count, negative for a fall and positive for a rise
percent change: the change as a percent of the starting count, with its sign
The change is the final count minus the starting count.
Dividing the change by the starting count states the change as a share of the start.
Multiplying by 100 turns that share into a percent.
To find the first reserve’s percent change, divide its change by its starting count, then multiply by 100.
The first reserve’s count fell from 240 raptors to 180 raptors. Calculate the percent change, with its sign.
So the first reserve’s percent change is −25 %.
The minus sign says the count went down. Keep the sign in front of the answer.
Now work the second reserve the same way: 400 raptors falling to 340.
The second reserve’s count fell from 400 raptors to 340 raptors. Calculate the percent change, with its sign.
So the second reserve’s percent change is −15 %.
The same loss of 60 birds is −25 % on the first reserve and −15 % on the second.
The percent change states in one number which loss was the larger share: the first reserve’s.
Now consider a lagoon where avocets, wading birds, are counted one year and again the next year: 250 avocets in the first year, and 320 in the second.
The count rose, so the change is positive.
The lagoon’s count rose from 250 avocets to 320 avocets. Calculate the percent change, with its sign.
So the avocets’ percent change is +28 %.
Here is a table of the three cases: the starting count, the final count, the change and the percent change.
The change is measured against where the count started.
So the starting count, never the final one, is what you divide by.
Suppose you divided the first reserve’s loss by its final count, 180, instead of its starting count, 240.
60 birds is a larger share of 180 birds than of 240 birds.
So the answer would come out larger than 25 %: a different number, and the wrong one.
What you are expected to know Calculate the percent change between two counts of a population, with its sign.
Suppose adders, snakes of open heathland, live on a heath. One spring 750 adders are counted on the heath, and the next spring 660 adders are counted.
Calculate the percent change in the adders’ count, with its sign.
Part 1. State the starting count of adders, the initial value.
Answer: 750 adders (tolerance ±0)
Part 2. State the change in the count, the final value minus the initial value, with its sign.
Answer: -90 adders (tolerance ±0)
Answer: -12 % (tolerance ±0)
Suppose zander, a freshwater fish, live in a reservoir. A survey counts 1 400 zander one summer and 1 876 zander the next summer.
Calculate the percent change in the zander’s count, with its sign.
Answer: 34 % (tolerance ±0)
Suppose corncrakes, ground-nesting birds, breed in a hay meadow. In one year 125 corncrakes are counted there, and in the next year 95 corncrakes are counted.
Calculate the percent change in the corncrakes’ count, with its sign.
Answer: -24 % (tolerance ±0)
Suppose a marsh’s count of snipe fell from 500 snipe to 300 snipe. A student says: “The change, 200 birds, is about 67 % of the final count, 300, so the percent change is about −67 %.”
Is the student correct?
- A. Yes: the change of 200 birds is compared with the final count, 300, so the percent change is about −67 %The change is measured against the starting count, 500, not the final count.
- B. ✓ No: the change of 200 birds is divided by the starting count, 500, so the percent change is −40 %
Why: The count started at 500 snipe.
The change is measured against where the count started.
So the change is divided by the starting count, not the final count.
The percent change is −40 %.
The table below gives two reserves’ counts of redshanks, wading birds, one year and again the next year. A student says: “Both reserves lost 30 birds, but Reserve J’s percent change is the bigger fall, because the same loss is a larger share of a smaller starting count.”
Is the student correct?
- A. No: a percent change depends on the size of the change alone, and the two reserves’ changes are equalA percent change is the change as a share of the starting count, and the two starting counts differ.
- B. ✓ Yes: 30 birds is a larger share of 300 than of 600, so Reserve J’s percent change is the bigger fall
Why: Reserve J lost 30 redshanks from a starting count of 300.
Reserve P lost 30 redshanks from a starting count of 600.
The same loss is the larger share of the smaller starting count.
So Reserve J’s fall is the bigger percent change: −10 % against −5 %.
Here is the table again: the two reserves’ raptors, and the lagoon’s avocets.
Both reserves lost 60 birds.
On the first reserve the loss is 25 % of the starting count, so its percent change is −25 %.
On the second reserve the same loss is 15 % of the starting count, so its percent change is −15 %.
Stated as percent changes with their signs, the two losses can be compared, whatever the size of the reserve.
53Numeric practice: percent change mixed practice
Suppose wheatears, small birds of open moorland, are counted on a moor: 860 wheatears one summer and 645 wheatears the next.
Calculate the change in the count, the final value minus the initial value, with its sign.
Answer: -215 wheatears (tolerance ±0)
Suppose a forest clearing’s count of nightjars rose from 36 pairs one summer to 45 pairs the next.
Which sign does the percent change carry?
- A. ✓ Positive
- B. NegativeA negative percent change means the count fell.
This count rose.
Why: The count rose from 36 pairs to 45 pairs.
A rise is a positive change.
So the percent change is positive.
Suppose stone loach, small bottom-living fish, live in a stream pool. One spring 64 stone loach are counted in the pool, and the next spring 96 are counted.
Calculate the percent change in the stone loach’s count, with its sign.
Answer: 50 % (tolerance ±0)
Suppose robins nest in a forest. One spring 160 robins are counted in the forest, and after a hard winter 104 robins are counted the next spring.
Calculate the percent change in the robins’ count, with its sign.
Answer: -35 % (tolerance ±0)
Suppose bitterns, herons of dense reeds, nest in a fen. One year 48 bitterns are counted in the fen, and the next year 84 bitterns are counted.
Calculate the percent change in the bitterns’ count, with its sign.
Answer: 75 % (tolerance ±0)
Suppose skylarks nest on farmland. One spring 475 skylarks are counted there, and the next spring 437 skylarks are counted.
Calculate the percent change in the skylarks’ count, with its sign.
Answer: -8 % (tolerance ±0)
APBIO-U08-L37 A limit changes, the population answers
Suppose an early-sprouting weed covers a field’s ground each spring, before the native plants come up. In one plot of the field a botanist pulls the weed out as soon as it appears.
By midsummer the native seedlings in that plot come up thick, while in the untouched plot they are few and small. What limited the natives, and what will their count do over the next few years?
Unit 8 · Ecology
1A limit changes, the population answers
Suppose killifish, small freshwater fish, breed in a tank, and the keeper feeds the tank the same amount of food each day. The count of killifish climbs and nears the tank’s K.
As the count nears K, what happens to births and deaths?
- A. Births rise and deaths fallEach killifish’s share of the food falls as the count climbs, so fewer are born and more die.
- B. ✓ Births fall and deaths rise
- C. Births and deaths both stay as they wereEach killifish’s share of the food falls as the count climbs, so births and deaths change.
Why: The keeper feeds the tank the same amount of food each day.
As the count climbs, each killifish’s share of the food falls.
So fewer killifish are born and more die, until births equal deaths.
Suppose two seed trays of one size sit under one lamp. One tray holds 20 lettuce seedlings and the other holds 80. The seedlings in each tray share the lamp’s light.
Which kind of limiting factor is the shortage of light?
- A. ✓ Density-dependent
- B. Density-independentThe shortage of light harmed a larger fraction of the crowded tray’s seedlings.
The fraction rose with the density.
Why: The same light is shared among 20 seedlings in one tray and 80 in the other.
In the crowded tray each seedling’s share is smaller, so a larger fraction die.
The fraction harmed rises with the density, so the shortage of light is density-dependent.
Suppose the grass of a dry plain feeds at most 140 guanacos, wild grazing animals of South America, so the plain’s K for guanacos is 140. A dry summer kills much of the grass, and the grass left feeds at most 70 guanacos. The herd stands at 130.
Predict what the herd’s count does over the following years.
- A. Rises toward 140The grass that fed 140 guanacos is gone; the grass left feeds 70.
- B. Stays at 130130 guanacos are more than the grass left can feed, so deaths outnumber births.
- C. ✓ Falls toward 70
Why: K is set by the plain’s grass.
The dry summer cut the grass, so K fell from 140 to 70 guanacos.
A herd of 130 is above the new K, so deaths outnumber births and the herd falls toward 70.
When a limit changes, how does a population answer?
First, name the factor: here the light and the ground space, which the weed takes first.
Then say which way the limit moved: pulling the weed eased it.
The natives’ count climbs.
As the natives crowd one another, each plant’s share of light and space shrinks.
So their growth slows, and the count levels off at a new, higher K.
Every prediction in this topic has those three parts: the factor, which way it moved, and the count’s answer.
Video: Watch: A limit changes, the population answers
The two plots are on screen, the weed covering one and the other bare, with native seedlings coming up thick. The native count of each plot is plotted year by year: the untouched plot’s points sit on the 25 gridline, and the weeded plot’s curve climbs and levels off on the 175 gridline, where a dashed line is drawn and labeled K.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L37a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L37a.mp4
Here are the two plots again, seen from above: each native plant is a dot, and the shaded ground is ground the weed covers.
Every March the weed comes up first, and it covers the ground of the untouched plot.
The native seedlings come up in April, under the weed, and most of them die because too little light reaches them.
By midsummer the untouched plot holds 25 native plants.
In the other plot the botanist pulls the weed as soon as it shows, so the native seedlings come up into bare ground and full light.
By midsummer the weeded plot holds 70 native plants.
The light and the ground space held the natives back in the untouched plot, because the weed took them first.
Light and space are resources the plants compete for: the more plants share a plot, the smaller each plant’s share.
So competition for light and space is a density-dependent factor.
Pulling the weed eased that limit.
More light and more bare ground reached each native seedling. So more seedlings survived.
A limit can move the other way, too: a change that takes light or space away from the plot tightens it.
The botanist counted the native plants in both plots every midsummer for eight years. Here are the counts as a table.
Here are the same counts plotted on the growth graph: population size N on the y-axis, time in years on the x-axis, one graph for each plot.
The untouched plot’s count stays near 25 for all eight years.
The weeded plot’s count climbs fast at first: 70, then 110, then 140.
Then the climb slows, and from year 7 the count sits on the 175 gridline.
The level a count settles at is its carrying capacity, K, drawn as a dashed line. The untouched plot’s K is 25, read from the 25 gridline, and the weeded plot’s K is 175, read from the 175 gridline.
The botanist pulled nothing but the weed, so why did the natives’ count stop at 175?
As the natives crowd one another, each plant’s share of light and space shrinks.
So fewer seedlings survive and more plants die, until births equal deaths, and the count stays steady.
As a population nears its carrying capacity each individual’s share of resources falls, births fall and deaths rise until they balance: negative feedback on growth.
The same factor, light and space, limits the natives again. This time the natives take it from one another, not from the weed.
The plot’s K for the natives was 25 with the weed and 175 without it.
K belongs to the place, for that species: when the weed stopped taking the plot’s light and space, the natives’ K rose.
Every prediction in this topic has three parts: the factor, which way it moved, and the count’s answer toward the new K.
For example, the botanist pulls the weed. The factor is the light and space, and pulling the weed eased it.
So K rises from 25 to 175 native plants, and the count climbs to the new K.
But now suppose the field’s farmer puts up a tall wooden fence along the weeded plot’s south side, and the fence shades half the plot. The factor is the light and space, and the fence tightened it.
So K falls from 175 to 125 native plants, and the count falls toward the new K.
And now suppose instead that a late frost one April kills 25 of the plot’s 175 natives. The frost takes plants, but it takes no light or space from the plot.
So K stays at 175, and the count climbs back to it.
Here are the fence and the frost on the growth graph, one above the other, with the question drawn above: did the change move K?
Some hazards kill a fraction of a population once and leave the place’s resources as they were: a late frost on the natives, a hailstorm on chicks.
Such a hazard moves the count for a season, and the count climbs back to the same K.
Other hazards also destroy the place’s resources: a fire that burns a reedbed, a drought that kills the grass. Such a hazard moves K as well.
Only a change in the place’s resources, or in a density-dependent factor, moves K.
What you are expected to know Predict how a population’s count changes when a resource or a density-dependent factor changes.
What you are expected to know Justify the prediction from the factor and the way it moved.
Suppose sea anemones, soft-bodied animals that fix themselves to hard surfaces, cover the wooden piles of a pier from top to bottom, and no bare wood is left for a young anemone to settle on. The harbor’s owners remove half the piles.
Predict what the count of anemones on the pier does over the following years.
- A. RisesThe pier has half as many piles, so it has less surface for anemones, not more.
- B. Stays the sameThe anemones on the removed piles are gone, and the piles left already had no bare wood for new ones.
- C. ✓ Falls
Why: Hard surface to settle on is the anemones’ limiting factor, and every pile was covered.
Removing half the piles took half that surface away, so the pier’s K fell.
The count falls toward the new K, and the piles left hold no more than before.
Suppose pied flycatchers, small birds that nest in tree holes, breed in a forest with few holes, and every hole holds a pair each spring. A bird group puts up 52 nest boxes in the forest.
Predict what the count of nesting pairs does over the following springs.
- A. Stays where it wasThe nest boxes add nest sites, so more pairs can nest than before.
- B. ✓ Rises, then levels off at a higher K
- C. Rises for as long as the boxes stand, and never levels offOnce every box and every hole holds a pair, nest sites limit the flycatchers again.
Why: Nest sites are the flycatchers’ limiting factor.
The boxes add nest sites, so the forest’s K rises and the count of pairs climbs.
Once every box and hole holds a pair, nest sites limit the flycatchers again, so the count levels off at the new K.
Suppose a tank of neon tetras, small shoaling fish, has held a steady count for a year, and the waste in the water is what limits them. The keeper fits a filter that removes the waste as fast as the tetras make it, and feeds the tank the same amount each day. A student says: “With the waste gone, nothing limits the tetras, so their count will climb for as long as the keeper feeds them.”
Is the student correct?
- A. Yes: with the waste gone, nothing limits the tetras, so the count climbs for as long as the keeper feeds themThe filter removes one limit, the waste; the tank’s food and space still limit the tetras.
- B. ✓ No: as the tetras crowd the tank, each tetra’s share of food and space shrinks, so the count levels off at a higher K
Why: The filter eases one limit: the waste.
As the count climbs, each tetra’s share of the food and the space shrinks.
So fewer fry survive and more tetras die, until births equal deaths.
The count levels off at a new, higher K.
A student reads that a botanist pulled the weed from one of two plots and left it in the other, and that the natives settled at a higher count in the weeded plot. The student says: “The plot’s K for the native plants is set by the light and the space the plot offers them, so pulling the weed raised the natives’ K.”
Is the student correct?
- A. ✓ Yes: K is set by the plot’s light and space for the natives, and the weed had been taking them, so pulling it raised K
- B. No: K belongs to the native plants themselves, the same in any plot, so pulling the weed cannot change itK belongs to the place, for that species: the same natives settled at a higher count once the weed was gone.
Why: K is set by the place’s resources for the species.
The weed took the plot’s light and space first.
Pulling the weed left that light and space to the natives, so their K rose.
Suppose goldcrests, tiny birds that eat insects, live in a spruce plantation, and their count has stayed near the plantation’s K for years. One winter a long cold spell kills 30 % of them. The plantation’s insects and nest sites are unchanged.
Predict what the count does over the following years.
- A. Keeps fallingThe cold spell is over, and the plantation’s insects and nest sites still feed and house as many goldcrests as before.
- B. Levels off at a new, lower KThe cold spell took goldcrests, but it took no insects or nest sites from the plantation, so K is unchanged.
- C. ✓ Climbs back to the old K
Why: K is set by the plantation’s insects and nest sites.
The cold spell killed goldcrests, but it took no insects or nest sites away, so K did not move.
The count is now below K, so births outnumber deaths and the count climbs back to the old K.
Here is a table comparing the three kinds of change: the factor, which way it moved, what K does, and what the count does.
Here are the two plots again: the weed left in one and pulled from the other, each native plant a dot.
The light and the ground space limited the natives, and the weed took them first.
Pulling the weed eased that limit, so the natives’ count climbed.
As the natives crowded one another, each plant’s share of light and space shrank, so the climb slowed.
The count leveled off at a new, higher K: 175 native plants in the weeded plot, where the untouched plot holds 25.
63Practice: predict and justify mixed practice
Suppose Tasmanian devils, meat-eating marsupials, live in a forest and feed on carrion, the bodies of dead animals. A disease that passes from devil to devil when they bite each other has kept the forest’s count near 26 devils for years. Vets then vaccinate every devil in the forest, and the disease stops passing between them.
(a) Identify the kind of limiting factor that the vaccination eased. (1 pt)
- Award 1 point for: density-dependent (accept: a disease passed by biting harms a larger fraction of a denser population).
(b) Describe how the count of devils changes over the following years. (1 pt)
- Award 1 point for: the count rises and then levels off at a higher level (accept: climbs to a new, higher carrying capacity).
(c) Explain why the count changes as you described in part (b) once the disease is gone. (2 pt)
So fewer young are born and more devils die, until births equal deaths.
So the count levels off at the K the forest’s carrion sets.
- Award 1 point for: as the count climbs, each devil’s share of the carrion shrinks (accept: the devils compete for the forest’s food; a density-dependent factor limits them again).
- Award 1 point for: so births fall and deaths rise until they are equal, and the count stays steady (accept: births equal deaths at the new K).
(d) Predict what happens to the count if the vaccination stops and the disease passes between devils again, and justify your prediction. (2 pt)
The disease passes when devils bite each other, so it harms a larger fraction of a denser population.
So the disease lowers the forest’s K, and the count, now above that K, falls toward it.
- Award 1 point for: the count falls (accept: falls toward a lower K; falls back toward 26).
- Award 1 point for: the disease is density-dependent, so it lowers K (accept: it passes by biting, so it harms a larger fraction of the crowded forest; deaths outnumber births above the new K).
APBIO-U08-P84 Practice questions: Topic 8.4
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one population’s growth one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. The formula sheet gives the growth equations, dN/dt = B − D, dN/dt = rmax N and dN/dt = rmax N ((K − N)/K), and percent change = (final − initial) / initial × 100.
Video: Watch first: The effect of density on populations, summed up
Population density; K read from the S-shaped curve; why growth slows near the top; density-dependent and density-independent limits; the braking term and dN/dt = r<sub>max</sub> N ((K − N)/K); fastest halfway to K; past K, and a moving K; percent change.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T84-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T84-summary.mp4
Suppose a net sample of 14 L of sea water holds 602 sprats, small fish.
Which of the following is the population density of the sprats in the sample?
- A. 0.02 sprats per L0.02 divides the volume by the count, 14 ÷ 602.
Population density is the count divided by the volume. - B. ✓ 43 sprats per L
- C. 588 sprats per L588 subtracts the volume from the count.
Population density is the count divided by the volume. - D. 8 428 sprats per L8 428 multiplies the count by the volume.
Population density is the count divided by the volume.
Why: Population density is the count divided by the volume of water.
population density = 602 ÷ 14 = 43 sprats per L.
Godwits, large wading birds, begin to winter on a new estuary. The graph below shows the count every winter for twelve years.
Which of the following is the estuary’s carrying capacity, K, for godwits?
- A. 40 godwits40 godwits is the first count, where the curve starts.
- B. 315 godwits315 godwits is half of the level the curve settles at.
K is the level itself. - C. ✓ 630 godwits
- D. 900 godwits900 godwits is the top of the y-axis.
The curve settles below it, on the 630 gridline.
Why: K is the level the count settles at.
From year 10 the curve is flat along the 630 gridline.
So K is 630 godwits.
Yellowhammers, small seed-eating birds, nest along one hedge, and the hedge’s seeds and nest sites stay as they were. The count of yellowhammers climbs and nears the hedge’s K.
As the count nears K, what happens to the births per yellowhammer and the deaths per yellowhammer each year?
- A. Births per bird rise and deaths per bird fallMore births per bird would need each bird to have more seed.
Each bird’s share is shrinking. - B. Births per bird and deaths per bird both keep their earlier valuesEach bird’s share of seed and nest sites shrinks as the count climbs, so the births and deaths per bird change.
- C. ✓ Births per bird fall and deaths per bird rise
- D. Births per bird fall and deaths per bird stay as they wereA bird short of seed dies sooner as well as raising fewer young.
Deaths per bird rise.
Why: The hedge’s seeds and nest sites are fixed.
As the count climbs, each yellowhammer’s share shrinks.
A bird short of seed raises fewer young and dies sooner.
So births per bird fall and deaths per bird rise, until births equal deaths.
Suppose greenfinches feed at a garden’s feeders all winter, and the gardener puts out the same weight of seed each day. By midwinter the seed is gone within an hour of being put out. The count of greenfinches at the feeders climbs through the autumn and then levels off.
Which of the following is the limiting factor for the greenfinches?
- A. ✓ The seed
- B. The winter coldCold kills a fraction of the flock whatever its size; it sets no level for the count.
The count settles where each bird’s share of seed is too small. - C. The perches on the feedersGreenfinches take turns at the perches and wait in the hedges between turns.
The seed is gone within an hour: the flock is short of seed, not of perches. - D. The greenfinches’ own nature: a flock stops growing at a set sizeA count settles where each bird’s share of a resource is too small, not at a size the birds carry in them.
More seed each day would raise the level.
Why: The gardener puts out the same weight of seed each day, and by midwinter it is gone within an hour.
As the count climbs, each greenfinch’s share of the seed shrinks.
So the seed holds the count back: the seed is the limiting factor.
Two hedges of the same length hold goldfinch nests. A hailstorm crosses both hedges on the same afternoon. The table below shows the nests in each hedge and how many the hail destroyed.
Which kind of limiting factor is the hail, and why?
- A. Density-dependent, because the hail destroyed more nests in the crowded hedgeA larger count alone does not decide.
18 of 72 and 6 of 24 are the same fraction, 25 %. - B. Density-dependent, because the hail destroyed a larger fraction of the crowded hedge’s nestsThe hail destroyed 25 % of the nests in both hedges.
The fraction did not rise with the density. - C. ✓ Density-independent, because the hail destroyed the same fraction of the nests in both hedges
- D. Density-independent, because the crowded hedge held three times as many nestsHow many nests a hedge held is not the ground.
The ground is that the fraction destroyed, 25 %, was the same in both hedges.
Why: Hedge J: 6 of 24 nests destroyed, 25 %.
Hedge P: 18 of 72 destroyed, 25 %.
The hail harmed the same fraction at both densities, so it is density-independent.
Redpolls, small finches, roost in two thickets of the same size on winter nights: a few dozen birds in one thicket and several hundred in the other. An infection that passes from bird to bird when they perch touching reaches both roosts. By spring it has reached a larger fraction of the crowded roost’s birds.
Which of the following explains why the infection reached a larger fraction of the crowded roost’s birds?
- A. The infection is stronger where the air is warmer, and a crowded roost is warmerThe same infection passes the same way in both roosts.
What differs is how many birds each bird touches. - B. The crowded roost holds more birds, so more birds are there to catch the infectionMore birds catching it would not by itself make the fraction larger.
The fraction is what rose. - C. The birds in the sparse roost are fitter, so they fight the infection offThe redpolls of the two roosts are the same kind of bird.
What differs between the roosts is how many birds each bird touches. - D. ✓ Each bird in the crowded roost perches touching more other birds every night
Why: The infection passes when birds perch touching.
In the crowded roost each redpoll touches more other birds every night.
So the infection passes to a larger fraction of the crowded roost’s birds.
Suppose an estuary’s mud holds food for no more than 2 320 dunlin, a small wading bird: K is 2 320 dunlin. In October the estuary holds 580 dunlin.
Which of the following is the braking term for the dunlin?
- A. 0.250.25 is N divided by K: the part of the estuary already filled, not the room left as a fraction of the whole.
- B. ✓ 0.75
- C. 44 is K divided by N.
The braking term divides the room left, K − N, by K. - D. 1 7401 740 dunlin is the room left, K − N: a count of birds, not a fraction of the whole.
Why: The room left is K − N: 2 320 − 580 = 1 740 dunlin.
The braking term is the room left divided by K.
1 740 ÷ 2 320 = 0.75.
A flock of pintails, a duck, on a lake has reached the lake’s carrying capacity: N equals K.
Which of the following does the logistic equation give for the flock’s dN/dt?
- A. ✓ 0 pintails per year
- B. r<sub>max</sub> Nr<sub>max</sub> N is the rate with no brake.
At K the braking term is 0, and any number multiplied by 0 is 0. - C. r<sub>max</sub> Kr<sub>max</sub> K would be the exponential rate for a flock of size K.
At K the braking term is 0, so dN/dt is 0. - D. A negative rateThe braking term is negative only above K.
At K it is 0, so dN/dt is 0.
Why: At K the room left, K − N, is 0, so the braking term is 0.
dN/dt is r<sub>max</sub> N multiplied by the braking term.
Any number multiplied by 0 is 0, so dN/dt is 0 pintails per year: the flock neither grows nor shrinks.
Suppose a shore’s seaweed and crevices can support no more than 2 120 sea slaters, shore-living relatives of woodlice: K is 2 120 sea slaters.
At which of the following counts does the sea slater population grow fastest?
- A. 530 sea slatersAt 530 sea slaters the brake is barely on, but few sea slaters are there to breed.
- B. ✓ 1 060 sea slaters
- C. 1 590 sea slatersAt 1 590 sea slaters many are there to breed, but the braking term is 0.25: the brake is nearly full on.
- D. 2 120 sea slatersAt 2 120 sea slaters the population is at K, so the braking term is 0 and the population does not grow.
Why: A logistic population grows fastest halfway to K.
Half of 2 120 sea slaters is 1 060 sea slaters.
So the population grows fastest at 1 060 sea slaters.
Suppose a reservoir’s fish feed no more than 76 goosanders, a fish-eating duck: K is 76 goosanders, and the flock has stood at 76 for years. The reservoir’s owners then stock it with fish every month, and the fish now feed no more than 152 goosanders.
Predict what the flock’s count does over the following years.
- A. Stays at 76 goosandersA count stays steady only at K.
The new K is 152, and 76 goosanders is below it. - B. Falls below 76 goosandersThe stocked fish are the goosanders’ food.
More food raises K, and the flock climbs toward it. - C. Rises with no limit, for as long as the stocking continuesThe stocked fish feed no more than 152 goosanders.
Once the flock reaches 152, the fish limit it again. - D. ✓ Rises toward 152 goosanders, the new K
Why: K is set by the reservoir’s fish, so the stocking raises K from 76 to 152 goosanders.
The flock, 76, is now below K, so births outnumber deaths.
The count rises toward the new K, 152.
Lugworms live buried in the sand of a beach. Suppose the food in one marked patch of sand can support no more than 2 040 lugworms: K is 2 040 lugworms. This spring the patch holds 510 lugworms, and the lugworms’ maximum per capita growth rate, r<sub>max</sub>, is 0.8 per year.
(a) Calculate the braking term for the lugworms this spring. (1 pt)
Frame braking term = (K − N)/K = (… − …)/… = …
Hint Which of the two counts is K, the most the patch can support, and which is N, the count now?
Answer: 0.75 (tolerance ±0)
- Award 1 point for: 0.75 (a fraction, no unit).
Slip Dividing N by K, 0.25. The braking term is the room left, K − N, as a fraction of K.
(b) Calculate r<sub>max</sub> N for the lugworms this spring. (1 pt)
Frame r<sub>max</sub> N = … × … = … lugworms per year
Hint r<sub>max</sub> is the number each lugworm adds per year with nothing holding it back; N is how many lugworms are adding.
Answer: 408 lugworms per year (tolerance ±0)
- Award 1 point for: 408 lugworms per year, with the unit (accept 408 per year).
Slip Adding 0.8 to 510. The two symbols written side by side are multiplied.
(c) Calculate dN/dt for the lugworms this spring, using your answers to parts (a) and (b). (1 pt)
Frame dN/dt = r<sub>max</sub> N × braking term = … × … = … lugworms per year
Hint The logistic equation multiplies the exponential part by the braking term.
Answer: 306 lugworms per year (tolerance ±0)
- Award 1 point for: 306 lugworms per year, with the unit (accept 306 per year).
Slip Dividing r<sub>max</sub> N by the braking term, 544. The braking term multiplies r<sub>max</sub> N.
(d) Predict how the number of lugworms the patch adds each year changes as the count climbs from 510 toward K, and explain your prediction. (1 pt)
Frame The yearly gain … and then …, because …
Hint Think of the two parts you multiplied in part (c): which one grows as the count climbs, and which one shrinks?
The gain peaks halfway to K, at 1 020 lugworms, so from 510 the gain rises before it falls.
- Award 1 point for: the number the patch adds each year rises at first, is largest halfway to K, at 1 020 lugworms, and then falls to 0 as the count reaches K, because r<sub>max</sub> N grows with the count while the braking term shrinks to 0 (accept: fastest growth at half of K, then the gain falls to 0 at K). ‘The number added falls as the count nears K’ on its own is the second half of the answer — it is true only once the count passes 1 020 — and is worth half of this point; the point wants the rise first.
Slip Predicting a gain that only falls as the count climbs. From 510 the count is below half of K, so the gain rises first; it falls once the count passes 1 020.
(e) Next spring the patch holds 816 lugworms. Calculate the percent change in the count from this spring to the next, with its sign. (1 pt)
Frame percent change = (final − initial)/initial × 100 = (… − …)/… × 100 = … %
Hint Which count is the starting count, the one you divide by?
Answer: 60 % (tolerance ±0)
- Award 1 point for: +60 % (accept 60 %; the sign positive).
Slip Dividing by the final count, 816, which gives 37.5 %. The change is measured against the starting count.
Stonechats, small perching birds, nest in the thorn scrub of a moor. Suppose the moor’s scrub shelters and feeds no more than 68 stonechat pairs: K is 68 pairs, and 65 pairs nest there this spring. In late summer a fire burns half of the scrub. No stonechat dies in the fire, but the scrub left shelters and feeds no more than 34 pairs. The burnt half stays bare in the years that follow.
(a) Identify the limiting factor that the fire changed for the stonechats. (1 pt)
- Award 1 point for: the scrub (accept: the nest sites; the shelter and food the scrub provides).
Slip Naming the fire itself. The fire is the event; the limiting factor is the resource it cut.
(b) Predict how the count of nesting pairs changes over the following years. (1 pt)
- Award 1 point for: the count falls (accept: falls toward the new K of 34 pairs).
Slip Predicting no change because no bird died in the fire. K fell, so the count is now above K and falls.
(c) Justify your prediction in part (b). (1 pt)
The 65 pairs are now above the new K.
So each pair’s share of nest sites and food is too small, and deaths outnumber births until the count reaches the new K.
- Award 1 point for: the count is above the new K, so each pair’s share of the scrub is too small and deaths outnumber births (accept: the braking term is negative above K; K fell because the scrub sets it).
Slip Justifying with the birds killed by the fire. None died; the resource fell, so K fell.
(d) A student says: “A fire is a density-independent factor, so it leaves K where it was, and the stonechats will nest at 68 pairs again within a few years.” Evaluate the student’s prediction. (1 pt)
Calling the fire a density-independent factor is fair: a fire kills the same fraction of birds whatever their density.
But this fire burned the scrub, the resource that sets K, so K fell from 68 to 34 pairs.
The scrub stays bare, so the count falls to about 34 pairs and settles there, not at 68.
- Award 1 point for: the judgement (the prediction is wrong) AND the ground (this fire destroyed the scrub that sets K, so K fell to 34 pairs and the count settles there while the scrub stays bare). Accept: the label is fair — a fire is density-independent — but the conclusion is wrong, because this fire took away the resource that sets K.
Slip Judging the prediction right because fires are listed as density-independent. Whether K moves depends on whether the hazard took a resource away.
APBIO-U08-T84 End-of-topic test: Effect of Density on Populations
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. The formula sheet gives the growth equations, dN/dt = B − D, dN/dt = rmax N and dN/dt = rmax N ((K − N)/K), and percent change = (final − initial) / initial × 100.
Ecologists count the muntjac, a small deer, in four woods. The table below gives each wood’s count and its area.
In which wood do the muntjac have the highest population density?
- A. Wood JWood J holds 42 muntjac on 7 km²: 6 muntjac per km².
Wood R holds 9 per km². - B. Wood MWood M holds 60 muntjac on 12 km²: 5 muntjac per km².
Wood R holds 9 per km². - C. ✓ Wood R
- D. Wood WWood W holds the most muntjac, 80, but on 20 km²: 4 muntjac per km², the lowest density of the four.
Why: Population density is the count divided by the area.
Wood J holds 6 muntjac per km² and Wood M 5 per km².
Wood R: 54 ÷ 6 = 9 per km². Wood W: 80 ÷ 20 = 4 per km².
The highest population density is Wood R’s.
Bullheads, small bottom-living fish, are released into a new stream pool and counted every month. The graph below shows the counts.
Which of the following is the pool’s carrying capacity, K, for bullheads?
- A. 30 bullheads30 bullheads is the count at month 0, where the curve starts, not the level it settles at.
- B. 420 bullheads420 bullheads is half of the level the curve settles at.
K is the level itself. - C. 690 bullheads690 bullheads is the count at month 10, while the curve is still climbing.
- D. ✓ 840 bullheads
Why: K is the level the count settles at.
From month 14 the curve is flat along the 840 gridline.
So K is 840 bullheads.
Prawns are bred in a tank that gets the same amount of food every day. The graph below shows the count of prawns every two weeks.
Which of the following explains why the curve flattens after week 16?
- A. The tank holds as many prawns as can fit, so no more prawns can be bornPrawns are still born near the top of the curve.
Births fall to match deaths, so the count stays steady. - B. ✓ Each prawn’s share of the food shrinks as the count rises, so fewer prawns are born and more die, until births equal deaths
- C. Each prawn keeps adding the same number of young, but a hazard from outside the tank kills a fixed fraction every dayA hazard from outside would kill the same fraction at any count.
The curve flattens because the count itself lowered each prawn’s share of food. - D. The prawns have used up all the food in the tank, so every prawn stops breeding and the count stays where it isThe tank gets the same food every day, so the food is not used up.
Each prawn’s share is smaller, and births fall until they equal deaths.
Why: The tank gets a fixed amount of food each day.
As the count rises, each prawn’s share shrinks.
A prawn short of food breeds more slowly and dies sooner, so births fall and deaths rise.
When births equal deaths the count stays steady, and the curve is flat.
Sandhoppers, small jumping crustaceans of the shore, are kept in four tanks of the same size at the same temperature. Each tank gets dead seaweed to eat and an area of damp sand to hide in. The table below gives what each tank gets and the count the sandhoppers settle at.
Which of the following is the limiting factor for the sandhoppers?
- A. ✓ The seaweed
- B. The damp sandTank J and Tank M get the same seaweed and different sand, and settle at the same count.
More sand adds no sandhoppers. - C. The seaweed and the damp sand togetherDoubling the sand changes nothing; doubling the seaweed doubles the count.
Only the seaweed holds the count back. - D. Neither: the sandhoppers’ own nature sets the countTank R settles at twice Tank J’s count when it gets twice the seaweed.
The count is set by the tank’s resources.
Why: Tank J and Tank M get the same seaweed and settle at the same count, though Tank M has twice the sand.
Tank R gets twice Tank J’s seaweed and settles at twice the count.
So the seaweed holds the count back: the seaweed is the limiting factor.
Two plantations of the same size hold young larch trees, one planted thinly and one planted densely. A fungal disease reaches both plantations in the same spring. The table below shows the saplings in each plantation and how many the fungus killed.
Which kind of limiting factor is the fungus, and why?
- A. Density-dependent, because the fungus killed more saplings in the dense plantationA larger count alone does not decide: the dense plantation holds twice as many saplings.
The sort compares the fractions. - B. ✓ Density-dependent, because the fungus killed a larger fraction of the dense plantation’s saplings
- C. Density-independent, because the fungus killed a fraction of the saplings in both plantationsEvery limiting factor kills a fraction in both plantations.
The sort asks whether that fraction rose with the density: 20 % against 40 %. - D. Density-independent, because a fungus spreads through a plantation whatever the spacing of its treesPlantation P lost 40 % of its saplings and Plantation J lost 20 %.
The fraction rose with the density.
Why: Plantation J: 33 of 165 saplings killed, 20 %.
Plantation P: 132 of 330 killed, 40 %.
The fungus killed a larger fraction of the denser plantation, so it is density-dependent.
Each of the following limits a population’s growth.
Which of the following is a density-independent factor?
- A. A disease that passes from one animal to another when the two touchA disease passed by touch reaches a larger fraction of a denser population, because each animal touches more others.
- B. A shortage of nest holes shared by all the birds of a woodIn a crowded wood a larger fraction of the pairs find no hole.
The fraction harmed rises with the density. - C. A hawk that hunts where its prey are easiest to findA hawk takes a larger fraction of its prey where the prey are dense.
The fraction harmed rises with the density. - D. ✓ A late frost that kills a fraction of a field’s seedlings
Why: A density-independent factor harms the same fraction at any density.
A frost reaches every seedling alike, however closely the seedlings stand.
So the frost kills the same fraction of a thin field and a dense field.
Buzzards hunt water voles on two meadows of the same size. The table below shows the voles on each meadow and how many the buzzards took in a week.
Which meadow lost the larger fraction of its voles, and why?
- A. ✓ Meadow P, because a buzzard finds a vole after a shorter search where voles are dense, and other buzzards gather there
- B. Meadow P, because it holds five times as many voles, so far more voles are there to be takenMore voles taken in all does not decide the fraction.
Meadow P lost 45 % of its voles and Meadow J 25 %. - C. Meadow J, because each vole there stands alone in the grass and is easier for a buzzard to seeMeadow J lost 7 of 28 voles, 25 %; Meadow P lost 63 of 140, 45 %.
The larger fraction was lost where the voles were dense. - D. Neither: a buzzard takes the same fraction of the voles on any meadow it hunts overMeadow J lost 25 % of its voles and Meadow P lost 45 %.
The fractions differ.
Why: Meadow J lost 25 % of its voles and Meadow P 45 %.
Where voles are dense, a buzzard finds one after a short search and hunts there again.
Other buzzards gather where the finding is easy.
So the buzzards take a larger fraction of the dense meadow’s voles.
Suppose a reef’s food and hiding places can support no more than 2 450 wrasse, a reef fish: K is 2 450 wrasse. The reef holds 490 wrasse.
Which of the following is the braking term for the wrasse?
- A. 0.20.2 is N divided by K: the part of the reef already filled, not the room left as a fraction of the whole.
- B. ✓ 0.8
- C. 55 is K divided by N.
The braking term divides the room left, K − N, by K. - D. 1 9601 960 wrasse is the room left, K − N: a count of fish, not a fraction of the whole.
Why: The room left is K − N: 2 450 − 490 = 1 960 wrasse.
The braking term is the room left divided by K.
1 960 ÷ 2 450 = 0.8.
Suppose the braking term for a flock of capercaillie, large grouse of pinewoods, is 0.25.
Which of the following is true of the flock?
- A. The flock is at 25 % of its KA braking term of 0.25 means 25 % of the whole is still free, so the flock fills the other 75 %.
- B. The flock has room for 25 % more birds than it holds nowThe 25 % is a fraction of K, the whole, not of the flock’s own count.
- C. ✓ The flock is at 75 % of its K
- D. The flock is above its KAbove K the room left is negative, so the braking term is negative.
0.25 is positive.
Why: The braking term is the room left as a fraction of the whole, K.
0.25 means 25 % of K is still free.
So the flock fills 75 % of K: it is at 75 % of its K.
An ecologist uses the logistic equation for the turnstones, a shore bird, on one estuary.
Which of the following does K stand for in the logistic equation?
- A. The number of turnstones on the estuary nowThe number of turnstones on the estuary now is N, the population size.
- B. The most turnstones each turnstone adds per year, with nothing holding the flock backThe most each turnstone adds per year is r<sub>max</sub>, the maximum per capita growth rate.
- C. The room left on the estuary as a fraction of the wholeThe room left as a fraction of the whole is the braking term, (K − N)/K.
- D. ✓ The most turnstones the estuary’s resources can support
Why: K is the carrying capacity.
The carrying capacity is the largest population a place’s resources can support.
So K is the most turnstones the estuary’s resources can support.
Which of the following is the logistic equation, as the formula sheet writes it?
- A. alone is the exponential equation.
It has no K, so the growth it describes never stops. - B. (N − K)/K is negative below K and positive above it.
The braking term, (K − N)/K, is positive below K and 0 at K. - C. ✓
- D. Without N, the rate does not grow with the population.
The logistic equation multiplies r<sub>max</sub> N by the braking term.
Why: The logistic equation is the exponential equation, r<sub>max</sub> N, multiplied by the braking term.
The braking term is the room left, K − N, divided by K.
So dN/dt = r<sub>max</sub> N ((K − N)/K).
Suppose a lagoon’s food can support no more than 2 720 smelt, a small fish: K is 2 720 smelt. The lagoon holds 680 smelt, and their r<sub>max</sub> is 0.9 per year.
Which of the following is dN/dt for the smelt?
- A. 153 smelt per year153 multiplies r<sub>max</sub> N by N divided by K, 0.25, the part of the lagoon already filled.
The braking term is the room left as a fraction, 0.75. - B. ✓ 459 smelt per year
- C. 612 smelt per year612 is r<sub>max</sub> N alone, 0.9 × 680.
The logistic equation multiplies it by the braking term, 0.75. - D. 1 836 smelt per year1 836 multiplies r<sub>max</sub> by the room left, K − N.
The equation multiplies r<sub>max</sub> N by the room left divided by K.
Why: Work the braking term first: (2 720 − 680) ÷ 2 720 = 0.75.
Then multiply r<sub>max</sub> N by it.
dN/dt = 0.9 × 680 × 0.75 = 459 smelt per year.
Arctic char, a cold-water fish, live in four lakes. The table below gives each lake’s carrying capacity for char and its population size now. Every lake’s char have the same r<sub>max</sub>, 0.1 per year.
Which lake’s char population adds the most fish this year?
- A. Lake JLake J’s braking term is 0.9, but only 460 char are there to breed: 0.1 × 460 × 0.9 = 41.4 char per year.
- B. Lake MLake M holds the most char below K, but its braking term is 0.1: 0.1 × 1 530 × 0.1 = 15.3 char per year.
- C. Lake RLake R’s 1 020 char are above its K of 850, so its braking term is negative and the population shrinks.
- D. ✓ Lake W
Why: A logistic population grows fastest halfway to K.
Lake W’s 1 100 char are half of its K, 2 200, so its braking term is 0.5.
dN/dt = 0.1 × 1 100 × 0.5 = 55 char per year, more than any other lake’s.
Sanderlings, small shore birds, are counted on one beach every winter for ten years. The beach’s food and roosting places stay as they were throughout. The table below gives the counts.
Which of the following is the best estimate of the beach’s carrying capacity, K, for sanderlings?
- A. 140 sanderlings140 sanderlings is the first count, before the flock had grown.
- B. 432 sanderlings432 is the mean of all ten counts, and the early counts pull it down.
K is the level the later counts return to. - C. ✓ 520 sanderlings
- D. 560 sanderlings560 sanderlings is the highest count, in year 5.
The count fell back from it, so the beach could not support 560 for long.
Why: K is the level a count settles at or returns to.
The count climbed to 560 in year 5, then fell back.
From year 6 the counts wander near 520 and end on it.
So the best estimate of K is 520 sanderlings.
Blennies, small rock-pool fish, are stocked into four tanks of the same size with the same food each day. The table below gives the number stocked in each tank and the count a year later. A fifth tank of the same kind is then stocked with 345 blennies.
Predict what the fifth tank’s count does over the following year.
- A. ✓ Falls toward 138 blennies
- B. Stays near 345 blenniesTank W was stocked above 138 and fell to 138.
A count above K falls toward K. - C. Rises above 345 blenniesAbove K each blenny’s share of food is too small, so deaths outnumber births.
The count falls. - D. Falls to 46 blenniesTank J was stocked with 46 and rose to 136.
The count moves toward 138 from either side, not to the smallest stocking.
Why: Every tank ends near 138 blennies, whether it was stocked below or above that number.
So 138 is the tanks’ K.
The fifth tank starts above K, so each blenny’s share of food is too small and deaths outnumber births.
The count falls toward 138.
Suppose whelks, sea snails, are counted on one shore: 1 300 whelks one spring and 1 040 whelks the next spring.
Which of the following is the percent change in the whelks’ count?
- A. −25 %−25 % divides the change by the final count, 1 040.
The change is divided by the starting count, 1 300. - B. ✓ −20 %
- C. −0.2 %−0.2 is the change divided by the starting count.
Multiplying by 100 gives the percent. - D. +20 %The count fell, so the percent change is negative.
Why: percent change = (final − initial) / initial × 100.
percent change = (1 040 − 1 300) / 1 300 × 100.
percent change = −260 / 1 300 × 100 = −20 %.
Suppose the insects along a stream feed no more than 330 wagtails: K is 330 wagtails. A flood scours away much of the streambed, and the insects left feed no more than 220 wagtails. The wagtails stand at 135.
Predict what the wagtails’ count does over the following years.
- A. Falls below 135 wagtailsK fell, but the count, 135, is still below the new K, 220.
Below K births outnumber deaths, and the count rises. - B. Stays at 135 wagtailsA count stays steady only at K.
135 wagtails is below the new K, 220. - C. Rises toward the old K, 330 wagtailsThe insects that fed 330 wagtails are gone.
The insects left feed 220, the new K. - D. ✓ Rises toward the new K, 220 wagtails
Why: K is set by the stream’s insects, so K fell from 330 to 220 wagtails.
The count, 135, is below the new K.
So the room left is positive, and births outnumber deaths.
The count rises toward 220.
Vendace, small lake fish, are eaten by one kind of large predatory fish in their lake, and their count has stayed near the lake’s K for years. Anglers then remove every one of the predatory fish. A student says: “With the predators gone, nothing limits the vendace, so their count will climb for as long as the lake exists.”
Which of the following is the best evaluation of the student’s claim?
- A. The claim is correct: with the predators gone, nothing holds the vendace backThe predators were one limit.
The lake’s food and space still limit the vendace once the count has climbed. - B. The claim is wrong: the vendace will stay at their old count, because K belongs to the vendaceK belongs to the lake, for vendace, and the predators were part of what held the count down.
With the predators gone, K rises. - C. ✓ The claim is wrong: the count climbs, then levels off at a new, higher K set by the lake’s food and space
- D. The claim is correct: the count climbs for ever, because a lake’s food is never used upAs the vendace crowd the lake, each fish’s share of food shrinks.
Births fall and deaths rise until they are equal.
Why: The predators were a density-dependent factor, and removing them eases one limit.
The count climbs.
As the vendace crowd the lake, each fish’s share of food and space shrinks, so births fall and deaths rise until they are equal.
So the count levels off at a new, higher K.
A fast-spreading floating weed covers the surface of a shallow lake each spring. Three kinds of native water plant grow in the lake too: water crowfoot, bur-reed and marsh marigold. The upper table below gives the months in which each kind of plant puts up new shoots. Botanists net off two plots of the lake shore of the same size. They leave Plot J untouched. In Plot M they skim the weed off the water every week from March to September. At the end of the summer they count every plant in each plot; the counts are in the lower table. The botanists skim Plot M again the next summer, and at the end of that summer Plot M holds more native plants than it did at the end of the first.
(a) Describe what limits the native water plants in Plot J. (1 pt)
Light and surface space are resources the plants compete for, so the more plants share the plot, the smaller each plant’s share: a density-dependent factor limits the natives.
- Award 1 point for: the weed takes the light and surface space (or nutrients) before the natives, and the natives are held back by that shortage of a shared resource (accept: competition for light / space; a density-dependent factor).
Slip Saying the weed ‘kills’ the natives. The weed takes a resource; the natives are short of light and space.
(b) Predict how the count of native water plants in Plot M changes over the summers after that, if the skimming continues. (1 pt)
- Award 1 point for: the count rises and then levels off (accept: climbs to a new, higher carrying capacity; an S-shaped curve).
Slip Predicting a rise with no levelling off. Once the natives crowd the plot, light and space limit them again.
(c) In a third netted plot, Plot R, the botanists remove every plant that comes up before the end of May, and then leave the plot alone. Determine which of the four kinds of plant are present in Plot R by the end of the summer. (1 pt)
Every kind puts up shoots after the end of May, so every kind comes up once the removal stops.
- Award 1 point for: all four kinds, including the floating weed (accept the list alone; accept with the reason that every kind keeps putting up shoots after May).
Slip Leaving the weed out. The weed’s shoots come up until the end of September, so it comes back after the removal stops.
(d) In one April a late frost kills 30 % of the native plants in Plot M. A student says: “The frost has left Plot M’s carrying capacity for the native plants where it was.” Evaluate the student’s claim. (1 pt)
The frost killed plants, but it took no light and no surface space from the plot.
K is set by the plot’s light and space, so K is unchanged.
The count is now below K, so births outnumber deaths and the count climbs back toward it.
- Award 1 point for: the judgement (the claim is right) AND the ground (the frost took no resource from the plot — the light and surface space that set K are as they were — so K stays; the count falls for a season and climbs back).
Slip Judging the claim wrong because 30 % of the plants died. A hazard that leaves the plot’s resources moves the count, not K.
Cockles, small shellfish, live buried in the sand of an estuary. Every spring for twelve years an ecologist counts the cockles in one marked square of sand. The estuary’s food and sand stay as they were throughout. The counts are in the table below, and the graph plots them.
(a) Make a claim about the carrying capacity, K, of the square for cockles. Give K to the nearest 50 cockles. (1 pt)
- Award 1 point for: K is about 1 750 cockles (accept any value from 1 700 to 1 800 cockles, given to the nearest 50).
Slip Claiming the highest count, 1 890, as K. The count fell back from it, so the square could not support 1 890 for long.
(b) Support your claim with evidence from the data. (1 pt)
When the count was above 1 750, in years 6 and 10, it fell back; when it was below, in years 8 and 9, it rose again.
The count keeps returning to about 1 750, so that level is K.
- Award 1 point for: the count returns to about 1 750 in the later years (years 7 to 11 lie within 100 of it), falling back when above it and rising when below it (accept: quoted counts from the later years).
Slip Quoting only the rise of the first years. The evidence for K is where the count settles or returns to, not how fast it climbed.
(c) Explain why the count fell between year 6 and year 7. (1 pt)
So each cockle’s share of the square’s food was too small.
Cockles short of food produced fewer young and more of them died, so deaths outnumbered births and the count fell.
- Award 1 point for: the count was above K, so each cockle’s share of food was too small and deaths outnumbered births (accept: the braking term was negative above K, so the count fell back toward K).
Slip Saying a hazard struck in year 6. The stimulus says the estuary stayed as it was; the count fell because it had climbed past K.
(d) The cockles’ maximum per capita growth rate, r<sub>max</sub>, is 0.7 per year. Using your value of K from part (a), calculate dN/dt for the cockles in year 0. (1 pt)
Answer: 231 cockles per year (tolerance ±5)
- Award 1 point for: about 231 cockles per year, with the unit (a K of 1 750 gives 231; accept any value from 228 to 233 cockles per year worked from the K claimed in part (a) — every K from 1 700 to 1 800 gives a value in that range).
Slip Leaving the braking term out, 0.7 × 440 = 308. The logistic equation multiplies r<sub>max</sub> N by the room left as a fraction of K.
APBIO-U08-L38 Who lives here, and how many of each
Imagine two rocky islands, each surveyed for a day. Each island has the same four species of shore animal.
On island A the counts are 25, 25, 25 and 25. On island B they are 70, 10, 10 and 10. Island A and island B each hold four species and 100 animals. Is one island more varied than the other, and how would you say so?
Unit 8 · Ecology
1A community: every species living in one place
An oak forest holds oaks, deer, beetles and fungi. All the living things of every species in the forest make one group.
Which level is that group?
- A. A populationA population is all the organisms of one species living in one place: the oaks alone.
- B. ✓ A community
- C. An ecosystemAn ecosystem is the community together with its non-living surroundings: the soil, the rain and the sunlight.
Why: The group holds every species in the forest: the oaks, the deer, the beetles and the fungi.
All the populations of every species living together in one place are a community.
How do ecologists describe a community, and what makes one community more diverse than another?
First they list which species are present and how many of each: the community’s species composition.
Then they ask two questions of the list.
How many species are there? That count is the species richness.
How evenly are the animals shared among the species? That share is the species evenness.
Island A and island B have the same species richness: four species each.
Island A shares its animals evenly, but one species dominates island B, so island A is the more diverse.
Diversity needs both richness and evenness.
Video: Watch: One shore, three groups
The shore of island A at low tide. First the flat winkles alone are ringed. Then every animal on the shore is ringed together. Then the rock, the seawater and the sunlight are added inside the ring. Each ring is named as it appears.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L38a.mp4
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Imagine the shore of island A at low tide. Flat winkles, velvet crabs, cushion stars and sea hares live among its rocks.
The flat winkles of the shore are one population, because they are one species living in one place.
The sea hares of the shore are one population too, because they are one species living in one place.
Now take every species on the shore together: the flat winkles, the velvet crabs, the cushion stars and the sea hares. Together they are the shore’s community, because they are every species living in one place.
Now add the rock the animals live on, the seawater and the sunlight. Rock, water and light are not living things.
So that larger group is not the community. The larger group is the shore’s ecosystem.
A population, a community and an ecosystem mean the same on this shore as in the oak forest.
Suppose a storm kills many of the shore’s sea hares one winter.
The velvet crabs scavenge the dead sea hares. So more velvet crabs are fed that winter, and more of them survive to breed.
By the next summer the shore holds fewer sea hares and more velvet crabs than before.
The shore’s species interact. What happens to one population changes the others.
That interaction has changed the shore’s community over time.
So a community is the interacting populations of different species living in one place, and it changes over time as those interactions play out.
The table below sorts the three groups on the shore: the flat winkles alone, every species together, and every species with the rock, the seawater and the sunlight.
A community means living things only. Add the rock, the seawater and the sunlight, and the group is the ecosystem.
What you are expected to know Decide whether a described group is one population, a community or an ecosystem.
All the house martins nesting under the eaves of one farmhouse.
Which level is that group?
- A. ✓ A population
- B. A communityA community is every species living in one place.
The house martins are one species. - C. An ecosystemAn ecosystem includes the non-living surroundings.
The house martins are living things of one species.
Why: The house martins are one species.
They all live in one place, the farmhouse’s eaves.
All the organisms of one species living in one place are a population.
The fish, insects, plants and bacteria of one flooded quarry, together with its water, its rock and the sunlight falling on it.
Which level is that group?
- A. A populationA population is one species in one place.
The group holds many species, and water, rock and light besides. - B. A communityA community is living things only.
The group includes the water, the rock and the sunlight. - C. ✓ An ecosystem
Why: The group holds every species in the quarry.
The group also holds the water, the rock and the sunlight, which are not living things.
The community together with its non-living surroundings is an ecosystem.
Every species living on one chalk hill: its grasses, its flowers, its insects and its birds.
Which level is that group?
- A. A populationA population is one species in one place.
The group holds grasses, flowers, insects and birds: many species. - B. ✓ A community
- C. An ecosystemAn ecosystem includes the non-living surroundings.
The group holds living things only.
Why: The group holds every species living on the hill.
The group holds living things only: no soil, no rain, no sunlight.
All the populations of every species living together in one place are a community.
All the corn buntings singing along one field margin.
Which level is that group?
- A. ✓ A population
- B. A communityA community is every species living in one place.
The corn buntings are one species. - C. An ecosystemAn ecosystem includes the non-living surroundings.
The corn buntings are living things of one species.
Why: The corn buntings are one species.
They all live in one place, the field margin.
All the organisms of one species living in one place are a population.
Every species living in one flooded quarry: its fish, its insects, its plants and its bacteria.
Which level is that group?
- A. A populationA population is one species in one place.
The group holds fish, insects, plants and bacteria: many species. - B. ✓ A community
- C. An ecosystemAn ecosystem includes the non-living surroundings.
The group holds living things only: the quarry’s water and rock are not in it.
Why: The group holds every species living in the quarry.
The group holds living things only.
All the populations of every species living together in one place are a community.
33Who lives here, and how many of each
Video: Watch: The counts become bars
The counts for island A and island B are written into one table, species by species. Then each count becomes a bar: four bars for island A, four bars for island B, side by side on the same scale.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L38b.mp4
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Imagine the surveyors on each island counting every shore animal they find in one day.
The table below gives the counts: the four species of shore animal, and the count of each species on island A and on island B.
Island A holds 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares. Island B holds 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares.
A list of which species are present in a community, and how many there are of each, is called the community’s .
Island A and island B have the same four species present. But the count of each species differs.
So the species compositions of island A and island B differ.
The two bar charts below draw the same counts: the species on the x-axis, the count of animals on the y-axis, one bar per species, and a gridline every 10 animals.
On island A every bar reaches the 25 gridline. On island B one bar reaches 70 and three bars reach 10.
The bars make the difference easy to see: island B’s flat winkles tower over its three other species.
What you are expected to know Describe a community’s species composition from a survey table: which species are present, and how many of each.
Suppose a student counts the birds that visit one garden’s bird table in an hour. The table below gives the count of each species.
Which species is the most numerous at the bird table?
- A. Collared dovesThe collared dove count is 6.
- B. SiskinsThe siskin count is 9.
- C. ✓ Tree sparrows
- D. WoodpigeonsThe woodpigeon count is 3, the smallest count in the table.
Why: The table gives 18 tree sparrows, 9 siskins, 6 collared doves and 3 woodpigeons.
The largest count is 18.
So the tree sparrows are the most numerous species.
Suppose a student nets the insects of one dew pond for an hour. The table below gives the count of each species.
Which of the following is the species composition of the dew pond’s insects?
- A. A total of 50 insects counted across the whole pondA total names no species and no count of each.
The species composition lists which species are present and how many of each. - B. Three species of insect counted across the whole pondA count of species names neither the species nor the count of each.
- C. ✓ 30 backswimmers, 12 whirligig beetles and 8 water scorpions
Why: The species composition is which species are present and how many of each.
The table lists three species, each with its count.
So the species composition is 30 backswimmers, 12 whirligig beetles and 8 water scorpions.
Suppose a student walks one woodland glade at noon and counts every butterfly seen. The table below gives the count of each species.
(a) Describe the species composition of the glade’s butterflies. (2 pt)
The counts are 27 ringlets, 14 brimstones, 9 orange-tips and 5 small coppers.
- Award 1 point for: naming the species present (all four: ringlets, brimstones, orange-tips, small coppers).
- Award 1 point for: the count of each species (27, 14, 9 and 5, each with its species).
48Quick quiz: species composition mixed practice
Suppose a student counts the plants in flower on one meadow plot in spring, and again on the same plot in late summer. The table below gives both counts.
(a) Describe how the species composition of the plot’s flowering plants changes from spring to late summer. (2 pt)
The counts change too: self-heal rises from 8 plants to 35, and cat’s-ear falls from 20 plants to 18.
- Award 1 point for: a change in which species are present (yellow rattle gone, or lady’s bedstraw appearing).
- Award 1 point for: a change in the count of a species present at both times (self-heal 8 to 35, or cat’s-ear 20 to 18).
Which of the following is a community’s species composition?
- A. How many individuals the survey counted in allA total count names no species.
- B. ✓ The species present, and how many there are of each
- C. How the counts changed between two surveysA change between two surveys compares two compositions; the composition itself is one survey’s list.
Why: The species composition is the list a survey gives.
The list says which species are present and how many there are of each.
51How many kinds, and how evenly shared
Video: Watch: Two questions of the list
Island A’s four level bars beside island B’s one tall bar and three short bars. The bars in each chart are counted. Then the heights in each chart are compared.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L38c.mp4
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Take the species compositions of island A and island B again. Ask two questions of each.
First: how many species are there? Island A holds four species, and island B holds four species.
The number of species in a community is called its .
Island A and island B have the same species richness: four.
Second: how equally are the animals shared among the species?
On island A each species holds 25 of the 100 animals: an equal share. On island B one species holds 70 of the 100 animals, and each of the other three holds 10: a lopsided share.
How equally a community’s individuals are shared among its species is called its .
Island A shares its animals equally, so island A has the greater species evenness. Island B’s shares are lopsided, so island B has the smaller species evenness.
Island A’s and island B’s bar charts return below.
Level bars mean equal shares and a greater species evenness. One tall bar over three short bars means lopsided shares and a smaller species evenness.
Now suppose a cove holds four species of shore animal: 97 flat winkles, 1 velvet crab, 1 cushion star and 1 sea hare.
The cove’s species richness is four, the same as island A’s.
But 97 of the cove’s 100 animals are flat winkles. Nearly every animal in the cove is a flat winkle.
So species richness alone does not say how varied a community is. The cove’s shares are as lopsided as they could be, and that matters as much as its four species.
How varied a community is depends on two things: its species richness and its species evenness. That variety is called the community’s .
More species make a community more diverse. More equal shares make a community more diverse.
Island A and island B have the same species richness, four species each.
Island A has the greater species evenness. So island A has the greater species diversity.
What you are expected to know Distinguish species richness, the number of species, from species evenness, how equally the individuals are shared among the species.
What you are expected to know Compare two communities’ species diversity using both species richness and species evenness.
The table below gives the count of each species of shore animal in two coves.
Which cove has the greater species richness?
- A. ✓ Cove J
- B. Cove PCove P’s list has four counts, so Cove P holds four species.
Cove J’s list has six.
Why: Species richness is the number of species.
Cove J’s list has six counts, so Cove J holds six species.
Cove P’s list has four counts, so Cove P holds four species.
The greater species richness is Cove J’s.
The table below gives the count of each species of shore animal in two inlets.
Which inlet has the greater species evenness?
- A. ✓ Inlet R
- B. Inlet MInlet M’s counts are 50, 30, 15 and 5: one species holds half the animals.
Inlet R’s counts are almost equal.
Why: Species evenness is how equally the animals are shared among the species.
Inlet R’s counts are 26, 25, 25 and 24: almost equal shares.
Inlet M’s counts are 50, 30, 15 and 5: lopsided shares.
The greater species evenness is Inlet R’s.
The table below gives the count of each species of shore animal on two islets.
Which islet has the greater species richness?
- A. Islet JIslet J’s list has two counts, so Islet J holds two species.
Islet R’s list has three. - B. ✓ Islet R
Why: Species richness is the number of species.
Islet R’s list has three counts, so Islet R holds three species.
Islet J’s list has two counts, so Islet J holds two species.
The greater species richness is Islet R’s.
The table below gives the count of each species of shore animal on two spits of shingle.
Which spit has the greater species evenness?
- A. Spit PSpit P’s counts are 60, 20, 10, 5 and 5: one species holds most of the animals.
Spit M’s counts are almost equal. - B. ✓ Spit M
Why: Species evenness is how equally the animals are shared among the species.
Spit M’s counts are 22, 20, 20, 19 and 19: almost equal shares.
Spit P’s counts are 60, 20, 10, 5 and 5: lopsided shares.
The greater species evenness is Spit M’s.
The table below gives the count of each species of shore animal on two capes.
Which cape has the greater species diversity?
- A. ✓ Cape R
- B. Cape JCape J holds three species with lopsided counts, 60, 30 and 10.
Cape R holds five species with almost equal counts.
Why: Species diversity depends on richness and evenness.
Cape R holds five species; Cape J holds three.
Cape R’s counts, 24, 22, 20, 18 and 16, are almost equal; Cape J’s, 60, 30 and 10, are lopsided.
So Cape R, richer and more even, has the greater species diversity.
A student reads that island J holds six species of shore animal and island A holds four, and says: “Island J has six species, so it must be more diverse than island A.”
Is the student correct?
- A. ✓ No: how evenly island J’s animals are shared among its six species is not known
- B. Yes: island J holds six species and island A holds four, so island J is the more diverseSpecies diversity depends on evenness as well as richness.
Island J’s counts are not given, so its evenness is not known.
Why: Island J holds six species, so its species richness is greater than island A’s.
But species diversity depends on both richness and evenness.
Island J’s counts are not given, so how evenly its animals are shared is not known.
So the student cannot say island J is the more diverse.
Island A and island B return: 25, 25, 25 and 25 animals of the four species of shore animal on island A, and 70, 10, 10 and 10 on island B.
The table below compares island A with island B: the species richness, the species evenness, and which island is the more diverse.
Island A and island B hold four species each. So their species richness is the same.
Island A shares its animals evenly; one species dominates island B. So island A is the more diverse island.
Diversity needs both richness and evenness.
84Quick quiz: species richness, species evenness, species diversity mixed practice
Suppose surveyors count the birds singing in two woods on one spring morning, Ash Wood and Elm Wood. The table below gives the count of each species in each wood.
(a) State the species richness of Ash Wood. (1 pt)
Answer: 4 species (tolerance ±0)
- Award 1 point for: 4 (accept four species).
(b) Identify which of Ash Wood and Elm Wood has the greater species evenness. (1 pt)
- Award 1 point for: Ash Wood.
(c) Determine which wood’s bird community is the more diverse. (1 pt)
- Award 1 point for: Ash Wood.
(d) Justify your answer to part (c). (1 pt)
So their species richness is the same.
Ash Wood’s counts, 24, 22, 20 and 18, are almost equal; Elm Wood’s counts, 60, 12, 8 and 4, are lopsided.
So Ash Wood has the greater species evenness.
Species diversity depends on both richness and evenness.
With the same richness the more even community, Ash Wood’s, is the more diverse.
- Award 1 point for: the two woods have the same species richness (four species each), so with the same richness the more even wood, Ash Wood, is the more diverse.
Which of the following is a community’s species richness?
- A. The number of individuals in the communityThe number of individuals is the size of the whole community, not how many kinds it holds.
- B. ✓ The number of species in the community
- C. The count of the community’s most numerous speciesThe count of one species is part of the species composition, not the richness.
Why: Species richness is how many kinds of organism the community holds.
So it is the number of species in the community.
Which of the following is a community’s species evenness?
- A. How many individuals the community holds in allA total count does not show how the individuals are shared among the species.
- B. How many individuals the community’s largest species holdsOne species’ count alone does not show how the rest are shared.
- C. ✓ How equally the individuals are shared among the species
Why: Species evenness compares the species’ counts with one another.
Equal counts make the species evenness greater; one dominant species makes it smaller.
So species evenness is how equally the individuals are shared among the species.
Which of the following is a community’s species diversity?
- A. ✓ How many species there are, and how equally the individuals are shared among them
- B. The total count of individuals of every species added togetherA total count of individuals is neither richness nor evenness.
- C. How many kinds there areHow many kinds there are is the species richness.
Diversity depends on evenness too.
Why: Species diversity depends on species richness and on species evenness.
So it is how many species there are, and how equally the individuals are shared among them.
89Mixed practice: who lives here, and how many of each mixed practice
Suppose a scree slope, a bank of loose broken rock on a mountainside, holds mosses, ferns, springtails and ring ouzels. A student groups them together with the slope’s stones, its rain and the sunlight falling on it.
Which level is the student’s group?
- A. A populationA population is one species in one place.
The group holds many species, and stones, rain and light besides. - B. A communityA community is living things only.
The group includes the stones, the rain and the sunlight. - C. ✓ An ecosystem
Why: The group holds every species on the slope.
The group also holds the stones, the rain and the sunlight, which are not living things.
The community together with its non-living surroundings is an ecosystem.
Suppose a moth trap left out in a garden for one night catches the moths listed in the table below.
Which species is the most numerous in the trap?
- A. Garden tigersThe garden tiger count is 6, the smallest count in the table.
- B. ✓ Large yellow underwings
- C. Silver YsThe silver Y count is 17.
Why: The table gives 41 large yellow underwings, 17 silver Ys and 6 garden tigers.
The largest count is 41.
So the large yellow underwings are the most numerous species.
The table below gives the count of each species of flowering plant along two field margins.
Which margin has the greater species richness?
- A. ✓ Margin J
- B. Margin RMargin R’s list has two counts, so Margin R holds two species.
Margin J’s list has five.
Why: Species richness is the number of species.
Margin J’s list has five counts, so Margin J holds five species.
Margin R’s list has two counts, so Margin R holds two species.
The greater species richness is Margin J’s.
Suppose two fishing lakes on one estate, the north lake and the south lake, each hold five species of fish. In the north lake the fish are shared almost equally among the five species; in the south lake most of the fish are one species. A student says: “The north lake is the more diverse, because its fish are shared more equally among the same number of species.”
Is the student correct?
- A. No: the same count of species means the same species diversity, whatever the shares of the fishThe two lakes have the same species richness, but the north lake has the greater species evenness, so the north lake is the more diverse.
- B. ✓ Yes: the north lake’s fish are shared more equally among the same five species, so it is the more diverse
Why: The two lakes each hold five species: the same species richness.
The north lake’s fish are shared almost equally, so its species evenness is greater.
Species diversity depends on both richness and evenness.
With the same richness, the more even lake is the more diverse: the north lake.
Every species living on one fallen beech trunk: the beetles, the woodlice, the fungi and the bacteria.
Which level is that group?
- A. A populationA population is one species in one place.
The group holds beetles, woodlice, fungi and bacteria: many species. - B. ✓ A community
- C. An ecosystemAn ecosystem includes the non-living surroundings.
The group holds living things only.
Why: The group holds every species living on the trunk.
The group holds living things only.
All the populations of every species living together in one place are a community.
The table below gives the count of each species of shore animal on two cays, small sandy islands.
Which cay has the greater species evenness?
- A. ✓ Cay J
- B. Cay RCay R’s counts are 70, 20 and 5: one species holds most of the animals.
Cay J’s counts are almost equal.
Why: Species evenness is how equally the animals are shared among the species.
Cay J’s counts are 33, 32 and 30: almost equal shares.
Cay R’s counts are 70, 20 and 5: lopsided shares.
The greater species evenness is Cay J’s.
Suppose a gardener lifts every plant pot in one garden and counts the spiders beneath. The table below gives the count of each species.
Which of the following is the species composition of the spiders under the pots?
- A. A total of thirteen spiders under the potsA total names no species and no count of each.
- B. Two species of spider under the potsA count of species names neither the species nor the count of each.
- C. ✓ 9 garden spiders and 4 wolf spiders
Why: The species composition is which species are present and how many of each.
Two species are present, each with its count.
So the species composition is 9 garden spiders and 4 wolf spiders.
Suppose surveyors sweep a net through the grass of two unsown strips of farmland on one farm, the upper strip and the lower strip, for the same time in each. The table below gives the count of each species of insect caught in each strip.
(a) State the species richness of the upper strip. (1 pt)
Answer: 4 species (tolerance ±0)
- Award 1 point for: 4 (accept four species).
(b) Identify which of the two strips has the greater species evenness. (1 pt)
- Award 1 point for: the upper strip.
(c) Determine which of the two strips has the greater species diversity. (1 pt)
- Award 1 point for: the upper strip.
(d) Justify your answer to part (c). (1 pt)
So the upper strip has the greater species richness.
The upper strip’s counts, 22, 20, 18 and 16, are almost equal; the lower strip’s, 44 and 6, are lopsided.
So the upper strip has the greater species evenness.
The decision rests on both: the upper strip is richer and more even.
- Award 1 point for: the decision rests on both species richness (four species against two) and species evenness (almost equal counts against 44 and 6).
Glossary
- species composition
- The list of which species are present in a community and how many there are of each, as a survey table gives it: for example, 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares.
- species richness
- The number of species in a community. Two communities with four species each have the same species richness, whatever their counts.
- species evenness
- How equally a community’s individuals are shared among its species. Equal counts make the species evenness greater; one dominant species makes it smaller.
- species diversity
- How varied a community is. It depends on both species richness, the number of species, and species evenness, how equally the individuals are shared among them.
APBIO-U08-L38B Draw the shares as a pie
Go back to the two rocky islands. Each island holds the same four species of shore animal. Island A holds 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares. Island B holds 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares.
A bar chart shows the four counts side by side. It does not show at a glance what share of the whole each species holds. How do you draw the shares so that one species holding most of the island is visible in one look?
Unit 8 · Ecology
1From counts to slices of a circle
Suppose a class nets 60 animals from a stream, and 12 of them are freshwater shrimps.
Calculate the freshwater shrimps’ percentage of the 60 animals, to the nearest whole percent.
Answer: 20 % (tolerance ±0)
Suppose an ecologist counts four species of longhorn beetle in a coppice.
Which of the following fits these four counts?
- A. ✓ A bar chart, one bar per species
- B. Points joined one to the next, one point per speciesPoints joined one to the next fit a measured amount on the x-axis, such as a concentration.
Four species are not a measured amount.
Why: The four species are separate categories: no amount runs from one species to the next.
Separate categories take a bar chart, one bar per category.
How do you draw shares of one whole?
Turn each count into a fraction of the total, then into a percentage.
Then cut a circle into slices, one slice per species.
A species with 25 % of the animals gets a slice that is 25 % of the circle.
A species with 70 % of the animals gets a slice that is 70 % of the circle.
A label on each slice gives its species and its percentage.
A circle cut this way is a pie chart.
A pie chart shows shares of one total only, never a change over time.
Video: Watch: The bars become a pie
Island B’s four bars stand side by side. Each count becomes its percentage of the 100 animals. A circle is cut into four slices, and the flat winkles’ slice covers 70 % of it. Each slice is labeled with its species and its percentage. Island A’s four equal bars become four equal slices beside it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L38Ba.mp4
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Island B holds 100 animals in all. That total is the whole that every share is taken from.
One species’ share of the whole is its count divided by the total count. Multiplied by 100, the share is a percentage.
One species’ percentage of the total: its count divided by the total count, multiplied by 100.
Island B holds 70 flat winkles among its 100 animals. Calculate the flat winkles’ percentage of the total.
The table below works the same calculation for all four species: the species, its count, its count as a fraction of the total, and its percentage.
Every animal on the island is in exactly one species. So the four percentages together make 100 %.
Both islands hold exactly 100 animals. So on these islands, each count is already its percentage.
That is a special case. Most surveys do not count exactly 100 animals, and then the division matters.
Cut a circle into four equal slices, and each slice is 25 % of the circle.
Two of the equal slices together are half the circle, 50 %. Three of them together are 75 %.
So a slice of 70 % reaches almost to the 75 % mark. A slice of 10 % is less than half of one 25 % slice.
Island B is drawn below as a circle. The slices start at the top and go clockwise in the table’s order: flat winkles, then velvet crabs, cushion stars and sea hares.
The flat winkles’ slice covers 70 % of the circle. The other three slices cover 10 % each.
Each slice carries its species and its percentage, written beside it.
A circle cut into slices, one per species, each slice’s size matching its share of the total, is called a .
A pie chart shows shares of one total. Every slice is a part of the same 100 animals.
A pie chart cannot show a change over time. A count that changes over time is drawn as a line graph, with time on the x-axis.
Island A holds 25 of each species. So each species’ share is 25 %, and its slice is 25 % of the circle.
Island A is the special case: equal counts give equal slices, so no slice stands out.
The two pies sit side by side below. One look at them shows which island is dominated by one species.
The table below lists what a finished pie chart has, row by row, beside island B’s pie.
A pie that fails any row of the table is drawn wrongly.
A slice whose size does not match its percentage misreads the counts. So does a slice with the wrong label.
What you are expected to know Construct a pie chart of a community’s species composition from its counts: one slice per species, sized to its percentage of the total, labeled with its species and its percentage.
Suppose a survey of a railway bank counts 40 plants of three species: 22 ragwort, 12 scabious and 6 mullein.
One species’ percentage of the total: its count divided by the total count, multiplied by 100.
Calculate the ragwort’s percentage of the 40 plants, to the nearest whole percent.
Part 1. Calculate the ragwort’s share of the total as a decimal fraction, to two decimal places.
Answer: 0.55 (tolerance ±0)
Answer: 55 % (tolerance ±0)
Suppose a survey of a railway bank counts 40 plants of three species: 22 ragwort, 12 scabious and 6 mullein. The three pie charts below, J, K and L, each claim to show these counts.
Which pie chart shows the railway bank’s counts correctly?
- A. ✓ Pie J
- B. Pie KPie K’s three slices are equal.
22 of 40 plants is 55 % of the total, and 55 % of the plants takes 55 % of the circle. - C. Pie LPie L’s largest slice is labeled mullein 15 %.
A species with 15 % of the plants covers 15 % of the circle, not the largest slice.
Why: 22 of 40 plants is 55 %, 12 of 40 is 30 % and 6 of 40 is 15 %.
So the ragwort’s slice covers 55 % of the circle, the scabious’s 30 % and the mullein’s 15 %.
Only pie J’s slice sizes match its labels.
Suppose a survey of a cemetery counts 60 plants of four species: 30 tormentil, 18 milkwort, 9 eyebright and 3 gentian. The pie chart below shows the four species as slices M, N, P and R, with no other labels. The slices are drawn in a shuffled order, different from the order of the species in the list.
Which slice is the tormentil’s?
- A. Slice MSlice M covers 15 % of the circle.
The tormentil is 30 of the 60 plants: half of them. - B. Slice NSlice N covers 5 % of the circle.
The tormentil is 30 of the 60 plants: half of them. - C. ✓ Slice P
- D. Slice RSlice R covers 30 % of the circle.
The tormentil is 30 of the 60 plants: half of them, not 30 %.
Why: The tormentil is 30 of the 60 plants: half of them, 50 %.
Half of the plants takes half of the circle.
Slice P covers half of the circle.
Suppose a survey of a rockery counts 40 animals of three species: 26 rove beetles, 12 oil beetles and 2 harvestmen.
One species’ percentage of the total: its count divided by the total count, multiplied by 100.
Calculate the rove beetles’ percentage of the 40 animals, to the nearest whole percent.
Answer: 65 % (tolerance ±0)
Suppose a survey of a rockery counts 40 animals of three species: 26 rove beetles, 12 oil beetles and 2 harvestmen. The three pie charts below, V, W and X, each claim to show these counts.
Which pie chart shows the rockery’s counts correctly?
- A. Pie VPie V draws each count as if it were a percentage of 100.
The total is 40 animals, so 26 animals is more than half of them. - B. Pie WPie W’s largest slice is labeled oil beetles 30 %.
The largest slice belongs to the species with the largest share, the rove beetles. - C. ✓ Pie X
Why: 26 of 40 animals is 65 %, 12 of 40 is 30 % and 2 of 40 is 5 %.
So the rove beetles’ slice covers 65 % of the circle, the oil beetles’ 30 % and the harvestmen’s 5 %.
Only pie X’s slice sizes match its labels.
Suppose a student counts the stag beetles in a walled garden each summer for four years: 8, 12, 15 and 20. The student says: “I will draw the four counts as a pie chart, one slice per year.”
Is the student correct?
- A. ✓ No: the counts are one species over four years, so a line graph shows them
- B. Yes: four counts make four slices of one circle, so a pie chart shows themA pie chart shows shares of one total.
Four years’ counts of one species are a change over time, not shares of one whole.
Why: A pie chart cuts one total into shares.
The four counts are not shares of one whole: they are the same species counted in four different years.
A count that changes over time is drawn as a line graph.
Suppose a student counts 90 beetles of three species in a kitchen garden: 45 tiger beetles, 30 dung beetles and 15 burying beetles. The student says: “The tiger beetles get half of the circle.”
Is the student correct?
- A. No: 45 is less than 100, so the tiger beetles’ slice is less than half of the circleThe share is the count divided by the total count, 90 beetles, not by 100.
- B. ✓ Yes: 45 of the 90 beetles is half of them, so the tiger beetles’ slice is half of the circle
Why: The total is 90 beetles, not 100.
45 of the 90 is half of them, 50 %.
Half of the total takes half of the circle.
The table below lists three graphs for counts, and the data each one fits.
The two rocky islands return below as pies: island A with 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares.
Island B holds 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares.
Island A’s pie is four equal slices of 25 %.
Island B’s pie has one slice covering 70 % of the circle and three small slices of 10 %.
One look at the two pies shows that one species holds most of island B.
49Quick quiz: pie chart mixed practice
Suppose a naturalist traps, counts and releases bank voles in a timber yard on the first day of each month for a year.
Which of the following fits these twelve counts?
- A. A pie chartA pie chart shows shares of one total.
Twelve monthly counts of one species are a change over time. - B. ✓ A line graph
Why: The twelve counts are one species counted at twelve times.
A count that changes over time is drawn as a line graph, with time on the x-axis.
Suppose a naturalist counts 120 craneflies of five species on a lit window in one night.
Which of the following fits these five species’ counts?
- A. ✓ A pie chart
- B. A line graphA line graph shows a count changing over time.
These five counts are five parts of one night’s count, not a change over time.
Why: The five counts are five parts of one total, the 120 craneflies.
Shares of one total are drawn as a pie chart, one slice per species.
Suppose a diver counts 80 squat lobsters of four species on one breakwater.
Which of the following fits these four species’ counts?
- A. ✓ A pie chart
- B. A line graphA line graph shows a count changing over time.
These four counts are four parts of one total, not a change over time.
Why: The four counts are four parts of one total, the 80 squat lobsters.
Shares of one total are drawn as a pie chart, one slice per species.
Suppose a survey of a grass roundabout counts 50 plants of three species. The table below shows the counts.
(a) Calculate the vetch’s percentage of the 50 plants, to the nearest whole percent. (1 pt)
Answer: 42 % (tolerance ±0)
- Award 1 point for: 42 %.
(b) Describe the vetch’s slice in a pie chart of these counts: its size and its label. (2 pt)
The slice is labeled vetch, 42 %.
- Award 1 point for: the slice covers 42 % of the circle (accept: a little less than half of the circle).
- Award 1 point for: the slice is labeled with the species, vetch, and its percentage, 42 %.
Which of the following is a pie chart?
- A. A bar for each species whose height shows that species’ countA bar for each species is a bar chart.
- B. ✓ A circle cut into slices sized to each species’ share of the total
- C. Points showing one count against time joined one to the nextPoints of one count against time joined one to the next are a line graph.
Why: A pie chart is a circle cut into slices.
Each slice is one species, and its size is that species’ share of the total.
Glossary
- pie chart
- A circle cut into slices, one per species, each slice’s size matching that species’ share of the total, and each slice labeled with its species and its percentage. It shows shares of one total, never a change over time.
APBIO-U08-L39 One number for diversity: what the formula does
Imagine again the two rocky islands, each surveyed for a day, each with the same four species of shore animal.
On island A the counts are 25, 25, 25 and 25. On island B they are 70, 10, 10 and 10. You can say A is more even. An ecologist wants one number for each island, so that fifty islands can be ranked. What kind of number rises when animals are shared evenly, and sinks when one species takes most of them?
Unit 8 · Ecology
1One share, and its square
Suppose 80 birds visit a garden feeder in one morning, and 36 of them are one species.
Which of the following is that species’ fraction of the 80 birds, written as a decimal?
- A. 0.360.36 is 36 out of 100, and the feeder had 80 birds, not 100.
- B. ✓ 0.45
- C. 2.22.2 comes from dividing the total by the count: the division the wrong way up.
Why: A fraction of a total is the count divided by the total.
36 out of 80 is 0.45.
Suppose two flowerbeds each hold the same five species of insect. In the first flowerbed the five counts are nearly equal. In the second flowerbed one species holds most of the insects.
Which flowerbed has the greater species evenness?
- A. ✓ The first flowerbed
- B. The second flowerbedIn the second flowerbed one species holds most of the insects: the counts are far from equal.
Why: Species evenness is how equally a community’s individuals are shared among its species.
The first flowerbed’s five counts are nearly equal.
So the first flowerbed has the greater species evenness.
How can one number stand for how diverse a community is?
Start with one species’ share of the whole.
On island B the big species holds 70 of the 100 animals: a share of 0.7.
Square the share: a big share squared stays big, and a small share squared becomes tiny.
So squaring makes the big species stand out even more.
Add up every species’ squared share, and a community dominated by one species gives a big sum.
Take that sum away from 1, and the more even community comes out with the higher number.
The formula sheet writes that number as , and it is called Simpson’s Diversity Index.
Video: Watch: One share, and its square
Island B’s counts are on screen: 70, 10, 10, 10. The big species’ count is divided by 100: the share is 0.7. The share is multiplied by itself, and the squared share 0.49 appears. The three small species’ shares are squared beside it. The four squared shares stack into one bar, and island A’s four equal squared shares stack beside it. The lopsided island’s bar stands taller.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L39a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L39a.mp4
Suppose you count island B again: 100 animals, and the big species is 70 of them.
A species’ share of the island is its own count divided by the count of all the animals. The line below shows the big species’ share.
island B’s big species: 70 animals out of 100 is a share of 0.7
The count of one species is written n. The count of all the animals is written N.
So a species’ share is : the fraction of all the animals that belong to that species.
Now multiply the share by itself, as the line below shows.
the big species’ share squared: 0.7 multiplied by itself is 0.49
A share multiplied by itself is called the share squared.
Each of island B’s three small species holds 10 of the 100 animals: a share of 0.1. That share squared is 0.01, as the line below shows.
one of island B’s small species: a share of 0.1, squared, is 0.01
The big species’ share is seven times the small species’ share.
The big species’ squared share is forty-nine times the small species’ squared share.
So squaring makes a big share stand out even more, and shrinks a small share almost to nothing.
On island A each of the four species holds 25 of the 100 animals: a share of 0.25. That share squared is 0.0625, as the line below shows.
one of island A’s species: a share of 0.25, squared, is 0.0625
Now suppose one species on an island of 100 animals grows: 10 of them, then 25, then 40, then 70. The table below compares its share and its squared share at each count.
From the first row to the last, the share grows seven times over. The squared share grows forty-nine times over.
So as one species comes to dominate, its squared share grows faster than its share.
Every species on an island has a squared share. Add the squared shares all up, one term for every species.
For island B the four squared shares add up to 0.52, as the line below shows.
island B: the four squared shares add up to 0.52
For island A the four squared shares add up to 0.25, as the line below shows.
island A: the four squared shares add up to 0.25
The squared shares are stacked up below, one bar per island. The height of a bar is that island’s sum.
Island B’s one big species pushes its sum up to 0.52. Island A’s four equal species keep its sum down at 0.25.
So a community dominated by one species has a big sum of squared shares. A community shared evenly has a small sum.
What you are expected to know Work out one species’ share of a community, , and square it.
What you are expected to know Say what happens to the sum of squared shares as one species comes to hold most of the organisms.
Island B holds 100 animals, and its big species is 70 of them. Calculate that species’ share of the island, and the share squared.
Suppose a survey of a park lake counts 250 fish of three species, and one species is 190 of them.
Calculate that species’ squared share of the lake’s fish, to four decimal places.
Part 1. Calculate that species’ share of the lake’s fish, to two decimal places.
Answer: 0.76 (tolerance ±0)
Answer: 0.5776 (tolerance ±0)
Suppose a survey of a spinney, a small wood, counts 50 birds of four species, and one species is 46 of them.
Calculate that species’ squared share of the spinney’s birds, to four decimal places.
Answer: 0.8464 (tolerance ±0)
Suppose a survey of a motorway verge counts 600 plants of six species, and one species is 84 of them.
Calculate that species’ squared share of the verge’s plants, to four decimal places.
Answer: 0.0196 (tolerance ±0)
Suppose two quays each hold 100 insects of the same three species. On the first quay the counts are 80, 10 and 10. On the second quay they are 40, 30 and 30.
Before any calculation, which quay has the bigger sum of squared shares?
- A. ✓ The first quay
- B. The second quayThe second quay’s biggest share is 0.4, and a share of 0.4 squared is far smaller than a share of 0.8 squared.
Why: On the first quay one species holds 80 of the 100 insects: a big share.
A big share squared stays big.
So the first quay has the bigger sum of squared shares.
Suppose two canal basins each hold 180 fish of the same five species. In the first basin every species is 36 fish. In the second basin the counts are 108, 18, 18, 18 and 18.
Before any calculation, which basin has the bigger sum of squared shares?
- A. The first basinThe first basin’s shares are all 0.2, and a share of 0.2 squared is tiny beside a share of 0.6 squared.
- B. ✓ The second basin
Why: In the second basin one species holds 108 of the 180 fish: a share of 0.6, a big share.
A big share squared stays big.
So the second basin has the bigger sum of squared shares.
Suppose two olive groves each hold 70 wild plants of the same two species beneath the trees. In the first grove the counts are 35 and 35. In the second grove they are 63 and 7.
Before any calculation, which grove has the bigger sum of squared shares?
- A. The first groveThe first grove’s two shares are both 0.5, and neither is a big share.
- B. ✓ The second grove
Why: In the second grove one species holds 63 of the 70 plants: a share of 0.9, a very big share.
A very big share squared stays very big.
So the second grove has the bigger sum of squared shares.
Suppose two towpaths each hold 100 birds of the same three species. On the first towpath the counts are 45, 45 and 10. On the second towpath they are 40, 40 and 20.
Before any calculation, which towpath has the bigger sum of squared shares?
- A. ✓ The first towpath
- B. The second towpathThe second towpath’s counts are the closer to equal, so its shares are the more even.
Why: On the first towpath two species hold 45 birds each and one holds only 10: the birds are shared less evenly.
Less even shares give a bigger sum of squared shares.
So the first towpath has the bigger sum.
Suppose a loch holds 240 fish of three species: 120, 60 and 60. Suppose a lochan, a small loch nearby, holds 60 fish of the same three species: 36, 12 and 12.
Which of the two has the bigger sum of squared shares?
- A. The lochThe loch’s biggest species is 120 of 240 fish: a share of 0.5, smaller than the lochan’s biggest share.
- B. ✓ The lochan
Why: A share is a species’ count divided by the count of all the fish, not the count alone.
The lochan’s biggest share is 36 of 60 fish: 0.6.
The loch’s is 120 of 240 fish: 0.5.
The bigger share gives the bigger sum.
44The formula sheet’s line
Suppose two species of insect live in a window box. One species has a share of 0.35 of the insects, and the other has a share of 0.65.
Which species has the larger squared share?
- A. The species with a share of 0.35Squaring a share smaller than 1 makes it smaller still, and 0.35 is the smaller share to start with.
- B. ✓ The species with a share of 0.65
Why: Squaring a bigger share gives a bigger square.
0.65 is the bigger share.
So the species with a share of 0.65 has the larger squared share.
Island A’s and island B’s sums return below: 0.25 for island A and 0.52 for island B. The lopsided island has the bigger sum.
An ecologist wants the number to come out higher for the more diverse island. So the ecologist takes the sum away from 1.
Video: Watch: The formula sheet’s line
The two stacked bars are on screen, with the sums 0.25 and 0.52 above them. Each sum is taken away from 1, and the more even island comes out higher. The formula sheet’s line appears, typeset, and each symbol is named beneath it: n, N and the sigma sign.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L39b.mp4
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Island A’s sum is smaller than island B’s. So 1 minus the sum is bigger for island A, and the more even island comes out with the higher number.
The formula sheet prints the whole recipe on one line, exactly as below. Beneath it is what each symbol means.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
n is the count of one species, and N is the count of all the organisms in the community.
So is that species’ share of the community.
The sigma sign, , says: add up one term for every species.
So the line says: square every species’ share, add the squares up, and take the sum away from 1.
A number worked out this way is called .
Simpson’s Diversity Index has no unit: it is a plain number.
The more evenly a community’s organisms are shared among its species, the higher the number comes out.
Now suppose every animal on an island belongs to one species. That species holds all the animals, so its share is 1.
Squaring that share of 1 changes nothing. There is no other species to add, so the sum is 1.
So an island of one species scores 0, the lowest value, as the line below shows.
one species only: its share is 1, its squared share is 1, the sum is 1, and the index is 0
Now suppose an island holds many species, all with equal counts. Each share is small.
So each squared share is tiny. The sum of those tiny squares is close to 0.
So the index is close to 1. The more species there are, the closer to 1 the index comes.
But every species has a share above 0. So the sum is never 0.
Simpson’s Diversity Index climbs toward 1 and never reaches it.
The table below compares three ways of sharing a community’s organisms: the sum of squared shares each gives, and the index.
An island of one species scores 0. An island where one species holds most scores low. An island shared evenly among many species scores close to 1.
What you are expected to know Read Simpson’s Diversity Index in its formula-sheet form: what n, N and the sigma sign stand for.
What you are expected to know Say what the index does: 0 for one species, climbing toward 1 for many equal species without reaching it.
The formula sheet prints Simpson’s Diversity Index as 1 minus a sum.
Which symbol in that line tells you to add up one term for every species?
- A. nn is the count of one particular species, a number to divide, not an instruction to add.
- B. NN is the count of all the organisms, the number every species’ count is divided by.
- C. ✓
Why: The sigma sign is the instruction to add up.
In this line it adds up one squared share for every species.
The formula sheet prints Simpson’s Diversity Index as 1 minus a sum.
Which symbol in that line stands for the total number of organisms of one particular species?
- A. ✓ n
- B. NN is the total number of organisms of all species, not of one species.
- C. The sigma sign is an instruction to add up, not a count.
Why: The sheet’s key reads: n is the total number of organisms of a particular species.
So one species’ count is n.
The formula sheet prints Simpson’s Diversity Index as 1 minus a sum.
Which symbol in that line stands for the total number of organisms of all species?
- A. nn is the count of one particular species only.
- B. ✓ N
- C. The sigma sign is an instruction to add up, not a count.
Why: The sheet’s key reads: N is the total number of organisms of all species.
So the count of every organism in the community is N.
Suppose two playing fields each hold 100 insects of the same four species. On the first field the counts are 28, 24, 24 and 24. On the second field they are 55, 15, 15 and 15.
Before any calculation, which field has the higher value of Simpson’s Diversity Index?
- A. ✓ The first field
- B. The second fieldOn the second field one species holds 55 of the 100 insects, a big share, so its sum of squared shares is the bigger.
Why: The first field’s counts are the closer to equal, so no species has a big share.
Even shares give a small sum of squared shares.
A small sum taken away from 1 leaves a higher index.
So the first field has the higher value.
A student counts 60 insects in a car park and finds that every one of them belongs to the same species. The student says: “Every insect is the same species, so Simpson’s Diversity Index for this car park is 1, the highest value it can take.”
Is the student correct?
- A. ✓ No: the one species has a share of 1, so the sum of squared shares is 1 and the index is 0, the lowest value
- B. Yes: every insect belongs to the one species, so the car park scores 1, the highest value the index can takeWith one species the share is 1, so the sum of squared shares is the whole of 1.
Nothing is left when the whole is taken away: the lowest value.
Why: The one species holds all 60 insects, so its share is 1.
Squaring that share changes nothing, and there is no other species to add, so the sum is 1.
Taking the whole sum away from 1 leaves nothing.
So the car park scores 0, the lowest value, not 1.
A student compares two gravel pits, flooded pits where gravel was once dug, that hold the same three species of water bird. In the first gravel pit the counts are 50, 25 and 25. In the second gravel pit they are 34, 33 and 33. The student says: “The second gravel pit scores the higher Simpson’s Diversity Index. Its birds are shared more evenly among the three species, so its sum of squared shares is the smaller.”
Is the student correct?
- A. No: the first gravel pit scores the higher index, since its biggest species holds the bigger shareA bigger share squared gives a bigger sum of squared shares, and a bigger sum taken away from 1 leaves a lower index.
- B. ✓ Yes: the second gravel pit’s more even shares give the smaller sum of squared shares, so its index is the higher
Why: In the second gravel pit the three shares are nearly equal.
So no share is big.
Even shares give a small sum of squared shares.
A small sum taken away from 1 leaves a higher index.
So the second gravel pit scores the higher index.
Suppose two jetties each hold 200 animals of four species. On the first jetty the biggest species is 180 of the 200 animals. On the second jetty the biggest species is 120 of the 200 animals.
Before any calculation, which jetty has the lower value of Simpson’s Diversity Index?
- A. ✓ The first jetty
- B. The second jettyOn the second jetty the biggest share is 0.6, smaller than the first jetty’s biggest share of 0.9, so its sum of squared shares is the smaller.
Why: On the first jetty one species holds 180 of 200 animals: a share of 0.9.
A share of 0.9 is most of the way to 1, so its square is big.
A big sum taken away from 1 leaves a low index.
So the first jetty has the lower value.
The two rocky islands return once more below, each with the same four species of shore animal.
Island A holds 25, 25, 25 and 25 of them. Island B holds 70, 10, 10 and 10.
On island B one species holds 70 of the 100 animals. That species’ squared share is 0.49 on its own.
So island B’s sum is at least 0.49. Island B’s Simpson’s Diversity Index is at most 0.51, as the line below shows.
island B’s big species alone puts 0.49 into the sum, so B’s index is at most 0.51
That is why one dominant species sinks the number, and why the even island comes out higher.
82Quick quiz: Simpson’s Diversity Index mixed practice
Ecologists count every organism in a community, species by species, and work out Simpson’s Diversity Index from the counts.
(a) State what happens to the value of Simpson’s Diversity Index as one species comes to hold more and more of the community’s organisms. (1 pt)
- Award 1 point for: the index falls (accept: gets lower, sinks toward 0).
(b) Explain why the index changes that way. (2 pt)
Its squared share grows faster than its share.
So the sum of the squared shares grows.
The index is 1 minus that sum.
So the index falls.
- Award 1 point for: the species’ share grows, and its squared share grows with it (accept: a bigger share squared is bigger), so the sum of squared shares grows.
- Award 1 point for: the index is 1 minus the sum of squared shares, so a bigger sum leaves a smaller index.
Which of the following is Simpson’s Diversity Index?
- A. The number of different species counted in the communityThe number of species is the community’s species richness, one input to diversity, not the index.
- B. The share of the community held by its commonest speciesThe commonest species’ share is one term in the working, not the whole index.
- C. ✓ 1 minus the sum of every species’ squared share of the community
Why: The formula sheet writes the index as 1 minus a sum.
The sum adds up every species’ share squared.
So Simpson’s Diversity Index is 1 minus the sum of every species’ squared share.
Glossary
- Simpson’s Diversity Index
- The number the formula sheet writes as Diversity Index = 1 minus the sum, over every species, of (n/N) squared, where n is the count of one species and N is the count of all the organisms. A plain number with no unit: 0 for a community of one species, climbing toward 1 (never reaching it) as organisms are shared more evenly among more species.
APBIO-U08-L39B Calculate the index, then say what it means
Go back to the two rocky islands, each with the same four species of shore animal. Island B first: 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares, 100 animals in all.
Island A holds 25 of each of the four species. What does Simpson’s Diversity Index give for island B, and what does it give for island A? And what does the gap between the two numbers say?
Unit 8 · Ecology
1Calculate the index
The formula sheet prints Simpson’s Diversity Index as 1 minus a sum, with one term for every species.
Which of the following does that sum add up?
- A. Every species’ count, nThe counts add up to N, the count of all the organisms, which is a number of animals, not a term of the index.
- B. Every species’ share, The shares of every species add up to 1, whatever the counts are, so a sum of shares could never tell two communities apart.
- C. ✓ Every species’ squared share,
Why: The formula sheet’s line squares each species’ share.
The sigma sign adds up one squared share for every species.
So the sum adds up every species’ squared share.
Suppose two roof gardens each hold the same three species of insect. On the first roof garden the squared shares add up to 0.30. On the second roof garden they add up to 0.45.
Which roof garden has the higher Simpson’s Diversity Index?
- A. ✓ The first roof garden
- B. The second roof gardenThe second roof garden’s sum of squared shares is the bigger, and a bigger sum taken away from 1 leaves a smaller index.
Why: Simpson’s Diversity Index is 1 minus the sum of squared shares.
The first roof garden’s sum is the smaller.
A smaller sum taken away from 1 leaves a higher index.
So the first roof garden has the higher index.
How do you calculate Simpson’s Diversity Index, and how do you read the answer?
Take one species at a time: write down its share of the community, then square the share.
Add up the squared shares, and take the sum away from 1.
Island B’s squared shares add up to 0.52. So island B’s index is 0.48.
Island A’s squared shares add up to 0.25. So island A’s index is 0.75.
Both islands hold four species. So island A scores higher only because its animals are shared more evenly.
A higher value of Simpson’s Diversity Index means more diverse, never better.
Video: Watch: Calculate the index
Island B’s counts are on screen: 70, 10, 10 and 10 out of 100. Each species’ share is squared on its own line. The four squared shares are added up, and the sum is taken away from 1: 0.48. Then island A’s four equal shares are worked the same way, line by line, to 0.75.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L39Ba.mp4
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The table below gives the two islands’ survey counts: the same four species of shore animal, and 100 animals on each island.
Each island’s index is worked from its own column, one species per line.
The formula sheet prints Simpson’s Diversity Index on one line, below. Every calculation on this page starts from that line.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Round as you go: each share to two decimal places, each squared share to four, and the index to two.
Island B is worked first.
Island B’s four shares are different. So its four lines of working are different.
Island B holds 100 animals: 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares. Calculate Simpson’s Diversity Index for island B.
Island B’s Simpson’s Diversity Index is 0.48. The index is a plain number with no unit.
Island A is the special case.
Island A’s four shares are all 0.25. So its four lines of working are all the same line.
Island A holds 100 animals: 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares. Calculate Simpson’s Diversity Index for island A.
Island A’s sum of squared shares is 0.25, the same number as each of its shares. The sum equals each of the shares only when every share is equal.
Do not expect that match on other communities.
Four steps give Simpson’s Diversity Index.
1 Write down each species’ count, n, and the total, N.
2 Square each species’ share, one line each.
3 Add the squared shares up.
4 Take the sum away from 1.
Write the index to two decimal places.
What you are expected to know Calculate Simpson’s Diversity Index from a community’s counts: one squared share per line, the sum, then 1 minus the sum, to two decimal places.
Suppose a survey of a slipway counts 20 animals of four species: 9, 7, 3 and 1.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate Simpson’s Diversity Index for the slipway, to two decimal places.
Part 1. Calculate the squared share of the species with 9 animals, to four decimal places.
Answer: 0.2025 (tolerance ±0)
Part 2. Add up the squared shares of all four species, to two decimal places.
Answer: 0.35 (tolerance ±0)
Answer: 0.65 (tolerance ±0)
Suppose a survey of a village pond counts 30 animals of three species: 15, 9 and 6.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate Simpson’s Diversity Index for the village pond, to two decimal places.
Answer: 0.62 (tolerance ±0)
Suppose a survey of a gravel drive counts 25 plants of two species: 20 and 5.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate Simpson’s Diversity Index for the gravel drive, to two decimal places.
Answer: 0.32 (tolerance ±0)
31Quick quiz: a share, a squared share, 1 minus the sum mixed practice
Suppose a survey of a pontoon counts 80 animals, and 44 of them are one species.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate that species’ share of the pontoon’s animals, to two decimal places.
Answer: 0.55 (tolerance ±0)
Suppose one species of insect holds a share of 0.95 of the insects in a polytunnel.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate that species’ squared share, to four decimal places.
Answer: 0.9025 (tolerance ±0)
Suppose the squared shares of a sandbank’s animals add up to 0.31.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate Simpson’s Diversity Index for the sandbank, to two decimal places.
Answer: 0.69 (tolerance ±0)
Suppose a survey of a barn roof counts 40 spiders, and 34 of them are one species.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate that species’ share of the barn roof’s spiders, to two decimal places.
Answer: 0.85 (tolerance ±0)
Suppose one species holds a share of 0.75 of the animals on a groyne, a low barrier of stone built out from a beach.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate that species’ squared share, to four decimal places.
Answer: 0.5625 (tolerance ±0)
37What the gap between two indices says
Suppose the edge of one footpath holds three species of plant, and the edge of a second footpath holds six species of plant.
Which footpath has the greater species richness?
- A. The first footpathThe first footpath holds three species, and three is the smaller number of species.
- B. ✓ The second footpath
Why: Species richness is the number of species in a community.
The second footpath holds six species, and the first holds three.
So the second footpath has the greater species richness.
Suppose two bowling greens each hold the same four species of worm. On the first bowling green one species holds most of the worms. On the second bowling green the four counts are nearly equal.
Which bowling green has the greater species evenness?
- A. The first bowling greenOn the first bowling green one species holds most of the worms: the counts are far from equal.
- B. ✓ The second bowling green
Why: Species evenness is how equally the organisms are shared among the species.
The second bowling green’s four counts are nearly equal.
So the second bowling green has the greater species evenness.
The two rocky islands return. Island A holds 25 of each of its four species of shore animal, and island B holds 70, 10, 10 and 10.
Island B’s Simpson’s Diversity Index is 0.48, and island A’s is 0.75. What does the gap between the two numbers say about the two islands?
The two islands’ squared shares are stacked up again below, one bar per island. A dashed box marks the gap from the top of each stack up to 1.
The index is that gap: take the sum of squared shares away from 1, and the gap is what is left.
Island B’s stack reaches 0.52, and the gap above it is 0.48. Island A’s stack reaches only 0.25, and the gap above it is 0.75.
Island A’s animals are shared out more equally among its four species than island B’s are. Island A’s higher value says exactly that: its animals are shared out more equally.
Both islands hold the same four species. So the two values differ only because the shares differ.
Island A has the greater species evenness. The two islands have the same species richness.
Now suppose a shore holds 100 animals of five species, 20 of each. Every species holds an equal share, just like on island A.
The line below works the shore’s index: 0.80.
the shore: five species with a share of 0.2 each, so five squared shares of 0.04; the sum is 0.20 and the index is 0.80
The shore scores higher than island A: 0.80 against 0.75.
The shares are equally even on the shore and on island A. So the shore’s higher value comes from its greater species richness: five species against four.
So a higher value of Simpson’s Diversity Index can mean more equally shared organisms, or more species, or both.
Simpson’s Diversity Index measures how varied a community is: how many species it holds, and how equally its organisms are shared among them.
The index does not say whether a community is healthy, or good, or better than another community.
Island B’s low value says that one species dominates island B. It does not say that island B is worse off.
The index is not a percentage of diversity either. Island A’s 0.75 does not mean island A is 75 % diverse.
The table below compares the two islands: species richness, species evenness and Simpson’s Diversity Index.
What you are expected to know Compare two communities by Simpson’s Diversity Index and state, in plain words, what the higher value means for species richness or species evenness.
Suppose two shingle banks hold the same three species of plant. The first shingle bank’s Simpson’s Diversity Index is 0.63, and the second shingle bank’s is 0.41.
Which of the following does the higher value say about the first shingle bank?
- A. More species of plant grow on the first shingle bank than on the secondBoth shingle banks hold the same three species, so neither has more species than the other.
- B. One species dominates the first shingle bank’s plants and holds most of themOne dominant species has a big squared share, so the sum is big and the value is low, and the first shingle bank has the higher value.
- C. ✓ The first shingle bank’s plants are shared more equally among the three species
Why: Both shingle banks hold the same three species, so their species richness is the same.
A higher value with the same richness comes from more equal shares.
So the first shingle bank’s plants are shared more equally among the three species.
Suppose a student surveys the plants of two fellsides. On the first fellside ten species grow, with equal counts of each; its Simpson’s Diversity Index is 0.90. On the second fellside two species grow, with equal counts of each; its index is 0.50.
(a) Explain what the difference between the two indices says about the two fellsides. (2 pt)
So the two fellsides have the same species evenness.
The first fellside holds ten species and the second holds two.
So the first fellside has the greater species richness.
So the first fellside’s higher value comes from its greater species richness, not from a difference in species evenness.
- Award 1 point for: the two fellsides have the same species evenness (accept: the plants are shared equally on both).
- Award 1 point for: the first fellside’s higher index comes from its greater species richness (accept: more species, ten against two).
A student works out Simpson’s Diversity Index for the plants along a bridleway and gets 0.77. The student says: “0.77 means the bridleway is 77 % diverse, so it is a better place for plants than a bridleway that scores 0.50.”
Is the student correct?
- A. ✓ No: the index is not a percentage, and a higher value means more diverse, not better
- B. Yes: a value of 0.77 means 77 % diverse, and the more diverse bridleway is the better oneThe formula sheet’s line gives a plain number with no unit, so 0.77 is not a percentage, and a higher value means more varied plants, not a better place.
Why: The index is a plain number with no unit.
So 0.77 does not mean the bridleway is 77 % diverse.
A higher value means more varied plants: more species, or more equal shares.
The index does not say which bridleway is the better place for plants.
A student says: “Two communities with the same number of species can have different values of Simpson’s Diversity Index, because the index also depends on how equally the organisms are shared among the species.”
Is the student correct?
- A. No: the index counts the species, so two communities with equally many species score the sameThe index squares each species’ share, so two communities with the same species but different shares get different sums.
- B. ✓ Yes: with the same number of species, more equal shares give a smaller sum of squared shares and a higher value
Why: The index is 1 minus the sum of squared shares.
With the same number of species, more equal shares give a smaller sum.
A smaller sum taken away from 1 leaves a higher value.
So two communities with the same number of species can score differently.
Suppose a sea loch’s Simpson’s Diversity Index for its animals falls from 0.72 to 0.33 over ten years, while the number of species in the sea loch stays the same.
Which of the following happened to the sea loch’s animals over the ten years?
- A. Every species’ count fell by the same fractionWhen every count falls by the same fraction, every share stays the same, so the index stays the same.
- B. The animals became more equally shared among the speciesMore equal shares give a smaller sum of squared shares and a higher value, and the sea loch’s value fell.
- C. ✓ One species came to hold most of the animals
Why: The number of species stayed the same, so the fall came from the shares.
A lower value with the same richness comes from less equal shares.
So one species came to hold most of the sea loch’s animals.
The two rocky islands return once more: island A with 25 flat winkles, 25 velvet crabs, 25 cushion stars and 25 sea hares.
Island B holds 70 flat winkles, 10 velvet crabs, 10 cushion stars and 10 sea hares.
Island B’s Simpson’s Diversity Index is 0.48, as the line below shows.
island B: the four squared shares add up to 0.52, and the index is 0.48
Island A’s Simpson’s Diversity Index is 0.75, as the line below shows.
island A: the four squared shares add up to 0.25, and the index is 0.75
The higher value says that island A’s animals are shared more equally among the same four species.
69Mixed practice: calculate the index, then say what it means mixed practice
A student calculates Simpson’s Diversity Index for the plants of a hedge bank with three species. The student adds up the three species’ shares, gets 1, takes 1 away from 1, and writes down 0.
Which step went wrong?
- A. ✓ The student added the shares instead of the squared shares
- B. The student should have divided the sum by the number of speciesThe formula sheet’s line has no division by the number of species: the sum of squared shares is taken away from 1.
- C. The student should have taken the sum away from 100The formula sheet’s line takes the sum away from 1, and the index runs from 0 toward 1, not toward 100.
Why: Every community’s shares add up to 1, whatever its counts are.
The formula sheet adds up the squared shares, not the shares.
So the student skipped the squaring, and a sum of 1 left an index of 0.
Suppose two churchyards score the same Simpson’s Diversity Index, 0.66. The first churchyard holds six species of plant, and the second churchyard holds three species of plant.
Which of the following must be true?
- A. The two churchyards have the same species evennessSix equal species would score higher than three equal species, so two churchyards scoring the same with six and three species cannot be equally even.
- B. ✓ The second churchyard’s plants are shared more equally among its species than the first churchyard’s
- C. The first churchyard’s plants are shared more equally among its species than the second churchyard’sSix species shared equally would score 0.83, so the first churchyard’s 0.66 means its shares are far from equal.
Why: Three species shared equally would score 0.67, so the second churchyard’s shares are almost equal.
Six species shared equally would score 0.83, so the first churchyard’s 0.66 means its shares are far from equal.
So the second churchyard’s plants are shared more equally.
Suppose a survey of a limestone pavement counts 360 plants of four species: 216, 72, 36 and 36.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
Calculate Simpson’s Diversity Index for the limestone pavement, to two decimal places.
Answer: 0.58 (tolerance ±0)
Suppose a student knows that an avenue holds 300 trees of four species, and that the most numerous species is 150 of them.
Which of the following does the student still need, to calculate Simpson’s Diversity Index for the avenue?
- A. The length of the avenueThe index is worked from counts of organisms; no length or area is in the formula sheet’s line.
- B. The count of trees in the avenue ten years agoThe index describes the community as it is now; no earlier count is in the formula sheet’s line.
- C. ✓ The counts of the other three species
Why: The index adds up one squared share for every species.
The student can work one species’ share, 150 out of 300, but not the other three.
So the student still needs the counts of the other three species.
Suppose a student records the fish in a marina each summer for five years, and Simpson’s Diversity Index for the fish rises from 0.40 to 0.70 over the five years. The student says: “The index rose, so the marina now holds more species of fish than it did five years ago.”
Is the student correct?
- A. ✓ No: a higher value can also come from the fish being shared more equally among the same species
- B. Yes: a higher value always means more species, because the index counts how many species there areA community keeps the same species and scores higher when one species stops dominating, because the shares become more equal.
Why: A higher value of Simpson’s Diversity Index means more species, or more equal shares, or both.
The same species of fish, shared more equally, would raise the value.
So the rise does not show that the marina holds more species.
A student says: “Simpson’s Diversity Index has no unit, because each share is a count divided by a count.”
Is the student correct?
- A. No: the index is a count of species, so its unit is species, just like N is a count of organismsThe index adds up squared shares, not species; a community of eight species can score lower than a community of three.
- B. ✓ Yes: a count divided by a count is a plain number, so the index is a plain number too
Why: A share is one species’ count divided by the count of all the organisms, so the units cancel.
A squared share, and a sum of squared shares, are plain numbers too.
So 1 minus that sum is a plain number with no unit.
Suppose Sand pit J and Sand pit R hold the same four species of bee. Sand pit J’s Simpson’s Diversity Index is 0.74, and Sand pit R’s is 0.51.
In which sand pit does one species hold the larger share of the bees?
- A. Sand pit JSand pit J has the higher value, and with the same four species the higher value comes from more equal shares.
- B. ✓ Sand pit R
Why: Both sand pits hold the same four species, so their species richness is the same.
Sand pit R has the lower value.
A lower value with the same richness comes from less equal shares.
So one species holds the larger share of the bees in Sand pit R.
Suppose a student sets a pitfall trap in each of two hop gardens, Hop garden J and Hop garden R, for the same time in each. In Hop garden J the trap catches 160 insects of three species: 96, 48 and 16. Hop garden R holds the same three species of insect, and its Simpson’s Diversity Index is 0.29.
n: the total number of organisms of a particular species (one species’ count)
N: the total number of organisms of all species (the count of every organism in the community)
: one species’ share of the community
(the sigma sign): add up one term for every species
Diversity Index: a number with no unit, from 0 up toward 1
(a) Calculate Simpson’s Diversity Index for Hop garden J, to two decimal places. (1 pt)
Answer: 0.54 (tolerance ±0)
- Award 1 point for: 0.54 (accept 0.54 with the working shown).
(b) Describe what the difference between the two indices says about how Hop garden R’s insects are shared among its three species. (1 pt)
So Hop garden R’s insects are shared less equally among the three species: one species holds most of Hop garden R’s insects.
- Award 1 point for: Hop garden R’s insects are shared less equally (accept: one species dominates Hop garden R; Hop garden R has the lower species evenness).
APBIO-U08-L40 Plus, minus, zero
Imagine four pairs of living things that share one place.
A tick feeds on a deer’s blood. A fungus and an alga grow together as one lichen. A bird nests in a tree, and the tree is neither helped nor harmed. Two species of warbler hunt the same insects in the same wood. In each pair, who gains and who loses?
Unit 8 · Ecology
1Who gains, who loses
Suppose an ecologist lists every species of animal and plant living on one stretch of savanna, and nothing else.
Which of the following is that list of living things?
- A. A populationA population is one species in one place.
The list holds every species on the savanna. - B. ✓ A community
- C. An ecosystemAn ecosystem is the community together with its non-living surroundings: the soil, the rain and the sunlight.
Why: The list holds every species living on the savanna, and only living things.
All the populations of every species living together in one place are a community.
One organism living inside another is called endosymbiosis.
Which of the following does the word symbiosis mean?
- A. Eating another organismSymbiosis says where two organisms live, not what one does to the other.
- B. Helping each otherSymbiosis says the two live together; it does not say that either one helps the other.
- C. ✓ Living together
Why: Endo means inside.
Symbiosis means living together.
So endosymbiosis is one organism living inside another.
How do two populations sharing one place affect each other?
Write a plus for a population that gains, and a minus for a population that loses. Write a zero for a population that is untouched.
Predation and parasitism are a plus for the eater and a minus for the eaten.
Competition is a minus for both populations.
Mutualism is a plus for both populations.
Commensalism is a plus for one population and a zero for the other.
The close, long-lasting ones, parasitism, mutualism and commensalism, are together called symbiosis.
Symbiosis means living together, not helping each other.
Five kinds in one table of signs is the classification.
Video: Watch: Who gains, who loses
A table with five empty rows is on screen: predation, parasitism, competition, mutualism, commensalism. A cheetah and a gazelle appear, and a plus and a minus fill the first row. A tick and a deer fill the second row with the same two signs. Two warbler species fill the third row with two minus signs. A fungus and an alga fill the fourth row with two plus signs. A bird and a tree fill the last row with a plus and a zero.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L40a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L40a.mp4
Two populations that share one place can affect each other.
When an interaction gives a population’s members food, shelter or safety, more of them survive and breed. That population gains.
Ecologists write a plus (+) for a population that gains.
When an interaction costs a population’s members food, blood or their lives, fewer of them survive and breed. That population loses.
Ecologists write a minus (−) for a population that loses.
When an interaction changes nothing for a population, that population is untouched.
Ecologists write a zero (0) for a population that is untouched.
Draw each population as a circle with its name inside.
Draw an arrow from one population to the other.
At the arrow’s head, write the sign of the effect on the population the arrow ends on.
So the sign at an arrow’s head always belongs to the population the arrow ends on. The sign never belongs to the population the arrow starts from.
Two populations affect each other both ways. So a pair of populations gets two arrows, one each way, and each arrow carries one sign.
What you are expected to know Read the sign at an arrow’s head as the effect on the population the arrow ends on.
Mistletoe is a plant that grows on a tree’s branch. In the drawing below, an arrow runs from the mistletoe to the trees. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which of the following does the drawing say?
- A. The trees harm the mistletoeThe arrow ends on the trees, so its sign is the effect on the trees.
- B. ✓ The mistletoe harms the trees
Why: The arrow runs from the mistletoe to the trees.
The sign at its head is a minus.
A minus at the head is a loss for the population the arrow ends on: the trees.
So the mistletoe harms the trees.
Orchids are plants, and some orchid species grow on a tree’s branch. In the drawing below, an arrow runs from the orchids to the trees. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which of the following does the drawing say?
- A. ✓ The orchids neither help nor harm the trees
- B. The trees neither help nor harm the orchidsThe arrow ends on the trees, so its sign is the effect on the trees, not on the orchids.
Why: The arrow runs from the orchids to the trees.
The sign at its head is a zero.
A zero at the head means the population the arrow ends on is untouched: the trees.
So the orchids neither help nor harm the trees.
For example, suppose cheetahs and gazelles live on one stretch of savanna. A cheetah catches a gazelle and eats it.
The cheetahs get a plus, because the cheetahs gain a meal. The gazelles get a minus, because the gazelles lose a member.
One population catching and eating members of another is called .
The eater is the predator, and the eaten is the prey.
And now suppose a tick lives on a deer’s skin and feeds on the deer’s blood for several days.
The ticks get a plus, because the ticks gain a meal of blood. The deer get a minus, because the deer lose blood.
One population living in or on another and feeding on it, while the other stays alive, is called .
The feeder is the parasite, and the population it lives on is the host.
Predation and parasitism give the same two signs: a plus for the eater and a minus for the eaten.
What separates them is whether the eaten member stays alive.
A cheetah kills the gazelle it eats. So the cheetah and the gazelle are a case of predation.
A tick feeds on a deer that walks away alive. So the tick and the deer are a case of parasitism.
But now suppose two species of warbler hunt the same insects in one wood.
Every insect one species eats is an insect the other species cannot eat.
Each warbler species gets a minus, because each species loses insects to the other.
Two populations needing the same food or space, so that each leaves less for the other, is called .
But now suppose a fungus and an alga grow together as one lichen, a crust on a rock or a tree trunk.
The alga in a lichen is a tiny organism that makes sugar from sunlight.
The alga makes sugar and shares it with the fungus. The fungus shelters the alga from drying out.
The fungus gets a plus, because the fungus gains sugar. The alga gets a plus, because the alga gains shelter.
An interaction in which both populations gain is called .
But now suppose a bird builds its nest in a tree.
The nest keeps the bird’s eggs off the ground, away from the animals that would eat them.
The nest is light, and the leaves that feed the tree sit above it. So the tree neither gains nor loses.
The birds get a plus, because the birds gain a safe place to breed. The trees get a zero, because the trees are untouched.
An interaction in which one population gains and the other is untouched is called .
Look again at the tick on the deer, the fungus with the alga, and the bird in the tree.
In each of these three pairs, the two populations live together for a long time. One population lives in or on the other.
A close, long-lasting interaction like these three is called .
Symbiosis means living together. It does not mean helping each other.
The tick lives on the deer and harms it. That harm is still symbiosis.
So parasitism, mutualism and commensalism are all kinds of symbiosis.
Predation is not symbiosis. The cheetah eats the gazelle in one chase.
Competition is not symbiosis. The two warbler species never live in or on each other.
What you are expected to know Give each population in an interaction its sign: a plus, a minus or a zero.
What you are expected to know Name the kind of interaction from its two signs, and say which kinds are symbiosis.
Pilot fish swim beside a shark and eat scraps from the shark’s meals. The shark eats as it did before.
Which sign does the shark get?
- A. a minus (−)The shark loses nothing: the pilot fish eat only the scraps.
- B. ✓ a zero (0)
Why: The pilot fish eat the scraps the shark leaves.
The shark eats as it did before.
A population that is untouched gets a zero.
A rattlesnake catches a young jackrabbit and eats it.
Which sign do the rattlesnakes get?
- A. a zero (0)The rattlesnake gains a meal, so the rattlesnakes are not untouched.
- B. ✓ a plus (+)
Why: The rattlesnake eats the jackrabbit.
The rattlesnakes gain a meal.
A population that gains gets a plus.
A tapeworm lives in a person’s gut for years and takes food from the meals the person eats.
Which sign do the people get?
- A. ✓ a minus (−)
- B. a zero (0)The tapeworm takes food the person ate, so the person loses food.
Why: The tapeworm takes food from the person’s meals.
The person loses food.
A population that loses gets a minus.
Suppose two species of seabird nest on one headland and catch the same small fish for their chicks. Every fish one species catches is one fish fewer for the other species.
Which sign does each of the two seabird species get?
- A. ✓ a minus (−)
- B. a plus (+)Each species loses fish to the other; neither gains from the other.
Why: Each species catches fish the other species would have caught.
Each species loses fish.
A population that loses gets a minus.
A cleaner fish picks parasites off the skin of a larger fish and eats them. The larger fish is rid of its parasites and swims off.
Which sign do the larger fish get?
- A. a minus (−)The larger fish loses nothing: the cleaner fish eats only its parasites, and the fish is rid of them.
- B. ✓ a plus (+)
Why: The cleaner fish eats the parasites off the larger fish’s skin.
The larger fish is rid of its parasites.
A population that gains gets a plus.
A bromeliad, a plant with stiff leaves, grows on a tree’s branch. It collects rainwater in its leaves and takes nothing from the tree, and the tree grows as it did before.
Which sign do the trees get?
- A. a minus (−)The bromeliad takes nothing from the tree, so the tree loses nothing.
- B. ✓ a zero (0)
Why: The bromeliad takes nothing from the tree.
The tree grows as it did before.
A population that is untouched gets a zero.
Plasmodium, a single-celled organism, lives inside a person’s red blood cells and feeds on them.
Which sign do the people get?
- A. ✓ a minus (−)
- B. a zero (0)Plasmodium feeds on the person’s red blood cells, so the person loses those cells.
Why: Plasmodium feeds on the person’s red blood cells.
The person loses red blood cells.
A population that loses gets a minus.
A clownfish lives among the stinging arms of a sea anemone, which keep the clownfish’s enemies away. The clownfish chases off the fish that nibble the anemone, and its droppings feed the anemone.
Which sign do the anemones get?
- A. a zero (0)The anemone gains: the clownfish drives off the fish that nibble it and feeds it, so the anemones are not untouched.
- B. ✓ a plus (+)
Why: The clownfish chases off the fish that nibble the anemone.
The clownfish’s droppings feed the anemone.
The anemones gain.
A population that gains gets a plus.
A cleaner fish picks parasites off the skin of a larger fish and eats them. The cleaner fish gets its food that way, and the larger fish is rid of its parasites.
Which kind of interaction is this?
- A. commensalismIn commensalism one population is untouched; here the cleaner fish gains food and the larger fish is rid of its parasites.
- B. ✓ mutualism
- C. parasitismIn parasitism one population loses; here neither the cleaner fish nor the larger fish loses.
Why: The cleaner fish gains food: a plus.
The larger fish is rid of its parasites: a plus.
An interaction in which both populations gain is mutualism.
Pilot fish swim beside a shark and eat scraps from the shark’s meals. The shark eats as it did before.
Which kind of interaction is this?
- A. ✓ commensalism
- B. mutualismIn mutualism both populations gain; here the shark eats as it did before.
- C. predationIn predation one population catches and eats members of the other; the pilot fish eat only scraps.
Why: The pilot fish gain scraps: a plus.
The shark eats as it did before: a zero.
An interaction in which one population gains and the other is untouched is commensalism.
A rattlesnake catches a young jackrabbit and eats it.
Which kind of interaction is this?
- A. competitionIn competition both populations lose; here the rattlesnake gains a meal.
- B. parasitismIn parasitism the eaten member stays alive; the rattlesnake kills the jackrabbit it eats.
- C. ✓ predation
Why: The rattlesnake catches the jackrabbit and eats it.
The jackrabbit does not stay alive.
One population catching and eating members of another is predation.
A tapeworm lives in a person’s gut for years and takes food from the meals the person eats. The person stays alive.
Which kind of interaction is this?
- A. commensalismIn commensalism the other population is untouched; here the person loses food.
- B. mutualismIn mutualism both populations gain; here the person loses food.
- C. ✓ parasitism
Why: The tapeworm lives inside the person and feeds on the person’s food.
The person loses food and stays alive.
One population living in another and feeding on it, while the other stays alive, is parasitism.
A student reads that a species of fern grows on the branches of a tree for years, takes nothing from the tree, and leaves the tree neither helped nor harmed. The student says: “The fern and the tree live together closely for years, so this is a symbiosis even though the tree gains nothing.”
Is the student correct?
- A. No: an interaction is a symbiosis only when both species gain, and the tree gains nothingMutualism is the kind of symbiosis in which both gain; a symbiosis with a zero in it is commensalism.
- B. ✓ Yes: the fern lives on the tree’s branches for years, so the two live together closely for a long time
Why: The fern lives on the tree’s branches for years.
The two populations live together closely for a long time, which is symbiosis.
The fern gains and the tree is untouched, so this symbiosis is commensalism.
A student reads that a species of louse lives its whole life in a swan’s feathers, feeding on the feathers, and that ecologists count this as a symbiosis. The student says: “Symbiosis means both species gain, so the swan must gain something from the lice.”
Is the student correct?
- A. ✓ No: the lice live on the swan for a long time, which is what symbiosis means, and they harm the swan
- B. Yes: ecologists only call an interaction a symbiosis when both species gain, so the swan gains tooThe lice eat the swan’s feathers, so the swan loses; a symbiosis can have a minus in it.
Why: Symbiosis means living together closely for a long time.
The lice live in the swan’s feathers for their whole lives: a symbiosis.
The lice eat the swan’s feathers, so the lice gain and the swan loses.
A symbiosis with a plus and a minus is parasitism.
The table below compares the five kinds of interaction: the sign each population gets, and one case of each.
Predation and parasitism share a plus and a minus. Competition is two minus signs.
Mutualism is two plus signs. Commensalism is a plus and a zero.
Look again at the four pairs of living things from the start of the lesson, each pair sharing one place.
The tick feeds on the deer’s blood. The tick gains and the deer loses: parasitism.
The fungus and the alga grow as one lichen. The fungus and the alga both gain: mutualism.
The bird nests in the tree. The bird gains and the tree is untouched: commensalism.
The two warbler species hunt the same insects. Both warbler species lose: competition.
Three of the four pairs live together closely for a long time. Those three are the tick and the deer, the fungus and the alga, and the bird and the tree.
Those three pairs are symbiosis.
89Quick quiz: predation, parasitism, competition, mutualism, commensalism, symbiosis mixed practice
Suppose a fungus grows around the roots of a tree and stays there for the life of the tree. The fungus takes sugar from the roots. The roots take up water and minerals that the fungus gathers from the soil around them.
(a) State the sign the fungus gets and the sign the tree gets. (1 pt)
The tree gets a plus.
- Award 1 point for: a plus for the fungus AND a plus for the tree (accept: + and +; both gain).
(b) Identify the kind of interaction between the fungus and the tree. (1 pt)
- Award 1 point for: mutualism.
- Accept the kind consistent with a wrong (a): award the point only for the kind of interaction that the student’s own two signs in (a) name.
(c) Explain why ecologists count this interaction as a symbiosis. (1 pt)
So the two populations live together closely for a long time.
A close, long-lasting interaction between two species is a symbiosis.
- Award 1 point for: the fungus and the tree live together closely (in contact) for a long time, which is what symbiosis means (accept: living together for years / for the tree’s life). Do not award for: both gain — that is why it is mutualism, not why it is a symbiosis.
One population catches and eats members of another population.
Which of the following is this interaction called?
- A. competitionIn competition both populations lose; here the eater gains.
- B. parasitismIn parasitism the eaten member stays alive while the parasite feeds on it.
- C. ✓ predation
Why: One population catching and eating members of another is predation.
The eater is the predator and the eaten is the prey.
Two populations need the same food or space, so each leaves less for the other.
Which of the following is this interaction called?
- A. ✓ competition
- B. mutualismIn mutualism both populations gain; here each population leaves less for the other, so both lose.
- C. predationIn predation one population eats members of the other; here neither eats the other.
Why: Each population leaves less food or space for the other.
Both populations lose.
Two populations needing the same food or space, so that each leaves less for the other, is competition.
One population lives in or on another and feeds on it, while the other stays alive.
Which of the following is this interaction called?
- A. commensalismIn commensalism the other population is untouched; here it is fed on and loses.
- B. ✓ parasitism
- C. predationIn predation the eaten member is killed; here it stays alive.
Why: The feeder lives in or on the other population and feeds on it.
The other population loses but stays alive.
That interaction is parasitism.
Two populations interact, and both populations gain.
Which of the following is this interaction called?
- A. commensalismIn commensalism one population is untouched; here both gain.
- B. ✓ mutualism
- C. parasitismIn parasitism one population loses; here both gain.
Why: Both populations gain, so both get a plus.
An interaction in which both populations gain is mutualism.
Two populations interact. One population gains, and the other is untouched.
Which of the following is this interaction called?
- A. ✓ commensalism
- B. mutualismIn mutualism both populations gain; here one is untouched.
- C. parasitismIn parasitism one population loses; here the other is untouched.
Why: One population gains: a plus.
The other is untouched: a zero.
An interaction in which one population gains and the other is untouched is commensalism.
Two populations live together, one in or on the other, for a long time. One may gain and the other lose, or both may gain, or one may be untouched.
Which of the following is this close, long-lasting interaction called?
- A. commensalismCommensalism is one kind of close, long-lasting interaction: the kind with a plus and a zero.
- B. mutualismMutualism is one kind of close, long-lasting interaction: the kind with two plus signs.
- C. ✓ symbiosis
Why: The two populations live together closely for a long time, whatever sign each population gets.
A close, long-lasting interaction like that is symbiosis.
Parasitism, mutualism and commensalism are its three kinds.
Glossary
- predation
- One population catching and eating members of another population. The eater is the predator and the eaten is the prey: a plus for the predator, a minus for the prey.
- parasitism
- One population living in or on another and feeding on it, while the other stays alive. The feeder is the parasite and the population it lives on is the host: a plus for the parasite, a minus for the host.
- competition
- Two populations needing the same food or space, so that each leaves less for the other: a minus for both.
- mutualism
- An interaction between two populations in which both gain: a plus for both.
- commensalism
- An interaction between two populations in which one gains and the other is untouched: a plus for one, a zero for the other.
- symbiosis
- A close, long-lasting interaction between two populations that live together, one in or on the other. Symbiosis means living together, not helping each other: parasitism, mutualism and commensalism are all kinds of symbiosis.
APBIO-U08-L40B One niche, two species
Photo: Barfooz, Wikimedia Commons, CC BY-SA 3.0.
Imagine two species of single-celled pond organism, both kinds of Paramecium, each grown alone in its own tube of water and food. Each species thrives alone.
Now put both species together in one tube with the same food. Within about two weeks one species is gone. Why can two species not share one tube when each does well on its own?
Unit 8 · Ecology
1What a species needs: its niche
Suppose two species of diving beetle hunt in one pond, and both species eat the same small pond animals. Each species leaves less food for the other, so both populations lose.
Which interaction is this?
- A. ✓ Competition
- B. MutualismIn mutualism both populations gain.
- C. PredationIn predation one population eats the other: the eater gains and the eaten loses.
Why: Both species of diving beetle eat the same food, and each leaves less for the other.
Both populations lose.
An interaction in which both populations lose is competition.
What happens when two species need the same things?
Everything a species uses and everything it lives in make up its niche.
Two species with the same niche in one place compete for everything.
The better competitor eats more of the food. So the other species dies out.
That outcome is competitive exclusion.
Two species do share a place when they divide the niche, each taking a different part of the resource.
Five warbler species feed in one spruce tree by hunting in five different parts of it. That dividing is niche partitioning.
Partitioning eases competition. It does not end it.
Video: Watch: One Paramecium and what it needs
One P. aurelia swims in its tube. The bacteria it eats are ringed and labeled its food. The water is labeled fresh water. A thermometer beside the tube is labeled the warmth it lives in. Then one ring is drawn around all three, and the ring is labeled its niche.
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Imagine one Paramecium aurelia, P. aurelia for short, swimming in its tube.
A Paramecium is one cell, covered in tiny beating hairs that row it through the water.
The Paramecium eats the bacteria in the tube’s water.
The Paramecium lives in fresh water, not salt water.
The Paramecium lives in water that is neither too cold nor too hot.
The bacteria and the water are things the Paramecium uses. The things a species uses, such as its food and its water, are its resources.
The saltiness of the water and its warmth are not things the Paramecium uses. They are the surroundings the Paramecium lives in: its conditions.
The resources a species uses and the conditions it lives in are together called its .
The table below sets out P. aurelia’s niche: the resources it uses and the conditions it lives in.
A niche is what a species uses and what it lives in. A niche is not a description of the species itself.
The Paramecium’s slipper shape is a feature of the Paramecium. The Paramecium does not use its shape or live in it.
So the shape is not part of its niche.
The count of Paramecium in the tube is the population size. A count is neither a resource nor a condition.
So the count is not part of the niche either.
What you are expected to know Decide whether a stated thing is part of a species’ niche.
A treecreeper is a small bird that picks insects out of cracks in tree bark. A student counts 14 treecreepers in one forest.
Is the count of treecreepers in the forest part of the treecreeper’s niche?
- A. YesA count of treecreepers is the population size, not a resource the bird uses or a condition it lives in.
- B. ✓ No
Why: A niche is the resources a species uses and the conditions it lives in.
A count of treecreepers is the size of the population.
The treecreeper neither uses the count nor lives in it, so the count is not part of its niche.
A treecreeper is a small bird that picks insects out of cracks in tree bark and eats them.
Are the insects in the bark part of the treecreeper’s niche?
- A. ✓ Yes
- B. NoThe insects are the treecreeper’s food, a resource it uses.
Why: A niche is the resources a species uses and the conditions it lives in.
The treecreeper eats the insects in the bark, so the insects are a resource it uses.
So the insects are part of its niche.
A treecreeper is a small bird that picks insects out of cracks in tree bark. The forest it lives in has cool summers and mild winters.
Is the warmth of the forest’s air part of the treecreeper’s niche?
- A. ✓ Yes
- B. NoThe warmth of the air is a condition the treecreeper lives in.
Why: A niche is the resources a species uses and the conditions it lives in.
The treecreeper lives in the forest’s air, cool in summer and mild in winter.
The warmth of that air is a condition it lives in, so the warmth is part of its niche.
A treecreeper is a small bird that picks insects out of cracks in tree bark with a thin, curved bill. A niche is the resources a species uses and the conditions it lives in.
Is the treecreeper’s thin, curved bill part of the treecreeper’s niche?
- A. YesThe bill is a feature of the bird, not a resource it uses or a condition it lives in.
- B. ✓ No
Why: A niche is the resources a species uses and the conditions it lives in.
The bill is a part of the treecreeper’s own body.
The treecreeper neither uses the bill as a resource nor lives in it, so the bill is not part of its niche.
A treecreeper is a small bird that picks insects out of cracks in tree bark. At night it roosts in a deep crack in the bark.
Is the crack the treecreeper roosts in part of the treecreeper’s niche?
- A. ✓ Yes
- B. NoThe crack is the treecreeper’s shelter, a resource it uses.
Why: A niche is the resources a species uses and the conditions it lives in.
The treecreeper roosts in the crack, so the crack is a shelter it uses: a resource.
So the crack is part of its niche.
32Quick quiz: niche mixed practice
Suppose a whinchat, a small bird, catches flies over rough grass, nests on the ground under low bushes and lives on hillsides with cool summers.
(a) Identify two parts of the whinchat’s niche from that description. (2 pt)
The cool summers of its hillside are a condition it lives in.
- Award 1 point for each part named, up to 2 points: the flies it eats (a resource); the low bushes it nests under (a resource); the cool summers, or the hillside, it lives in (a condition).
Which of the following is a species’ niche?
- A. The number of individuals of the species living in one placeA count of individuals is the population size.
- B. The shape and size of the species’ bodyShape and size are features of the organism, not things it uses or lives in.
- C. ✓ The resources the species uses and the conditions it lives in
Why: A niche is everything a species uses and everything it lives in.
So it is the resources the species uses and the conditions it lives in.
35Two species in one tube
A population’s count climbs, slows and then levels off at its carrying capacity, K.
What is the carrying capacity, K?
- A. ✓ The largest population that the place’s resources can support
- B. The number of individuals born in one yearThe births in one year are a rate, not the level the count settles at.
- C. The time the population takes to doubleA doubling time is a time, not a count of individuals.
Why: The count levels off because the place’s resources can feed no more individuals.
The level it settles at is the carrying capacity, K: the largest population that the place’s resources can support.
Video: Watch: Two species, one tube
P. aurelia is counted alone for 16 days, and its points climb and level off. P. caudatum is counted alone, and its points climb and level off lower. Then both species are put into one tube. The P. aurelia points climb and level off. The P. caudatum points climb for four days, then fall to zero.
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Imagine P. aurelia alone in a tube of water and bacteria, and P. caudatum alone in a second tube of the same water and the same bacteria.
The ecologist Georgy Gause grew the two species this way and counted the cells in a fixed sample of each tube’s water over 16 days.
Gause’s counts are drawn as a graph below: time in days on the x-axis, the number of cells in the sample on the y-axis, a gridline every 2 days and every 50 cells.
The solid line is P. aurelia and the dashed line is P. caudatum.
Alone, P. aurelia’s count climbs and levels off near 250 cells, read against the 250 gridline.
Alone, P. caudatum’s count climbs and levels off just above the 50 gridline.
Each species alone reaches its own carrying capacity, K: the bacteria in its tube can feed no more cells.
Now put P. aurelia and P. caudatum together in one tube of water and bacteria.
The counts from the shared tube are drawn below on the same axes. The solid line is P. aurelia and the dashed line is P. caudatum.
P. aurelia’s count climbs and levels off a little above the 200 gridline.
P. caudatum’s count climbs a little by day 4, staying below the 50 gridline. Then the count falls, and by day 16 it sits on the 0 gridline.
Within about two weeks P. caudatum is gone from the shared tube, though it thrived alone.
P. aurelia and P. caudatum eat the same bacteria and live in the same water: P. aurelia and P. caudatum have the same niche.
Every bacterium P. aurelia eats is a bacterium P. caudatum cannot eat. So P. aurelia and P. caudatum compete for every bacterium in the tube.
P. aurelia eats more of the bacteria than P. caudatum does. So P. aurelia is the better competitor.
Each P. caudatum gets less food than it needs. So fewer P. caudatum divide, and more P. caudatum die.
So P. caudatum’s count falls day by day, until no P. caudatum is left.
When two species with the same niche live in one place, the better competitor wins the food and the other species dies out. That outcome is called .
Competitive exclusion is the reason two species with the same niche cannot both persist in one place.
What you are expected to know Predict, from a graph of two species grown together, which species is excluded.
What you are expected to know Explain why two species with the same niche cannot both persist in one place.
The graph below shows Gause’s counts of P. aurelia and P. caudatum grown together in one tube of water and bacteria. The solid line is P. aurelia and the dashed line is P. caudatum.
Which species’ count falls to zero?
- A. P. aureliaThe solid line, P. aurelia, climbs and levels off above the 200 gridline.
- B. ✓ P. caudatum
Why: The dashed line is P. caudatum.
The dashed line stays below the 50 gridline, then falls.
By day 16 the dashed line sits on the 0 gridline.
So P. caudatum’s count falls to zero.
Suppose two species of water boatman, small swimming insects, live in one cattle trough. Both species eat only the algae growing on the trough’s sides. By the end of the summer one species is gone.
(a) Explain why one species of water boatman died out. (3 pt)
So the two species of water boatman compete for the algae.
The better competitor eats more of the algae.
Each water boatman of the other species gets less food than it needs, so fewer breed and more die.
So that species’ count falls until none is left: competitive exclusion.
- Award 1 point for: the two species have the same niche (the same food in the same place), so they compete.
- Award 1 point for: the better competitor eats more of the algae, so the other species gets too little food (fewer breed, or more die).
- Award 1 point for: the other species’ count falls to zero; competitive exclusion (accept the outcome described without the name).
Suppose a baker puts two species of firebrat, small insects that eat flour, into one tub of flour. Both species eat only the flour. One species breeds faster and eats more of the flour than the other.
Predict what happens to the slower-breeding species’ count over the following months.
- A. The count rises to the tub’s carrying capacityReaching the tub’s carrying capacity needs enough flour for every firebrat; the faster breeder eats more of the flour, so the slower breeder gets too little.
- B. The count levels off at the same level as the faster breeder’sThe faster breeder eats more of the flour, so the slower breeder cannot reach the same level.
- C. ✓ The count falls to zero
Why: Both species of firebrat eat only the flour in one tub: the same niche.
The faster breeder eats more of the flour: it is the better competitor.
Each slower-breeding firebrat gets too little food, so fewer breed and more die.
So the slower breeder’s count falls to zero: competitive exclusion.
62Quick quiz: competitive exclusion mixed practice
(a) State what competitive exclusion is. (1 pt)
- Award 1 point for: two species with the same niche in one place, and the better competitor drives the other out (the other species dies out, or disappears).
Which of the following is competitive exclusion?
- A. One species in a place eats another species living there and the eaten species dies outOne species eating another is predation, not competition.
- B. Two species with different niches share one place and compete so little that both persistTwo species with different niches compete little, so neither drives the other out.
- C. ✓ Two species with the same niche share one place and the better competitor drives the other out
Why: Competitive exclusion happens between two species with the same niche in one place.
The better competitor gets more of the resources, and the other species dies out.
65Sharing the tree
Video: Watch: Five warblers, one spruce
One spruce tree is drawn. Five warblers appear one at a time, each in its own part of the tree: the top twigs, the upper outer branches, the middle outer branches, the middle inner branches, and the bottom branches with the ground. Each feeding zone is ringed as its bird arrives. Then the narrow strip where two zones touch is shaded.
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Imagine five species of warbler, small songbirds that eat insects, all feeding in one spruce tree, an evergreen with needles.
All five species eat insects, and all five live in the same spruce forests. Their niches overlap.
Yet all five species persist there year after year. No species excludes the others.
The ecologist Robert MacArthur watched where in the tree each species fed.
The Cape May warbler feeds at the top of the tree, on the outer twigs.
The Blackburnian warbler feeds in the upper branches, on the outside of the tree, below the Cape May warbler.
The black-throated green warbler feeds in the middle of the tree, on the outer branches.
The bay-breasted warbler feeds in the middle of the tree too, but on the inner branches, near the trunk.
The yellow-rumped warbler feeds on the bottom branches and on the ground beneath the tree.
Each species hunts the insects of a different part of the tree.
So the five species compete less for insects: a Cape May warbler at the top and a yellow-rumped warbler on the ground rarely hunt the same insect.
When competing species each use a different part of a shared resource, that dividing of the resource is called .
The Paramecium species and the warblers differ in one thing.
P. aurelia and P. caudatum ate the same bacteria in the same water. So one excluded the other.
Each warbler species takes the insects of a different part of the tree. So all five persist.
Niche partitioning eases competition. It does not end it.
The feeding zones touch at their edges, and insects fly from one zone into the next.
So two warbler species whose zones touch still hunt some of the same insects. They still compete, though less than two species in one zone would.
What you are expected to know Explain how competing species share one place by niche partitioning, each using a different part of a shared resource.
What you are expected to know Judge a claim that dividing a resource ends the competition for it.
The table below gives the part of one spruce tree where each of five warbler species does most of its feeding.
Which two species feed at the same height in the tree?
- A. ✓ Bay-breasted warbler and black-throated green warbler
- B. Blackburnian warbler and yellow-rumped warblerThe Blackburnian warbler feeds in the upper branches and the yellow-rumped warbler at the bottom.
- C. Cape May warbler and bay-breasted warblerThe Cape May warbler feeds at the top and the bay-breasted warbler in the middle.
Why: The height column reads top, upper, middle, middle, bottom.
The two middle rows are the black-throated green warbler and the bay-breasted warbler.
So those two species feed at the same height, and the distance column keeps them apart: outer branches for one, inner branches for the other.
Five species of warbler eat insects in the same spruce tree, and all five species persist year after year.
Which of the following is the reason no species excludes the others?
- A. The five species eat different kinds of food and so never competeAll five species eat insects: the food is the same, and the tree is divided instead.
- B. ✓ The five species feed in different parts of the tree and so compete less for insects
- C. The tree holds more insects than the five species can eat and so the insects never grow scarceThe tree’s insects do grow scarce; the five species compete for them, but less, because each feeds in a different part of the tree.
Why: All five species eat insects, so their niches overlap.
Each species feeds in a different part of the tree.
So two species rarely hunt the same insect, and the five compete less.
Less competition means no species drives another out.
A student reads that five species of warbler each feed in a different part of one spruce tree, and says: “Once the warblers divide the tree, they stop competing for insects.”
Is the student correct?
- A. ✓ No: the five species compete less, but their zones touch and insects move between them, so they still compete
- B. Yes: each species feeds in its own part of the tree, so no two species ever hunt the same insectTwo zones that touch share their edges, and insects fly between zones, so two species still hunt some of the same insects.
Why: Dividing the tree gives each species its own feeding zone, so the five compete less.
But the zones touch at their edges, and insects fly from one zone into the next.
So two species still hunt some of the same insects.
Niche partitioning eases competition; it does not end it.
Suppose two species of bee-fly drink nectar from the same flowers in one garden. One species feeds in the morning and the other in the afternoon.
Predict how much the two species of bee-fly compete for nectar, compared with two species that feed at the same time of day.
- A. MoreFeeding at different times means the two species rarely visit a flower at the same moment, so they compete less, not more.
- B. The sameTwo species feeding at the same time compete for every flower; two species feeding at different times share the flowers by time.
- C. ✓ Less
Why: The two species of bee-fly use the same nectar at different times of day.
The time of feeding is part of the niche, so the two have divided it: niche partitioning.
Each flower is shared by time.
So the two species compete less than two feeding at the same time.
The two species of Paramecium return: P. aurelia and P. caudatum, each thriving alone in its own tube of water and bacteria.
Put together in one tube with the same bacteria, P. caudatum is gone within about two weeks.
P. aurelia and P. caudatum ate the same bacteria in the same water: one niche.
P. aurelia ate more of the bacteria. So P. caudatum starved out: competitive exclusion.
The five warblers in one spruce avoid that fate. Each species feeds in a different part of the tree.
So the five compete less. That dividing is niche partitioning.
The table below compares two outcomes: the same niche in one place, and a divided niche.
98Quick quiz: niche partitioning mixed practice
(a) State what niche partitioning is. (1 pt)
- Award 1 point for: competing species each use a different part of a shared resource (a different part of the place, a different food, or a different time), so competition is eased.
Which of the following is niche partitioning?
- A. ✓ Competing species each use a different part of a shared resource
- B. One species drives a competitor with the same niche out of a placeOne species driving out a competitor with the same niche is competitive exclusion.
- C. Two species in one place eat different foods and never competeSpecies that never compete have not divided a shared resource; partitioning eases competition between species that do share one.
Why: Niche partitioning is the dividing of a shared resource.
Each competing species uses a different part of it, so the species compete less.
So it is competing species each using a different part of a shared resource.
101Mixed practice: one niche, two species mixed practice
Suppose a great tit, a small bird, is one of 40 great tits counted in a forest, and it feeds caterpillars to its chicks.
Which of the following is part of the great tit’s niche?
- A. The great tit’s black and yellow feathersFeathers are a feature of the bird, not a resource it uses or a condition it lives in.
- B. ✓ The caterpillars it feeds to its chicks
- C. The count of 40 great tits in the forestA count of great tits is the population size, not a resource or a condition.
Why: A niche is the resources a species uses and the conditions it lives in.
The caterpillars are the food the great tit uses: a resource.
So the caterpillars are part of its niche.
Suppose two species of scale insect, tiny sap-sucking insects, settle on one blackthorn hedge, and both species suck sap from the same young twigs. One species draws more sap than the other.
Predict what happens to the two species of scale insect over the following years.
- A. Both species die outThe species that draws more sap gets enough food to persist.
- B. Both species persist at half the count each would reach aloneThe species that draws more sap does not share the sap equally; the other species gets too little.
- C. ✓ One species dies out and the other persists
Why: Both species of scale insect use the same sap from the same twigs: the same niche.
The species that draws more sap is the better competitor.
The other species gets too little food, so fewer breed and more die.
So the other species dies out: competitive exclusion.
Suppose two species of cyclops, tiny crustaceans, live in one rainwater cistern and eat the same food. The graph below shows the count of each species over 12 weeks: the solid line is species J and the dashed line is species M.
Which species is being excluded?
- A. Species JThe solid line, species J, climbs and levels off above the 150 gridline.
- B. ✓ Species M
Why: The dashed line is species M.
The dashed line climbs a little, then falls.
By week 12 the dashed line sits on the 0 gridline.
So species M is being excluded.
A student reads that two species of mason bee gather nectar from the same garden’s flowers, one species flying in spring and the other in summer, and says: “The two species can both persist in the garden, because they take the nectar at different times.”
Is the student correct?
- A. No: two species that use the same flowers have the same niche, so one must exclude the otherThe time a species feeds is part of its niche; feeding in different seasons divides the niche, so the two niches are not the same.
- B. ✓ Yes: the two species divide the nectar by time, so they compete little and both persist
Why: The two species of mason bee use the same nectar in different seasons.
The season a species flies in is part of its niche, so the two have divided it: niche partitioning.
Spring flowers feed one species; summer flowers feed the other.
So the two compete little, and both persist.
Suppose two species of nematode, tiny worms, live in the soil of one flowerpot, and both species eat only the same bacteria. After a month only one species is left.
Which of the following is the reason the other species is gone?
- A. ✓ The better competitor ate more of the bacteria and starved the other species out
- B. The better competitor hunted and ate the worms of the other speciesBoth species of nematode eat bacteria; neither eats worms.
- C. The pot’s soil had room for one species of worm and not for twoRoom in the soil is not the resource the two species share; the bacteria are.
Why: Both species of nematode eat only the same bacteria in one pot: the same niche.
So the two species compete for every bacterium.
The better competitor eats more of the bacteria.
The other species gets too little food, so it starves out: competitive exclusion.
Suppose two species of alderfly larva live in one beck, a small stream, and both species eat the same small animals. One species hunts under the stones of the fast water, the other in the mud of the slow pools, and both species persist year after year.
Which of the following names the way the two species of alderfly larva share the beck?
- A. Competitive exclusionIn competitive exclusion one species dies out; both species of alderfly larva persist.
- B. ✓ Niche partitioning
- C. PredationIn predation one species eats the other; the two species of alderfly larva eat the same small animals.
Why: Both species of alderfly larva eat the same food in one stream, so they compete.
Each species hunts in a different part of the stream: the fast water or the slow pools.
So the two use different parts of the shared resource and compete less.
That is niche partitioning.
Suppose two species of insect eat the same food in one meadow.
Which of the following would let both species persist in the meadow?
- A. Breeding faster than the other speciesBreeding faster makes one species the better competitor, which excludes the other.
- B. Eating more of the food than the other speciesEating more of the food makes one species the better competitor, which excludes the other.
- C. ✓ Feeding at a different time of day from the other species
Why: Two species with the same niche in one place compete for everything, and one excludes the other.
Feeding at a different time of day divides the niche.
Each species then takes a different part of the shared food, so the two compete less: niche partitioning.
So both species persist.
Suppose two species of leafcutter bee nest in the same clay bank and cut pieces from the leaves of the same plants to line their nests. The bank has room for far more nests than the bees dig. Only the leaves are scarce. A student counts both species every summer for five summers. One species’ count rises each summer. The other species’ count falls each summer and reaches zero in the fifth summer.
(a) Identify the interaction between the two species of leafcutter bee while both were present. (1 pt)
- Award 1 point for: competition (both species use the same resource, the leaves).
(b) Explain why the second species’ count fell to zero. (2 pt)
The first species cut more of the leaves, so it was the better competitor.
Each bee of the second species had too little leaf to line its nest, so fewer of its young survived each summer.
So the second species’ count fell to zero: competitive exclusion.
- Award 1 point for: the two species had the same niche (the same leaves from the same plants in one place), so the better competitor took more of the leaves.
- Award 1 point for: the second species got too little of the resource, so fewer of its young survived each summer and its count fell to zero (competitive exclusion).
(c) Predict how the outcome would differ if the two species had cut leaves from different plants, and justify your prediction. (2 pt)
Cutting leaves from different plants divides the shared resource: niche partitioning.
So the two species would compete less, and neither would drive the other out.
- Award 1 point for: both species persist (neither dies out).
- Award 1 point for: cutting leaves from different plants divides the resource (niche partitioning), so the two species compete less and neither excludes the other.
Glossary
- niche
- The resources a species uses and the conditions it lives in: for P. aurelia, the bacteria it eats and the water it swims in, and how fresh and how warm that water is.
- competitive exclusion
- The outcome when two species with the same niche live in one place: the better competitor gets more of the resources, and the other species dies out. P. caudatum grown with P. aurelia was gone within about two weeks.
- niche partitioning
- Competing species each using a different part of a shared resource, such as five warbler species each feeding in a different part of one spruce tree. Partitioning eases competition; it does not end it.
APBIO-U08-L41 Lynx and hare
Photos: Keith Williams, Wikimedia Commons, CC BY 2.0 (resized); National Park Service / Jim Peaco, Wikimedia Commons, public domain (resized).
For ninety years, between 1845 and 1935, a fur-trading company in Canada bought pelts from trappers: the pelts of snowshoe hares, and the pelts of the Canada lynx that hunt them. Ecologists plotted both yearly counts on one graph.
The hare count climbs and crashes about every ten years. The lynx count climbs and crashes too, and lags a little behind the hares. The hare count climbs higher than the lynx count ever does. So the graph gives each animal its own y-axis. Which line belongs to which axis, and why does the lynx line lag behind the hare line?
Unit 8 · Ecology
1Two lines, two y-axes
Suppose a line graph plots the depth of snow in a garden, measured each morning, against the day. The points are joined one to the next.
To read the snow depth on day 6, which of the following do you do?
- A. Find the highest point of the line and read its depth from the y-axisThe highest point is the deepest snow of the whole graph, not the depth on day 6.
- B. Find 6 on the y-axis, go across to the line, then down to the x-axisThe day is on the x-axis, so a value on the y-axis cannot be the day.
- C. ✓ Find day 6 on the x-axis, go up to the line, then across to the y-axis and read the depth
Why: The day was set, so day 6 sits on the x-axis.
The depth was measured, so it is read from the y-axis.
Go up from day 6 to the line, then across to the y-axis: that reading is the depth on day 6.
How do you read two populations on one graph?
Each line is read against its own y-axis. The hare line is read against the left axis, and the lynx line against the right.
The two scales are never compared with each other.
Video: Watch: Two lines, two y-axes
The hare pelts are plotted first, against a left y-axis numbered to 160 thousand. Then the lynx pelts are plotted against a right y-axis numbered to 80 thousand. One value is read from each line at 1875: about 100 thousand hare pelts, about 30 thousand lynx pelts.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L41a.mp4
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The Canada lynx, a wild cat of the northern forests, hunts the snowshoe hare. The photographs below show a lynx on snow and a hare in its white winter coat.
A pelt is one animal’s skin with the fur still on it. For those ninety years the fur-trading company’s records give a yearly count of hare pelts and of lynx pelts.
So each year’s pelt count stands in for the number of animals trapped that year. More hares in the forest meant more hare pelts bought.
The graph below plots both pelt counts against the year on the x-axis, 1845 to 1935.
The solid line is the hare pelts. It is read against the left y-axis, which is numbered from 0 to 160 thousand pelts.
The dashed line is the lynx pelts. It is read against the right y-axis, which is numbered from 0 to 80 thousand pelts.
The legend at the top names the two lines. Each y-axis carries the name of its animal.
Both lines climb and crash, over and over. The hare line reaches its highest point, about 150 thousand pelts, in 1863.
The lynx line reaches its highest point, about 80 thousand pelts, in 1886.
The lynx count never passes 80 thousand pelts. Plotted against the hare axis, the lynx line would use only the lower half of the frame.
Its own axis reaches only 80 thousand. So the lynx line fills the frame, and its rises and falls are as easy to read as the hare line’s.
A graph with two y-axes, one scale for each of two quantities plotted against the same x-axis, is called a .
Take the year 1875. The graph below rings the point where each line crosses the 1875 gridline.
Go up the 1875 gridline to the solid hare line. Read across to the left axis: about 100 thousand hare pelts.
Go on up the 1875 gridline to the dashed lynx line. Read across to the right axis: about 30 thousand lynx pelts.
The 30 thousand gridline of the right axis is the same line as the 60 thousand gridline of the left axis.
Read the lynx line against the left axis by mistake, and 30 thousand lynx pelts becomes 60 thousand: twice the true count.
So a lynx count is read from the right axis only, and a hare count from the left axis only. The heights of the two lines are never compared with each other.
What you are expected to know Read a graph with two y-axes: match each line to its axis, and read a value from each line against its own axis.
The graph below shows the years 1865 to 1870 of the pelt records, with one of the two lines drawn.
Against which y-axis is the line read?
- A. The left y-axisThe legend names the dashed line as the Canada lynx pelts, and the right y-axis carries the lynx pelts.
- B. ✓ The right y-axis
Why: The legend says the dashed line is the Canada lynx pelts.
The right y-axis is titled Canada lynx pelts.
So the dashed line is read against the right y-axis.
The graph below shows the years 1893 to 1898 of the pelt records, with one of the two lines drawn.
Against which y-axis is the line read?
- A. The left y-axisThe legend names the dashed line as the Canada lynx pelts, and the right y-axis carries the lynx pelts.
- B. ✓ The right y-axis
Why: The legend says the dashed line is the Canada lynx pelts.
The right y-axis is titled Canada lynx pelts.
So the dashed line is read against the right y-axis.
The graph below shows the years 1850 to 1855 of the pelt records, with one of the two lines drawn.
Against which y-axis is the line read?
- A. ✓ The left y-axis
- B. The right y-axisThe legend names the solid line as the snowshoe hare pelts, and the left y-axis carries the hare pelts.
Why: The legend says the solid line is the snowshoe hare pelts.
The left y-axis is titled snowshoe hare pelts.
So the solid line is read against the left y-axis.
The graph below shows the years 1910 to 1915 of the pelt records, with one of the two lines drawn.
Against which y-axis is the line read?
- A. ✓ The left y-axis
- B. The right y-axisThe legend names the solid line as the snowshoe hare pelts, and the left y-axis carries the hare pelts.
Why: The legend says the solid line is the snowshoe hare pelts.
The left y-axis is titled snowshoe hare pelts.
So the solid line is read against the left y-axis.
The graph below shows the years 1925 to 1930 of the pelt records, with one of the two lines drawn.
Against which y-axis is the line read?
- A. The left y-axisThe legend names the dashed line as the Canada lynx pelts, and the right y-axis carries the lynx pelts.
- B. ✓ The right y-axis
Why: The legend says the dashed line is the Canada lynx pelts.
The right y-axis is titled Canada lynx pelts.
So the dashed line is read against the right y-axis.
The graph below shows the years 1880 to 1890 of the pelt records, both lines, with a gridline at every year.
Read the lynx pelts in 1884 from the graph, in thousands, to the nearest 5 thousand. Answers within 5 thousand are accepted.
Part 1. Read the hare pelts in 1884 from the graph, in thousands, to the nearest 5 thousand. Answers within 5 thousand are accepted.
Answer: 50 thousand hare pelts (tolerance ±5)
Answer: 45 thousand lynx pelts (tolerance ±5)
32Quick quiz: dual-y-axis graph mixed practice
Suppose an ecologist plots the yearly counts of gyrfalcons, large falcons of the far north, and of the willow grouse they hunt, on the dual-y-axis graph below. The grouse count is plotted against the left y-axis and the gyrfalcon count against the right. A student reads the grouse count in year 6 from the right y-axis and writes down 20 grouse.
(a) Explain why the student’s reading is wrong. (2 pt)
Each line is read against its own y-axis only.
So the grouse count must be read from the left y-axis, and 20 is the gyrfalcon count for year 6, not a grouse count.
- Award 1 point for: each line is read against its own y-axis (accept: the two axes carry different scales).
- Award 1 point for: the grouse count is read from the left y-axis, so the right-axis number is not a grouse count.
Which of the following is a dual-y-axis graph?
- A. ✓ A graph with two y-axes, one scale for each of two quantities plotted against the same x-axis
- B. A graph with two x-axes, one span of time for each of two quantities plotted against the same y-axisThe two quantities share one x-axis, the time; it is the y-axis that is doubled.
- C. A graph whose one y-axis is split into two equal halves, one half for each of two quantitiesEach quantity gets a whole y-axis of its own, with its own scale, not half of one axis.
Why: Two quantities are plotted against the same x-axis.
Each quantity gets its own y-axis, with its own scale.
A graph drawn that way is called a dual-y-axis graph.
35Why the lynx follows the hare
A fisher, a large weasel-like hunter of the northern forests, catches, kills and eats a porcupine.
Which of the following gives the sign of this interaction for each population?
- A. Fishers −, porcupines +The fisher eats: it gains, so its sign is a plus, not a minus.
- B. Fishers +, porcupines +The porcupine is killed and eaten: it loses, so its sign is a minus, not a plus.
- C. ✓ Fishers +, porcupines −
Why: One population catches and eats members of another: that interaction is called predation.
The eater gains, so the fisher population gets a plus.
The eaten loses, so the porcupine population gets a minus.
A limiting factor takes a larger fraction of a population when the population is denser.
Which of the following is that kind of limiting factor called?
- A. ✓ Density-dependent
- B. Density-independentA density-independent factor takes the same fraction of a population, however dense the population is.
Why: The factor takes a larger fraction of the denser population.
A larger fraction when denser means density-dependent.
Why do a predator and its prey rise and fall in turn?
More hares mean more food for the lynx. So the lynx count climbs after the hare count.
Many lynx kill many hares. So the hare count falls.
With few hares the lynx starve. So the lynx count falls.
With few lynx, few hares are killed. So the hares recover, and the cycle repeats.
So the lynx peaks follow the hare peaks.
Video: Watch: Why the lynx follows the hare
One cycle of the graph plays through. The hare line climbs, and the lynx line climbs after it. The hare line crashes, and the lynx line crashes after it. Then the hares recover, and the lag between the two peaks is marked.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L41b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L41b.mp4
Take the cycle of the years 1919 to 1932. The graph below shades those years and rings the hare peak of 1923 and the lynx peak of 1926.
In 1919 the hare line reads about 10 thousand pelts and the lynx line about 3 thousand pelts. Both animals were scarce.
From 1919 to 1923 the hare count climbs, to about 80 thousand pelts.
Each hare is food for a lynx. So more hares mean that more lynx are fed.
So the lynx count climbs after the hare count, to about 55 thousand pelts in 1926.
Many lynx kill many hares. So from 1923 the hare count falls, to about 5 thousand pelts in 1926.
With few hares, fewer lynx are fed. So the lynx count falls, to about 7 thousand pelts in 1930.
With few lynx, few hares are killed. So the hare count climbs again, to about 80 thousand pelts in 1932.
The hare peak came in 1923 and the lynx peak in 1926. The lynx count lagged three years behind the hare count.
On this graph, in every cycle after 1850 the lynx peak comes with or after the hare peak, never before it. In most cycles it comes one to four years after.
The hare peaks come about ten years apart: 1863, 1875, 1886, 1895, 1904, 1912, 1923 and 1933. So the whole cycle repeats about every ten years.
1 More hares feed more lynx: the lynx count climbs.
2 Many lynx kill many hares: the hare count falls.
3 Few hares starve the lynx: the lynx count falls.
4 Few lynx let the hares recover.
Predation is not the whole story: crowding among the hares is a density-dependent factor too.
In a crowded year the hares raise fewer young. Fewer young also drive the hare count down.
What you are expected to know Explain the predator–prey cycle read from the graph, and say why the lynx peaks lag the hare peaks.
Suppose the snowshoe hare count in a northern forest peaks this year.
Over the next year, which of the following does the lynx count do?
- A. The lynx count fallsMany hares this year mean much food for the lynx, so the lynx count does not fall yet.
- B. ✓ The lynx count climbs
- C. The lynx count stays levelMany hares mean more lynx are fed, so the lynx count does not stay level.
Why: A hare peak means many hares.
Many hares are much food for the lynx.
So more lynx are fed, and the lynx count climbs after the hare peak.
Suppose cougars, large wild cats, hunt bighorn sheep on a mountain range. The bighorn count climbs for four years to a peak in 2014, then crashes over the next three years. The cougar count peaks in 2016, then falls.
(a) Explain why the cougar count peaks in 2016, and then falls. (3 pt)
So more cougars are fed, and the cougar count climbs after the bighorn count.
Many cougars kill many bighorn sheep, so the bighorn count crashes.
With few bighorn sheep, fewer cougars are fed.
So the cougar count falls after the bighorn count.
- Award 1 point for: more bighorn sheep mean more food, so more cougars are fed and the cougar count climbs after the bighorn count.
- Award 1 point for: many cougars kill many bighorn sheep, so the bighorn count crashes.
- Award 1 point for: with few bighorn sheep, fewer cougars are fed (accept: the cougars starve), so the cougar count falls.
The graph below shows the years 1920 to 1930 of the pelt records. A student says: “The two lines cross just after 1923, so at that moment the company bought as many lynx pelts as hare pelts.”
Is the student correct?
- A. ✓ No: each line is read against its own y-axis, so the two counts differ where the lines cross
- B. Yes: where two lines cross on a graph, the two counts they show are equal at that momentWhere the lines cross, the hare line reads about 76 thousand on the left axis and the lynx line about 38 thousand on the right axis.
Why: The hare line is read against the left y-axis.
The lynx line is read against the right y-axis.
Where the lines cross, the hare count is about 76 thousand and the lynx count about 38 thousand.
So the two counts are not equal there.
Suppose wolves and moose on an island are counted every year for twenty years. The wolves hunt the moose. The moose count peaks in the eighth year.
In which of the following does the wolf count peak?
- A. ✓ After the eighth year
- B. In the eighth yearThe wolves are fed by the moose, so the wolf count is still climbing when the moose count peaks.
- C. Before the eighth yearThe wolf count climbs only after more moose have fed more wolves, so it cannot peak first.
Why: More moose mean more food for the wolves.
So the wolf count climbs after the moose count.
The moose count peaks in the eighth year.
So the wolf count peaks after the eighth year.
The ninety years of pelt records are below once more, 1845 to 1935. Snowshoe hare pelts are read against the left y-axis, and Canada lynx pelts against the right.
Read each line against its own axis: in 1875, about 100 thousand hare pelts on the left axis, and about 30 thousand lynx pelts on the right.
The lynx peaks follow the hare peaks. More hares feed more lynx.
Many lynx then kill many hares. The hare count falls.
With few hares the lynx starve. The hares recover.
69Mixed practice: two y-axes, one cycle mixed practice
Suppose a survey counts ruffed grouse and pairs of goshawks in one forest every year for twenty years. The graph below plots the ruffed grouse against the left y-axis and the goshawk pairs against the right.
About how many goshawk pairs were counted in year 13?
- A. 5 pairsThe dashed goshawk line crosses the year-13 gridline at the 10 gridline of the right axis, not the 5.
- B. ✓ 10 pairs
- C. 200 pairs200 is the left axis’s number on that gridline, and the left axis carries the ruffed grouse.
Why: The goshawk pairs are read against the right y-axis.
The dashed line crosses the year-13 gridline at the right axis’s 10 gridline.
So about 10 goshawk pairs were counted in year 13.
A student says: “When the hare count crashes, the lynx count falls soon after, because fewer hares means less food for the lynx.”
Is the student correct?
- A. No: the lynx count falls first, and the fall in lynx is what makes the hare count crashThe hares crash because many lynx kill many hares; the lynx fall afterwards, when the hares are scarce.
- B. ✓ Yes: with fewer hares, fewer lynx are fed, so the lynx count falls after the hare count
Why: Hares are the lynx’s food.
When the hare count crashes, fewer lynx are fed.
So the lynx count falls after the hare count.
Suppose the count of a prey animal in a valley doubles over three years, and the count of the predator that hunts it climbs over the same years.
Which of the following is the reason the predator count climbs?
- A. The predators raise more young so that the prey count is brought back downA predator does not raise young to bring another count down; each predator eats and raises young when it is fed.
- B. More prey means more space for the predators to live inThe prey are the predators’ food, not their space: more prey gives more meals, not more room.
- C. ✓ More prey means more food for the predators, so more predators are fed
Why: The prey are the predators’ food.
More prey means more food.
So more predators are fed, and the predator count climbs.
Suppose the ruffed grouse count in a forest has crashed to its lowest in ten years, and the goshawk count is now falling.
Over the next few years, which of the following does the ruffed grouse count do?
- A. ✓ The ruffed grouse count climbs
- B. The ruffed grouse count keeps fallingWith few goshawks, few grouse are killed, so the grouse count does not keep falling.
- C. The ruffed grouse count stays at its lowestFew goshawks kill few grouse, and the grouse that remain raise young, so the count does not stay at its lowest.
Why: The goshawk count is falling, so fewer goshawks hunt the grouse.
Fewer grouse are killed.
So the ruffed grouse count climbs again.
Suppose an ecologist has twenty years of counts for two species in one valley. One species’ count lies between 400 and 6 000 animals. The other species’ count lies between 4 and 60 animals.
Which of the following is the reason to plot the second species against its own y-axis?
- A. Two y-axes let the two counts be compared with each other by the heights of their two linesThe two axes carry different scales, so the heights of the two lines are never compared.
- B. Two y-axes let the larger count fit inside the frame instead of going off the top of the graphThe larger count fits on one axis already; the second axis is for the smaller count.
- C. ✓ On the first species’ axis, the second species’ line would lie flat along the bottom, unreadable
Why: The first axis must reach 6 000 to fit the larger count.
On that axis a count of 60 sits almost on the x-axis: a flat line along the bottom.
A second y-axis numbered to 60 stretches that line so it can be read.
The graph below shows the years 1900 to 1910 of the pelt records, both lines, with a gridline at every year.
In which year does the hare count peak?
- A. 1903In 1903 the solid hare line is still climbing steeply.
- B. ✓ 1904
- C. 1905In 1905 the solid hare line has already turned down.
Why: The hare pelts are the solid line.
The solid line is highest at the 1904 gridline.
So the hare count peaks in 1904.
A student looks at the ninety-year graph and says: “The lynx count and the hare count rise and fall in the same years, so the two animals must be competing for the same food.”
Is the student correct?
- A. ✓ No: the lynx eat the hares, so the lynx count follows the hare count
- B. Yes: two counts that rise and fall together must share one food supplyThe lynx hunt the hares: the hare count feeds the lynx count, and the lynx peaks come with or after the hare peaks.
Why: The lynx eat the hares.
More hares feed more lynx, so the lynx count climbs after the hare count.
Two counts that rise and fall in turn like this show a predator and its prey, not two competitors.
Suppose fennec foxes, small foxes of the desert, hunt jerboas, small hopping rodents, across a stretch of desert. Both animals are counted every spring. In one spring the jerboa count reaches its highest in ten years.
(a) Predict what the fennec fox count does by the next spring. (1 pt)
- Award 1 point for: the fennec fox count climbs (accept: rises; increases; peaks after the jerboas).
(b) Justify your prediction. (2 pt)
A high jerboa count gives the foxes more food.
So more foxes are fed, and the fox count climbs after the jerboa count.
- Award 1 point for: more jerboas mean more food for the fennec foxes.
- Award 1 point for: so more foxes are fed and the fox count climbs after the jerboa count (accept: the predator lags the prey).
- Accept reasoning consistent with a wrong prediction in (a): award the second point only for a food link that leads to the direction the student predicted.
Glossary
- dual-y-axis graph
- A graph with two y-axes, one scale for each of two quantities plotted against the same x-axis. Each line is read against its own y-axis, and the two scales are never compared with each other.
APBIO-U08-L41B Draw the two-axis graph
Suppose wolves and moose on an island are counted every year for twenty years. The moose counts lie between 700 and 2 400 moose. The wolf counts lie between 12 and 48 wolves.
On one y-axis that fits the moose, the wolf line lies flat along the bottom of the graph. Not one of its rises and falls can be read. How do you draw the moose count and the wolf count so that each line can be read?
Unit 8 · Ecology
1Build the graph, one axis at a time
Suppose a student counts one population at the end of every year for ten years, and then graphs the ten counts.
Which quantity goes on the x-axis?
- A. ✓ Time, in years
- B. Population size, in individualsThe population size was measured, so it goes on the y-axis.
Why: The years were set: the count was made at the end of every year.
The quantity that was set goes on the x-axis.
So time, in years, goes on the x-axis.
A dual-y-axis graph carries two lines, one for each of two populations, and two y-axes, one at the left and one at the right.
Against which y-axis is each line read?
- A. Both lines against the left y-axisThe left y-axis carries one population’s scale only; the other population’s line has a y-axis of its own.
- B. Each line against whichever y-axis is nearer to itNearness decides nothing: the legend and the axis titles say which y-axis carries which line.
- C. ✓ Each line against its own y-axis
Why: Each population has its own y-axis, with its own scale.
The legend and the axis titles say which line belongs to which y-axis.
So each line is read against its own y-axis, and the left scale is never compared with the right scale.
How do you put two populations of very different size on one graph?
Time goes on the x-axis.
The moose take the left y-axis, numbered from 0 to 2 500 moose.
The wolves take the right y-axis, numbered from 0 to 50 wolves.
Each y-axis title names the species, the quantity and its unit.
A legend names the moose line and the wolf line.
Drawn this way, the wolf peaks show up as clearly as the moose peaks.
So the lag between a moose peak and the wolf peak after it can be read.
Video: Watch: Build the graph, one axis at a time
The twenty years go on the x-axis. The left y-axis is numbered from 0 to 2 500 moose, and the moose points go on. The right y-axis is numbered from 0 to 50 wolves, and the wolf points go on. The points are joined, and the legend names the moose line and the wolf line.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L41Ba.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L41Ba.mp4
The survey’s counts are in the table below: the year, the moose counted and the wolves counted.
The smallest moose count is 700 moose, in year 12. The largest is 2 400 moose, in year 8.
The smallest wolf count is 12 wolves, in year 5. The largest is 48 wolves, in year 10.
Try one y-axis first, numbered from 0 to 2 500, a gridline every 500.
Every moose count sits clear of the x-axis. So the moose line can be read.
Every wolf count is below 50 wolves. So every wolf point sits under the first gridline, 500, pressed against the x-axis.
The wolf line lies flat. Its rises and falls cannot be read.
A count of 48 wolves needs a scale that ends near 50, not at 2 500. So the wolves need a y-axis of their own.
The island’s moose and wolves were counted once a year: the year was set. So the year goes on the x-axis, from year 0 to year 20, a gridline every year.
The x-axis title is year of the survey.
The moose take the left y-axis, as the hares did on the graph of hares and lynx.
The largest moose count is 2 400 moose. So the left y-axis is numbered from 0 to 2 500, a round number just above 2 400, with a gridline every 500.
Its title names the species, the quantity and its unit: moose population size, N (individuals).
The wolves take the right y-axis, as the lynx did on that graph.
The largest wolf count is 48 wolves. So the right y-axis is numbered from 0 to 50 wolves, a round number just above 48, with a gridline every 10.
Its title names the wolves the same way: wolf population size, N (individuals).
The 10-wolf gridline of the right y-axis is the same line as the 500-moose gridline of the left y-axis. One set of gridlines serves the left y-axis and the right y-axis.
Each line is read against its own y-axis only.
In year 1 the survey counted 1 000 moose. Go up the year-1 gridline to the 1 000 gridline of the left y-axis, and mark the point there.
In year 2 the survey counted 1 500 moose. Go up the year-2 gridline to the 1 500 gridline of the left y-axis, and mark the point there.
In year 1 the survey counted 30 wolves. Go up the year-1 gridline to the 30 gridline of the right y-axis, and mark the point there.
In year 2 the survey counted 24 wolves. That point sits between the 20 gridline and the 30 gridline of the right y-axis, a little under halfway up.
The other 36 points go on the same way: each count above its year, at its height on its own y-axis.
Join the moose points, each to the next, with straight lines: the moose line is solid. Join the wolf points the same way: the wolf line is dashed.
A legend at the top names the two lines: solid line, moose; dashed line, wolves.
Now the wolf line fills the frame. Its peaks, 48 wolves in year 10 and 44 wolves in year 19, are as clear as the moose peaks, 2 400 moose in year 8 and 2 100 moose in year 17.
Each wolf peak comes two years after a moose peak.
Each line has a y-axis of its own. So the lag can be read.
What you are expected to know Construct a dual-y-axis graph from two series of different range: time on the x-axis, each series against its own y-axis, numbered to fit it and titled with species, quantity and unit, the points joined, and a legend.
Suppose cottontails, a kind of rabbit, and the red foxes that hunt them are counted every year for twelve years on a ranch. The cottontail counts lie between 300 and 3 600 cottontails. The red fox counts lie between 8 and 70 red foxes.
Which of the following scales fits the red foxes’ y-axis?
- A. 0 to 40 red foxes, a gridline every 5The largest red fox count is 70.
On a scale that ends at 40, every count above 40 sits above the top of the axis. - B. ✓ 0 to 80 red foxes, a gridline every 10
- C. 0 to 4 000 red foxes, a gridline every 500On a scale that ends at 4 000, every red fox count sits under the first gridline, 500, pressed against the x-axis.
Why: The largest red fox count is 70.
A round number just above 70 is 80.
So the red foxes’ y-axis is numbered from 0 to 80 red foxes, a gridline every 10, and every count fits inside it.
Suppose cottontails, a kind of rabbit, and the red foxes that hunt them are counted every year for twelve years on a ranch. The graph frame below carries the cottontails’ y-axis at the left, 0 to 4 000, and the red foxes’ y-axis at the right, 0 to 80. In year 5 the survey counted 40 red foxes. Four rings, lettered J, K, M and P, mark four positions on the frame.
Which ring marks the point for the red foxes in year 5?
- A. Ring JRing J sits at 40 on the left y-axis, the cottontails’ axis.
The red foxes are plotted against the right y-axis. - B. Ring KRing K sits above year 4.
The count of 40 red foxes was made in year 5. - C. Ring MRing M sits on the 50 gridline of the right y-axis.
The count was 40 red foxes. - D. ✓ Ring P
Why: The red foxes are plotted against the right y-axis.
Go up the year-5 gridline to the 40 gridline of the right y-axis.
Ring P sits there.
Suppose cottontails, a kind of rabbit, and the red foxes that hunt them are counted every year for twelve years on a ranch. The graph frame below carries the cottontails’ y-axis at the left, 0 to 4 000, and the red foxes’ y-axis at the right, 0 to 80. In year 5 the survey counted 2 500 cottontails. Three rings, lettered R, S and T, mark three positions on the frame.
Which ring marks the point for the cottontails in year 5?
- A. ✓ Ring R
- B. Ring SRing S sits on the 2 000 gridline of the left y-axis.
The count was 2 500 cottontails. - C. Ring TRing T sits above year 6.
The count of 2 500 cottontails was made in year 5.
Why: The cottontails are plotted against the left y-axis.
Go up the year-5 gridline to the 2 500 gridline of the left y-axis.
Ring R sits there.
Suppose waxwings, small crested songbirds, and the merlins, small falcons, that hunt them are counted every year for ten years on a stretch of birch scrub. The waxwing counts lie between 200 and 1 800 waxwings. The merlin counts lie between 4 and 36 merlins. Three graphs of the counts, lettered F, G and H, are drawn below.
Which graph shows the waxwing count and the merlin count so that each line can be read?
- A. Graph FIn graph F the right y-axis carries no numbers, and the dashed merlin line lies flat along the x-axis: the merlins were plotted against the left y-axis, the waxwings’ axis.
- B. Graph GIn graph G the right y-axis is numbered to 400.
Every merlin count is below 40, so the merlin line sits pressed against the x-axis. - C. ✓ Graph H
Why: Graph H plots the waxwings against the left y-axis, 0 to 2 000, and the merlins against the right y-axis, 0 to 40.
The right y-axis ends just above the largest merlin count, 36.
So the merlin line fills the frame, and each line can be read.
A student drawing the island’s graph of moose and wolves says: “The largest moose count is 2 400, so I will number the left y-axis from 0 to 5 000, to leave plenty of room above the moose line.”
Is the student correct?
- A. ✓ No: the top of a scale is a round number just above the largest count, so the left y-axis is numbered from 0 to 2 500
- B. Yes: a scale that ends well past the largest count leaves the moose line room, so 0 to 5 000 fits the mooseOn a scale to 5 000, the moose line uses only the lower half of the frame, and its rises and falls shrink by half.
Why: The largest moose count is 2 400.
A round number just above 2 400 is 2 500.
A scale to 5 000 squashes the moose line into the lower half of the frame.
So the left y-axis is numbered from 0 to 2 500.
A student drawing the island’s graph of moose and wolves says: “The largest wolf count is 48, so a right y-axis numbered from 0 to 50 shows every wolf count clearly.”
Is the student correct?
- A. No: the left y-axis and the right y-axis carry the same scale, so the right y-axis is numbered to 2 500 like the leftThe left y-axis and the right y-axis carry different scales.
On a scale to 2 500, every wolf count sits under the first gridline. - B. ✓ Yes: 50 is a round number just above 48, so a right y-axis to 50 fits every wolf count and the wolf line fills the frame
Why: The largest wolf count is 48.
A round number just above 48 is 50.
So a right y-axis numbered from 0 to 50 fits every wolf count, and the wolf line fills the frame.
Suppose brown hares and the eagle owls that hunt them are counted every spring for fifteen years. A student plots the hare count and the owl count against one y-axis, as the graph below shows. The largest owl count is 34 owls.
(a) Identify the feature of the student’s graph that stops the owl counts from being read. (1 pt)
Every owl count is below 40, so the owl line lies flat along the x-axis, under the first gridline.
- Award 1 point for: the owls are plotted against the hares’ y-axis (0 to 1 500), so the owl line lies flat along the bottom of the frame / under the first gridline.
(b) Describe how the student should redraw the graph so that the owl counts can be read: which axis the owls take, its scale and its title. (3 pt)
Its scale is numbered from 0 to 40 owls, a round number just above the largest owl count, 34, with a gridline every 5.
Its title names the species, the quantity and its unit: eagle owl population size, N (individuals).
- Award 1 point for: a second y-axis at the right of the graph, for the owls.
- Award 1 point for: a scale whose top is a round number just above the largest owl count, 34 (accept 0 to 35 or 0 to 40), with a gridline at a stated interval.
- Award 1 point for: a title naming the species, the quantity and its unit (eagle owl population size, N (individuals); accept eagle owls counted (individuals)).
The table below lists seven features of a finished dual-y-axis graph, each lined up against the island’s graph.
A graph that fails any row of the table is drawn wrongly.
The island’s graph returns below: twenty years of moose and wolves, the moose numbered from 0 to 2 500 on the left y-axis and the wolves from 0 to 50 on the right y-axis.
Each line is read against its own y-axis. The wolf line is as clear as the moose line.
The wolf peaks, in years 10 and 19, come two years after the moose peaks, in years 8 and 17.
54Quick quiz: which axis, which scale mixed practice
A dual-y-axis graph carries a left y-axis numbered to 3 000 individuals and a right y-axis numbered to 60 individuals. One series of counts lies between 6 and 55 individuals.
Which y-axis takes this series?
- A. The left y-axisOn the left y-axis, numbered to 3 000, every count below 60 sits pressed against the x-axis.
- B. ✓ The right y-axis
Why: The largest count is 55.
The right y-axis ends at 60, just above 55.
So the series takes the right y-axis.
A series of counts lies between 3 and 27.
Which of the following scales fits this series?
- A. ✓ 0 to 30, a gridline every 5
- B. 0 to 300, a gridline every 50On a scale to 300, every count below 30 sits under the first gridline, 50, pressed against the x-axis.
- C. 0 to 3 000, a gridline every 500A scale to 3 000 ends a hundred times above the largest count, 27, so the line lies flat along the x-axis.
Why: The largest count is 27.
A round number just above 27 is 30.
So the scale from 0 to 30 fits every count, and the line fills the frame.
A dual-y-axis graph carries a left y-axis numbered to 3 000 individuals and a right y-axis numbered to 60 individuals. One series of counts lies between 350 and 2 700 individuals.
Which y-axis takes this series?
- A. ✓ The left y-axis
- B. The right y-axisThe right y-axis ends at 60.
Every count above 60 would sit above the top of that axis.
Why: The largest count is 2 700.
The left y-axis ends at 3 000, just above 2 700.
So the series takes the left y-axis.
A series of counts lies between 90 and 760.
Which of the following scales fits this series?
- A. 0 to 80, a gridline every 10On a scale to 80, every count sits above the top of the axis.
- B. 0 to 700, a gridline every 100On a scale to 700, the largest count, 760, sits above the top of the axis.
- C. ✓ 0 to 800, a gridline every 100
Why: The largest count is 760.
A round number just above 760 is 800.
So the scale from 0 to 800 fits every count.
A series of counts lies between 40 and 360.
Which of the following scales fits this series?
- A. 0 to 40, a gridline every 5On a scale to 40, every count above 40 sits above the top of the axis.
- B. ✓ 0 to 400, a gridline every 50
- C. 0 to 4 000, a gridline every 500On a scale to 4 000, every count below 400 sits under the first gridline, 500, pressed against the x-axis.
Why: The largest count is 360.
A round number just above 360 is 400.
So the scale from 0 to 400 fits every count, and the line fills the frame.
APBIO-U08-L42 When the wolves came back
Photos: Doug Smith / National Park Service, Wikimedia Commons, public domain (resized); Jan Kronsell, Wikimedia Commons, public domain (resized).
In 1995, wolves were released into Yellowstone National Park, in the western United States, after about seventy years without them.
Wolves eat elk. In the years that followed, the elk were fewer, and the elk that remained kept away from the open streambanks. Along those streams, young willows that the elk had grazed began to grow again. How did adding one animal at the top of the food chain change the trees at the bottom?
Unit 8 · Ecology
1A change at the top runs down the chain
A goanna, a large lizard, catches a ship rat and eats it. Write a plus for a population that gains and a minus for a population that loses.
Which of the following gives the sign each population gets?
- A. Goannas −, ship rats +The goanna gains a meal, so the goannas get a plus.
- B. Goannas +, ship rats +The ship rat is killed and eaten, so the ship rats lose: a minus.
- C. ✓ Goannas +, ship rats −
Why: The goanna catches the ship rat and eats it.
The goannas gain a meal: a plus.
The ship rats lose a member: a minus.
One population catching and eating members of another is predation.
Suppose an island’s food web has each arrow going from the eaten organism to the organism that eats it. One arrow goes from the wetas, large flightless insects, to the moreporks, small owls.
Which of the following does that arrow say?
- A. ✓ The moreporks eat the wetas
- B. The wetas eat the moreporksThe arrow starts at the eaten organism and ends at the eater, so the wetas are the eaten.
Why: Each arrow is drawn from the eaten organism to the eater, the way the energy goes.
This arrow starts at the wetas and ends at the moreporks.
So the moreporks eat the wetas.
How does a change in one population reach a population two steps away?
Wolves eat elk. So more wolves means fewer elk, and the elk that remain keep away from the open streambanks.
Elk eat young willows. So fewer elk means more young willows.
The change passed from the top of the food chain down two levels: a trophic cascade.
Draw the populations as circles joined by arrows, each arrow carrying the sign of its effect. Then the drawing predicts the change for any chain of three before anyone counts them.
Video: Watch: A change at the top runs down the chain
Three circles stand in a column: wolves, elk, willows. The wolf circle grows, and an arrow with a minus is drawn down to the elk circle, which shrinks. A second arrow with a minus is drawn down to the willow circle, which grows. The word up appears beside the wolves, down beside the elk and up beside the willows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L42a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L42a.mp4
A gray wolf hunts in a pack and eats large animals. An elk is a large deer that eats grass and the young shoots of trees and bushes.
The photographs below show a gray wolf lying in snow and an elk grazing.
In Yellowstone, wolves catch and eat elk. So the wolves are the predator and the elk are the prey.
The wolves get a plus, because the wolves gain a meal. The elk get a minus, because the elk lose members.
The drawing below shows the wolves’ effect on the elk. An arrow runs from the wolves to the elk, with a minus at its head.
In 1995, wolves were released into the park. The wolves killed elk, and over the years that followed the elk count fell.
The elk that remained changed where they grazed. To avoid the wolves, the elk no longer grazed the open streambanks and riverbeds.
Willows are trees and bushes that grow along streams.
Elk eat the young willow shoots. So few young willows grow tall where many elk graze.
The elk get a plus, because the elk gain food. The willows get a minus, because the willows lose their young shoots.
With fewer elk on the streambanks, fewer young willows were eaten. So young willows grew along the streams again.
Put the two steps together. The wolves eat the elk, and the elk eat the young willows: a food chain of three levels.
The wolves went up. The wolves eat elk, so the elk went down.
The elk eat young willows, so the willows went up.
The change started with the wolves, at the top of the food chain. The change passed down to the elk, one level below, and then to the willows, two levels below.
A change in one population that passes down the food chain through two or more trophic levels is called a .
A cascade is water falling step by step down a slope. A trophic cascade is a change falling step by step down the trophic levels.
Each step of a trophic cascade has a direction: up or down.
Along a minus arrow, the direction flips. More wolves means fewer elk, and fewer elk means more willows.
Two flips bring the direction back. So the wolves and the willows move the same way, and the elk move the opposite way.
A change in one population passes down the chain: predator up, prey down, the prey’s food up.
Suppose someone asks what the wolves’ return does to the willows. “The willows will be affected” is not a prediction, because it does not say which way the willows go.
A prediction names the direction: more young willows grow. An answer that names no direction earns nothing.
What you are expected to know Predict how a change in one population passes down two or more trophic levels, naming the direction at each level.
Wolves were released into Yellowstone National Park in 1995. Wolves eat elk.
Over the years that followed, which of the following did the elk count do?
- A. The elk count roseThe wolves killed elk, so the elk count could not rise.
- B. ✓ The elk count fell
Why: Wolves eat elk.
More wolves killed more elk.
So the elk count fell.
In Yellowstone, wolves eat elk, and elk eat the young willow shoots along the streams. Wolves were released into the park in 1995.
Over the years that followed, which of the following did the young willows do?
- A. Fewer young willows grewFewer elk grazed the streambanks, so fewer young willows were eaten.
- B. ✓ More young willows grew
Why: More wolves killed more elk, so the elk count fell.
Fewer elk ate fewer young willow shoots.
So more young willows grew.
Suppose wildcats return to a forest where they had been hunted out. Wildcats eat wood mice. Wood mice eat bilberry seedlings, the young plants of a small bush. After the wildcats return, more bilberry seedlings survive.
(a) Explain why more bilberry seedlings survive after the wildcats return. (2 pt)
Frame Wildcats eat wood mice, so …
So there are fewer wood mice.
Wood mice eat bilberry seedlings, so fewer wood mice eat fewer seedlings.
So more bilberry seedlings survive.
- Award 1 point for: more wildcats kill more wood mice, so there are fewer wood mice (the direction must be stated).
- Award 1 point for: fewer wood mice eat fewer bilberry seedlings, so more seedlings survive (the direction must be stated). Do not award: ‘the seedlings are affected’ with no direction.
Suppose sea eagles return to a bay where they had died out. Sea eagles eat brant, small geese, and brant graze the bay’s seagrass. A student is asked to predict what the sea eagles’ return does to the seagrass. The student writes: “The seagrass will be affected, so that is my prediction.”
Is the student correct?
- A. ✓ No: a prediction names the direction, and here more seagrass grows
- B. Yes: the student has said that the seagrass changes, and that is the prediction“Affected” does not say which way the seagrass goes, so it predicts nothing.
Why: More sea eagles eat more brant, so the brant count falls.
Fewer brant graze less seagrass.
So more seagrass grows.
A prediction names that direction; “affected” names none.
Sea otters eat sea urchins. Sea urchins destroy kelp if they are left alone. Now imagine the sea otters along a stretch of coast are gone.
Which of the following does the kelp do?
- A. ✓ Less kelp grows
- B. More kelp growsWith no otters eating them, the sea urchins multiply, and more urchins destroy more kelp.
Why: The sea otters are gone, so fewer sea urchins are eaten.
The sea urchin count rises.
More sea urchins destroy more kelp.
So less kelp grows.
39Quick quiz: trophic cascade mixed practice
(a) State what a trophic cascade is. (1 pt)
- Award 1 point for: a change in one population passing down the food chain through two or more trophic levels (accept: predator up, prey down, the prey’s food up).
A change in one population passes down the food chain through two or more trophic levels.
Which of the following is this called?
- A. a food chainA food chain is the line of organisms, each eaten by the next; it is not a change passing along that line.
- B. ✓ a trophic cascade
- C. predationPredation is one population catching and eating members of another: one link, not a change passing down two or more.
Why: The change passes down through two or more trophic levels.
A change passing down the food chain like that is a trophic cascade.
42Draw the chain, then predict from it
Dodder is a plant with no leaves that winds around another plant and takes sugar from it. In the drawing below, an arrow runs from the dodder to the alfalfa, a crop plant. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which of the following does the drawing say?
- A. The alfalfa harms the dodderThe arrow ends on the alfalfa, so its sign is the effect on the alfalfa.
- B. ✓ The dodder harms the alfalfa
Why: The arrow runs from the dodder to the alfalfa.
The sign at its head is a minus.
A minus at the head is a loss for the population the arrow ends on: the alfalfa.
So the dodder harms the alfalfa.
How do you draw a chain of populations so that the drawing predicts?
Draw each population as a circle, the eater above what it eats.
Draw one arrow from each eater down to what it eats, and write the sign of its effect at the arrow’s head.
Then read down from the population that changed: at every minus arrow, the direction flips.
Video: Watch: Draw the chain, then predict from it
Three empty circles appear in a column and are labeled wolves, elk, willows. An arrow is drawn from the wolves down to the elk, and a minus is written at its head. A second arrow is drawn from the elk down to the willows, with a minus at its head. A finger starts at the wolves with the word up. The finger moves down the first arrow and writes down beside the elk. Then the finger moves down the second arrow and writes up beside the willows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L42b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L42b.mp4
Take the three Yellowstone populations: wolves, elk and young willows.
1 Draw each population as a circle with its name inside, the eater above what it eats. So the wolves go at the top, the elk in the middle and the willows at the bottom.
2 Draw one arrow from each eater down to what it eats. One arrow runs from the wolves to the elk, and one from the elk to the willows.
A food-web arrow runs from the eaten to the eater, the way the energy goes. This chain’s arrows run the other way, from the eater to the eaten: the way each effect goes.
3 At each arrow’s head, write the sign of the eater’s effect on the eaten. The wolves harm the elk, a minus; the elk harm the willows, a minus.
A pair of populations gets two arrows, one each way.
A chain drawing keeps only the eater’s arrow down to the eaten. That arrow is the one the change travels along.
Start at the population that changed and read down the arrows.
The wolves went up. The arrow from the wolves to the elk carries a minus, so the elk went down.
The arrow from the elk to the willows carries a minus, so the willows went up.
Now imagine the wolves gone from the park again. Read down the same drawing: wolves down, so elk up, so willows down.
The drawing predicts both ways, and it needs no count of any population.
What you are expected to know Represent interactions as circles joined by signed arrows, and predict from the drawing which way a third population goes when one is removed.
Now suppose sea otters eat sea urchins along a coast, and sea urchins destroy kelp if they are left alone. The drawing in the two checks below has the three circles and the two arrows, with no signs yet.
Sea otters eat sea urchins. In the drawing below, an arrow runs from the sea otters to the sea urchins. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which sign goes at the head of the arrow from the sea otters to the sea urchins?
- A. ✓ a minus (−)
- B. a plus (+)The otters eat the urchins, so the urchins lose members; a plus would say the urchins gain.
Why: The sea otters eat the sea urchins.
The sea urchins lose members.
A population that loses gets a minus, at the head of the arrow that ends on the urchins.
Sea urchins destroy kelp if they are left alone. In the drawing below, an arrow runs from the sea urchins to the kelp. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which sign goes at the head of the arrow from the sea urchins to the kelp?
- A. ✓ a minus (−)
- B. a plus (+)The urchins destroy the kelp, so the kelp loses; a plus would say the kelp gains.
Why: The sea urchins destroy the kelp.
The kelp loses.
A population that loses gets a minus, at the head of the arrow that ends on the kelp.
The finished drawing carries a minus at the head of each arrow.
Now imagine the sea otters gone. Read down from the otters: otters down, so urchins up, so kelp down.
The drawing gives the prediction before anyone counts an urchin.
Suppose dholes, wild dogs of Asia, eat chital, a deer. Chital eat the young shoots of teak trees. Three students drew the chain below, in drawings J, K and M. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which drawing represents the two sentences correctly?
- A. Drawing JDrawing J puts a plus at the chital: it says the dholes help the chital.
- B. ✓ Drawing K
- C. Drawing MDrawing M draws its upper arrow from the chital to the dholes: it says the chital harm the dholes.
Why: Dholes eat chital, so the arrow runs from the dholes down to the chital with a minus at its head.
Chital eat young teak, so the arrow runs from the chital down to the teak with a minus at its head.
Drawing K has both.
Now suppose secretary birds and honey badgers both eat black rats on a farm, and black rats eat cassava, a root crop.
A model of the four populations is drawn in the two checks below. The sign of each effect sits at the arrow’s head.
The model below draws four populations on a farm: secretary birds and honey badgers both eat black rats, and black rats eat cassava, a root crop. The sign at an arrow’s head is the effect on the population the arrow ends on. Now imagine the secretary birds are gone.
Which of the following does the black rat count do?
- A. The black rat count fallsFewer black rats are eaten when one of their eaters is gone, so the black rat count rises.
- B. ✓ The black rat count rises
Why: The arrow from the secretary birds to the black rats carries a minus.
The secretary birds go down, and along a minus arrow the black rats go the opposite way.
So the black rat count rises.
The model below draws four populations on a farm: secretary birds and honey badgers both eat black rats, and black rats eat cassava, a root crop. The sign at an arrow’s head is the effect on the population the arrow ends on. Now imagine the secretary birds are gone.
Which of the following does the cassava do?
- A. More cassava growsMore black rats eat more cassava, so less cassava grows.
- B. The cassava will be affected“Affected” does not say which way the cassava goes, so it predicts nothing.
- C. ✓ Less cassava grows
Why: The secretary birds go down.
Along the minus arrow to the black rats, the rats go up.
Along the minus arrow from the rats to the cassava, the cassava goes down.
So less cassava grows.
In 1995, wolves were released into Yellowstone National Park after about seventy years without them.
The wolves killed elk. So the elk count fell, and the elk that remained kept away from the open streambanks.
Fewer elk ate fewer young willow shoots. So young willows grew along the streams again.
Drawn as a chain, each arrow carries a minus at its head. The change reads down the arrows: wolves up, elk down, willows up.
The same drawing for sea otters, sea urchins and kelp gives the same reading. Losing the otters lets the urchins rise and the kelp fall.
78Mixed practice: cascades and chains mixed practice
Suppose jackals eat dik-diks, small antelope, and dik-diks eat the young shoots of acacia trees. Now imagine the jackals are gone from a stretch of grassland.
Which of the following do the young acacia trees do?
- A. ✓ Fewer young acacia trees grow
- B. More young acacia trees growWith no jackals, more dik-diks eat more young shoots, so fewer young trees grow.
Why: The jackals are gone, so fewer dik-diks are eaten and the dik-dik count rises.
More dik-diks eat more young acacia shoots.
So fewer young acacia trees grow.
Sand cats eat sandgrouse, ground birds, and sandgrouse eat the seeds of desert grasses. The drawing below shows sand cats, sandgrouse and desert grasses as a chain, with the sign of each effect at the arrow’s head. The sign at an arrow’s head is the effect on the population the arrow ends on. Now imagine sand cats return to a desert where they had died out.
Which of the following does the desert grass do?
- A. ✓ More desert grass grows
- B. Less desert grass growsMore sand cats eat more sandgrouse, so fewer sandgrouse eat grass seeds, and more grass grows.
Why: The sand cats go up.
Along the minus arrow to the sandgrouse, the sandgrouse go down.
Along the minus arrow to the desert grasses, the grasses go up.
So more desert grass grows.
A student reads that after wolves returned to Yellowstone National Park, young willows grew along the streams again. The student says: “The wolves changed the willows, although wolves eat elk and leave the willows uneaten, because the change passed through the elk.”
Is the student correct?
- A. No: a population can only change a population it eats, so the wolves changed the elk aloneA change passes down the chain level by level, so it reaches populations the wolves never eat.
- B. ✓ Yes: the wolves changed the elk, and the elk changed the willows, so the wolves’ effect reached the willows
Why: Wolves eat elk, so more wolves means fewer elk.
Elk eat young willows, so fewer elk means more young willows.
The wolves’ effect passed through the elk to the willows: a trophic cascade.
Suppose harpy eagles, large forest eagles, eat spider monkeys, and spider monkeys eat the young leaves of cecropia trees. A student says: “If the harpy eagles die out, the spider monkeys will die out too, because the whole chain falls together.”
Is the student correct?
- A. ✓ No: with fewer harpy eagles, fewer spider monkeys are eaten, so the spider monkey count rises
- B. Yes: every population in a chain rises and falls with the population at the topAlong a minus arrow the direction flips: a predator’s fall is its prey’s rise.
Why: Harpy eagles eat spider monkeys.
Fewer harpy eagles eat fewer spider monkeys.
So the spider monkey count rises; it does not fall with the eagles.
Suppose fruit bats eat the young leaves of breadfruit trees, and the trees lose those leaves. A student draws the two populations below to show the bats’ effect on the trees. The sign at an arrow’s head is the effect on the population the arrow ends on. Judge the arrow as the student drew it.
Which of the following is true of the student’s drawing?
- A. ✓ The arrow ends on the wrong population
- B. The sign at the arrow’s head is wrongThe trees feed the bats, so a plus is right for the arrow as drawn.
The fault is the arrow’s end: the arrow shows the trees’ effect on the bats. - C. The drawing is correctThe arrow ends on the bats, so its plus is the trees’ effect on the bats.
The bats’ effect on the trees needs an arrow ending on the trees.
Why: The drawing must show the bats’ effect on the trees, so its arrow must end on the trees.
The student’s arrow ends on the bats.
For the arrow as drawn, the plus is right: the trees feed the bats.
So the arrow ends on the wrong population.
Three events are described.
Which of the following is a trophic cascade?
- A. A cat catches a bird and eats itOne population eating another is predation: one link, and no change passing down the chain.
- B. Two bird species hunt the same insects in one wood, and each leaves fewer insects for the otherTwo populations needing the same food is competition: one link, and no change passing down the chain.
- C. ✓ More moreporks eat more wetas, so more of the plants the wetas eat survive
Why: The change passes from the moreporks to the wetas, then to the plants: two levels down the food chain.
A change passing down two or more trophic levels is a trophic cascade.
Suppose a dry grassland lies in Australia. Dingoes, wild dogs, eat wallaroos, large kangaroos. Wallaroos graze spinifex, a spiky grass. The model below draws the three populations, with the arrows but no signs. The sign at an arrow’s head is the effect on the population the arrow ends on.
(a) Identify the sign that belongs at the head of the arrow from the dingoes to the wallaroos. (1 pt)
- Award 1 point for: a minus (accept: −; negative).
(b) Predict what happens to the spinifex if the dingoes are removed from the grassland. (1 pt)
- Award 1 point for: less spinifex / the spinifex decreases (the direction must be stated). Do not award: ‘the spinifex is affected’ with no direction.
(c) Justify your prediction. (1 pt)
More wallaroos graze more spinifex.
So less spinifex grows.
- Award 1 point for: fewer wallaroos are eaten so the wallaroo count rises, AND more wallaroos graze more spinifex (both links, each with its direction).
- Accept reasoning consistent with a wrong prediction in (b): award the point only for two links, each with its direction, that lead to the direction the student predicted.
Glossary
- trophic cascade
- A change in one population that passes down the food chain through two or more trophic levels: predator up, prey down, the prey’s food up.
APBIO-U08-L43 Do the two plots really differ?
Suppose a grassland is split into twenty plots. Cattle graze ten of the plots. The other ten plots are fenced off. No cattle graze them. An ecologist counts the plant species in every plot.
On average, a grazed plot has 8.2 species. On average, an ungrazed plot has 12.4. Each mean is drawn as a bar with an error bar of two standard errors either side. The grazed bar’s error bar is ±0.9 species. The ungrazed bar’s error bar is ±1.1 species. Is the gap between 8.2 and 12.4 real, or could it be chance?
Unit 8 · Ecology
1Do the two error bars overlap?
Suppose one plot of grassland has 210 plants of 12 species.
What is the plot’s species richness?
- A. 210 plants210 is the number of plants, not the number of species.
- B. 210 plants shared equally among the speciesHow the plants are shared among the species is the plot’s species evenness.
- C. ✓ 12 species
Why: The number of species in a community is called its species richness.
The plot has 12 species.
So its species richness is 12 species.
Suppose the mean of a set of plots is 9.0 species per plot. The mean’s error bar represents ±2SE, as its legend says.
What does the error bar show?
- A. How spread out the individual plot counts wereA standard-deviation bar shows the spread of the individual counts.
A ±2SE bar shows where the true mean is likely to lie. - B. ✓ The range the true mean is likely to lie in
Why: Two standard errors either side of a mean is the range the true mean is likely to lie in.
The legend says the bar is ±2SE.
So the error bar shows the range the true mean is likely to lie in.
In Unit 3, two mean catalase rates each carried a ±2SE error bar. The two error bars were clear of each other, with a gap between them.
What did that show about the two rates?
- A. ✓ The difference between the two rates was very unlikely to be chance
- B. The two rates were the same at the two temperaturesA gap between two ±2SE error bars shows the two true means are very unlikely to be the same.
- C. The data showed nothing until many more tubes had been testedTwo ±2SE error bars with a gap between them already show a real difference.
More tubes are not needed.
Why: Each ±2SE error bar is the range its true mean is likely to lie in.
The two error bars had a gap between them.
So the two true means were very unlikely to be the same.
So the difference between the two rates was very unlikely to be chance.
Suppose one community’s Simpson’s Diversity Index is 0.52 and a second community’s is 0.79.
Which community is the more diverse?
- A. The community scoring 0.52A lower value of Simpson’s Diversity Index means less diverse.
- B. Neither: the index measures how even the shares are, not how diverse the community isSimpson’s Diversity Index measures how varied a community is: how many species, and how equally the organisms are shared among them.
- C. ✓ The community scoring 0.79
Why: A higher value of Simpson’s Diversity Index means more diverse, never better.
0.79 is the higher value.
So the community scoring 0.79 is the more diverse.
How do you tell whether two community measurements really differ?
Look at the error bars, exactly as you did for the enzyme graphs in Unit 3.
If the two error bars do not overlap, the difference is unlikely to be chance.
If the two error bars overlap, you cannot tell. An overlap does not show that the two are equal.
On the grassland, the ungrazed bar reaches from 11.3 to 13.5 species. The grazed bar reaches from 7.3 to 9.1 species.
The two error bars are clear of each other. So grazing very likely lowers the number of plant species in a plot.
Video: Watch: Do the two error bars overlap?
The grazed mean and the ungrazed mean are drawn as two bars. Each error bar extends two standard errors above and below its mean. The top of the grazed bar is compared with the bottom of the ungrazed bar. The gap between them is marked. So the difference is very unlikely to be chance.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L43a.mp4
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Cattle graze ten plots of the grassland. The other ten plots are fenced off, and no cattle graze them.
An ecologist counts the plant species in every plot. On average, a grazed plot has 8.2 species and an ungrazed plot has 12.4.
Each mean has its standard error: 0.45 species for the grazed plots and 0.55 species for the ungrazed plots.
Each mean is drawn as a bar with an error bar of two standard errors either side. The legend says so: error bars represent ±2SE.
The grazed mean is 8.2 species per plot. Its standard error of the mean, SE, is 0.45 species. Between which values does its ±2SE error bar reach?
The ungrazed bar is worked the same way. The table below compares the two sets of plots: mean, standard error, two standard errors, and the error bar’s two ends.
The graph below draws the two means as bars, each with its ±2SE error bar, on gridlines every 1 species. The two ends of each error bar are printed beside it.
Compare the top of the lower bar with the bottom of the higher bar.
The grazed bar reaches from 7.3 to 9.1 species and the ungrazed bar from 11.3 to 13.5 species; both represent ±2SE. Do the error bars overlap, and what does that show?
The graph below marks the gap. The grazed bar’s top sits at 9.1 species, the ungrazed bar’s bottom at 11.3.
The clear space between the two error bars is 2.2 species.
Each ±2SE error bar is the range its true mean is likely to lie in. The two ranges share no value.
So the two true means are very unlikely to be the same. So the difference between 8.2 and 12.4 species is very unlikely to be chance.
Grazing very likely lowers the number of plant species in a plot.
Suppose a hayfield is split into sixteen plots. Eight plots are cut for hay each June.
Eight plots are left uncut all summer.
On average, a cut plot has 9.6 species and an uncut plot has 10.4. The graph below draws the two means with their ±2SE error bars.
The cut bar reaches from 8.4 to 10.8 species. The uncut bar reaches from 9.4 to 11.4 species.
The uncut bar’s bottom, 9.4 species, sits below the cut bar’s top, 10.8 species. So the two error bars overlap.
Each true mean could lie anywhere in its bar. So both true means could lie in the shared range, from 9.4 to 10.8 species, at the same value.
So the difference between 9.6 and 10.4 species could be chance. So these data cannot tell whether cutting changes the number of species.
Overlapping error bars mean “cannot tell”, not “the same”. The two true means may still differ.
With more plots, each standard error shrinks. So each error bar shortens.
Shorter error bars might then show a difference.
The rule is for error bars of ±2SE. So read the legend first.
What you are expected to know Decide from two means with ±2SE error bars whether two community measurements differ: bars apart, the means differ; bars overlapping, the data cannot tell.
Suppose twelve plots on a river’s floodplain flood each spring, and twelve plots sit just above the flood line. The graph below shows the mean number of plant species per plot for each set, with ±2SE error bars.
Which of the following do the two error bars show?
- A. The two means differThe above-the-flood-line bar’s bottom, 6.2 species, sits below the flooded bar’s top, 6.7 species.
So the error bars overlap. - B. ✓ The data cannot tell
Why: The flooded bar’s top is 6.7 species.
The above-the-flood-line bar’s bottom is 6.2 species.
6.2 sits below 6.7.
So the two error bars overlap.
So the data cannot tell whether the two means differ.
Suppose eight stretches of a stream are wooded and eight stretches are treeless. The graph below shows the mean number of insect species caught per net in each set, with ±2SE error bars.
Which of the following do the two error bars show?
- A. ✓ The two means differ
- B. The data cannot tellThe treeless bar’s top, 5.0 species, sits below the wooded bar’s bottom, 6.9 species.
There is clear space between the error bars.
Why: The treeless bar’s top is 5.0 species.
The wooded bar’s bottom is 6.9 species.
5.0 sits below 6.9.
So the two error bars do not overlap.
So the difference is very unlikely to be chance.
The two means differ.
Suppose nine log piles in a forest are new and nine are old. The graph below shows the mean number of beetle species per pile for each set, with ±2SE error bars.
Which of the following do the two error bars show?
- A. ✓ The two means differ
- B. The data cannot tellThe new-pile bar’s top, 11.0 species, sits below the old-pile bar’s bottom, 11.6 species.
The gap is small but real.
Why: The new-pile bar’s top is 11.0 species.
The old-pile bar’s bottom is 11.6 species.
11.0 sits below 11.6.
So the two error bars do not overlap, even though the gap is small.
So the two means differ.
Suppose a student visits a schoolyard on eight mornings and a vacant lot, an empty piece of city land, on eight mornings. On each visit the student lists the bird species seen. The graph below shows the mean number of bird species per visit for each place, with ±2SE error bars.
Which of the following do the two error bars show?
- A. The two means differThe vacant-lot bar’s bottom, 18 species, sits below the schoolyard bar’s top, 19 species.
So the error bars overlap.
The distance between the two means does not decide the question. - B. ✓ The data cannot tell
Why: The schoolyard bar’s top is 19 species.
The vacant-lot bar’s bottom is 18 species.
18 sits below 19.
So the two error bars overlap.
So the data cannot tell whether the two means differ, however far apart the two means sit.
Suppose twelve plots of cropland are in crop and twelve plots are left fallow, unplanted for the year. The graph below shows the mean number of plant species per plot for each set, with ±2SE error bars.
Which of the following do the two error bars show?
- A. ✓ The two means differ
- B. The data cannot tellThe in-crop bar’s top, 3.4 species, sits below the fallow bar’s bottom, 3.8 species.
Overlap is about position, not the length of the bars.
Why: The in-crop bar’s top is 3.4 species.
The fallow bar’s bottom is 3.8 species.
3.4 sits below 3.8.
So the two error bars do not overlap.
So the two means differ.
The table below compares the two cases: error bars that do not overlap, and error bars that overlap.
Suppose fifteen plots of chaparral, a dry shrubby hillside, burned five years ago, and fifteen plots burned twenty years ago. The graph below shows the mean number of plant species per plot for each set, with ±2SE error bars. A student says: “The error bars overlap, so the two sets of plots have the same species richness.”
Is the student correct?
- A. ✓ No: overlapping error bars mean the data cannot tell whether the two differ
- B. Yes: overlapping error bars show that the two means are the sameOverlapping error bars leave the question open.
The two true means may still differ.
More plots might show it.
Why: Each ±2SE error bar is the range its true mean is likely to lie in.
The two bars overlap.
So the two true means could be the same, or could differ.
So the data cannot tell whether the two sets of plots differ.
“Cannot tell” is not “the same”.
Suppose six plots of a highway median, the strip of grass between the two directions of traffic, are mowed every two weeks, and six plots are mowed once a year. The graph below shows the mean number of plant species per plot for each set, with ±2SE error bars. A student says: “The two error bars are clear of each other, so mowing often very likely lowered the number of plant species.”
Is the student correct?
- A. No: the two error bars are different lengths, so the two means cannot be compared with each otherOverlap is about position, not length.
The often-mowed bar’s top, 4.8 species, sits below the once-a-year bar’s bottom, 7.4 species. - B. ✓ Yes: the two error bars are clear of each other, so mowing often very likely lowered the species count
Why: The often-mowed bar’s top is 4.8 species.
The once-a-year bar’s bottom is 7.4 species.
4.8 sits below 7.4.
So the two ±2SE error bars do not overlap.
So the difference is very unlikely to be chance.
Mowing often very likely lowered the number of plant species.
Suppose an ecologist sets eight insect traps in each of two shelterbelts, lines of trees planted to slow the wind. For each trap the ecologist calculates Simpson’s Diversity Index for the insects caught. The graph below shows the two mean indices, with ±2SE error bars.
Which of the following do the data show?
- A. The north shelterbelt is the more diverseThe north bar’s top, 0.63, sits below the south bar’s bottom, 0.68.
The south belt has the higher index. - B. ✓ The south shelterbelt is the more diverse
- C. The data cannot tell which shelterbelt is the more diverseThe north bar’s top, 0.63, sits below the south bar’s bottom, 0.68.
The error bars do not overlap, so the data do show a difference.
Why: The north bar’s top is 0.63.
The south bar’s bottom is 0.68.
0.63 sits below 0.68.
So the two error bars do not overlap.
So the difference is very unlikely to be chance.
A higher index means more diverse.
So the south shelterbelt is the more diverse.
The twenty grassland plots are below once more. On average an ungrazed plot has 12.4 species and a grazed plot 8.2, each mean with its ±2SE error bar.
The ungrazed bar reaches from 11.3 to 13.5 species and the grazed bar from 7.3 to 9.1 species. The two error bars are clear of each other.
So the difference between 8.2 and 12.4 species is very unlikely to be chance. Grazing very likely lowers the number of plant species in a plot.
APBIO-U08-L44 What the interaction does to the counts
Suppose a grass brought from another continent is spreading through a valley of native bunchgrass, a grass that grows in clumps.
Where the newcomer grass grows, aphids, small insects that suck sap from grass stems, breed faster. The aphids carry a virus that kills only the native grass. Someone proposes releasing ladybugs, small beetles that eat aphids. What is happening to the native grass, and what will the ladybugs do?
Unit 8 · Ecology
1Name it, follow it, predict it
Suppose two species of duiker, small forest antelope, eat the same fallen fruit under one stand of trees, and there is never enough fruit for both. Write a plus for a population that gains, a minus for one that loses and a zero for one that is untouched.
Which sign does each duiker species get?
- A. ✓ Each species gets a minus (−)
- B. Each species gets a plus (+)Each species loses fruit to the other, so neither species gains.
Why: Each species eats fruit the other species needs.
So each species leaves less fruit for the other.
Two populations needing the same food, so that each leaves less for the other, is competition.
Competition is a minus for both populations.
Suppose two species of guinea fowl, ground birds, scratch for the same seeds in one clearing and roost in the same trees. The two species have the same niche.
Which of the following happens over the following years?
- A. Both species persist, each at half its countTwo species with the same niche compete for every seed; the better competitor takes the seeds, so the two do not share the clearing.
- B. ✓ The better competitor drives the other out
- C. Both species die outThe better competitor wins the seeds and persists; only the other species dies out.
Why: The two species have the same niche, so they compete for every seed.
The better competitor wins the seeds.
Each bird of the other species gets less food than it needs, so fewer breed and more die.
So the other species dies out: competitive exclusion.
Suppose fishing cats, wild cats that hunt in water, eat the fish of a crater lake, and those fish graze the lake’s algae. Now imagine the fishing cats are gone.
Which of the following does the algae do?
- A. More algae growsWith no fishing cats, more fish survive, and more fish graze more algae.
- B. The algae will be affected“Affected” does not say which way the algae goes, so it predicts nothing.
- C. ✓ Less algae grows
Why: The fishing cats are gone, so fewer fish are eaten.
The fish count rises.
More fish graze more algae.
So less algae grows: a trophic cascade, read down the chain.
Suppose an ecologist counts the bird species on ten cocoa farms shaded by tall trees and ten cocoa farms with no shade trees. The table below gives each mean with its ±2SE, so each error bar’s two ends can be worked out.
Which of the following do the two error bars show?
- A. ✓ The two means differ
- B. The data cannot tellThe shaded farms’ bar reaches down to 11.2 species and the unshaded farms’ bar reaches up to 8.7 species, so the two error bars do not overlap.
Why: The shaded farms’ error bar reaches from 11.2 to 14.4 species.
The unshaded farms’ error bar reaches from 6.3 to 8.7 species.
8.7 sits below 11.2, so the two error bars do not overlap.
So the difference is very unlikely to be chance: the two means differ.
How do you explain a change in the counts, and predict the next one?
Name the interaction. The two grasses compete for water, light and soil: a minus for both.
The aphids’ virus harms only the native grass: a minus for the native grass.
Follow the chain. The newcomer grass feeds more aphids, more aphids carry more virus, and the native grass falls.
Predict the next link with a direction word. Name who eats whom, then read down the chain to the native grass.
Every step names who does what to whom, and which way the count moves.
Video: Watch: Name it, follow it, predict it
Three circles stand on screen: newcomer grass, aphids, native grass. An arrow with a minus is drawn from the newcomer grass to the native grass. An arrow with a plus is drawn from the newcomer grass to the aphids. An arrow with a minus is drawn from the aphids to the native grass. A finger starts at the newcomer grass with the word up. It follows each arrow, writes up beside the aphids, and writes down beside the native grass.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-L44a.mp4
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The native bunchgrass grows in clumps, with bare soil between the clumps. Each clump takes its water and its light from that soil and that open ground.
The newcomer grass seeds into the bare soil between the clumps. Where a newcomer plant grows, it takes the water, the light and the soil that a native clump would have used.
Every newcomer plant leaves less water, light and soil for the native grass. Every native clump leaves less for the newcomer grass.
Two populations needing the same water, light and soil, so that each leaves less for the other, is competition. Each grass gets a minus.
A competing pair gets two arrows, one each way.
The change travels from the newcomer grass to the native grass. So the model keeps that arrow.
The model starts with three circles: the newcomer grass, the aphids and the native grass.
An arrow from the newcomer grass to the native grass carries a minus at its head. The newcomer grass harms the native grass.
Aphids suck sap from grass stems. On the newcomer grass the aphids breed faster than on the native grass.
So where the newcomer grass grows, there are more aphids. The newcomer grass feeds the aphids: a plus at the aphids.
This arrow runs from the grass to the aphids: from the eaten to the eater.
The change travels from the grass to the aphids. So the model keeps that arrow.
The aphids carry a virus. When an aphid sucks sap from a native grass stem, the virus passes into that plant.
The virus kills many of the native grass plants it infects. The newcomer grass carries the virus and is unharmed.
So more aphids means more native grass plants infected and killed. The aphids harm the native grass: a minus at the native grass.
Start at the population that changed. The newcomer grass went up.
Along the minus arrow from the newcomer grass to the native grass: newcomer up, so native down.
Along a plus arrow, the direction stays the same. Along a minus arrow, the direction flips.
Along the plus arrow to the aphids: newcomer up, so aphids up. Along the minus arrow from the aphids to the native grass: aphids up, so native down.
Both routes end on the native grass with a down. So the native grass count falls.
The native grass loses in two ways: to competition and to the aphids’ virus.
Now imagine the newcomer grass is pulled out of one plot of the valley.
Read down from the newcomer grass: newcomer down, so native up.
Newcomer down, so aphids down, so native up.
A prediction names the direction: in that plot, the native grass count rises. “The native grass will be affected” names no direction and earns nothing.
A new population joins the model the same way. Draw its circle, draw its arrow to the population it eats, write the sign at the head, and read down.
What you are expected to know Explain, from counts of two populations, how a named interaction drives the change in each.
What you are expected to know Predict the next change in a count with a direction word, and justify it link by link.
Suppose a grass brought from another continent is spreading through a valley of native bunchgrass. Where the newcomer grass grows, aphids breed faster. The aphids carry a virus that kills only the native grass. An ecologist marks ten plots where the newcomer grass has spread and ten plots of equal size where it has not. In every plot the ecologist counts the native grass plants, and the aphids on 100 native grass stems. The table below gives each mean with its ±2SE. Someone proposes releasing ladybugs, small beetles that eat aphids, into the valley.
(a) Describe the difference in the number of native grass plants between the two kinds of plot. (1 pt)
The two ±2SE error bars, 14 to 22 and 41 to 51, do not overlap.
So the difference is very unlikely to be chance.
- Award 1 point for: fewer native grass plants where the newcomer grass grows (the direction must be stated; accept: 18 against 46 plants per plot). Accept with or without: the ±2SE error bars do not overlap, so the difference is very unlikely to be chance.
(b) Explain how the newcomer grass changes the number of native grass plants in the valley. (1 pt)
The newcomer grass also feeds more aphids.
More aphids carry more virus to the native grass.
The virus kills many of the native plants it infects, so more native plants die.
- Award 1 point for ONE of the following: the newcomer grass takes water, light or soil the native grass needs, so fewer native plants can grow (competition); OR the newcomer grass raises the aphid count, so more virus reaches the native grass and more native plants die.
(c) Predict the effect of releasing ladybugs on the number of native grass plants. (1 pt)
- Award 1 point for: the native grass count rises / increases (the direction must be stated). Do not award: ‘the native grass will be affected’ with no direction.
(d) Justify your prediction in part (c). (1 pt)
Fewer aphids carry less virus to the native grass.
Fewer native grass plants are infected and killed.
So the native grass count rises.
- Award 1 point for: ladybugs eat aphids so the aphid count falls, AND fewer aphids carry less virus to the native grass (both links, each with its direction). Do not award: ‘predators help the grass’ with no aphid or virus link.
- Accept reasoning consistent with a wrong prediction in (c): award the point only for both links, each with its direction, leading to the direction the student predicted.
Back in the valley, the grass brought from another continent spreads through the native bunchgrass.
The two grasses compete for water, light and soil, so the native grass loses to the newcomer grass.
The newcomer grass feeds more aphids, more aphids carry more virus, and the virus kills more native grass.
Ladybugs eat aphids. So the ladybugs join the model as a circle above the aphids.
The ladybugs’ arrow ends on the aphids, with a minus at its head.
Read down from the ladybugs: ladybugs up, so aphids down, so native grass up.
Less virus reaches the native grass. So the native grass count rises.
49Mixed practice: interactions and counts mixed practice
Suppose ten fields are sown with a cover crop over winter and ten fields are left bare. In spring an ecologist counts the plant species in each field. The table below gives each mean with its ±2SE, so each error bar’s two ends can be worked out.
Which of the following do the two error bars show?
- A. The two means differThe cover-crop bar reaches up to 6.4 species and the bare bar reaches down to 4.6 species, so the two error bars overlap.
- B. ✓ The data cannot tell
Why: The cover-crop fields’ error bar reaches from 3.8 to 6.4 species.
The bare fields’ error bar reaches from 4.6 to 7.4 species.
4.6 sits below 6.4, so the two error bars overlap.
So the data cannot tell whether the two means differ.
Suppose a diver counts the fish of five species on two shipwrecks, the shallow wreck and the deep wreck. The table below gives the counts.
On which wreck is the species evenness greater?
- A. ✓ The shallow wreck
- B. The deep wreckOn the deep wreck one species holds 90 of the 100 fish, so its fish are shared very unequally among the five species.
Why: Species evenness is how equally a community’s individuals are shared among its species.
The shallow wreck’s five species number 24, 22, 20, 18 and 16: nearly equal shares.
On the deep wreck one species holds 90 of the 100 fish.
So the species evenness is greater on the shallow wreck.
Suppose a lake holds four fish species. Over ten years one species’ share rises from 0.50 to 0.80 of all the fish, and the other three species’ shares shrink. The species richness stays at four.
Which of the following does the lake’s Simpson’s Diversity Index do?
- A. The index risesA share that grows is squared into a bigger term, so the sum of squared shares grows and the index falls.
- B. The index stays the sameThe shares changed, and the big share grew.
A bigger share squared is a bigger term, so the sum of squared shares grew and the index fell. - C. ✓ The index falls
Why: The big species’ share rises from 0.50 to 0.80 of all the fish.
A big share squared stays big, so the sum of squared shares grows.
The index is 1 minus that sum.
So the index falls.
Oxpeckers are birds that ride on buffalo. The oxpeckers pick the ticks off the buffalo’s skin and eat them, and the buffalo are rid of the ticks that were drinking their blood. Write a plus for a population that gains, a minus for one that loses and a zero for one that is untouched.
Which sign does each population get?
- A. Oxpeckers +, buffalo 0The buffalo are rid of the ticks that were drinking their blood, so the buffalo are not untouched.
- B. Oxpeckers +, buffalo −The oxpeckers eat the ticks, not the buffalo, so the buffalo lose nothing.
- C. ✓ Oxpeckers +, buffalo +
Why: The oxpeckers eat the ticks: the oxpeckers gain a meal, a plus.
The buffalo are rid of the ticks that were drinking their blood: the buffalo gain, a plus.
An interaction in which both populations gain is mutualism.
Suppose two species of bulbul, fruit-eating birds, eat the fruit of the same trees along one river. One species, with a small bill, takes the small fruits. The other species, with a larger bill, takes the large fruits.
Which of the following happens to the two species over the following years?
- A. ✓ Both species persist
- B. One species drives the other outThe two species take fruits of different sizes from the same trees, so they compete little and neither is excluded.
Why: Each species takes a different part of the trees’ fruit: the small fruits, or the large.
So the two species compete little for food.
Neither species drives the other out.
Both species persist: niche partitioning.
Suppose francolins, ground birds, and the black-footed cats that hunt them are counted on one plain every year for twenty years. Each cat peak comes about a year after a francolin peak. A student says: “The cat peaks come after the francolin peaks because more francolins feed more cats, so the cat count climbs after the francolin count.”
Is the student correct?
- A. No: the two counts are of different sizes, so their peaks are unrelatedThe two counts differ in size, but each cat peak follows a francolin peak by a year, so the peaks are linked.
- B. ✓ Yes: more francolins feed more cats, so the cat count climbs after the francolin count
Why: More francolins feed more cats.
So the cat count climbs after the francolin count.
So each cat peak comes after a francolin peak.
More francolins feeding more cats is where the predator–prey cycle starts.
Suppose rock pythons, large snakes, eat bushbuck, a forest antelope, and bushbuck browse the seedlings of mahogany trees. The drawing below shows the three populations as a chain, with the sign of each effect at the arrow’s head. The sign at an arrow’s head is the effect on the population the arrow ends on. Now imagine rock pythons return to a forest where they had been hunted out.
Which of the following do the mahogany seedlings do?
- A. ✓ More mahogany seedlings survive
- B. Fewer mahogany seedlings surviveMore rock pythons eat more bushbuck, so fewer bushbuck browse the seedlings, and more seedlings survive.
Why: The rock pythons go up.
Along the minus arrow to the bushbuck, the bushbuck go down.
Along the minus arrow to the mahogany seedlings, the seedlings go up.
So more mahogany seedlings survive.
Suppose a grass brought from another continent spreads through a valley of native bunchgrass, and the two grasses compete for water, light and soil. A student says: “The two grasses interact closely for years, so their competition is a symbiosis.”
Is the student correct?
- A. ✓ No: symbiosis is two populations living in or on each other, and neither grass lives in or on the other
- B. Yes: any interaction between two populations that lasts for years is a symbiosisNeither grass lives in or on the other.
So their competition is not a symbiosis.
Why: Symbiosis is a close, long-lasting interaction in which two populations live in or on each other.
The two grasses take the same water, light and soil, but neither lives in or on the other.
So their competition is not a symbiosis.
APBIO-U08-P85 Practice questions: Topic 8.5
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one community’s Simpson’s Diversity Index one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. The formula sheet gives Simpson’s Diversity Index, Diversity Index = 1 − Σ(n/N)², where n is the total number of organisms of a particular species and N is the total number of organisms of all species.
Video: Watch first: Community ecology, summed up
Population, community and ecosystem; species composition, richness, evenness and diversity; the pie chart; Simpson’s Diversity Index — what the formula does, how to calculate it and what a difference in it means; the five interactions and symbiosis; niche, competitive exclusion and niche partitioning; the dual-y graph and the predator–prey cycle; trophic cascades and the signed-arrow model; the ±2SE overlap rule.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U08-T85-summary.mp4
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A biologist lists every species of bird, plant and insect living in one gorge, together with the gorge’s rock, its stream and the sunlight that reaches its floor.
Which of the following is that group?
- A. A populationA population is every member of ONE species in one place.
The list holds every species in the gorge, and its rock, stream and sunlight. - B. A communityA community is every species living in one place: living things only.
The list adds the rock, the stream and the sunlight. - C. ✓ An ecosystem
- D. The gorge’s non-living surroundingsThe rock, the stream and the sunlight are the non-living surroundings.
The list holds every living species as well.
Why: A community is every species living in one place.
An ecosystem is the community together with its non-living surroundings.
The list holds every species in the gorge and its rock, stream and sunlight.
So the group is an ecosystem.
Suppose surveyors count the plants of four ravines, narrow steep-sided valleys, and list the count of each species in each ravine. The table below gives the four lists.
Which ravine has the greatest species diversity?
- A. Ravine QRavine Q’s four counts are almost equal, but Ravine U holds five species shared almost as equally.
More species with equal shares is more diverse. - B. ✓ Ravine U
- C. Ravine YRavine Y holds the most species, six, but one species holds 90 of its 100 plants.
Diversity needs evenness as well as richness. - D. Ravine ZRavine Z’s 100 plants are all one species: no variety at all.
Why: Species diversity depends on richness and evenness.
Ravine U holds five species shared almost equally: 19, 21, 20, 20 and 20.
Ravine Y holds six species, but 90 of its 100 plants are one species.
Ravine Q holds four species, Ravine Z one.
So Ravine U is the most diverse.
Suppose a survey of the shrubs of one canyon counts 60 plants of four species: 30 saltbush, 15 buckthorn, 9 wild rose and 6 dogwood. The pie chart below shows the four species as slices Q, U, Y and Z, with no other labels. The slices are drawn in a shuffled order, different from the order of the species in the list.
Which slice is the buckthorn’s?
- A. Slice QSlice Q covers 15 % of the circle.
The buckthorn is 15 of the 60 plants: 25 % of them. - B. Slice USlice U covers 50 % of the circle: half of it.
The buckthorn is 15 of the 60 plants: 25 % of them. - C. Slice YSlice Y covers 10 % of the circle.
The buckthorn is 15 of the 60 plants: 25 % of them, not 10 %. - D. ✓ Slice Z
Why: The buckthorn is 15 of the 60 plants: 25 % of them.
25 % of the plants takes 25 % of the circle.
Slice Z covers 25 % of the circle.
Suppose a student counts 80 insects in a dovecote, a tower built for pigeons, and finds that every one of them belongs to the same species.
Which of the following is Simpson’s Diversity Index for the dovecote’s insects?
- A. ✓ 0
- B. 0.50.5 would need the sum of squared shares to be 0.5.
One species holding every insect has a share of 1, and 1 squared is 1. - C. 11 is the value the index climbs toward with many equal species.
One species alone gives the sum 1, and 1 − 1 is 0. - D. It cannot be worked out until the species is namedThe index is worked from counts and shares.
The species’ name is not in the formula sheet’s line.
Why: The one species holds all 80 insects, so its share is 1.
Squaring that share changes nothing, and there is no other species to add, so the sum is 1.
Taking 1 from 1 leaves 0, the lowest value the index can take.
Suppose a diver records the fish of one reef flat, the shallow shelf of a coral reef, each year for ten years. Six species of fish are present every year. Simpson’s Diversity Index for the fish rises from 0.44 to 0.82 over the ten years.
Which of the following happened to the reef flat’s fish over the ten years?
- A. More species of fish came to live on the reef flatSix species were present every year, so the number of species did not change.
- B. ✓ The fish became more equally shared among the six species
- C. One species came to hold most of the fishOne species holding most of the fish has a big squared share, so the sum is big and the index is low.
The index rose. - D. Every species’ count fell by the same fractionWhen every count falls by the same fraction, every share stays the same, so the index stays the same.
Why: The number of species stayed at six, so the rise came from the shares.
A higher value with the same richness comes from more equal shares.
So the fish became more equally shared among the six species.
A kinkajou, a small tree-living mammal of the tropical Americas, drinks nectar from the flowers of the balsa tree at night. As it moves from flower to flower it carries pollen between them, and the flowers set seed. Write a plus for a population that gains, a minus for one that loses and a zero for one that is untouched.
Which of the following gives the sign each population gets?
- A. Kinkajous +, balsa trees 0The kinkajou carries pollen between the balsa flowers, so the flowers set seed.
The balsa trees gain, so their sign is a plus. - B. Kinkajous +, balsa trees −The balsa trees lose a little nectar, and their flowers set seed.
Setting seed is a gain, so their sign is a plus. - C. Kinkajous 0, balsa trees +The kinkajou drinks the nectar: it gains a meal.
Its sign is a plus, not a zero. - D. ✓ Kinkajous +, balsa trees +
Why: The kinkajou drinks nectar: the kinkajous gain a meal, a plus.
The kinkajou carries pollen between the flowers, so the balsa trees set seed: the trees gain, a plus.
An interaction in which both populations gain is mutualism.
Suppose two species of pipistrelle, small bats, hunt the same small flies along one escarpment, a long steep slope. One species hunts at dusk and the other after midnight.
Compared with two species that hunt at the same time of night, how much do the two species of pipistrelle compete for the flies?
- A. MoreTwo species that hunt at different times of night rarely chase the same fly.
So they compete less, not more. - B. The sameTwo species hunting at the same time compete for every fly.
Two species hunting at different times take different flies. - C. ✓ Less
- D. Not at all: they never hunt at the same timeThe flies a species eats at dusk are flies the other species cannot eat after midnight.
Dividing a resource eases competition; it does not end it.
Why: The two species use the same flies at different times of night.
The time of hunting is part of the niche, so the two have divided it: niche partitioning.
Each night’s flies are shared by time.
So the two species compete less than two hunting at the same time.
Suppose redwings, thrushes that arrive in one wood each winter, and firecrests, tiny songbirds, are counted in the wood every winter for twelve winters. The redwing counts lie between 600 and 4 000 redwings. The firecrest counts lie between 9 and 45 firecrests. A student plots both series against the winters, one series against a left y-axis and the other against a right y-axis.
Which population takes the right y-axis, and why?
- A. ✓ The firecrests, because their counts are the smaller and the right y-axis is numbered to fit them
- B. The firecrests, because the right y-axis always carries the species counted secondThe order of counting decides nothing.
Each y-axis is numbered to fit the series it carries. - C. The redwings, because the larger counts need the taller axis and the right y-axis is the tallerThe two y-axes are drawn the same height.
Each axis is numbered to fit its own series, so no axis is the taller. - D. Either population, because the two y-axes carry one scaleThe two y-axes carry different scales: one numbered to fit the redwings, the other to fit the firecrests.
Why: Each line is read against its own y-axis, numbered to fit its series.
The redwings take the left y-axis, numbered to 4 000, as the prey series did on the hare and lynx graph.
So the firecrests take the right y-axis, numbered to 50, just above their largest count, 45.
Which of the following is a trophic cascade?
- A. A dogfish, a small shark, catches a whiting and eats itOne population catching and eating members of another is predation: one link, and no change passing down the chain.
- B. Two species of skate hunt the same shellfish in one bay, and each leaves fewer for the otherTwo populations needing the same food is competition: one link, and no change passing down the chain.
- C. ✓ More dogfish eat more whiting, so more of the brown shrimps the whiting eat survive
- D. Booklice feed on the fungus growing under a tree’s bark, and the tree is untouchedOne population gaining while the other is untouched is commensalism: one link, and no change passing down the chain.
Why: The change passes from the dogfish to the whiting, then to the brown shrimps: two levels down the food chain.
A change passing down two or more trophic levels is a trophic cascade.
Suppose a birdwatcher visits the ditches of two polders, flat land reclaimed from the sea behind a dike, twelve times each: an old polder and a new polder. On each visit the birdwatcher lists the bird species seen. The table below gives each polder’s mean number of bird species per visit with its ±2SE, so each error bar’s two ends can be worked out.
Which of the following do the two error bars show?
- A. The two means differThe old polder’s bar reaches up to 7.5 species and the new polder’s down to 6.6.
6.6 sits below 7.5, so the two error bars overlap. - B. ✓ The data cannot tell
- C. The two means are the sameOverlapping error bars leave the question open.
The two true means may still differ. - D. The new polder holds more bird speciesThe distance between the two means does not decide the question.
The bars overlap, so the difference could be chance.
Why: The old polder’s error bar reaches from 5.3 to 7.5 species.
The new polder’s reaches from 6.6 to 9.2 species.
6.6 sits below 7.5, so the two error bars overlap.
So the data cannot tell whether the two means differ.
Suppose a botanist counts the plants growing on one crag, a rocky outcrop, and finds 50 plants of four species. The table below gives the counts. The formula sheet gives , where n is the total number of organisms of a particular species and N is the total number of organisms of all species.
(a) Calculate the share of the crag’s plants held by the most numerous species, to two decimal places. (1 pt)
Frame share = n ÷ N = … ÷ … = …
Hint n is the count of that one species; N is the count of every plant on the crag.
Answer: 0.38 (tolerance ±0)
- Award 1 point for: 0.38 (a fraction, no unit).
Slip Dividing the total by the count, 50 ÷ 19. A share is one species’ count divided by the count of all the plants.
(b) Calculate that species’ squared share, to four decimal places. (1 pt)
Frame squared share = share × share = … × … = …
Hint Multiply the share by itself, not by 2.
Answer: 0.1444 (tolerance ±0)
- Award 1 point for: 0.1444 (no unit).
Slip Doubling the share, 0.76. A squared share is the share multiplied by itself.
(c) Calculate the sum of the four species’ squared shares, to four decimal places. (1 pt)
Frame sum of the squared shares = … + … + … + … = …
Hint Square each of the other three shares the same way, then add all four squares up.
Answer: 0.2856 (tolerance ±0)
- Award 1 point for: 0.2856 (no unit).
Slip Adding the shares instead of the squared shares, which always gives 1. The formula sheet adds up the squared shares.
(d) Calculate Simpson’s Diversity Index for the crag, to two decimal places. (1 pt)
Frame Diversity Index = 1 − … = …
Hint Take the sum of the squared shares away from 1, then round.
Answer: 0.71 (tolerance ±0)
- Award 1 point for: 0.71 (no unit; accept a value worked consistently from the sum in part (c)).
Slip Taking the sum away from 100. The index runs from 0 toward 1, not toward 100.
(e) A second crag holds the same four species of plant, and its Simpson’s Diversity Index is 0.53. Describe what the difference between the two crags’ indices says about how the second crag’s plants are shared among its four species. (1 pt)
Frame The second crag’s plants are …
Hint Both crags hold the same four species, so which of richness and evenness is left for the two indices to differ by?
- Award 1 point for: the second crag’s plants are shared less equally (accept: one species dominates the second crag; the second crag has the smaller species evenness). Accept a description consistent with the index calculated in part (d), provided it is compared with 0.53.
Slip Saying the second crag holds fewer species. Both crags hold the same four species; with the same richness a lower index comes from less equal shares.
Suppose kookaburras, large birds of the kingfisher family, hunt dunnarts, mouse-sized marsupials, on one stretch of Australian grassland, and the dunnarts eat soldier beetles. The drawing below shows the three populations as a chain, with the arrows drawn but no signs. The sign at an arrow’s head is the effect on the population the arrow ends on.
(a) Identify the sign that belongs at the head of the arrow from the kookaburras to the dunnarts. (1 pt)
- Award 1 point for: a minus (accept: −; negative).
Slip Writing a plus because the kookaburras gain a meal. The sign at the head is the effect on the dunnarts, which the arrow ends on: they lose members.
(b) Suppose kookaburras return to the grassland after years away. Predict what happens to the soldier beetles over the following years. (1 pt)
- Award 1 point for: more soldier beetles / the soldier beetle count rises (the direction must be stated). Do not award: ‘the soldier beetles are affected’ with no direction.
Slip Predicting fewer soldier beetles because a hunter has come back. The kookaburras never eat soldier beetles; the change reaches the beetles through the dunnarts.
(c) Justify your prediction in part (b). (1 pt)
Fewer dunnarts eat fewer soldier beetles.
So more soldier beetles survive.
- Award 1 point for: more dunnarts are eaten so the dunnart count falls, AND fewer dunnarts eat fewer soldier beetles (both links, each with its direction).
- Accept reasoning consistent with a wrong prediction in (b): award the point only for two links, each with its direction, that lead to the direction the student predicted.
Slip Justifying with the kookaburras alone. The point wants the dunnart link and the beetle link, each with its direction.
(d) A student says: “Removing the kookaburras will change the soldier beetle count, although no kookaburra ever eats a soldier beetle.” Evaluate the student’s claim. (1 pt)
Kookaburras eat dunnarts, and dunnarts eat soldier beetles.
So a change in the kookaburras passes down the chain through the dunnarts to the soldier beetles, which the kookaburras never eat.
- Award 1 point for: the judgement (the claim is right) AND the ground (the change passes down the chain level by level — kookaburras eat dunnarts and dunnarts eat soldier beetles — so it reaches a population the kookaburras never eat).
Slip Judging the claim wrong because the kookaburras never touch a soldier beetle. A change passes down the chain through the population in between.
APBIO-U08-T85 End-of-topic test: Community Ecology
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. The formula sheet gives Simpson’s Diversity Index, Diversity Index = 1 − Σ(n/N)², where n is the total number of organisms of a particular species and N is the total number of organisms of all species.
A biologist studies the living things on the wooden pilings of one wharf: wracks, which are seaweeds, dog whelks, which are sea snails, and the sea bass that hunt among them.
Which of the following groups is the pilings’ community?
- A. All the dog whelks living on the pilingsAll the dog whelks are one species in one place: a population.
A community holds every species living in the place. - B. Every species living on the pilings, together with the pilings’ wood, the seawater and the sunlightThe wood, the seawater and the sunlight are not living things.
A community together with its non-living surroundings is an ecosystem. - C. The pilings’ wood, the seawater and the sunlight around themThe wood, the seawater and the sunlight are the non-living surroundings.
A community is living things only. - D. ✓ Every species living on the pilings: the wracks, the dog whelks and the sea bass
Why: A community is all the populations of every species living together in one place.
The wracks, the dog whelks and the sea bass are every species on the pilings, and all are living things.
So that group is the pilings’ community.
Suppose a student walks one stretch of downland, dry chalk grassland, at noon and counts every butterfly seen. The table below gives the count of each species.
Which of the following is the species composition of the downland’s butterflies?
- A. ✓ 28 fritillaries, 12 hairstreaks and 5 skippers
- B. 45 butterflies counted in allA total names no species and gives no count of each.
- C. Three species of butterflyA count of species is the species richness.
It names neither the species nor the count of each. - D. Fritillaries, the most numerous speciesThe most numerous species is one entry of the list, not the whole list.
Why: The species composition is which species are present and how many there are of each.
The table lists three species, each with its count.
So the species composition is 28 fritillaries, 12 hairstreaks and 5 skippers.
Suppose a botanist counts the flowering plants of the same five species in two water meadows, Meadow Q and Meadow U. The table below gives the count of each species in each meadow.
Which meadow has the greater species diversity, and why?
- A. Meadow Q, because it holds more species than Meadow UMeadow Q’s list has five counts and Meadow U’s has five: the same species richness.
- B. ✓ Meadow Q, because its plants are shared more equally among the same five species
- C. Meadow U, because its most numerous species holds more plants than any species in Meadow QOne species holding most of the plants makes the shares lopsided.
Lopsided shares are a smaller species evenness. - D. Neither: the two meadows hold the same five species, so their species diversity is the sameSpecies diversity depends on evenness as well as richness.
The two meadows share their plants very differently.
Why: Both meadows hold five species: the same species richness.
Meadow Q’s counts, 20, 19, 18, 17 and 16, are almost equal; Meadow U’s, 70, 10, 5, 3 and 2, are lopsided.
So Meadow Q has the greater species evenness.
With the same richness, the more even meadow is more diverse.
Suppose a survey of the grass bank of a levee, a raised bank built to hold back a river’s floods, counts 60 plants of four species: 27 knapweed, 18 wild carrot, 9 chicory and 6 teasel. The four pie charts below, Q, U, Y and Z, each claim to show these counts.
Which pie chart shows the levee’s counts correctly?
- A. Pie QPie Q’s four slices are equal.
27 of 60 plants is 45 % of the total, and 45 % of the plants takes 45 % of the circle. - B. Pie UPie U draws each count as if it were a percentage of 100.
The total is 60 plants, so 27 plants is almost half of them. - C. ✓ Pie Y
- D. Pie ZPie Z’s largest slice is labeled teasel 10 %.
A species with 10 % of the plants covers 10 % of the circle, not the largest slice.
Why: 27 of 60 is 45 %, 18 of 60 is 30 %, 9 of 60 is 15 % and 6 of 60 is 10 %.
So the four slices cover 45 %, 30 %, 15 % and 10 % of the circle.
Only pie Y’s slice sizes match its labels.
Suppose one species’ share of the birds counted on an airfield rises from 0.13 one year to 0.39 ten years later.
Which of the following gives that species’ squared share, , at the start and ten years later?
- A. 0.0169, then 0.05070.0507 is 3 × 0.0169: the squared share scaled up like the share.
0.39 × 0.39 = 0.1521, nine times 0.0169. - B. ✓ 0.0169, then 0.1521
- C. 0.13, then 0.390.13 and 0.39 are the shares themselves.
The squared share is each share multiplied by itself. - D. 0.26, then 0.780.26 and 0.78 are the shares doubled.
A squared share is the share multiplied by itself, not by 2.
Why: The squared share is the share multiplied by itself.
0.13 × 0.13 = 0.0169 and 0.39 × 0.39 = 0.1521.
The share grew three times over; the squared share grew nine times over, 3 × 3.
So a dominating species’ squared share grows faster than its share.
Suppose a survey of a saltpan’s edge counts 60 plants of three species, 24 of them samphire. A student works out Simpson’s Diversity Index for the plants from the formula sheet’s line.
Which of the following is N in that line, for this survey?
- A. 0.40.4 is 24 divided by 60: the samphire’s share, .
- B. 33 is the number of species: the number of terms the sum adds up, not N.
- C. 2424 is the samphire’s count, n: the total number of organisms of one particular species.
- D. ✓ 60
Why: The formula sheet’s key reads: N is the total number of organisms of all species.
The survey counted 60 plants in all.
So N is 60.
Suppose two oxbow lakes, old river bends cut off from their river, hold the same five species of fish. The first lake’s Simpson’s Diversity Index is 0.78 and the second lake’s is 0.47.
Which of the following does the difference between the two values say about the first lake?
- A. ✓ The first lake’s fish are shared more equally among the five species
- B. More species of fish live in the first lake than in the secondBoth lakes hold the same five species, so neither has more species than the other.
- C. One species holds most of the first lake’s fishOne dominant species has a big squared share, so the sum is big and the index is low.
The first lake has the higher index. - D. The first lake is 78 % diverse, so it is the better place for fishThe index is a plain number with no unit, not a percentage of diversity.
A higher value means more diverse, never better.
Why: Both lakes hold the same five species, so their species richness is the same.
A higher value with the same richness comes from more equal shares.
So the first lake’s fish are shared more equally among the five species.
A sundew is a bog plant with sticky hairs on its leaves. Small flies land on the leaves and stick fast, and the sundew digests them.
Which kind of interaction is this?
- A. commensalismIn commensalism one population gains and the other is untouched.
Here the flies are killed and eaten. - B. mutualismIn mutualism both populations gain.
Here the flies lose their lives. - C. parasitismIn parasitism the eaten member stays alive while the parasite feeds on it.
The sundew kills the flies it digests. - D. ✓ predation
Why: The sundew catches the flies and digests them.
The sundew gains food, and the flies lose their lives.
One population catching and eating members of another is predation.
A bullfinch is a plump songbird with a stout, rounded bill and a pink breast. In one valley a student counts 40 bullfinches. Each spring they eat the buds of the valley’s fruit trees.
Which of the following is part of the bullfinch’s niche?
- A. Its stout, rounded billThe bill is a part of the bullfinch’s own body.
The bird neither uses it as a resource nor lives in it. - B. ✓ The buds of the fruit trees it eats
- C. The count of 40 bullfinches in the valleyA count of bullfinches is the population size, not a resource the bird uses or a condition it lives in.
- D. Its pink breast feathersFeathers are a feature of the bird, not a resource it uses or a condition it lives in.
Why: A niche is the resources a species uses and the conditions it lives in.
The bullfinch eats the buds of the fruit trees: the buds are a resource it uses.
So the buds are part of its niche.
Suppose two species of parakeet nest in one wood, and each species nests only in the hollows of the wood’s old trees. The wood holds few hollows. Whenever both species want one hollow, the larger species drives the smaller species out, so the larger species’ pairs take nearly all the hollows each spring.
Predict what happens to the count of the smaller species over the following years.
- A. ✓ It falls to zero
- B. It levels off at half the wood’s carrying capacityThe two species do not share the hollows equally.
The larger species takes a hollow whenever both want it. - C. It levels off at the same count as the larger species’ countThe larger species takes nearly all the hollows, so the smaller species cannot raise as many young.
- D. It rises to the wood’s carrying capacityReaching the wood’s carrying capacity needs a hollow for every pair.
The larger species takes nearly all the hollows, so few pairs of the smaller species breed.
Why: Both species need the same hollows in one wood: the same niche.
The species that takes a hollow whenever both want it is the better competitor.
Fewer pairs of the smaller species find a hollow, so fewer breed each year.
So its count falls to zero: competitive exclusion.
Suppose two species of gurnard, bottom-living fish, hunt the same small crustaceans in one bay. One species hunts over the sand at about 10 m deep, the other over the sand at about 40 m deep, and both species persist year after year.
Which of the following names the way the two species of gurnard share the bay?
- A. competitive exclusionIn competitive exclusion one species dies out.
Both species of gurnard persist. - B. mutualismIn mutualism both populations gain from each other.
The two species of gurnard take food from the same supply. - C. ✓ niche partitioning
- D. predationIn predation one species eats the other.
The two species of gurnard eat the same small crustaceans.
Why: Both species eat the same small crustaceans in one bay, so they compete.
Each species hunts at a different depth: about 10 m or about 40 m.
So the two use different parts of the shared resource and compete less.
That is niche partitioning.
Suppose ragworms, which live in the mud of a shore, and the whimbrels, wading birds that eat them, are counted on one stretch of shore each autumn for ten years. The graph below plots the ragworms per square of mud against the left y-axis and the whimbrels against the right y-axis.
About how many whimbrels were counted in year 7?
- A. ✓ 25 whimbrels
- B. 30 whimbrelsThe dashed whimbrel line crosses the year-6 gridline at the right axis’s 30, not the year-7 gridline.
- C. 500 whimbrels500 is the left y-axis’s reading halfway between the right axis’s 20 and 30 gridlines, and the left y-axis carries the ragworms.
- D. 620 whimbrels620 ragworms per square is the solid ragworm line’s reading in year 7, on the left y-axis.
Why: The legend names the dashed line as the whimbrels, and the right y-axis is titled whimbrels.
The dashed line crosses the year-7 gridline halfway between the right axis’s 20 and 30.
So about 25 whimbrels were counted in year 7.
Suppose ling, large fish of deep water, hunt Norway pout, small fish, and both are counted off one stretch of coast every year for twenty years. In one year the Norway pout count reaches its highest in ten years.
Over the next year, which of the following does the ling count do?
- A. The ling count fallsMany Norway pout this year are much food for the ling, so the ling count does not fall yet.
- B. The ling count falls, then climbsThe ling are fed first.
Their count falls only later, when many ling have eaten the Norway pout down. - C. The ling count stays levelMany Norway pout mean more ling are fed, so the ling count does not stay level.
- D. ✓ The ling count climbs
Why: A Norway pout peak means many Norway pout.
Many Norway pout are much food for the ling.
So more ling are fed, and the ling count climbs after the Norway pout peak.
Suppose bleak, small river fish, and the burbot that hunt them are counted in one river every year for twelve years. The bleak counts lie between 320 and 2 800 bleak. The burbot counts lie between 9 and 58 burbot. A student plots both series against the years, the bleak against a left y-axis and the burbot against a right y-axis.
Which of the following pairs of scales fits the two y-axes?
- A. Left 0 to 30 000, a gridline every 5 000; right 0 to 60, a gridline every 10On a left axis numbered to 30 000, every bleak count sits under the first gridline, 5 000, pressed against the x-axis.
- B. ✓ Left 0 to 3 000, a gridline every 500; right 0 to 60, a gridline every 10
- C. Left 0 to 3 000, a gridline every 500; right 0 to 600, a gridline every 100On a right axis numbered to 600, every burbot count is below 60, so the burbot line sits pressed against the x-axis.
- D. Left 0 to 3 000, a gridline every 500; right 0 to 3 000, a gridline every 500On a right axis numbered to 3 000, the burbot line lies flat along the bottom of the frame.
The burbot need a scale that ends near their largest count.
Why: A y-axis ends at a round number just above the largest count.
The largest bleak count is 2 800, so the left y-axis is numbered to 3 000.
The largest burbot count is 58, so the right y-axis is numbered to 60.
Every right gridline lies on a left one.
Suppose a disease killed off the serotines, large bats, of one valley, and now serotines return to the valley. Serotines eat chafers, large beetles, and chafer grubs eat the roots of the valley’s grass.
Which of the following happens to the valley’s grass over the following years?
- A. ✓ More grass grows
- B. Less grass growsMore serotines eat more chafers, so fewer chafer grubs eat grass roots.
The grass grows more, not less. - C. The grass is unchangedA change passes down the chain level by level, so it reaches the grass the serotines never eat.
- D. More grass grows at first, then the grass returns to the amount it had beforeThe serotines stay in the valley, so fewer chafer grubs eat the grass roots than before.
So the grass keeps its gain year after year.
Why: The serotines go up.
Serotines eat chafers, so the chafers go down.
Chafer grubs eat grass roots, so with fewer grubs the grass goes up.
So more grass grows: a trophic cascade read down the chain.
Horseflies bite gemsbok, large antelope, and drink their blood; the gemsbok lose blood and live on. Four students each drew the two populations as circles joined by two arrows, one each way, in drawings Q, U, Y and Z below. The sign at an arrow’s head is the effect on the population the arrow ends on.
Which drawing represents the interaction correctly?
- A. Drawing QDrawing Q puts a minus at the horseflies: it says the gemsbok harm the horseflies, whose meal of blood is a gain.
- B. Drawing UDrawing U puts the plus at the gemsbok and the minus at the horseflies: each sign sits at the wrong end.
- C. Drawing YDrawing Y puts a plus at the gemsbok: it says the horseflies help the gemsbok, which lose blood.
- D. ✓ Drawing Z
Why: The horseflies gain a meal of blood, so the arrow that ends on the horseflies carries a plus.
The gemsbok lose blood, so the arrow that ends on the gemsbok carries a minus.
Drawing Z has both.
Suppose four stretches of one seawall were built 5, 20, 50 and 100 years ago. On each stretch a student counts the plant species in ten squares of 1 m². The graph below shows each stretch’s mean number of plant species per square, with ±2SE error bars; the two ends of each error bar are printed beside it.
Which of the following pairs of stretches do the data show to differ in their mean number of plant species?
- A. The 5-year-old and 50-year-old stretches onlyThe 5-year-old bar, 2.2 to 3.8, also sits clear of the 20-year-old bar, 4.4 to 6.4, and of the 100-year-old bar, 5.0 to 7.0.
- B. The 20-, 50- and 100-year-old stretches, each with each of the other twoThe 20-, 50- and 100-year-old bars all share the range 5.0 to 6.4.
Overlapping bars cannot tell. - C. ✓ The 5-year-old stretch and each of the other three stretches
- D. Every pair of stretchesThe 50-year-old bar, 5.0 to 7.4, and the 100-year-old bar, 5.0 to 7.0, overlap.
That pair’s data cannot tell.
Why: The 5-year-old bar reaches from 2.2 to 3.8 species.
The other three bars start at 4.4, 5.0 and 5.0, all above 3.8: no overlap, so those three means differ from the 5-year-old mean.
The 20-, 50- and 100-year-old bars overlap one another, so the data cannot tell those pairs apart.
Suppose a student studies two species of grain beetle, species Q and species U, whose larvae eat stored grain. In the first treatment the student puts 10 adults of species Q into one jar of cracked wheat and 10 adults of species U into a second jar of the same size with the same amount of cracked wheat. In the second treatment the student puts 5 adults of each species into a third jar of the same size with the same amount of cracked wheat. The jars are kept in the same room, and the adults of each species are counted every four weeks. The table below gives the counts.
(a) Identify the species that is excluded when the student keeps the two species together in one jar. (1 pt)
- Award 1 point for: species U (accept: the species whose count falls to 0 in the shared jar).
Slip Naming species Q because its count is lower together than alone. Species Q still climbs in the shared jar; species U falls to 0.
(b) The student claims that in the third jar each species takes cracked wheat the other species needs. Support the claim with evidence from the table. (1 pt)
- Award 1 point for any ONE of the following: species Q grows more slowly together than alone (80 against 190 adults at week 8); species Q reaches a lower count together than alone (420 against 500 adults by week 24); species U grows more slowly together than alone (45 against 120 adults at week 8); species U reaches a lower count together than alone, or dies out together (0 against 300 adults by week 24).
- The evidence must compare the same species alone and together.
Slip Quoting the counts of one jar alone. Evidence for the claim compares each species alone with the same species together.
(c) Identify the sign, a plus (+), a minus (−) or a zero (0), that each species’ population gets from the other in the third jar. (1 pt)
- Award 1 point for: a minus for both species (competition harms both populations).
Slip Giving species Q a plus because it wins. Species Q reaches 420 with species U present against 500 alone: it loses too.
(d) Explain why, by week 24, the third jar holds adults of one species only, although each species thrives alone. (1 pt)
The better competitor eats more of the wheat.
Each larva of the other species gets less food than it needs, so fewer of them survive to breed and more die.
So the other species’ count falls until none is left: competitive exclusion.
- Award 1 point for: the two species have the same niche (the same food in the same jar), so the better competitor takes more of the food and the other species gets too little (fewer breed or more die) and its count falls to zero (accept the outcome described without the name competitive exclusion).
Slip Saying one species eats the other. Neither species eats the other; both eat the wheat, and one takes more of it.
Suppose bush dogs, small wild dogs of the South American jungle, hunt pacas, large rodents, in one jungle reserve. Pacas eat the seeds that fall from the reserve’s trees, so where pacas are many, few seeds are left to sprout into seedlings. Both animals are counted every year for twenty years. The graph below plots the pacas against the left y-axis and the bush dogs against the right y-axis.
(a) Describe the pattern in the timing of the bush dog peaks relative to the paca peaks. (1 pt)
- Award 1 point for: the bush dog peaks come after the paca peaks (accept: about two years after; the bush dog count lags the paca count).
Slip Saying the two counts peak in the same years. The dashed bush dog line peaks two gridlines to the right of each solid paca peak.
(b) Explain how the paca count and the bush dog count affect each other over one rise and fall. (1 pt)
So more bush dogs are fed and survive to breed, and the bush dog count climbs.
Many bush dogs kill many pacas.
So the paca count falls.
- Award 1 point for: more pacas feed more bush dogs, so the bush dog count climbs, AND many bush dogs then kill many pacas, so the paca count falls (both links, each with its direction).
Slip Stopping at ‘the bush dogs have more food’. The point wants the bush dog count’s direction and then the paca count’s.
(c) Suppose trappers remove every bush dog from the reserve. The drawing below shows the three populations as a chain, with the arrows drawn but no signs. The sign at an arrow’s head is the effect on the population the arrow ends on. Predict what happens to the number of tree seedlings in the reserve over the following years. (1 pt)
- Award 1 point for: fewer seedlings (the number of seedlings falls; the direction must be stated). Do not award: ‘the seedlings will be affected’ with no direction.
Slip Predicting more seedlings because the bush dogs are gone. The bush dogs never ate seedlings; the change reaches the seedlings through the pacas.
(d) Justify your prediction in part (c). (1 pt)
More pacas eat more of the fallen seeds.
So fewer seeds are left to sprout, and fewer seedlings grow.
- Award 1 point for: fewer pacas are killed so the paca count rises, AND more pacas eat more of the fallen seeds so fewer seedlings sprout (both links, each with its direction).
- Accept the two links written as the signs at the arrow heads, a minus at the pacas and a minus at the seedlings, with the direction of each change stated.
- Accept reasoning consistent with a wrong prediction in (c): award the point only for two links, each with its direction, that lead to the direction the student predicted.
Slip Justifying with the bush dogs alone. The point wants the paca link and the seed link, each with its direction.
Suppose a student surveys the fish in two sluice pools, pools held back by the gates of one drain: Pool Q and Pool U. Both pools hold the same four species of fish. The table below gives Pool Q’s counts. Pool U’s Simpson’s Diversity Index is 0.37.
(a) Calculate the squared share, , of Pool Q’s most numerous species, to four decimal places. (1 pt)
Answer: 0.3364 (tolerance ±0)
- Award 1 point for: 0.3364 (no unit).
(b) Calculate Simpson’s Diversity Index for Pool Q, to two decimal places. (1 pt)
Answer: 0.59 (tolerance ±0)
- Award 1 point for: 0.59 (no unit; accept a value worked consistently from the squared share in part (a) with the other three squared shares right).
(c) Describe what the difference between the two pools’ indices says about how Pool U’s fish are shared among its four species. (1 pt)
So Pool U’s fish are shared less equally among the four species: one species holds most of Pool U’s fish.
- Award 1 point for: Pool U’s fish are shared less equally (accept: one species dominates Pool U; Pool U has the smaller species evenness). Accept a description consistent with the index calculated in part (b), provided it is compared with 0.37.
Slip Saying Pool U holds fewer species. Both pools hold the same four species; with the same richness a lower index comes from less equal shares.
(d) A student says: “Pool U’s index of 0.37 means that 37 % of Pool U’s fish belong to one species.” Evaluate the student’s claim. (1 pt)
The index is 1 minus the sum of every species’ squared share.
So 0.37 is a plain number with no unit, worked from every species’ share; it is not the share of any one species and not a percentage.
- Award 1 point for: the judgement (the claim is wrong) AND the ground (the index is 1 minus the sum of the squared shares of every species — a plain number, not one species’ share or a percentage).
- Accept as the ground: the index combines every species’ share, so one species’ share is not read from it.
Slip Judging the claim right because 0.37 looks like 37 %. The index is a plain number with no unit; no single share can be read from it.
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