Campbell Biology with MasteringBiology, 9E Jane B. Reece Solution Manual.pdf (22)

5. In calculating the resting and action potentials of the axon, we generally don’t worry about the concentration differences of calcium or chloride ions. Explain.

The concentration differences are set up by active transport of the ions across the neuron membranes. The same is true for all such concentration differences. In general, we are not concerned with the concentration differences of calcium and chloride ions because the neuron membrane is essentially impermeable to these under normal conditions.

6. What effects would the following changes in extracellular K+ concentration be likely to have on the resting membrane potential of neurons in a human? a. Change extracellular K from 5 mM to 2 mM EK+ = 62/1 × log(2/140) = -113 mV This would hyperpolarize the cells, making it more difficult to generate an action potential. b. Change extracellular K from 5 mM to 10 mM EK+ = 62/1 × log(20/140) = -52 mV The normal potassium equilibrium potential is -90 mV. Changing the external potassium concentration to 20 mM would most likely change the equilibrium potential enough to bring it above threshold. As a result, all neurons in the body would fire and death would be the result. (Note: Death by lethal injection often involves sedatives and muscle relaxers in combination with intravenous delivery of a high concentration of potassium.)

Activity 48.2 How do neurons function to transmit information?

Working in groups of three or four, construct a dynamic (claymation-type) model of the transmission of an action potential along a neuron and then across a synapse to generate an action potential in a postsynaptic neuron.

When developing and explaining your model, be sure to include definitions or descriptions of the following terms and structures.

Neurons or Parts of Neurons Ions Gates dendrite

K+

voltage-gated ion channels

axon

Na

Na+ gates or channels

cell body

negative organic ions

K+ gates or channels

synaptic vesicles

Ca++

Ca++ gates or channels

presynaptic neuron postsynaptic neuron

Building the Model • Using chalk on a tabletop or a marker on a large sheet of paper, draw the membranes of two neurons and the synaptic region between them. Each neuron should each be at least 4 inches across and a foot or more in length.

• Identify the axon, cell body, and dendrite(s) on your drawing.

• Make the ions and gates from playdough or cutout pieces of paper. Indicate the placement of gates in Ch 44_56-18

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the membranes and ions inside the membrane versus outside the membranes.

• Include a key for your model that indicates how ions and gates are differentiated from each other. You may use color coding for the ions and gates.

• Start the model by “initiating” an action potential in the axon hillock.

• Indicate how this action potential is propagated along the axon and how it can lead to production of an action potential in the postsynaptic neuron.

• When you have completed your model, explain it to another student or to your instructor.

Use your understanding of how action potentials are generated and propagated to answer the questions.

1. All cells maintain an ionic (and therefore electrical) potential difference across their membranes. In most cells, this potential difference is between -50 and -100 mV. That is, the inside of the cells is more negative than the outside by 50 to 100 mV. Although all cells in the body maintain this potential difference across their membranes, only certain cells (for example, neurons) are capable of generating action potentials. a. How is this potential difference across the cell membrane generated?

Sodium-potassium pumps in cell membranes actively pump Na+ ions out of cells and K+ ions into cells.

In general, for every three Na+ ions pumped out of a cell, two K+ ions are pumped into the cell. As a result, the inside of the cell contains fewer positively charged ions than the outside; that is, the inside of the cell is more negative than the outside of the cell. This difference in ion distribution can be measured as a voltage or potential difference across the cell membrane. For the squid axon, the potential difference is -70 millivolts (mV). b. What characteristics of membranes allow cells to concentrate or exclude ions?

Cell membranes are selectively permeable. Ions such as Na+ and K+ cannot diffuse across the phospholipid bilayer. Ions must either be actively transported or diffuse through specific ion channels in the membranes. The ratio of Na+ to K+ ion channels can also affect the relative concentrations of the ions in cells. c. What is it about neurons (nerve cells) that make their properties different from those of other cells? In other words, what enables nerve cells to produce action potentials?

All cells generate a potential difference across their membranes. However, only nerve cells have voltage-gated ion channels that open and close in response to changes in the potential difference across the membrane. Postsynaptic neurons also contain chemically gated ion channels that open and close in response to neurotransmitters released from presynaptic neuron terminals. d. How is an action potential started and propagated?

Refer to Figure 48.11 of Campbell Biology, 9th edition, and keep in mind that all of the following responses are recorded at a single site on the axon.

• Local depolarizations in the neuron membrane open some of the Na+ gates and Na+ ions diffuse into the axon. When the potential difference at that site on the membrane reaches the threshold potential (-55 mV for the squid axon), the Na+ activation gates at the site all open and Na+ ions diffuse in (down the concentration gradient). During this time, the K+ gates remain closed.

• When the potential difference across the membrane at this site on the axon reaches +30 mV, the Na+ inactivation gates close and the K+ channels open.

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• K+ ions move out of this region of the axon down the concentration gradient, and this region or site is repolarized.

• When the potential difference reaches -70 mV (resting potential), the K+ gates are stimulated to close. The K+ gates are slow in closing, however, which produces the characteristic undershoot.

During this time, both the Na+ activation and inactivation gates are closed. This is the refractory period. When the Na+ inactivation gates are closed, the neuron is insensitive to depolarization. This limits the number of action potentials that can be generated per unit time. The refractory period also helps explain why action potentials normally travel in only one direction along the axon. e. Is any direct or indirect energy input required to generate an action potential? If so, when and where is the energy used?

Other than whatever energy is required to generate a depolarizing stimulus, no direct energy input is required to generate an action potential. Energy is required indirectly, however, to fuel the action of the sodium-potassium pumps, which generate the potential difference across the membrane. f. What happens in time and space (along the axon) once an action potential begins?

In part d, we looked at what happened at a single site on the axon as it was depolarized to generate an action potential. At the height of the action potential (+30 mV), the inside of the axon is more positive than the outside, at least at this one point. The Na+ ions that enter the axon at this site diffuse in all directions from their point of entry. As they diffuse, they change the potential difference across the membrane. The next site on the axon reaches threshold potential and its Na+ activation gates open. An action potential is generated. The Na+ ions that enter at this site diffuse in all directions and depolarize the next part of the axon, generating an action potential. This repeats along the length of the axon.

(Note: Although the Na+ ions diffuse in all directions along the inside of the axon, the action potential is generated in only one direction. The refractory period prevents the generation of an action potential in the opposite direction.) g. What factors ultimately limit the ability of the nervous system to respond (that is, to continue to generate impulses)?

Ultimately, the operation of the sodium-potassium pump limits the ability of the system to respond. The sodium-potassium pump establishes the potential difference (-70 mV, for example) across the neuron membrane. In experimental situations, isolated neurons are used to generate and record action potentials. When the sodium-potassium pumps in the neuron membrane are poisoned (or are nonfunctional), the axon is still able to generate action potentials (in some cases up to several hundred action potentials). This occurs because the number of Na+ and K+ ions that move across the membrane during an action potential is very small (on the order of eight or ten ions per gate per action potential).

On the other hand, the Na+ and K+ ion concentrations inside and outside the neuron are measured in millimoles. (See Figure 48.11.) If a molar solution of K+ ions contains 6.02 × 1023 K+ ions per liter, then a millimolar solution contains 6.02 × 1020 K+ ions per liter. As a result, many action potentials can be generated before the potential difference across the membrane is reduced enough to prevent their generation.

2. If an axon is stimulated in the middle of its length, nervous signals (action potentials) will move out from the point of stimulus in both directions. Normally, however, nerve signals move in only one direction along neurons. Explain.

Normally, the refractory period prevents the action potential from moving in more than one direction.

When an axon is artificially stimulated at any point along its length, however, no refractory period exists, so action potentials are generated in both directions.

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3. Whether or not an action potential is generated in a postsynaptic neuron depends on a number of factors.

What are they? What ultimately determines whether or not an action potential is generated in the postsynaptic neuron?

Whether or not an action potential is generated in a postsynaptic neuron depends on • the type of neurotransmitter—that is, whether it is excitatory or inhibitory, • the number and/or frequency of excitatory versus inhibitory inputs to the postsynaptic neuron per unit time, and • how close the synapses are to the axon hillock.

If the combination of all these inputs depolarizes the postsynaptic neuron to threshold level, an action potential is generated.

4. Diffusion is efficient over only very short distances. In fact, as you can see in this table, diffusion is efficient only for distances of about 1 to 100 µm.

Diffusion Distance (µm)

Time Required for Diffusion

1

0.5 msec

10

50 msec

100

5 sec

1,000 (1 mm)

8.3 min

a. How wide is a synapse?

The average synapse is between 10 and 20 nm wide. b. If a synapse were two times as wide, what effect would it have on the transmission of nerve signals from one neuron to the next? How would this change affect the response time of an organism?

If the synapse were two times as wide, the time for diffusion would increase by the square, or four times. It would take longer for the organism to respond. (Note: Although the response time would be somewhat longer, it would not necessarily be four times longer. Response time is also affected by the time required for a given neurotransmitter vesicle to release its contents into the synaptic cleft as well as by the need for the neuron to integrate or sum all of the inputs it receives.)

5. If you examine neuron transmission within an organism, you discover that every action potential generated is stereotyped; for example, every action potential reaches the same maximum height and the same minimum height. In addition, the generation of action potentials is an all-or-none phenomenon. That is, once the potential difference across the membrane reaches threshold, an action potential will be generated.

Given this, how does the nervous system signal differences in intensity of signal?

Intensity of signal is generally determined by the number of action potentials fired per unit time along a given neuron, the total number of neurons firing per unit time, or both.

Activity 48.3 What would happen if you modified a particular aspect of neuron function?

In the following questions, test your understanding of the various parts of the nervous system by asking yourself what would happen if a certain part was damaged. What would the system still be able to do? What would it be Reece_IG

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unable to do?

1. Some nerve gases and insect poisons work by destroying acetylcholine esterase. Acetylcholine esterase is normally present in acetylcholine synapses and acts to degrade acetylcholine. What is likely to happen to nervous transmission in insects exposed to this type of insect poison?

In the central nervous system (of vertebrates), acetylcholine can cause inhibition or excitation of the postsynaptic neuron. Which it does depends on the type of postsynaptic receptor present. In neuromuscular junctions (voluntary muscle), acetylcholine is excitatory. In heart muscle, the effect of acetylcholine is to reduce the rate and strength of heart muscle contractions. If acetylcholine esterase is destroyed, any acetylcholine released into a synapse will remain there. As a result, the postsynaptic neuron will be continuously stimulated by acetylcholine to either depolarize (if its receptor is excitatory) or hyperpolarize (if its receptor is inhibitory). For example, after an action potential is initiated in a muscle fiber, the muscle will contract continuously. The heart rate will slow. Inability to control the contraction and relaxation of muscles (such as the diaphragm) will quickly result in death.

2. The pufferfish (fugu) contains the poison tetrodotoxin. Some shellfish produce a paralytic poison called saxotoxin. Both of these poisons block the Na+ channels in neurons. What specific effects could these toxins have on neuron function?

If the Na+ channels were blocked, axons would be unable to generate or propagate action potentials.

3. A type of spider (the funnel-web spider) produces a toxin that blocks the Ca+ channels. a. Can a neuron exposed to this toxin fire an action potential? Explain.

Neurons exposed to this toxin should still be able to generate action potentials. b. Can a neuron transmit a signal across the synapse using neurotransmitters? Explain.

In chemical synapses, an influx of Ca2+ is required to cause the synaptic vesicles to fuse with the membrane of the presynaptic neuron. If these Ca2+ channels are blocked, no neurotransmitters are released into the synapse. Therefore, the signal cannot be transmitted across the synapse.

4. You isolate a section of a squid giant axon and arrange an experiment so that you can change the solution bathing the axon. You insert an electrode into the axon and place another electrode outside the cell so that you can measure the potential across the cell membrane. With the axon bathed in normal extracellular fluid, you observe a resting potential of -70mV and action potentials, when stimulated, that reach +55 mV. mM concentration of each ion Normal concentrations

Experimental concentration in (a)

Ion

Inside neuron

Outside neuron

Inside neuron

Outside neuron

Na+

50

440

50

440

K+

400

20

400

40

a. You change the solution bathing the neuron by increasing the K+ concentration to 40 mM. What effect will this have on the neuron? For example, will it depolarize the membrane and make it easier to start an action potential? Will it hyperpolarize the membrane and make it more resistant to starting an action potential? Or will it have no effect? Explain your answer.

To answer this question, keep these facts in mind: Ch 44_56-22

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• The axon’s permeability to K+ ions is much greater than its permeability to Cl- ions, which is much greater than to Na+ ions.

• At rest, the concentrations of negative ions inside and outside the axon are approximately equal.

• The equilibrium potential for each ion is dependent both on its concentration gradient and on the electrical gradient.

K+ ions cannot be added without simultaneously adding a counterbalancing anion—for example, Cl-.

As a result, the addition of KCl adds both K+ and Cl- ions to the outside of the membrane. This addition disrupts both the K+ and Cl- equilibrium potentials. The axon is much more permeable to K+ ions (than to Cl- or Na+ ions). As a result, K+ ions move more quickly into the axon (moving down the electrical gradient) to reestablish its equilibrium potential. Therefore, the result will be to depolarize the axon and make it more likely to generate an action potential. b. What would happen if, instead of adding more K+ to the outside, you added more Na+ to the fluid bathing the neuron? Explain.

As in part a, Na+ ions cannot be added without adding a counterbalancing anion—for example, Cl-. The addition disrupts the Na+ and Cl- equilibrium potentials. As before, the axon is more permeable to Cl(than to Na+), so Cl- ions would move into and hyperpolarize the axon.

In both parts a and b, the sodium-potassium pump will ultimately restore the Na+ and K+ gradients.

Notes to Instructors Chapter 49 Nervous Systems What is the focus of these activities?

Maintaining homeostasis is central to maintaining life. Control of homeostasis is the function of both the nervous system and the endocrine system. In Chapter 45 (Hormones and the Endocrine System) students learned that much of endocrine control is based on feedback loops as well as antagonistic action of hormones.

In this activity, students examine how the nervous system is set up to control not only the gross movements of organisms (somatic nervous system) but also the finer homeostatic responses of smooth muscle, cardiac muscle, and glands that support homeostasis (autonomic nervous system).

What is this activity designed to do?

This activity allows students to sort out and organize the functions of the peripheral nervous system. In doing this it also helps them to understand how interactions between the sympathetic and parasympathetic divisions of the autonomic system maintain homeostasis.

Activity 49.1 How is our nervous system organized?

Neurons are single cells composed of processes called dendrites, which carry information to the neuron’s cell body, and an axon, which carries information away from the cell body. In simple terms, nerves are bundles of dendrites, or axons or mixed axons and dendrites. Ganglia are collections of nerve cell bodies and brains are collections of ganglia. Processing and integration of the information occurs as a result of interaction within and between ganglia.

1. Responding to stimuli requires at a minimum: a. reception of a stimulus which triggers b. an action potential in a sensory neuron which synapses with Reece_IG

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c. a motor neuron which causes a response in d. a muscle or gland To coordinate activity an interneuron is often found between the sensory and motor neurons. The interneuron can function to send information to the central nervous system and as a result lead to a response by the whole organism.

In the space below, draw this minimum system. Be sure to name each neuron and label each of its parts.

[Unnumbered Figure Comes Here] 2. The blink response is one type of simple reflex response that occurs, for example, in response to water splashing up out of the sink or to a snowball aimed at your face. When these events happen, you blink and only afterwards do you “realize” why you blinked. a. What would you have to add to your diagram above to allow you to “realize” why you blinked?

You would need to add an interneuron or another synapse to the sensory axon that led to an interneuron that would carry the information to the brain to allow you to “realize” what happened. b. Why does it take you longer to “realize” why you blinked than it takes for the blink reaction to occur?

The simple reflex (sensory to motor neuron) cuts out the “middle man,” the interneuron, which allows reception and interpretation functions of the brain to occur. This means that the protective response (the blink) can occur very quickly. Most such reflexes are designed to protect or ensure the livelihood of the organism.

3. What composes the central nervous system (CNS) and how does the CNS differ in general form and function from the peripheral nervous system (PNS)?

The CNS is the integration and processing center. As such, it contains large collections of cell bodies and interneurons.

The PNS contains the sensory neurons, which carry information from the sensors to the CNS and motor neurons, which carry information from the CNS to the effectors.

4. The peripheral nervous system is subdivided into the somatic, autonomic and enteric nervous systems. a. What is the general function of each?

PNS subdivision:

General function:

Somatic nervous system

Carries signals to and from skeletal muscles, primarily in response to external stimuli.

Autonomic nervous system

Functions to regulate the internal environment by controlling function of smooth and cardiac muscle and glands.

Enteric nervous system

Controls secretion and function of the digestive tract (including peristalsis), the pancreas, and the gall bladder. While it can function on its own to do this, it is normally regulated by interaction between the sympathetic and parasympathetic divisions of the autonomic system.

b. The autonomic system is again further subdivided into the sympathetic and parasympathetic systems.

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How do the general functions of each of these divisions interact to maintain homeostasis?

Autonomic nervous system divisions: Sympathetic

Functions: See Figure 49.8 also.

Responses by the sympathetic nervous system are often called “fight or flight” responses. These tend to up-regulate or increase conversion of glycogen to glucose for energy generation; to cause secretion of epinephrine; and to increase heart rate, blood pressure, and flow of blood to skeletal muscles and brain.

Parasympathetic

The parasympathetic system acts antagonistically to the sympathetic system for the most part. When stimuli to the sympathetic system (e.g., stress, danger) are removed, the parasympathetic system promotes “calming” activities (e.g., reduced heart rate and blood pressure and increased digestion and glycogen production). One exception to this is in control of reproductive organs, where the sympathetic and parasympathetic systems complement each other (or work in concert).

49.1 Test Your Understanding Fibromyalgia affects about 2% of the U.S. population. Its primary symptoms are chronic pain, difficulty sleeping and fatigue that is not relieved by sleep. Other reported symptoms include: jaw clenching at night, dryness of mouth and eyes, an increased need to urinate, irritable bowels, muscle numbness and tingling, and headaches.

1. Several researchers have proposed that fibromyalgia results from failed regulation of part of the autonomic nervous system. If this is true, failed regulation of which division of the autonomic nervous system could account for these symptoms? Explain your reasoning.

The sympathetic system is responsible for arousal and energy generation while the paraympathetic system acts antagonistically to the sympathetic system. As a result, students may think that overactivity of the parasympathetic system could produce the symptoms of fibromyalgia. Symptoms not accounted for by action of the parasympathetic system would be dry mouth, muscle problems, and headaches.

2. To test their ideas, the researchers recorded heart rate of both controls and individuals with fibromyalgia for 24-hour periods of normal activity. They ob-served that patients with fibromyalgia showed greatly elevated heart rates both while active and asleep. Does the addition of this information change or support the hypothesis you developed in 1 above?

Given this information, the primary symptoms appear to be the result of hyperactivity of the sympathetic system. Hyperactivity of the sympathetic system could elevate the heart rate and blood pressure and lead to difficulty sleeping and chronic fatigue. Prolonged exposure to high levels of adrenaline associated with hyperactivity of the sympathetic system also leads to a reduced stress response. The researchers propose that this may be the cause of the irritable bowel and bladder as well as limb numbness and dryness of eyes and mouth.

Notes to Instructors Chapter 50 Sensory and Motor Mechanisms What is the focus of these activities?

Skeletal muscles are made up of bundles of muscle fibers (muscle cells). Each muscle fiber contains numerous myofibrils arranged parallel to one another. As a result, the degree of contraction of a given skeletal muscle is Reece_IG

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determined by which of its muscle fibers contract at any given time. Still, the maximum contraction that can be achieved by a skeletal muscle is almost always about 70% of its uncontracted or relaxed length. This is determined by the structure of the sarcomere. To understand muscle function, students need to understand both the overall structure and function of muscles and their subcellular structure.

What are the particular activities designed to do?

Activity 50.1 How does sarcomere structure affect muscle function?

The activity is designed to help students review and understand the basic structure and function of the sarcomere and how these constrain muscle function.

Activity 50.2 What would happen if you modified particular aspects of muscle function?

These questions are designed to have students test their understanding of muscle function by asking themselves what would happen if some part of the muscular system were damaged. What would the system still be able to do? What would it be unable to do?

Answers

Activity 50.1 How does sarcomere structure affect muscle function?

Working in groups of three or four, use playdough or cutout pieces of paper to construct a dynamic model of the sarcomere and demonstrate the sliding-filament model for muscle contraction.

When developing and explaining your model, be sure to include definitions or descriptions of the following terms and structures. thick filament thin filament actin myosin Z line H zone cross-bridges sarcomere contracted length extended length troponin complex tropomyosin Ca++ Alternatively, assign at least three people to model different parts of the sarcomere and demonstrate its action in muscle contraction. Assign a different role to each person, and assign specific actions to each role.

When you have completed your model, explain it to another student or to your instructor.

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Use your understanding of the model to answer the questions.

1. How are sarcomeres arranged within muscle fibers (and therefore within muscles)?

Sarcomeres are arranged linearly in myofibrils. Myofibrils are arranged linearly in muscle fibers. Muscle fibers are organized into muscles. Within a given muscle, the muscle fibers are all oriented in the same direction. (See Figure 50.26 of Campbell Biology, 9th edition.)

2. Describe the sliding-filament model of muscle action. For example, what interactions power the movement of filaments in association with each other? Include troponin complex, Ca++ binding sites, tropomyosin, myosin binding site, and actin in your discussion.

Refer to Figures 50.27, 50.28, and 50.29. In the sarcomere, thick myosin filaments lie between thin actin filaments. The actin filaments are attached to the Z lines of the sarcomere. The thick filaments contain myosin heads, which alternate between low- and high-energy states. In its low-energy state, a myosin head is bound to an ATP molecule. The ATP is hydrolyzed to generate the myosin head’s high-energy state. The actin filaments are composed of two strands of actin intertwined to form a helix. Long strands of tropomyosin are wound around the actin helix. At rest, the tropomyosin strands block the myosin binding sites on the actin. Action potentials in the sarcoplasmic reticulum and T tubules cause the sarcoplasmic reticulum to release Ca2+ into the cytoplasm of the myofibrils. When Ca2+ binds to the troponin complexes (which are situated at intervals along the tropomyosin strands), they change configuration. The configuration change alters the position of the tropomyosin strands and exposes the myosin binding sites.

The myosin heads hydrolyze ATP and pull the actin filaments toward the center of the sarcomere—that is, toward the M line. The muscle contracts.

3. How do your answers to questions 1 and 2 help explain why, at maximum contraction, muscle measures 70% of its extended (uncontracted) length?

During contraction, the actin filaments are pulled toward the center of the sarcomere by myosin heads on the thick filaments. At maximum contraction, the sarcomere width is equal to the length of the thick filaments. In the relaxed sarcomere, the length of the thick filaments is about 70% of the sarcomere width from Z line to Z line. You can check this by measuring the widths of the relaxed sarcomere and the contracted sarcomere in Figure 50.27.

4. Review Figure 50.30 and the associated text in Campbell Biology, 9th edition. Then answer the next questions. a. Acetylcholine, an excitatory neurotransmitter, is the neurotransmitter released into the neuromuscular junction. This release can trigger an action potential along the length of the muscle membrane. Describe the process involved.

A motor neuron synapses with a muscle fiber at a neuromuscular junction. If an action potential is generated in the neuron, the motor end plate of the neuron releases acetylcholine. This triggers an action potential in the muscle cell, which travels into the T tubule system. As noted in question 2, this causes the release of Ca2+ from the sarcoplasmic reticulum, which then triggers contraction of the sarcomere. b. If acetylcholine always produces an excitatory response in the neuromuscular junction, how can we regulate which muscles (in the arm, for example) are contracted and which are extended at any given time? For example, how can we bend the arm only partially?

Muscles are arranged into motor units. The contraction of each motor unit is controlled by separate neurons. The degree of response depends on how many motor units have been stimulated to contract.

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c. Severe calcium deficiency can lead to a reduction in bone mass. It can also have serious effects on the functioning of the nervous system and on the action of muscles. Explain what role(s) calcium plays in the activity of muscles.

As noted, it is the release of Ca2+ from the sarcoplasmic reticulum that ultimately causes the muscles to contract. In cases of severe calcium deficiency, muscles can become weak, or incapable of strong contractions.

50.1 Test Your Understanding The level of musculature in various organs is directly related to the amount of work done by that organ or organ part.

A few examples are noted below. a. Give two examples of different parts of organ b. For each example, explain the advantage of the systems where this is evident. (Use two different difference in musculature for each system. organ systems.)

Examples in the circulatory system include:

This is advantageous because:

• Heart—Atria are much less muscular than ventricles.

• Heart—Right ventricle is much less muscular than left ventricle.

• Arteries contain more muscle than veins.

• The atria pump blood a very short distance into the ventricles. The ventricles must pump blood greater distances.

• The right ventricle pumps blood to the lungs, a shorter distance than the ventricle , which pumps blood through theleft body .

• Arteries carry blood under higher pressure than veins.

An example in the digestive system:

This is advantageous because the stomach acts

• The stomach wall is more muscular than the esophagus.

to churn the ingested food , which in breaking it into smaller pieces (functions aidedassists by action of stomach acid ).

The primary of the esophagus—to undergo peristalsis and transfer food to the stomach—

require much of the less musculature. (The peristaltic activity Activity 50.2 What would happen if you modified particular aspects of muscle ? eat upside esophagus explains whyfunction you can Test your understanding of muscle function by asking yourself what would happen if a specific part was down .) damaged. What would the system still be able to do? What would it be unable to do?

In each of the following situations, indicate whether the proposed answers are true or false. If false, indicate why it is false.

1. A toxin in a newly discovered bacterial strain causes irreversible inactivation of the acetylcholine receptor.

If you were infected with this organism, the symptoms would include: T/F a. Uncontrollable muscular contractions Ch 44_56-28

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Explain: False—If the acetylcholine receptors were inactivated, no action potentials would be generated in muscle cells or T tubule systems. Therefore, no contractions would occur.

T/F b. Muscular paralysis Explain: True—As noted in part a, inactivation of the acetylcholine receptors would prevent muscle contractions.

T/F c. Loss of membrane potential of muscle fibers Explain: False—The sodium-potassium pumps are responsible for maintaining the membrane potential in muscle fibers. Poisoning the acetylcholine receptors should not affect these pumps.

2. In an experiment, a physiologist was able to destroy some but not all of the motor neurons to a specific skeletal muscle. Given what you know about motor units, what is the result of this action?

T/F a. There is no effect because motor neurons are redundant; multiple neurons innervate each muscle fiber as backup.

Explain: False—Each motor unit is innervated by a specific motor neuron (and its branches). As a result, if you destroyed its motor neuron, the motor unit would be unable to respond.

T/F b. The muscle will contract faster if the physiologist destroyed inhibitory motor neurons.

Explain: False—No inhibitory motor neurons innervate motor units in skeletal muscle.

T/F c. The muscle will in general, contract more weakly because fewer fibers are capable of contracting.

Explain: True.

T/F d. The ability of the muscle to contract will not change because the electrical potential will spread from stimulated fibers to adjacent fibers.

Explain: False—In skeletal muscle, each muscle cell in a motor unit receives separate innervation from the branches of the motor neuron associated with that motor unit.

Notes to Instructors Chapter 51 Animal Behavior What is the focus of this activity?

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other words, an organism’s behavior occurs directly (or proximally) as a response to current stimuli and indirectly (or ultimately) as the result of the process of natural selection or evolution of the species as a whole.

What is this particular activity designed to do?

Activity 51.1 What determines behavior?

In this activity, students are given specific examples of types of behavior. Then they are asked to propose: •

what questions they would need to ask to determine proximate causation for each behavior,

•

what questions they would need to ask to determine ultimate causation for each behavior, and

•

what kinds of experiments or investigations they would propose to answer the questions.

Answers

Activity 51.1 What determines behavior?

Note: A wide range of questions can be asked to determine proximate versus ultimate causation in each of the examples. A few representative questions are provided below. Similarly, a wide range of experiments or investigations can be proposed.

1. Some plant species (for example, many orchids) rely on a single species of insect for pollination. If the insect species dies out, so will the plant species. In a through c, refer only to the behavior of the insect species. a. What questions would you need to ask to determine proximate causation for this species-specific insect behavior?

What are the specific characteristics of the orchid and the insect?

Does some feature(s) of the morphology of the insect limit the types of flowers that it can pollinate?

Is this specific insect species unable to pollinate other flowers?

What factors (color pattern, fragrance, type of nectar, and so on) draw the insect species to only this type of orchid? b. What questions would you need to ask to determine ultimate causation for this insect behavior?

Why would this type of tight symbiotic relationship between orchid and pollinator evolve? c. What kinds of experiment(s) or investigation(s) you would propose to answer at least one of the questions in parts a and b?

What factor(s) draw the insect species to the flower?

You would need to test for each factor separately. You could place individual insects in test chambers.

To test for nectar preference, put small samples of nectar from the orchid and from several other orchid species into the chamber. Then observe the insects and record their responses to the different types of nectar. You can determine whether or not the insects are attracted by nectar from only the one species of orchid and how often (per unit time) they sample each type of nectar.

To test for fragrance preference, remove the nectar-containing parts of various flowers, including the orchid. Then put the orchid into a sealed box that has a small opening in one side. Do the same for another species of flower. Connect one end of a Y tube to one box and the other to the second box.

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Place the insect in the vertical arm of the Y tube and record how often it moves to one box or the other.

Repeat the procedure many times. You can randomize which box (right side or left side of Y tube) contains the orchid. In some experiments, use no orchid.

2. Many bird species that are common in the northern United States during spring and summer fly south in the fall to overwinter and feed in Central or South America. In the spring, they return to states in the northern United States to breed. a. What questions would you need to ask to determine proximate causation for this behavior?

How do birds “know” when to fly south versus north? For example: What environmental factors (for example, day length or temperature) stimulate the birds to flock and fly south in the fall? What environmental factors stimulate the birds to fly north in the spring? What, if any, physiological changes occur to trigger these flights?

How do birds navigate (find their way) on these long flights? b. What questions would you need to ask to determine ultimate causation for this behavior?

Why do birds fly south? Why would this type of behavior evolve? c. What kinds of experiment(s) or investigation(s) you would propose to answer at least one of the questions in parts a and b?

What environmental factors stimulate flocking behavior for the flight south?

You could test the effects of changes in day length on bird behavior. For example, maintain different flocks of birds in separate artificial habitats that allow you to regulate day length. In one habitat, maintain the flock on normal day-length cycles. In others, alter the day-length cycles by speeding up or slowing down the approach of shorter day lengths. You can then observe any differences in the behaviors of the different flocks of birds.

3. Many species of animals engage in complex courtship rituals. Among these species is the bower bird of Australia. Male bower birds construct elaborate structures or bowers from twigs, leaves and moss and decorate them with colorful objects, such as berries and shells. The bowers and the dances the males perform are designed to attract female bower birds for mating. After mating, the females fly away to build a nest and raise the offspring. The males remain at their bowers and try to attract additional mates. These birds can live for up to 17 years. Males are territorial and build their bowers in the same location each year.

In studying their behavior, researchers have noted that about 25% of the females “shop around” going from one bower to another before deciding on a mate. The other 75% appear to go directly to a single bower to mate. These behaviors cannot be observed in captivity. a. What questions would you need to ask to determine proximate causation for the bower building behavior?

What characteristics of the environment (etc.) trigger the bower building response in the male? b. What questions would you need to ask to determine the proximate causation for the female choice?

What characteristics of the bower or the male’s dance trigger the mating response in the female? c. What questions would you need to ask to determine ultimate causation for these behaviors?

Why did this elaborate mating ritual develop? d. What kinds of experiment(s) or investigation(s) would you propose to answer at least one of the Reece_IG

Ch 44_56-31

questions in either parts a or b?

What characteristics of the environment trigger the male bower building behavior? Because these birds cannot be studied in captivity, you would need to do a detailed analysis of the environmental characteristics in the days/weeks leading up to the behavior and compare these from year to year to determine key factors.

To determine what characteristics of the bower attract females you could record the characteristics of the bowers that attract the most females and do an analysis of similarities among these. You would need to do the same to compare the characteristics and dance patterns of the males that were most successful.

There are obviously many more questions that students could ask. Some would have to do with the life span of the birds and their ability to learn over time not only which males are considered the best mates but also the location of these males, which could explain why many females fly directly to specific males’ bowers (etc.)

Notes to Instructors Chapter 52 An Introduction to Ecology and the Biosphere What is the focus of this activity?

Ecology is the scientific study of the interactions of organisms and their environments. These interactions can be studied at a variety of levels, ranging from the individual organism to populations, communities, ecosystems, landscapes, biomes, and ultimately the entire biosphere. As a result, studying ecology requires that we integrate our understanding of the biology of organisms (for example, cell structure and function, genetics, evolution, physiology) with our understanding of the physical and chemical factors that affect the world and its climates.

What is this particular activity designed to do?

Activity 52.1 What factors determine climate?

In this activity, students apply their understanding of the factors that produce the biomes and climates on Earth to propose how these factors would affect environmental conditions on a hypothetical continent on Earth.

Answers Activity 52.1 What factors determine climate?

The map on page 343 shows a hypothetical continent on Earth. Assume that biomes and climates on this continent are produced by the same factors that produce biomes and climates on Earth’s real continents. Use this map to answer the questions in this activity. Where needed, draw the required features directly on the map.

1. a. On the map of the hypothetical continent, indicate the location(s) of the biomes listed in the table below.

To do this, draw approximate boundary lines to delimit each biome type, and then label each delimited area with the type of biome it contains.

Use the information in Figures 52.3, 52.10, and 52.15 of Campbell Biology, 9th edition, to answer these questions. b. In the table indicate the annual temperature and precipitation ranges for each biome.

Biome type

Annual temperature and precipitation ranges

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a. Tropical rain forest(s)

In general, tropical rain forests lie between the equator and about 25° north and south latitudes, where the annual mean precipitation exceeds 150 cm and the annual mean temperature exceeds about 23°C.

b. Temperate grasslands

Temperate grasslands tend to lie in the interior of continents or on the east coast in South America, between 25° and 45° north or south latitudes. Annual precipitation is between 30 and 100 cm, with periodic droughts common.

Temperature ranges between -10°C and 30°C.

c. Chaparral

The chaparral biome tends to lie between 25° and 45° north and south latitudes, where the general direction of the trade winds is away from the coast and the annual mean precipitation averages 30 to 50 cm. Average winter temperatures range between 10 and 12°C and average summer temperatures can reach 30°C with a maximum of 40°C.

d. Temperate forest(s)

Temperate forests tend to lie between 25° and 60° north and south latitudes, where the annual mean precipitation ranges from 70 to 200 cm and the annual temperature range is from below zero to 30°C.

e. Tundra

Tundra lies above 60° north and south latitude (Note: On maps of the Earth, there is no tundra in the southern hemisphere because there are no land masses between 60° and 90° south latitude. Antarctica lies above 90° south latitude for the most part). Annual precipitation ranges from 20 to 60 cm, and temperatures range from below -30°C up to about 10°C in summer.

2. Atmospheric circulation is driven primarily by differential heating of Earth’s surface. More heat is delivered near the equator than near the poles. This seems to explain the northward and southward flows of air. What introduces the eastward and westward components into air movement? (Hint: Review Figure 52.3, Global Air Circulation and Precipitation Patterns and Global Wind Patterns, in Campbell Biology, 9th edition.)

As indicated in Figure 52.3, “as Earth rotates on its axis, land near the equator moves faster than that at the poles, deflecting the winds from the vertical paths” introduced by differential heating. This deflection creates more easterly flow patterns between the equator and 30° north and south latitudes and more westerly flow patterns between 30° and 60° north and south latitudes.

[Unnumbered Figure Comes Here] 3. Use your understanding of global air circulation and wind patterns to draw arrows on the map of the hypothetical continent indicating: a. The direction of prevailing winds at points W1, W2, and W3 b. The direction of flow of surface currents in the ocean at points O1, O2, and O3

(Hint: Note on the map in Figure 52.3 that surface currents in the ocean follow the major wind systems at the surface.)

4. Are the surface winds at the given points warming or cooling as they move? Explain.

Point on the map

Are the winds warming or cooling as they move?

Explanation

a. X 1

The winds are warming

The general wind pattern in this area is from the southwest (westerlies). These winds tend to pick up Reece_IG

Ch 44_56-33

moisture as they move from 30° to 60° N latitude. b. X2

The winds in this region tendThis region tends to be under the influence of cooling easterly winds. These winds tend to pick up moisture as to be cooling. they flow from 30° to 0° N latitude.

c. X3

The winds are warming

The general wind pattern in this region is westerly.

These winds tend to pick up moisture as they flow from 30° to 60° S latitude.

5. What biomes or vegetation types would most likely be found at the given points? (Assume all are at sea level or low altitudes.)

Point on the map

Most likely type of biome or vegetation

A

This region would most likely contain coniferous forest.

B

This region would most likely be a temperate grassland that will graduate into temperate forest as it moves east.

This would most likely be tropical forest, but could be savanna depending on local water conditions.

This is most likely a region of tropical rainforest.

C D

6. Would the climate at point E be relatively wet or dry? Explain.

This region would be relatively dry. It lies in the rain shadow of the mountains. In addition it lies at about 30° N latitude, a region of very drying winds.

7. What would the direction of the prevailing winds be at Earth’s surface at point X3?

The prevailing winds at this point should be westerly.

8. In the United States, temperate forest extends from the East Coast westward for about 1,200 miles (to the Mississippi River Valley). From there, the forest begins to thin out toward the west into oak savannas (or temperate woodland), and it finally gives way to open grassland (Great Plains). The grasslands extend 1,000 miles westward to the foothills of the Rocky Mountains. a. Why does grassland replace forest west of the Mississippi River?

This area is under the influence of the westerlies (wind flow pattern), which carry moist air from the Pacific Ocean onto the North American continent. When these winds reach the Rocky Mountains, they rise and cool. This causes the moisture in the air to condense and fall as rain on the western slopes of the Rockies. The wind that reaches the eastern slopes is dry, so much of the land in this “rain shadow area” is desert. As the winds continue overland, they pick up some water; however, the area west of the Mississippi receives less rainfall than areas farther east. Grasslands are found in such regions where the mean annual precipitation is between about 30 and 80 cm. b. What is the rain shadow effect?

As noted in part a, the rain shadow effect occurs on the leeward side of mountain ranges. When the air currents encounter the mountains, they rise. As air rises, it cools and the moisture in it condenses and falls as rain. As a result, the air that reaches the other side of the mountain is dry, so the leeward side of the mountain is dry.

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c. Draw a rain shadow somewhere in the southern hemisphere of the map.

The rain shadow is always on the leeward side of the mountain. Therefore, in the southern hemisphere, the rain shadow should be on the western side of the mountains if the site is between 0° and 30° south latitude but on the eastern side if the site is between 30° and 60° south latitude.

9. What biome would exist at point F? Point F is at the transition between temperate grassland and temperate forest. It is difficult to determine which biome would exist in this exact location, and therefore it is likely a transition zone.

10. How are the general characteristics of plants (for example, morphology) influenced by climate? In other words, explain what effects climate has on the types of plants that grow in an area.

As indicated in Figure 52.10, the type of biome is strongly influenced by both the annual mean temperature of the region and the annual mean precipitation. For example, when both are relatively high, you find temperate rain forests. Plants in these regions do not experience extremes in temperature or water over the year. When mean annual temperature is high and mean annual precipitation is low, deserts are formed and the plants are those that can survive in this type of climate. Refer to Figure 52.12 for examples of the types of plants that would be found in the various types of biomes.

11. Refer to Figure 52.10 in Campbell Biology, 9th edition. In general, how is the distribution of major ecosystems or biomes related to climate? If you know the mean annual precipitation and the mean annual temperature of an area, would you be able to accurately predict the type of biome that could exist there?

Explain.

As noted above, the type of biome or ecosystem is determined by the mean annual temperature and the mean annual precipitation in an area. However, how the precipitation and temperature are distributed over the year also has a significant impact on the type of biome. For example, is the temperature relatively constant over the course of the year or does it range between high summer temperatures and very low winter temperatures? Other factors—for example, type of soil—can also affect the type of biome.

12. Why isn’t Earth’s climate uniform? To answer this, summarize the major factors that can produce differences in climate from place to place.

The primary factors affecting the type of climate are the annual amount of solar energy received and the annual amount of water received. The amount of solar energy received per year in a given area is directly related to its location on the globe and the fact that the earth is tilted 23 degrees on its axis. As a result, the equator receives approximately the same amount of solar energy throughout the year. On the other hand, more northern areas in the United States will experience different seasons over the course of the year. The major wind patterns are set up by differential heating at the equator. The major precipitation patterns are a function of these wind patterns.

52.1 Test Your Understanding 1. A large asteroid hits the Earth, and alters its axis tilt from 23.5° to 10°. a. What effect(s) would this have on seasons in the northern hemisphere?

With less tilt, there would be less seasonality. Overall, the seasons would be warmer and less distinctly different from one another. The sun’s rays over the course of the year would be more direct, and the day length would tend to be less variable in summer versus winter. Days in the summer would still be longer than days in the winter, but the difference in average day length would not be as great. b. What effect(s) would this have on biome distribution in the Northern Hemisphere? (Refer to Figure Reece_IG

Ch 44_56-35

52.10 in Campbell Biology, 9th edition.)

As the annual mean temperature increased, we would see much less tundra and boreal/coniferous forest.

Over time, the land that was previously occupied by the tundra and boreal/coniferous forest would become temperate forest, grassland, or desert, depending on location and precipitation. A catastrophic event such as this would likely lead to mass extinction events and it would take a very long time for the biomes and flora/fauna to reposition/establish themselves on the Earth.

2. If you travel across the United States on a line from North Carolina to Southern California, you will find that deciduous forest extends from the East Coast westward for approximately 1,200 miles to about the valley of the Mississippi River. There it begins to thin out toward the west into oak savannas and finally gives way entirely to open grassland (Great Plains), which extends another 1,000 miles westward up into the foothills of the Rocky Mountains. If you cross over the mountains you find chaparral in Southern California.

T/F The single most important factor affecting the type of biome present in each of these major regions of the United States is rainfall.

True

3 to 6. The type of vegetation that will grow in a particular region of the earth is strongly affected by: T/F 3. Extremes of temperature (high and low) experienced over the year in that region.

True T/F 4. Altitude of that region.

True T/F 5. Seasonal availability of water.

True T/F 6. Longitudinal location of that region.

False Notes to Instructors Chapter 53 Population Ecology What is the focus of these activities?

To study ecology, we often use mathematical models and statistical analysis to estimate population size and distribution. We have also used mathematical models to make predictions about what may happen to the population in the future. Students should keep in mind, however, that any model is limited by the assumptions used in building it. At best, models provide us with a generalized idea of what could happen if the specific assumptions of the model are met.

What are the particular activities designed to do?

Activity 53.1 What methods can you use to determine population density and distribution?

The questions used in this activity are designed to help students review and understand some of the basic methods used in determining population density and distribution.

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Activity 53.2 What models can you use to calculate how quickly a population can grow?

The questions are designed to help students review and understand how various mathematical models can be used in studying populations.

Answers

Activity 53.1 What methods can you use to determine population density and distribution?

1. To measure the population density of chipmunks occupying a particular park, you sample several quadrats and capture 50 chipmunks. You mark each of them with a small dot of red paint on their backs, and then release them. The next day, another 50 chipmunks are captured. Among the 50, you find 10 that are marked. a. Using the mark–recapture formula below, how many chipmunks do you estimate the population contains?

Number of recaptures in second catch/ Total number in second catch = Number marked in first catch/Total population N If 10/50 = 50/N, then N = 250 chipmunks. b. What effect would each of the following discoveries have on your estimate? i. You later discover that you sampled the one area of the park that was most favored by the chipmunks.

If the area was favored by the chipmunks, you most likely captured more chipmunks in this area than you would have caught in other areas of the park. As a result, your estimate of the population size would be smaller than if you had sampled more of the park. ii.You later discover that the chipmunks were licking the mark off of each others’ backs.

Some of the recaptured chipmunks that were counted as unmarked could have been among those that had their marks licked off. As a result, your estimate of population size would be higher than if the marks had not been licked off. For example, if 10 of those captured had licked off their marks, the actual recapture ratio would be 20/50. Then the population estimate would change to 125 (rather than 250) chipmunks in the population. iii. You later discover that the marked chipmunks are easier to see and therefore more susceptible to predation.

If more marked chipmunks are captured by predators, you will recapture fewer marked chipmunks than you should. For example, if you mark 100 and 50 are eaten, the probability of your recapturing any of the marked chipmunks is reduced by 50%. Therefore, the population size you calculate will also be reduced by up to 50%. As a result, you will not be able to accurately quantify the population size. c. How could you modify your sampling program to ensure that you make more accurate estimates of population size?

At a minimum, you should: i. sample several areas of the park.

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ii. use a marking system that cannot be removed easily. iii. use a marking system that is not easily detected by predators.

2. Refer to the two proposals for the distribution of a tree species below.

[Unnumbered Figure Comes Here] a.

What type of distribution is represented in each of the proposals?

Distribution 1: = clumped Distribution 2: = random b. Given these two possible distributions, what factors do you need to consider in setting up a sampling plan for the area? Propose sampling strategies and the results you would get if organisms were distributed as in 1 vs. 2 above. For each sampling strategy proposed, indicate how you will know if you have chosen both an appropriate quadrat size and number of quadrates to provide you a good representation of both the size of the population and the actual distribution of organisms within the sampling area.

The answer to this question will depend on the size of the total area relative to the size of the sample quadrats and the number of quadrats sampled per distribution. The following gives an overview of the results you would get for each type of distribution and different sampling strategies.

Assume that you divide each area into four quadrats and then sample only quadrat D in both areas.

A

B

C

D

If you count the total number of trees in quadrat D in both distributions 1 and 2, you discover that quadrat D in distribution 1 contains 10 trees, while quadrat D in distribution 2 contains 5 trees. These differences would indicate that the density of trees in distribution 1 may be higher than the density in distribution 2. It doesn’t help you determine how the trees are distributed, however.

Now, assume that you divide each area into 16 quadrats.

A1

A2

B1

B2

A3

A4

B3

B4

C1

C2

D1

D2

C3

C4

D3

D4

If you sample the four quadrats in section D in each area (1 and 2), your data would look something like this: Quadrat number:

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Number of trees Area 1

Area 2

D1

1

2

D2

6

1

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D3

3

1

D4

0

1

These data indicate that areas 1 and 2 contain not only different densities but different distributions of trees as well.

If you divide the areas further into 64 quadrats, you would get a better idea of the exact distribution of the trees. However, if your goal is to determine relative distributions rather than absolute distributions, you don’t need this much detail. In other words, the size and number of quadrats that need to be sampled depend on the goals of your study.

3. The following table provides the numbers of deaths per acre per year resulting from two different agents of mortality applied to grasshopper populations of different densities. a. Fill in the mortality rates for agents A and B in the table below.

Grasshopper population density (individuals/acre)

Deaths per year per acre

Mortality rate (%)

Agent A

Agent B

Agent A

Agent B

100

4

0

4%

0%

1,000

40

25

4%

2.5%

10,000

400

500

4%

5%

100,000

4,000

50,000

4%

50%

b. Graph the data below.

[Unnumbered Figure Comes Here] c. Which of the two agents of mortality (A or B) is operating in a density-independent manner? Explain your answer.

Regardless of the population size per acre, agent A kills only 4% of the population. In other words, it is working in a density-independent manner. Agent B is acting in a density-dependent manner because as the density changes, the percentage of individuals killed by agent B also changes. In this example, as the density of grasshoppers increases, the percentage of grasshoppers killed per acre also increases. d. Which of the two agents of mortality (A or B) is likely to act as a factor stabilizing the size of the grasshopper population? Explain your answer.

No matter how large the population gets, agent A removes only 4%. In contrast, agent B removes an increasing percentage of the grasshoppers as the population size increases. Therefore, agent B is more likely to act to stabilize the size of the population over time.

53.1 Test Your Understanding A researcher has recently discovered three species of parasites (A, B, and C) that infect developing salamanders. He suspects that one or more of these species cause fatalities during salamander development and that the death rate varies with salamander population density. To test his hypothesis, the researcher sets up a series of experiments. In each experiment, he varies the density of the salamander populations. At the start of Reece_IG

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each experiment, he infects 5% of each test population with a single parasite species and then measures the mortality/death rate after four weeks. To establish a baseline mortality rate, he sets up a control experiment that differs only in that no parasite is introduced at any density. The data table and graph below relate the mortality rate of salamanders (caused by the three different parasites) to the original density of the developing salamander population. a. Given the data presented in the graph, indicate whether each of the parasite species (A, B, and C) is acting in a density-dependent or density-independent fashion. Explain your answers.

Exp. 1

Exp. 2

Exp. 3

Control

Control

A B C Density Density Density MortalityRate MortalityRate MortalityRate MortalityRate

Density

20

0.1

20

0.82

20

0.4

20

0.1

40

0.2

40

0.74

40

0.25

40

0.096

80

0.4

80

0.76

80

0.49

80

0.12

160

0.6

160

0.8

160

0.35

160

0.15

320

0.85

320

0.76

320

0.36

320

0.092

[Unnumbered Figure Comes Here] Parasite A is acting in a density-dependent fashion, because parasite A causes progressively higher mortality rates as salamander density increases.

Parasite B produces greater mortality than the control. However, the percent mortality is fairly consistent over the range of population densities. As a result, it is acting in a density-independent fashion.

Parasite C, while it produces greater mortality than that shown in the control, also does not appear to act in a density-dependent fashion.

Line D is the control, showing approximately 10% mortality regardless of population density. b. Looking at the data for parasites A and B, develop an argument to indicate which is more likely to cause extinction of the salamander population. Explain your reasoning.

Because of its overall high mortality rate at all salamander densities, parasite B may decrease the population size such that it cannot rebound. In contrast, parasite A has lower mortality at lower population densities. As a result you are likely to see an oscillation in population size and parasite mortality over time.

Only at very high population densities (>320) is parasite A likely to cause extinction.

Activity 53.2 What models can you use to calculate how quickly a population can grow?

1. In the simplest population growth model (dN/dt = rN). a. What do each of the terms stand for?

Term dN Ch 44_56-40

Stands for Change in the number of individuals Reece_IG

dt

Time interval during which the change occurred

r

Per capita population growth rate = per capita birth rate minus per capita death rate

N

Initial population size

b. What type of population growth does this equation describe?

Exponential growth of a population (J-shaped curve) c. What assumptions are made to develop this equation?

Unlimited space (habitat) and resources

2. Population growth may also be represented by the model, dN/dt = MmaxN[(K – N)/K]. a. What is K? K is the carrying capacity or the number of individuals that can be sustained over time. K is a function of the environment. b. If N = K, then what is dN/dt?

When N = K, dN/dt = 0; that is, there is no change in the population size over time. c. Describe in words how dN/dt changes from when N is very small to when N is large relative to K. When N is very small, the population is growing exponentially. The actual number of new individuals added is relatively small, however, because N is so small. As N nears the value of K/2, dN/dt approaches its maximum. As N approaches K, dN/dt decreases. d. What assumptions are made to develop this equation?

The logistic model assumes that resources and space limit the growth of a population and that these factors determine the maximum number of individuals that can be sustained over time.

3. You and your friends have monitored two populations of wild lupine for one entire reproductive cycle (June year 1 to June year 2). By carefully mapping, tagging, and censusing the plants throughout this period, you obtain the data listed in the chart.

Parameter

Population A

Population B

Initial number of plants

500

300

Number of new seedlings established

100

30

Number of the initial plants that died

20

100

a. Calculate the following parameters for each population.

Parameter

Population A

Population B

B (births during time interval)

100

30

D (deaths during time interval)

20

100

100/500=0.20

30/300=0.10

b (per capita birth rate)

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m (per capita death rate) r (per capita rate of increase)

20/500=0.04

100/300=0.33

0.20-0.04=0.16

0.10-0.33=-0.23

b. Given the initial population size and assuming that the population is experiencing exponential growth at growth rate r, what will the number of plants be in each population in 5 years? (Use the initial population size as time 0, and compute to time 5.)

Population A: 1,050 Population B: 81

4. Using the exponential growth formula, you can determine the amount of time it will take for a population to double in size if you know rmax or r. Doubling time is equal to: log102 / log10 (1 + r) Alternatively this value can be estimated by using the formula:

70 divided by the percent increase per unit time (as a whole number) = doubling time per unit time or 70/r = approximate doubling time.

Using either of these formulas—the exponential growth formula or the approximate doubling rate formula—calculate the following. a. If the population of a country is growing at 2% per year, how many years will it take for the population to double?

Doubling time = 70/2 per year = 35 years. b. If your bank account is growing at a rate of 1% per year, how many years will it take for it to double?

Doubling time = 70/1 per year = 70 years.

5. You are studying the growth of a particular strain of bacteria. You begin with a tiny colony on a Petri plate.

One day later, you determine that the colony grew and exactly doubled in size. A calculation showed that if the colony continued to grow at the same (constant) rate, it would cover the entire plate in 30 days.

(Assume that colony size is directly proportional to the number of individual bacteria.) a. What is value of r?

See question 4.

You can calculate r using the formula log102/log10 (1 + r).

Alternatively, r can be estimated by using the formula: 70 divided by the percent increase per unit time (as a whole number) equals the doubling time per unit time. Therefore, if 70/r = 1 day, then r must equal approximately 0.70, or 70%. b. On what day would the bacteria cover half the plate?

If the bacteria colony doubles in size every day and completely covers the plate on day 30, then it must cover half the plate on day 29.

6. You collect data on birth and mortality in three populations of grasshoppers, and you calculate the following birth and death rates for these populations. Both populations are experiencing exponential growth: Ch 44_56-42

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b

d

Population

0.90

0.80

A

0.45

0.35

Population

0.15

0.05

B true or false?

Are the following statements Population F a. Population A is growing at the fastest rate.

C T b. Population C has the lowest death rate.

F c. Population C is growing at the slowest rate.

T d. All populations are growing at the same rate.

7. In a herd of bison, the number of calves born in 1992, 1993 and 1994 was 55, 80, and 70, respectively. In which year was the birth rate greatest?

You cannot answer this question unless you also know the population size for each of those years.

8. A population of pigeons on the west side of town has a per capita annual growth rate of 0.07. A separate population of pigeons on the east side of town has a per capita annual growth rate of 0.10. If the populations are both growing exponentially and both are censused the following year, in which of the populations will dN/dt be greatest?

Again, you cannot answer this question because you don’t know how large the populations are initially. If the two population sizes are equal, or if the population on the east side is larger than the population on the west side, then dN/dt will be greater on the east side. However, if the population on the west side is large and the population on the east side is much smaller, then dN/dt of the west-side population could be the greater of the two.

9. Suppose you have a “farm” on which you grow harvest, and sell edible freshwater fish. The growth of the fish population is logistic. You want to manage your harvest to maintain maximum yields (that is, the maximum rate of production) from your farm over a number of years.

As a fisheries manager, you are responsible for deciding how many walleye can be harvested without destabilizing the population. a. Below is a data table showing the walleye population in a typical pond on your fish farm over 24 weeks. Draw a graph showing how population size in the pond changes through time.

Time 1 (weeks) Population 100 Size Time 13 (weeks) Population 188 Size

2

3

4

5

6

7

8

9

10

11

12

101

102

103

104

106

110

115

125

140

155

172

14

15

16

17

18

19

20

21

22

23

24

201

209

217

221

225

229

233

235

237

238

239

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b. How large should you let the population get before you harvest? Identify the point on your graph and explain why.

The rate of production will be fastest at the point on the curve where the slope is the steepest. This would be at about point C (the midpoint) on the graph. dN/dt is the measure of the rate of change (that is, production) of individuals in the population. When you graph N versus t, the slope corresponds to dN/dt. When the population size (N) is very small, then [(K - N)/K] is large. However, there are so few individuals in the population that the gain in population numbers, dN/dt, is also relatively small. When the population size (N) is very large, then there are many individuals but [(K – N)/K] is small. The rate of population growth (production measured as dN/dt) is at a maximum when N is equal to 1/2 K. c. Assume the carrying capacity for your pond is 250 individuals. Check your answer in part b by using the data in the chart and computing the change in the population size (dN/dt) when the population is at several different levels relative to its carrying capacity. Use K = 250 and rmax = 0.20.

Population size (N)

(K – N)/K

dN/dt

25 (low)

0.1

0.5

50 (moderately low)

0.8

8

125 (half K)

0.5

12.5

200 (moderately high)

0.2

8

250 (high)

0

0

10. A rabbit population has the following life table.

Age class

Number of deaths Mortality rate

0–1

Number of survivors 100

10

0.10

Number of offspring per reproducing pair 0

1–2

90

30

0.33

1.5

2–3

60

30

0.50

2.0

3–4

30

24

0.80

2.5

4–5

6

6

1.0

0

a. Fill in the missing data in the table. b. Owing to a good food supply and small predator population, the rabbit population is growing by leaps and bounds. The rabbits call a meeting to discuss population control measures. Two strategies are proposed: • Delay all rabbit marriages until age class 2–3 (rabbits never breed until after marriage).

• Sterilize all rabbits in age class 3–4.

Which of the proposed strategies will be more effective in slowing population growth? Explain your reasoning and show your calculations.

In the first two years of their lives, each reproducing pair of rabbits will have an average of 1.5 offspring. For a cohort of 100 rabbits born in year 0–1, 45 pairs of rabbits (the 90 survivors) will each Ch 44_56-44

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have 1.5 offspring during year 1–2; they will add about 67.5 rabbits to the population. When this same cohort reaches age class 3–4, it will consist of 15 pairs, each of which will produce an average of 2.5 offspring for a total of 37.5 rabbits. Therefore, the population size will be more limited by delaying rabbit marriages until age class 2–3.

Notes to Instructors Chapter 54 Community Ecology What is the focus of this activity?

Campbell Biology, 9th edition, defines community ecology as “the study of how interactions between species affect community structure and organization.” Biologists recognize that the organisms in communities are continuously evolving in association with one another. This implies that community structure and organization are also continuously evolving. This activity focuses on some of the key principles and factors that can affect the structure and organization of communities.

What is the particular activity designed to do?

Activity 54.1 What do you need to consider when analyzing communities of organisms?

Activity 54.2 What effects can disturbance have on a community?

These activities are designed to help students recognize that studying community ecology requires the consideration and integration of a number of ecological principles.

Activity 54.3 How can distance from the mainland and island size affect species richness?

This activity allows students to explore the theory of island biogeography and the effects distance from the mainland and island size have on species numbers.

All of these activities give students practice in data analysis and graphical representation and interpretation.

Answers

Activity 54.1 What do you need to consider when analyzing communities of organisms?

Understanding problems in community ecology most often requires integration of a number of ecological principles.

For questions 1 to 5, analyze the situations described. Then explain which of the following ecological principles could be active in each particular situation.

Ecological principles: coevolution realized niche fundamental niche tolerance competitive exclusion resource partitioning

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character displacement topdown vs. Bottom-up controls trophic structure or trophic levels

keystone

species

competition disturbance

1. A small clan of hyenas killed an antelope. While they were feeding on the carcass, two female lions approached, growled at the hyenas, and chased them away from the carcass.

The kill is an example of predation by the hyenas. The action of the lionesses is an example of competition. One species is actively keeping another away from a resource.

2. Two species of closely related swallows live in England. The black swallow lives in coniferous forests, and the yellow swallow lives in deciduous forests. In Ireland, where the black swallow has never been introduced, only the yellow swallow is present and it lives in both coniferous and deciduous forests.

This is an example of how a population’s realized (actual) niche can differ from its fundamental or ecological niche. When the black swallow is not present, the yellow swallow is capable of living in both deciduous and coniferous forests. In England, where both the black swallow and the yellow swallow are found, the yellow swallow is limited to deciduous forests. You would need more information to determine whether this is also a case of competitive exclusion or resource partitioning. For example, the birds’ distribution in England could be the result of resource partitioning if black sparrows are known to live in both deciduous and coniferous forests when yellow sparrows are not present. The distribution could be the result of competitive exclusion if black swallows live only in coniferous forests.

3. In a woodland community, three species of rodents coexist: voles, field mice and shrews. All three species eat seeds and nuts. Each species has a preference for seeds of the most appropriate sizes for their teeth and mouths. However, all three species compete for the same kinds of nuts. An owl species also lives in this woodland community. The owl preys on all three rodent species. During one particular year, a parasite that causes pneumonia in birds is introduced into the community. This dramatically reduces the owl population, which remains low for several years as a result. Following the initial reduction in the owl population, there is a dramatic increase in the population of field mice and a dramatic decrease in the populations of both voles and shrews.

The owl is a keystone species. Because the vole and shrew populations decline and the mouse population increases following removal of the owls, the mice are obviously better competitors for the available resources.

4. In 1962, five mute swans escaped from captivity and began a breeding population in Chesapeake Bay. Today, there are over 4,000 mute swans living in the bay. Each year they eat approximately 10.5 million pounds of aquatic grasses. These grasses provide habitat for waterfowl and crustaceans, improve water quality, decrease erosion, and increase dissolved oxygen concentrations in the bay. The birds are also aggressively territorial, and have been known to trample nests of other birds (e.g., least terns and black skimmers) and drive away native birds like tundra swans and black ducks from feeding and roosting areas.

This is an example of competition for territory and resources as well as an example of disturbance, in this case caused by the cascading effects of non-native species introduction. Disturbance of the trophic structure is occurring as the primary producers are decimated, and cannot perform their other ecological functions (oxygenation, cover, sediment stabilization).

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This could also be considered a form of top-down control gone wrong. Here, introduction of the mute swans caused loss of the aquatic grasses that supported the system. Their loss led to the destruction of the system.

5. Known as the “Hawaiian woodpecker,” the `akiapola`au (aki-a-pul-a-ow) is found only in montane mesic old-growth koa/`ohi` forests, and only on the Big Island (Hawaii). It has a distinctive beak that is like a multiple-use tool.

[Unnumbered Figure Comes Here] The short straight lower mandible is used to peck holes in the wood and the long curved upper mandible is used to probe for insects and larvae. Males have larger beaks than females and feed on the trunks of trees.

Females feed higher on branches and twigs. ‘Akiapola`au are thought to have the lowest reproduction rate for a small bird—only one chick per year, which is cared for by the parents for 6 months or more. The decline in their numbers appears to correspond with the introduction of rats, cats and logging on the island.

This species has a fairly specialized niche in old-growth forest. As a result, it is suffering loss of niche from logging. Introduced species are preying on the birds and may also be competing for habitat.

Significant reduction in numbers implies the birds have a limited tolerance range. Character displacement (beak size) in males and females exists; this allows resource partitioning (feeding on different parts of trees). Relating this example back to previous chapters, the parental care aspect of this bird indicates that it is K-selected, which is unusual for animals of such small size.

6. A researcher collected data on an experiment she conducted on two desert islands. The islands were of similar size, climate, and species composition and richness, and were the same distance from the mainland.

Originally, the same species of snake was present on both islands (A and B). In her experiment, the researcher removed the species of snake from island A. For comparison, the species of snake was NOT removed from island B. She then recorded the number of animal species on each island over a period of 24 months. Her data are presented in the table below. a. Graph the data.

# Species Time (months) 1

A

B

36

36

2

38

37

3

35

34

4

33

35

5

31

36

6

32

38

7

29

40

8

29

36

9

26

38

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10

24

39

11

20

36

12

18

37

13

18

35

14

13

34

15

11

36

16

10

38

17

9

40

18

10

38

19

10

36

20

8

38

21

9

39

22

10

36

23

9

37

24

8

35 [Unnumbered Figure Comes Here]

b. Construct a hypothesis that explains the difference between the numbers of species present on island A versus island B over the 24 month period.

One reasonable hypothesis would be that the snake was acting as a keystone species on both islands.

When it was removed from island A, another competitor/predator dominated and drove many of the other species on the island to extinction. c. Propose an experimental design to test your hypothesis. Explain the reasoning behind your design.

One possible design would be to remove the snake from island B, reintroduce these snakes to island A, and again collect data on species diversity over time. You would ideally collect data on not only species diversity, but also population size and species richness. This would allow you to better understand which competitor species is being held in check by the snake. d. What would you expect to find as a result of your experiment? Describe your expected results and draw them on the graph below.

Over time you would expect to see the numbers of species on island A increase and the number of species on island B decrease.

Possible graph:

[Unnumbered Figure Comes Here]

Activity 54.2 What effects can a disturbance have on a community?

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Disturbances rarely affect only one species of a community. More often disturbances that have direct effects on a single species produce a cascade of additional effects on other species.

For questions 1 and 2, analyze the ecological situations described. Then answer the questions that follow each situation.

1. A particularly popular island vacation site is home to many species of orchids. The primary income generating businesses on the island are tourism and orchid sales.

Disturbance: To make the island more attractive to visitors, a local politician suggested that the island be periodically “fogged” with insecticides.

Response: This fogging reduced the insect numbers. It also appeared to reduce the number of birds and new growths of orchids.

(Hint: Examine the ideas of limits of tolerance, coevolution, predation, and bioaccumulation.) a. Is it likely that the disturbance directly caused the response?

Fogging with insecticides is likely to have directly reduced the number of insects. However, the reduction in the number of birds and new growths of orchids may not have been directly caused by the pesticides. b. What other factors might be involved?

Many pesticides are fat-soluble. The amount of pesticide directly encountered by the birds may have been within their limits of tolerance. As insect eaters, however, they may have accumulated much more of the pesticide over time. Any one insect contains only a small amount of the pesticide, but over time, the birds consume many insects and the pesticide in each accumulates in the birds’ adipose (fat) tissues.

Most orchids are pollinated by insects. General fogging for insect pests would also kill the insects that pollinate the orchids and result in less new growth in the orchid population. Many plants have coevolved with specific insect species. If their pollinator dies out, the orchid species dies out as well. c. How could you test different factors for their effect on the response? For example, what experiments could you set up?

Different types of experiments could be set up. One possible experiment would test the effect of direct spraying on a population of birds that had no prior exposure to the pesticides. Groups of birds would need to be maintained in a facility that allows you to control what they eat and to observe their behavior. At the beginning of the experiment, all the birds would be directly sprayed with pesticides at a level consistent with possible exposure in the wild. For the remainder of the experiment, half of the birds would be fed insects that contained no pesticides. The other half would be fed insects sprayed with pesticide. Immediately before spraying and immediately after spraying, you would measure the level of pesticides in specific tissues in the birds. You would continue to measure the levels of pesticide in these tissues at set time intervals. The behavior and levels of pesticides in both sets of captive birds could be compared to each other and to those of birds in the wild. d. What would you expect to find if the other factors you proposed affected the response?

If direct spraying caused the deaths in the birds, you would expect to see equal death rates in both populations of captive birds. If bioaccumulation caused the deaths, you would expect to see higher numbers of deaths in the birds fed insects sprayed with pesticide. You would also expect to see a relatively constant level of pesticide in the tissues of the birds fed unsprayed insects and increasing levels of pesticide in the tissues of birds fed sprayed insects. When you compared the levels of Reece_IG

Ch 44_56-49

pesticides in the tissues, you would expect to see a strong correlation between the levels of pesticides in the birds fed sprayed insects and in the birds from the wild.

2. Australia and New Zealand are home to a very wide range of marsupials (for example, kangaroos and other pouched mammals). Until colonization by foreign traders and other developments, placental mammals were not found in these areas.

Disturbance: Following colonization, the rabbit was introduced to Australia.

Response: The rabbit multiplied rapidly and ultimately became a pest species, doing considerable damage to both crop and natural plants.

(Hint: Examine the ideas of competition, competitive exclusion, predation, and coevolution.) a. Is it likely that the disturbance directly caused the response?

The introduction of the rabbit is the most likely disturbance. The response is the considerable damage to crops and natural plants caused by the uncontrolled rabbit populations. There is a direct cause-andeffect relationship between the introduction of the rabbit and the rabbits’ effects on crops and other plants. b. What other factors might be involved?

In areas where the rabbit is a native species, it is generally not a pest because the rabbit has coevolved with its predators (including disease organisms), its food organisms, and its competitors. The situation is different when a species is artificially introduced into a new habitat. In the new habitat, there may be no predators or disease organisms to keep the introduced species’ population under control. The introduced species may be a better competitor for available food sources, so it is likely to become a pest. c. How could you test different factors for their effect on the response? For example, what experiments could you set up?

The government of Australia tested the hypothesis that lack of disease organisms was a central cause for the explosion in the rabbit population. In 1950, the government imported rabbit myxovirus and infected the wild populations. Significant decreases in the rabbit population were reported. As a result, there were significant increases in the populations of not only native plants but also native animals such as the kangaroo that depend on the native plants. (More recently, rabbit calcivirus, RCD, has been used to keep some of the more resistant rabbit populations under control.) d. What would you expect to find if the other factors you proposed affected the response?

If lack of disease organisms was a factor in the overpopulation of rabbits, you would expect that the introduction of disease organisms would significantly reduce the population (which did occur).

3. In the boreal forest of Canada, wildfires are important disturbance factors. A single wild fire seldom burns the whole forest. Instead it burns large patches or stands and leaves others untouched. Following a wildfire in a black spruce forest, there is usually a predictable regrowth of the vegetation, starting with ground lichens and small spruce seedlings. As the spruce trees grow and form a closed-crown canopy, feather mosses (Bryophytes) are found in an increasing proportion on the forest floor. In some cases, the peat moss Sphagnum outcompetes the feather mosses and eventually dominates the ground cover. Because wildfires occur naturally about every 10 years, a forest stand can sometimes burn before Sphagnum dominance is reached, and the whole process repeats. a. Graph “Tree Biomass vs. Time” over a 100-year period: Ch 44_56-50

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i.

at the stand scale (the stand is a particular part of the forest that burned) and

ii.

at the landscape scale (composed of many forest stands).

•

Assume, at the stand scale, that when a stand burns, all trees in that stand die.

• Assume, at the landscape scale, that there is one fire every 10 years, and that each fire burns a different stand.

The graphs should look something like this (X-axix is approx 100 yrs):

[Unnumbered Figure Comes Here] b. Explain how (and why) the graphs differ.

In the scenario presented, even though there are large variations in tree biomass at the stand scale, the variations are damped at the landscape scale, and at the landscape scale the overall biomass remains fairly constant.

Activity 54.3 How can distance from the mainland and island size affect species richness?

1. Island biogeography theory attempts to explain the patterns of species richness and turnover on islands as a function of the size of the island and its distance from the mainland. a. What are the basic tenets of this theory? (Refer to Figure 54.27 in Campbell Biology, 9th edition.)

The theory states that species richness (number of species) on islands is determined by both the size of the island and its distance from a mainland (source of species). Among the basic tenets are the following: Larger islands tend to have higher immigration rates and lower extinction rates than smaller islands. As a result, larger islands have more species than smaller islands. If two islands are the same size, they have equal rates of extinction. However, the one closer to the mainland has a higher rate of immigration and therefore a larger number of species. b. In the graph below, curves A and A’ represent the historic immigration and extinction rates, respectively, for an island off the coast of South America. Given these data, what is the equilibrium number of species for this island?

The equilibrium species number would be approximately 8.

[Unnumbered Figure Comes Here] c. If a land bridge or an isthmus forms and connects the island to the mainland, which two curves would best represent the resulting immigration and extinction rates for the island?

Line C would represent the new rate of immigration and line A’ would remain the rate of extinction.

Rate of extinction is a factor of island size (and therefore of habitat availability); as a result it is generally not affected by distance from the mainland. d. What would be the new equilibrium number of species?

The new equilibrium species number would be between 10 and 11.

2. A research scientist comes back from a study of a group of islands in the South Pacific. He says that his data show that in some cases, the smaller islands he investigated had higher numbers of lizard species on them than did the larger islands. He suggests these data indicate that we should reexamine the whole theory Reece_IG

Ch 44_56-51

of island biogeography. a. What additional information would you like to have about the group of islands he studied?

Some of the questions that could be asked include the following.

•

How far are the islands from the mainland?

•

What types of habitat are available on the islands?

•

How did the scientist determine how many species each island had?

b. Is there any way his data could be accounted for using the current theory of island biogeography?

Explain your reasoning.

Given the theory of island biogeography, it is possible that a smaller island may have more species than a larger island if the smaller island is much closer to a mainland than the larger island is.

Notes to Instructors Chapter 55 Ecosystems What is the focus of this activity?

In Activity 53.2, students compared exponential growth and logistic growth models. On one hand, the exponential growth model assumes no limits on the availability of resources. On the other hand, the logistic growth model assumes that each population has a carrying capacity, an upper limit on its population size that is determined by its environment. In Activity 55.1, students investigate various factors that can affect a population’s carrying capacity.

What is this particular activity designed to do?

Activity 55.1 What limits do available solar radiation and nutrients place on carrying capacities?

This activity is designed to help students understand basic energy relationships in ecosystems, including how the availability of energy can limit population growth.

Answers

Activity 55.1 What limits do available solar radiation and nutrients place on carrying capacities?

1. Energy transformations in a community can be diagrammatically represented as a trophic structure. a. Diagram a simple trophic structure for a grassland community.

Figure 55.11 on page 1,226 of Campbell Biology, 9th edition, shows an idealized pyramid of net production and may be used as an example of the trophic structure for a grassland community. b. At what trophic level(s) would humans belong in your diagram? Explain.

Humans, in general, have an omnivorous diet. Much like the mouse in Figure 55.11, humans consume both other animals and plants. Humans who are strictly vegetarians and eat only food derived from plants would be classified as primary consumers. However, if part of their diet is from plants and part from animals, which are primary consumers, then humans are classified as secondary consumers. (This demonstrates one of the difficulties associated with using very simple models. Not all organisms fit neatly into only one of the categories.)

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2. Energy is lost at every step in the trophic structure where energy is transferred—for example, from primary producers to primary consumers. It is estimated that only 1% of the solar energy striking plants is converted to chemical energy, only 50–90% of gross primary productivity (GPP) becomes net primary productivity (NPP), and only about 10% of the energy available at each tropic level transfers to the next level (as biomass). a. Restate the preceding information in your own words. Be sure to explain what GPP and NPP are and how they relate to the percentage of energy transfers from one trophic level to another.

Gross primary productivity (GPP) is a measure of the total amount of light energy used in photosynthesis per unit time. It is usually expressed in joules per square meter per year (J/m2/yr). Net primary productivity (NPP) equals GPP minus the energy the photosynthetic organisms use for cellular respiration. NPP can similarly be expressed in J/m2/yr, or it can be expressed as the total biomass of primary producers added per year to an area. Biomass is usually expressed in g/m2/yr. (The assumption is that each gram of biomass contains a given amount of energy, which can be released by oxidation.)

The information given in the question states that, of the sunlight that reaches plants, only 1% is converted to chemical energy. This is a measure of GPP. Only 50 to 90% of this 1%—that is, 0.5 to 0.9%—of the energy that reaches plants becomes actual biomass, or NPP. If all the NPP were consumed by primary consumers, you would expect to see 10% of it end up in biomass of secondary consumers. That is, 0.05 to 0.09% of the sunlight that reaches plants is actually converted into biomass of primary consumers. b. Where does the “lost energy” go? Is it correct to say that energy is lost in these transfers? If not, what would be a better way of expressing this?

The energy that is “lost” in these transfers is energy used by the organisms in their own metabolism.

For example, a primary consumer needs to eat 10 g of plant food to gain 1 g of biomass. The other 9 g of plant food are oxidized (via cellular respiration) to produce the ATP needed to build more biomass and to maintain the life functions of the organism.

3. Campbell Biology, 9th edition, indicates that each day the earth is “bombarded by about 1022 joules of solar radiation. (1 J = 0.239 cal).” It has also been estimated that primary producers on Earth collectively create about 170 billion tons of organic material per year. Do the following calculation to determine how these two values compare. a. Convert 1022 joules/day into kilocalories per year.

1022 J/day × 0.239 cal/J = 0.239 × 1022 cal/day = 2.39 × 1021 cal/day = 2.39 × 1018 kcal/day × 365 days/yr = 8.7 × 1023 kcal/yr b. Convert 170 billion tons of organic material into kilocalories. To do this, assume all of the 170 billion tons is glucose. Also, assume tons are metric tons. (Note: 180 grams of glucose when burned in a calorimeter gives off 686 Kcal of energy.)

170 × 109 metric tons/yr × 1,000 g/kg × 1,000 kg/metric ton × 686 kcal/mole = 170 × 1015 g/yr of biomass × 1 mole of glucose/180 g = 9.4 × 1014 moles/yr × 686 kcal/mole Reece_IG

Ch 44_56-53

= 6.5 × 1017 kcal/yr c. Given your answers to parts a and b, what percentage of the total incoming solar energy is captured as biomass (glucose)?

6.5 × 1017 kcal/yr/8.7 × 1023 kcal/yr = 7 × 10 -5 = 0.00007 = 0.007% d. Measurements of photosynthetic conversion of the energy in sunlight to biomass indicate that at best, plants can convert about 75% of the energy they absorb as sunlight into biomass. The other 25% is used to support metabolism. What could account for the apparently low efficiency of photosynthesis relative to total incoming solar energy that you calculated in part c?

Much of Earth is covered by water. In fact, only about 29% of Earth is land. Of that land, only about 11% is arable (suitable for agriculture). Some of the arable lands have been paved over (as streets, for example) or covered with buildings and are not available for agriculture. In addition, most of the arable land is in the temperate regions, where crops can grow only part of the year. This alone reduces the amount of biomass produced by at least 50%. Other factors that can affect productivity include variations in weather patterns from year to year, which can produce droughts or floods that cause crop loss and therefore loss of the energy stored in them.

4. If we can calculate the earth’s total primary productivity, we can use this value to develop an estimate of the total number of humans Earth can support. a. How much energy in kilocalories per year would it take to support the Earth’s current human population if all individuals weighed about 75 kg and each required 2,000 kcal per day? Assume 6 × 109 humans currently inhabit Earth.

2,000 kcal/day/human × 365 days/yr = 7.3 × 105 kcal/yr/human × 6 × 109 humans on Earth = 4.4 × 1015 kcal/yr consumed by the human population b. How does the estimate of the energy contained in the 170 billion tons of organic material (from part b in question 3) compare to the amount of energy required to support the current human population (from part a in question 4)?

If we accept the assumptions made in part a, then:

4.4 × 1015 kcal/yr/6.5 × 1017 kcal/yr = 6.8 × 10-1 or 68% of the total biomass is consumed by humans. c. Is it reasonable to assume that all of the primary productivity on Earth is available to support humans?

If not, what else do you need to consider?

The plant biomass on Earth ultimately supports or nourishes all the animals, the fungi, and many of the prokaryotes. If all the humans alive today were consuming 2,000 kcal per day, only 32% of the total biomass would be available for the other organisms.

5. You have been monitoring the net primary productivity (NPP) of a grassland area for several years. Over the years, NPP increased initially and then leveled off. You suspect that the availability of a nutrient is limiting productivity. a. Design an experiment to determine whether there is a limiting nutrient.

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To test whether a particular nutrient is limiting NPP, you could subdivide the area into a number of smaller plots. Treat some plots by adding more of the nutrient, and leave others untreated. Then measure the NPP produced on each type of plot. If the nutrient is limiting, you should find higher NPPs on treated plots than on untreated plots. You can repeat the experiment with a number of different nutrients. b. Given your design, what results would you expect if only one of the nutrients you test is limiting?

If your test plots are each set up to test addition of a single nutrient, you would expect to see an increase in growth only on the plot supplemented with the limiting nutrient. c. What results would you expect if more than one of the nutrients are limiting? Would this affect your design in any way?

If your test plots are each set up to test addition of a single nutrient, you would likely see no difference in growth across all plots if more than one nutrient is limiting. At this point you could start testing combinations of nutrients or other factors (see d below). d. What factor(s) other than nutrient limitation might cause NPP to level off?

Among other things, plant growth can be limited by water availability, physical or chemical conditions related to the density of plants per unit area, and the presence or absence of symbiotic mycorrhizal fungi.

55.1 Test Your Understanding 1 to 3. Two species of hawks are found at various times of the year in the United States. Hawk species A eats mice and squirrels (both of which eat seeds). Hawk species B eats snakes that feed on mice and squirrels. Given what you know about trophic structures, if other characteristics of the two hawk species are similar, which of the following are likely to be true? Explain your reasoning.

A = Likely to be true.

B = Not likely to be true.

1. For a given feeding area, the number of hawk species A that could be supported is likely to be smaller than the number of hawk species B that could be supported.

B—Not likely to be true. Species A, feeding on primary consumers, will have on average a 10-fold greater food base available than species B, which feeds on snakes, which are predators and thus secondary consumers.

2. Reduction in the available primary productivity in the feeding area would tend to have a lesser effect on the hawk species A than on hawk species B. B—Not likely to be true. Assume primary productivity is reduced from 100 kg of biomass to 50 kg. Prior to the loss, the feeding area would support 10 kg of primary consumers (e.g., mice and squirrels). That 10 kg would be used to support 1 kg of secondary consumers (e.g., snakes and hawk A). Therefore, less than 0.1 kg of hawk B would be supported. The reduction of primary productivity biomass to 50 kg would support 5 kg of primary consumers (on average) and 0.5 kg of snakes and hawk A. As a result, less than 0.05 kg of hawk B could be supported. In other words, the percent reduction for both populations would be about the same.

3. If the seeds consumed by the mice and squirrels in this feeding area were contaminated with heavy metals, you would expect to find much lower concentrations of heavy metals in hawk species A than in hawk Reece_IG

Ch 44_56-55

species B. A—Likely to be true. Hawk A would need to eat 10 g of mice and squirrels for every g of hawk mass added. Hawk B would have to eat 10 g of snake that ate 100 g of mice and squirrels for every g of hawk mass added. As a result, heavy metals, which are not metabolized but accumulate through the food chain, would be found in much higher concentrations in hawk species B. Notes to Instructors Chapter 56 Conservation Biology and Restoration Ecology What is the focus of this activity?

Many species of both plants and animals are rapidly going extinct as a result of habitat reduction or loss. Efforts to conserve individual species and species diversity require an understanding of the species’ physiology and behavior as well as its community ecology, evolution, and population genetics.

What is this particular activity designed to do?

Activity 56.1 What factors can affect the survival of a species or community?

Students are asked to consider and integrate a wide range of principles to address questions in conservation biology.

Answers

Activity 56.1 What factors can affect the survival of a species or community?

Making decisions to preserve communities requires an understanding and integration of many factors. Assume you work for the U.S. government and manage a large national forest. You are told that to maintain the economy in the area, the government has agreed to allow foresters to remove half a million acres of trees from a million-acre parcel. This parcel is almost square.

You have asked your staff to do an analysis of two possible methods for implementing the plan: Proposal I: Split the million acres into two parcels of a half million acres each and allow the foresters to harvest all trees on one of these parcels.

Proposal II: Divide the million-acre tract into 50 parcels of 10,000 acres each and allow the foresters to cut half the trees in each parcel.

1. List some of the ecological advantages and disadvantages of each proposal.

(Note: Each proposal has many possible advantages and disadvantages. The following are just a few of these.)

Proposal I Advantages

Proposal II Disadvantages

Advantages

• Fewer roads would be • The regeneration rate needed. for trees would be slower because seed • There would be less sources are more edge habitat relative to Ch 44_56-56

Disadvantages

• All cut areas would be • There would be more roads and therefore closer to seed sources, more edge and more and natural disturbance. regeneration would be

Reece_IG

interior habitat.

distant.

faster.

• The result may be invasion of more exotic species.

• Replanting efforts

• Interior species may be • More soil nutrients • Fewer soil nutrients favored more. would be lost from the would be lost due to land. runoff.

• Harvesting and would beefficient more diffuse and less . replanting would be • Complete loss of some • Disturbed animal more focused and communities could species may be able to therefore could occur result if animal or move to adjacent uncut more efficiently with plant communities areas. less disturbance to the were distributed other half-million • Species that roam over heterogeneously. acres. a wide geographic range may benefit.

2. Given the following characteristics of various animal species described below, which of the forestrycutting proposals would be more likely to ensure continued success of each animal species? Explain your answers. a. C. arnivora is a secondary and occasionally a tertiary consumer or carnivore. Behaviorally, it roams over about 20 square miles of “home range” in search of food—for example, rabbits (herbivores) and foxes (carnivores).

One square mile contains 2.8 × 107 square feet; 20 square miles contains 56 × 107 square feet; 1 acre contains 2.6 × 105 square feet. Therefore, a 1 million-acre forest contains 26 × 1010 square feet, or 26 × 1010/2.8 × 107, or about 9.3 × 103 square miles. As a result, this area could support as many as 4.6 × 102 C. arnivora (if these were the only carnivores).

Half of the forest (proposal I) contains about 4.6 × 103 square miles. This area could support a maximum of 2.3 × 102 C. arnivora.

The half of the forest left in proposal II is patchy. Although patches would be contiguous, the patchiness of the environment may have an effect on its desirability to predators.

In general, the effects of reducing the forest size on this species depend not only on the absolute area remaining but also on the size of the existing population, the minimum number of individuals required to maintain the population long term, the response of the carnivore to disturbance, and other factors. b. R. odentia is a small rodentlike herbivore that exists in small numbers in the forest. Its preferred food and habitat are found along the edges of the forest and are composed primarily of herbaceous (nonwoody) annual plants which produce tender shoots in the spring and plentiful seeds later in the year.

Proposal II would produce more edge habitat than proposal I. If the success of this species is directly related to the amount of edge habitat, then it would be more successful if proposal II were adopted. c. P. redatoria is a predatory bird which feeds on small rodents and occasionally on snakes and other reptiles. It can range over large distances looking for food. It nests in hollows that occur naturally in a particular species of tree. These hollows are not found in trees under 20 years of age but are commonly found in trees 40 years and older.

If we assume that 40-year-old trees are randomly distributed throughout the forest, then both proposals would eliminate about half of these trees. In other words, both proposals would reduce the available Reece_IG

Ch 44_56-57

nesting sites for this species by half. The immediate effect on the population would depend on the existing population size. If the number of P. redatoria in the population and the number of available nest sites are currently about equal, then half of the P. redatoria would lose their nest sites. They would not be able to reproduce and would eventually die without leaving offspring. If the P. redatoria population is much smaller than the number of available nesting sites, then there may be little immediate effect on the population. However, if this plan is repeated in 20 years and the other half of the forest is cut, the P. redatoria species would have no nesting sites left and would disappear in this region.

3. Conservation biologists have debated extensively which is better: many, small reserves or a few large ones. a. What types of factors should be considered in making judgments about size and location of reserves?

What is better will vary from species to species. Some species roam over wide ranges and may do better if many small reserves are connected into one large one. Some species can survive in only oldgrowth forest areas that have appropriate nesting sites. Other species are much more successful in open areas or edge habitat. In other words, the requirements of the species to be preserved should be of primary importance in making decisions about the size and location of reserves. b. Some ecologists argue that we should be concerned about preserving the largest number of species.

Others argue that we should be most concerned with saving those species judged to be of unusual importance. Develop an argument to support one of these viewpoints. Your argument should be based on our existing understanding of biology in general and ecology in particular.

Many logical arguments can be proposed. In developing your argument, you should consider the habitat and energy (food) requirements of the plant and animal species as well as any behavioral requirements of the species.

4. Based on your answers to questions 1–3, which of the proposals for removing half a million acres from the national forest in question 1 would you recommend? Explain your reasoning.

Again, a number of possible arguments can be developed to support either proposal I or proposal II. Your argument should include a comparison of the advantages and disadvantages of each proposal as they relate to both the plant and animal communities in the forest.

Ch 44_56-58

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Appendix A

An Introduction to Data Analysis and Graphing for a PC

This appendix provides a quick introduction to the use of some of the data analysis and graphing capabilities of one of the more commonly used computer programs, Excel. This and other similar software systems make it much easier to graph and analyze large data sets.

1.

TO SET UP EXCEL FOR DATA ANALYSIS AND GRAPHING •

Open Excel and activate the “Data Analysis” tool. o Under the “Tools” pull-down menu, look for “Data Analysis.” o If you don’t see “Data Analysis,” look for “Add-Ins.” Click on “Data Analysis” or “Add-Ins” and select “Analysis ToolPack”; then click “OK.” o If you did a custom install instead of a complete install of Excel, you may not find “Analysis ToolPack” under “Add-Ins.” If this is the case, go to your original installation CD to add this feature.

•

Enter the data to be analyzed or graphed in columns on the Excel spreadsheet. o If you want to graph height versus arm length, enter the data in two columns with the data you want to appear on the y-axis (independent variable) of a graph in the right-hand column. o

The data to appear on the x-axis (dependent variable) should be in the adjacent left-hand column.

o An example is on the next page.

Note: If you are using Excel 2007 to find and activate Add-ins: •

Click on the “Office” button in the upper left-hand corner.

•

Then click on the “Excel options” button at the bottom of the pop-up window.

•

Select “Add-ins” from the left-hand column.

•

Select/highlight “Analysis ToolPak” in the list of Inactive Application Add-ins, then click “Go.”

•

Check the box in front of “Analysis Tool Pak” in the window that appears and click “OK.”

•

Now click on the “Data” tab at the top of the screen. You should see a new “Data Analysis” pulldown menu at the right top of the screen.

A

B

C

Gender

Arm Length (cm)

Height (cm)

3

f

73.7

170.2

4

m

76.7

177.8

5

m

71.1

170.2

1 2

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App A-1

2.

f

66

165.1

7

f

68.6

172.7

8

m

76.7

172.7

9

m

78.7

177.8

10

m

81.3182.9

TO SORT THE DATA

•

In the upper left-hand corner of the Excel spreadsheet, click the box between Row 1 and Column A. The whole table should be highlighted when you do this.

Pull down the “Data” menu at the top of the screen and click “Sort.”

•

If you have headers or titles on your columns, click the  in front of “Header row.”

•

Select the column you want to sort by (e.g., Gender).

•

• •

3.

6

Select either “Ascending” if you want the sort to go from low to high (or A to Z) or “Descending” for high to low (or Z to A).

Click “OK.”

TO GET DESCRIPTIVE/SUMMARY STATISTICS FOR THE DATA a. Pull down the “Tools” menu at the top of the Excel spreadsheet and click on “Data Analysis.” b. Click on “Descriptive Statistics” and then “OK.” • • • •

In the window that appears, for input range click on the icon to the right of the “Input Range:” box.

Highlight (select) the data in the Excel spreadsheet that you want analyzed. For example, select all of the arm length data (males plus females).

Click on the chart icon next to “Input Range” again to return to the Descriptive Statistics window.

To indicate where to place the output, click on the  in front of “Output Range:” and then click on the chart icon to the right of the “Output Range:” box. o To select the location for the data to appear, click on any open box in the Excel spreadsheet. o Click the chart icon next to “Output Range” again to return to the Descriptive Statistics window.

•

In the Descriptive Statistics window select: Summary Statistics ▪ Confidence Level for Mean ▪

• App A-2

Click “OK.” A box containing the summary statistics will appear. For example: Reece_IG

Arm Length

Definitions:

Mean

74.1

The average of all the values.

Standard Error

1.849421

Median

75.2

Mode

76.7

Standard Deviation

5.230952

An estimate of the standard deviation of the sample mean.

The number that lies in the middle of the set of numbers.

The most frequent number in the set.

The square root of the variance.

Sample Variance

27.36286

A measure of how far from the average the data points are.

A measure of how peaked or flat the distribution is.

A measure of the degree of symmetry of the distribution.

The difference between the lowest and highest values in the set.

The smallest value in the set.

Kurtosis

–0.99518

Skewness

–0.29163

Range

15.3

Minimum

66

The largest value in the set.

Maximum

81.3

The total of all values in the set.

Sum

592.8

Count

8

The total number of values in the set.

Two data sets may have the same mean but very different amounts of variability. To indicate the amount of variability that exists in a given set of data, the arithmetic mean for those data is often expressed as the mean  the standard deviation. Here, the smaller the standard deviation is, the smaller the variance is in the treatment group. The smaller the variance is, the more confidence we have that the data in our sample set represent a good estimate of the values for the larger population.

4. TO GRAPH THE DATA a. To make a linear graph of x vs. y values: •

Click on the “Chart Wizard” icon at the top of the Excel spreadsheet.

•

Click on “XY (Scatter).”

•

Click “Next.”

•

Click the chart icon to the left of “Data range:” o Select all of the data in the two columns (Arm length [cm] and Height [cm]).

Reece_IG

App A-3

o Note: If your data are not in side-by-side columns with the y data in the right-hand column, you can click on the “Series” tab instead of the “Data range:” tab. Click “Add” and then enter the data separately for the x and y-axes by highlighting the columns separately here.

•

Click “Next” and select the “Titles” tab to add titles for the chart and the x and y-axes.

•

Click “Next” and select where you want the chart to appear (on the same sheet or a new sheet).

Then click “Finish.”

•

On the graph that appears, click on one of the data points to highlight it, then right click (on PC computers) and select “Add trendline.” o Under the “Type” tab, select “Linear.” o Under the “Options” tab, select: o “Display equation on chart” and “Display R-squared value on chart.” o Click “OK.”

•

For the data in the previous table you should get the following graph.

[Unnumbered figures come here] The formula y = 0.9409x + 103.95 is in the format y = mx + b. When the value of m is positive it indicates that there is a direct relationship between x and y; in this case it indicates that as x increases, y increases. (If m were negative, it would indicate an inverse relationship—as x increases, y decreases.)

For this specific graph, if it were reasonable to extrapolate the line to a point where x = 0, the line would intercept the y axis at b, or 103.95 cm. With this as the zero point for x on the line, the y values would increase from this point by m, or 0.9409 for every unit change in x.

The R2 value has no units and a range of 0 to 1. An R2 value of zero means there is no linear relationship between x and y values. On the other hand, high values of R2 indicate a strong correlation between the x and y values. High R2 values are also a good indicator of statistical significance. b. To make a histogram of the data: •

Select a set of data to use (e.g., heights for males only). (See previous section “To Sort the Data.”)

•

Make a new column of data containing heights increasing by 5 cm each—for example, 160, 165, 170, 175, 180, 185 cm. (This is called your Bin Range.)

•

Go to the “Tools” pull-down menu and click on “Data Analysis,” then on “Histogram.” Then click “OK.” o Click on the chart icon to the right of the “Input Range:” box. o Select/highlight the height data for males only (include a column heading). o Click the chart icon again to return to the Histogram window. o Click on the chart icon to the right of the “Bin Range:” box. o Select/highlight the column containing heights increasing by 5 cm each. o Click on the chart icon again to return to the Histogram window.

App A-4

Reece_IG

o Click on the  in front of “Output Range:” and then click on the chart icon to the right of the “Output Range:” box. o To select the location for the data to appear, click on any open box in the Excel spreadsheet. o Click the chart icon again to return to the Histogram window. o In the Histogram window select:

 Chart Output o Click “OK.” The frequency data and histogram should appear.

Histograms are often developed to help determine whether or not the data appear to be distributed normally. In this case, only five values (the five heights for males) are included—enough to determine how the heights among these five individuals are distributed but not enough to make any general statements about the total population of males.

Note: If the graph of the data approximates a bell-shaped curve or normal distribution, the standard deviation takes on additional meaning. For normal data distributions, about 68% of the data points will fall within one standard deviation (on either side) of the mean. Ninety-five percent of the data will fall within two standard deviations of the mean; and 99% will fall within three standard deviations of the mean.

5. TO DETERMINE STATISTICAL SIGNIFICANCE The statistical significance of the difference in means can be determined by using a statistical test. The null hypothesis (H0) is generally assumed for such tests. A simple null hypothesis would state that there is no difference in x (some characteristic) between two populations (e.g., there is no difference in arm length between males and females). This null hypothesis further assumes that any differences observed between the two populations are the result of chance variation or sampling error alone. An alternative hypothesis (Ha) would be that there is a difference or relationship between the populations. The alternative hypothesis is supported when the null hypothesis is rejected.

The Data Analysis function of Excel allows you to perform a number of different statistical tests, including t tests and analysis of variance (ANOVA). These tests are designed to show whether the observed difference in means, between groups or between control and test groups, is due to random chance/variability. A significance level of 0.05 or 5% means that statistically there is only a 5% chance that the difference observed is due to chance alone. A significance level of 0.05 is usually accepted as evidence that the treatment produced a significant effect. A significance level of 0.01, or 1%, is considered highly significant. It would mean that only 1% of the time would such a difference be likely to occur as a result of chance alone. In these cases, the null hypothesis is rejected.

Reece_IG

App A-5

Appendix B

An Introduction to Data Analysis and Graphing for a Mac

This appendix provides a quick introduction to the use of some of the data analysis and graphing capabilities of one of the more commonly used computer programs, Excel. This and other similar software systems make it much easier to graph and analyze large data sets.

1.TO SET UP EXCEL FOR DATA ANALYSIS AND GRAPHING • Open Excel and activate the “Data Analysis” tool. o Under the “Tools” pull-down menu, look for “Data Analysis.” o If you don’t see “Data Analysis,” look for “Add-Ins.” Click on “Data Analysis” or “Add-Ins” and select “Analysis ToolPack”; then click “OK.” o If you did a custom install instead of a complete install of Excel, you may not find “Analysis ToolPack” under “Add-Ins.” If this is the case, go to your original installation CD to add this feature.

• Enter the data to be analyzed or graphed in columns on the Excel spreadsheet. o If you want to graph height versus arm length, enter the data in two columns with the data you want to appear on the y-axis (independent variable) of a graph in the right-hand column. o The data to appear on the x-axis (dependent variable) should be in the adjacent left-hand column. o For example: A

B

C

Gender

Arm Length (cm)

Height (cm)

3

f

73.7

170.2

4

m

76.7

177.8

5

m

71.1

170.2

6

f

66

165.1

7

f

68.6

172.7

8

m

76.7

172.7

9

m

78.7

177.8

10

m

81.3

182.9

1 2

2. TO SORT THE DATA •

In the upper left-hand corner of the Excel spreadsheet, click the box between Row 1 and Column A. The whole table should be highlighted when you do this.

• Pull down the “Data” menu at the top of the screen and click “Sort.”

Reece_IG

App B-1

• If you have headers or titles on your columns, click the  in front of “Header row.”

• Select the column you want to sort by (e.g., Gender).

• Select either “Ascending” if you want the sort to go from low to high (or A to Z) or “Descending” for high to low (or Z to A).

• Click “OK.”

3. TO GET DESCRIPTIVE/SUMMARY STATISTICS FOR THE DATA a. Pull down the “Tools” menu at the top of the Excel spreadsheet and click on “Data Analysis.” b. Click on “Descriptive Statistics” and then “OK.” •

In the window that appears, for input range click on the icon to the right of the “Input Range:” box.

•

Highlight (select) the data in the Excel spreadsheet that you want analyzed. For example, select all of the arm length data (males plus females).

• •

Click on the icon to the right of “Input Range” again to return to the Descriptive Statistics window.

To indicate where to place the output, click on the  in front of “Output Range:” and then click on the icon to the right of the “Output Range:” box. o To select the location for the data to appear, click on any open box in the Excel spreadsheet. o Click the icon next to “Output Range” again to return to the Descriptive Statistics window.

•

In the Descriptive Statistics window select:

 Summary Statistics  Confidence Level for Mean •

Click “OK.” A box containing the summary statistics will appear. For example:

Arm Length

Definitions:

Mean

74.1

The average of all the values.

Standard Error

1.849421

Median

75.2

An estimate of the standard deviation of the sample mean.

The number that lies in the middle of the set of numbers.

Mode

76.7

Standard Deviation

5.230952

Sample Variance

27.36286

Kurtosis

–0.99518

App B-2

The most frequent number in the set.

The square root of the variance.

A measure of how far from the average the data points are.

A measure of how peaked or flat

Reece_IG

the distribution is.

Skewness

–0.29163

Range

15.3

Minimum

66

Maximum

81.3

Sum

592.8

Count

8

A measure of the degree of symmetry of the distribution.

The difference between the lowest and highest values in the set.

The smallest value in the set.

The largest value in the set.

The total of all values in the set.

The total number of values in the set.

Two data sets may have the same mean but very different amounts of variability. To indicate the amount of variability that exists in a given set of data, the arithmetic mean for those data is often expressed as the mean  the standard deviation. Here, the smaller the standard deviation is, the smaller the variance is in our set of results or data. The smaller the variance is, the more confidence we have that the data in our sample set represent a good estimate of the values for the larger population.

4. TO GRAPH THE DATA a. To make a linear graph of x vs. y values: •

Click on the “Chart Wizard” icon at the top of the Excel spreadsheet.

•

Click on “XY (Scatter).”

•

Click “Next.”

•

Click the icon to the right of “Data range:” o Select all of the data in the two columns (Arm length [cm] and Height [cm]). o Note: If your data are not in side-by-side columns with the y data in the right-hand column, you can click on the “Series” tab at the top of the screen that pops up (instead of the “Data range” tab). Click “Add” and then enter the data separately for the x- and y-axes by highlighting the columns separately here.

•

Click “Next” and select the “Titles” tab to add titles for the chart and the x- and y-axes.

•

Click “Next” and select where you want the chart to appear (on the same sheet or a new sheet).

Then click “Finish.”

•

On the graph that appears, click on one of the data points to highlight it, then click on the “Chart” pull-down menu at the top of the screen and select “Add trendline.”

 Under the “Type” tab, select “Linear.”  Under the “Options” tab, select:  “Display equation on chart” and “Display R-squared value on chart.”

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App B-3

 Click “OK.” •

For the data in the previous table you should get the following graph.

The formula y = 0.9409x + 103.95 is in the format y = mx + b. When the value of m is positive it indicates that there is a direct relationship between x and y; in this case it indicates that as x increases, y increases. (If m were negative, it would indicate an inverse relationship—as x increases, y decreases.)

For this specific graph, if it were reasonable to extrapolate the line to a point where x = 0, the line would intercept the y-axis at b, or 103.95 cm. With this as the zero point for x on the line, the y values would increase from this point by m, or 0.9409 for every unit change in x.

The R2 value has no units and a range of 0 to 1. An R2 value of zero means there is no linear relationship between x and y values. On the other hand, high values of R2 indicate a strong correlation between the x and y values. High R2 values are also a good indicator of statistical significance. b. To make a histogram of the data: • • •

Select a set of data to use (e.g., heights for males only). (See the previous section “To Sort the Data.”)

Make a new column of data containing heights increasing by 5 cm each for example, 160, 165, 170, 175, 180, 185 cm. (This is called your Bin Range.)

Go to the “Tools” pull-down menu and click on “Data Analysis,” then on “Histogram.” Then click “OK.” o Click on the icon to the right of the “Input Range:” box

 Select/highlight the height data for males only (include a column heading).  Click the chart icon again to return to the Histogram window. o Click on the icon to the right of the “Bin Range:” box.

 Select/highlight the column containing heights increasing by 5 cm each.  Click on the icon again to return to the Histogram window. o

Click on the  in front of “Output Range:” and then click on the icon to the right of the “Output Range:” box. 

To select the location for the data to appear, click on any open box in the Excel spreadsheet.

 Click the icon again to return to the Histogram window. o In the Histogram window select:

 Chart Output o Click “OK.” The frequency data and histogram should appear.

Histograms are often developed to help determine whether or not the data appear to be distributed normally. In this case, only five values (the five heights for males) are included—enough to determine how the heights among these five individuals are distributed but not enough to make any general statements about the total population of males.

App B-4

Reece_IG

Note: If the graph of a large data set approximates a bell-shaped curve or normal distribution, the standard deviation takes on additional meaning. For normal data distributions, about 68% of the data points will fall within one standard deviation (on either side) of the mean. Ninety-five percent of the data will fall within two standard deviations of the mean; and 99% will fall within three standard deviations of the mean.

5. TO DETERMINE STATISTICAL SIGNIFICANCE The statistical significance of the difference in means can be determined by using a statistical test. The null hypothesis (H0) is generally assumed for such tests. A simple null hypothesis would state that there is no difference in x (some characteristic) between two populations (e.g., there is no difference in arm length between males and females). This null hypothesis further assumes that any differences observed between the two populations are the result of chance variation or sampling error alone. An alternative hypothesis (Ha) would be that there is a difference in arm length (or other characteristic) between the populations. The alternative hypothesis is supported when the null hypothesis is rejected.

The Data Analysis function of Excel allows you to perform a number of different statistical tests, including t tests and analysis of variance (ANOVA). These tests are designed to show whether the observed difference in means, between groups or between control and test groups, is due to random chance/variability. A significance level of 0.05 or 5% means that statistically there is only a 5% chance that the difference observed is due to chance alone. A significance level of 0.05 is usually accepted as evidence that the treatment produced a significant effect. A significance level of 0.01, or 1%, is considered highly significant. It would mean that only 1% of the time would such a difference be likely to occur as a result of chance alone. In these cases, the null hypothesis is rejected.

Reece_IG

App B-5