38.4: Muscle Contraction and Locomotion
Skills to Develop
- Classify the different types of muscle tissue
- Explain the role of muscles in locomotion
Muscle cells are specialized for contraction. Muscles allow for motions such as walking, and they also facilitate bodily processes such as respiration and digestion. The body contains three types of muscle tissue: skeletal muscle, cardiac muscle, and smooth muscle (Figure \(\PageIndex{1}\)).
Skeletal muscle tissue forms skeletal muscles, which attach to bones or skin and control locomotion and any movement that can be consciously controlled. Because it can be controlled by thought, skeletal muscle is also called voluntary muscle. Skeletal muscles are long and cylindrical in appearance; when viewed under a microscope, skeletal muscle tissue has a striped or striated appearance. The striations are caused by the regular arrangement of contractile proteins (actin and myosin). Actin is a globular contractile protein that interacts with myosin for muscle contraction. Skeletal muscle also has multiple nuclei present in a single cell.
Smooth muscle tissue occurs in the walls of hollow organs such as the intestines, stomach, and urinary bladder, and around passages such as the respiratory tract and blood vessels. Smooth muscle has no striations, is not under voluntary control, has only one nucleus per cell, is tapered at both ends, and is called involuntary muscle.
Cardiac muscle tissue is only found in the heart, and cardiac contractions pump blood throughout the body and maintain blood pressure. Like skeletal muscle, cardiac muscle is striated, but unlike skeletal muscle, cardiac muscle cannot be consciously controlled and is called involuntary muscle. It has one nucleus per cell, is branched, and is distinguished by the presence of intercalated disks.
Skeletal Muscle Fiber Structure
Each skeletal muscle fiber is a skeletal muscle cell. These cells are incredibly large, with diameters of up to 100 µm and lengths of up to 30 cm. The plasma membrane of a skeletal muscle fiber is called the sarcolemma . The sarcolemma is the site of action potential conduction, which triggers muscle contraction. Within each muscle fiber are myofibrils —long cylindrical structures that lie parallel to the muscle fiber. Myofibrils run the entire length of the muscle fiber, and because they are only approximately 1.2 µm in diameter, hundreds to thousands can be found inside one muscle fiber. They attach to the sarcolemma at their ends, so that as myofibrils shorten, the entire muscle cell contracts (Figure \(\PageIndex{2}\)).
The striated appearance of skeletal muscle tissue is a result of repeating bands of the proteins actin and myosin that are present along the length of myofibrils. Dark A bands and light I bands repeat along myofibrils, and the alignment of myofibrils in the cell causes the entire cell to appear striated or banded.
Each I band has a dense line running vertically through the middle called a Z disc or Z line. The Z discs mark the border of units called sarcomeres , which are the functional units of skeletal muscle. One sarcomere is the space between two consecutive Z discs and contains one entire A band and two halves of an I band, one on either side of the A band. A myofibril is composed of many sarcomeres running along its length, and as the sarcomeres individually contract, the myofibrils and muscle cells shorten (Figure \(\PageIndex{3}\)).
Myofibrils are composed of smaller structures called myofilaments . There are two main types of filaments: thick filaments and thin filaments; each has different compositions and locations. Thick filaments occur only in the A band of a myofibril. Thin filaments attach to a protein in the Z disc called alpha-actinin and occur across the entire length of the I band and partway into the A band. The region at which thick and thin filaments overlap has a dense appearance, as there is little space between the filaments. Thin filaments do not extend all the way into the A bands, leaving a central region of the A band that only contains thick filaments. This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments. Both the Z disc and the M line hold myofilaments in place to maintain the structural arrangement and layering of the myofibril. Myofibrils are connected to each other by intermediate, or desmin, filaments that attach to the Z disc.
Thick and thin filaments are themselves composed of proteins. Thick filaments are composed of the protein myosin. The tail of a myosin molecule connects with other myosin molecules to form the central region of a thick filament near the M line, whereas the heads align on either side of the thick filament where the thin filaments overlap. The primary component of thin filaments is the actin protein. Two other components of the thin filament are tropomyosin and troponin. Actin has binding sites for myosin attachment. Strands of tropomyosin block the binding sites and prevent actin–myosin interactions when the muscles are at rest. Troponin consists of three globular subunits. One subunit binds to tropomyosin, one subunit binds to actin, and one subunit binds Ca 2+ ions.
Link to Learning
View this animation showing the organization of muscle fibers.
Sliding Filament Model of Contraction
For a muscle cell to contract, the sarcomere must shorten. However, thick and thin filaments—the components of sarcomeres—do not shorten. Instead, they slide by one another, causing the sarcomere to shorten while the filaments remain the same length. The sliding filament theory of muscle contraction was developed to fit the differences observed in the named bands on the sarcomere at different degrees of muscle contraction and relaxation. The mechanism of contraction is the binding of myosin to actin, forming cross-bridges that generate filament movement (Figure \(\PageIndex{4}\)).
When a sarcomere shortens, some regions shorten whereas others stay the same length. A sarcomere is defined as the distance between two consecutive Z discs or Z lines; when a muscle contracts, the distance between the Z discs is reduced. The H zone—the central region of the A zone—contains only thick filaments and is shortened during contraction. The I band contains only thin filaments and also shortens. The A band does not shorten—it remains the same length—but A bands of different sarcomeres move closer together during contraction, eventually disappearing. Thin filaments are pulled by the thick filaments toward the center of the sarcomere until the Z discs approach the thick filaments. The zone of overlap, in which thin filaments and thick filaments occupy the same area, increases as the thin filaments move inward.
ATP and Muscle Contraction
The motion of muscle shortening occurs as myosin heads bind to actin and pull the actin inwards. This action requires energy, which is provided by ATP. Myosin binds to actin at a binding site on the globular actin protein. Myosin has another binding site for ATP at which enzymatic activity hydrolyzes ATP to ADP, releasing an inorganic phosphate molecule and energy.
ATP binding causes myosin to release actin, allowing actin and myosin to detach from each other. After this happens, the newly bound ATP is converted to ADP and inorganic phosphate, P i . The enzyme at the binding site on myosin is called ATPase. The energy released during ATP hydrolysis changes the angle of the myosin head into a “cocked” position. The myosin head is then in a position for further movement, possessing potential energy, but ADP and P i are still attached. If actin binding sites are covered and unavailable, the myosin will remain in the high energy configuration with ATP hydrolyzed, but still attached.
If the actin binding sites are uncovered, a cross-bridge will form; that is, the myosin head spans the distance between the actin and myosin molecules. P i is then released, allowing myosin to expend the stored energy as a conformational change. The myosin head moves toward the M line, pulling the actin along with it. As the actin is pulled, the filaments move approximately 10 nm toward the M line. This movement is called the power stroke, as it is the step at which force is produced. As the actin is pulled toward the M line, the sarcomere shortens and the muscle contracts.
When the myosin head is “cocked,” it contains energy and is in a high-energy configuration. This energy is expended as the myosin head moves through the power stroke; at the end of the power stroke, the myosin head is in a low-energy position. After the power stroke, ADP is released; however, the cross-bridge formed is still in place, and actin and myosin are bound together. ATP can then attach to myosin, which allows the cross-bridge cycle to start again and further muscle contraction can occur (Figure \(\PageIndex{5}\)).
Link to Learning
Watch this video explaining how a muscle contraction is signaled.
Art Connection
Which of the following statements about muscle contraction is true?
- The power stroke occurs when ATP is hydrolyzed to ADP and phosphate.
- The power stroke occurs when ADP and phosphate dissociate from the myosin head.
- The power stroke occurs when ADP and phosphate dissociate from the actin active site.
- The power stroke occurs when Ca 2+ binds the calcium head.
Link to Learning
View this animation of the cross-bridge muscle contraction.
Regulatory Proteins
When a muscle is in a resting state, actin and myosin are separated. To keep actin from binding to the active site on myosin, regulatory proteins block the molecular binding sites. Tropomyosin blocks myosin binding sites on actin molecules, preventing cross-bridge formation and preventing contraction in a muscle without nervous input. Troponin binds to tropomyosin and helps to position it on the actin molecule; it also binds calcium ions.
To enable a muscle contraction, tropomyosin must change conformation, uncovering the myosin-binding site on an actin molecule and allowing cross-bridge formation. This can only happen in the presence of calcium, which is kept at extremely low concentrations in the sarcoplasm. If present, calcium ions bind to troponin, causing conformational changes in troponin that allow tropomyosin to move away from the myosin binding sites on actin. Once the tropomyosin is removed, a cross-bridge can form between actin and myosin, triggering contraction. Cross-bridge cycling continues until Ca 2+ ions and ATP are no longer available and tropomyosin again covers the binding sites on actin.
Excitation–Contraction Coupling
Excitation–contraction coupling is the link (transduction) between the action potential generated in the sarcolemma and the start of a muscle contraction. The trigger for calcium release from the sarcoplasmic reticulum into the sarcoplasm is a neural signal. Each skeletal muscle fiber is controlled by a motor neuron, which conducts signals from the brain or spinal cord to the muscle. The area of the sarcolemma on the muscle fiber that interacts with the neuron is called the motor end plate . The end of the neuron’s axon is called the synaptic terminal, and it does not actually contact the motor end plate. A small space called the synaptic cleft separates the synaptic terminal from the motor end plate. Electrical signals travel along the neuron’s axon, which branches through the muscle and connects to individual muscle fibers at a neuromuscular junction.
The ability of cells to communicate electrically requires that the cells expend energy to create an electrical gradient across their cell membranes. This charge gradient is carried by ions, which are differentially distributed across the membrane. Each ion exerts an electrical influence and a concentration influence. Just as milk will eventually mix with coffee without the need to stir, ions also distribute themselves evenly, if they are permitted to do so. In this case, they are not permitted to return to an evenly mixed state.
The sodium–potassium ATPase uses cellular energy to move K + ions inside the cell and Na + ions outside. This alone accumulates a small electrical charge, but a big concentration gradient. There is lots of K + in the cell and lots of Na + outside the cell. Potassium is able to leave the cell through K + channels that are open 90% of the time, and it does. However, Na + channels are rarely open, so Na + remains outside the cell. When K + leaves the cell, obeying its concentration gradient, that effectively leaves a negative charge behind. So at rest, there is a large concentration gradient for Na + to enter the cell, and there is an accumulation of negative charges left behind in the cell. This is the resting membrane potential. Potential in this context means a separation of electrical charge that is capable of doing work. It is measured in volts, just like a battery. However, the transmembrane potential is considerably smaller (0.07 V); therefore, the small value is expressed as millivolts (mV) or 70 mV. Because the inside of a cell is negative compared with the outside, a minus sign signifies the excess of negative charges inside the cell, −70 mV.
If an event changes the permeability of the membrane to Na + ions, they will enter the cell. That will change the voltage. This is an electrical event, called an action potential, that can be used as a cellular signal. Communication occurs between nerves and muscles through neurotransmitters. Neuron action potentials cause the release of neurotransmitters from the synaptic terminal into the synaptic cleft, where they can then diffuse across the synaptic cleft and bind to a receptor molecule on the motor end plate. The motor end plate possesses junctional folds—folds in the sarcolemma that create a large surface area for the neurotransmitter to bind to receptors. The receptors are actually sodium channels that open to allow the passage of Na + into the cell when they receive neurotransmitter signal.
Acetylcholine (ACh) is a neurotransmitter released by motor neurons that binds to receptors in the motor end plate. Neurotransmitter release occurs when an action potential travels down the motor neuron’s axon, resulting in altered permeability of the synaptic terminal membrane and an influx of calcium. The Ca 2+ ions allow synaptic vesicles to move to and bind with the presynaptic membrane (on the neuron), and release neurotransmitter from the vesicles into the synaptic cleft. Once released by the synaptic terminal, ACh diffuses across the synaptic cleft to the motor end plate, where it binds with ACh receptors. As a neurotransmitter binds, these ion channels open, and Na + ions cross the membrane into the muscle cell. This reduces the voltage difference between the inside and outside of the cell, which is called depolarization. As ACh binds at the motor end plate, this depolarization is called an end-plate potential. The depolarization then spreads along the sarcolemma, creating an action potential as sodium channels adjacent to the initial depolarization site sense the change in voltage and open. The action potential moves across the entire cell, creating a wave of depolarization.
ACh is broken down by the enzyme acetylcholinesterase (AChE) into acetyl and choline. AChE resides in the synaptic cleft, breaking down ACh so that it does not remain bound to ACh receptors, which would cause unwanted extended muscle contraction (Figure \(\PageIndex{6}\)).
Art Connection
The deadly nerve gas Sarin irreversibly inhibits acetycholinesterase. What effect would Sarin have on muscle contraction?
After depolarization, the membrane returns to its resting state. This is called repolarization, during which voltage-gated sodium channels close. Potassium channels continue at 90% conductance. Because the plasma membrane sodium–potassium ATPase always transports ions, the resting state (negatively charged inside relative to the outside) is restored. The period immediately following the transmission of an impulse in a nerve or muscle, in which a neuron or muscle cell regains its ability to transmit another impulse, is called the refractory period. During the refractory period, the membrane cannot generate another action potential. . The refractory period allows the voltage-sensitive ion channels to return to their resting configurations. The sodium potassium ATPase continually moves Na + back out of the cell and K + back into the cell, and the K + leaks out leaving negative charge behind. Very quickly, the membrane repolarizes, so that it can again be depolarized.
Control of Muscle Tension
Neural control initiates the formation of actin–myosin cross-bridges, leading to the sarcomere shortening involved in muscle contraction. These contractions extend from the muscle fiber through connective tissue to pull on bones, causing skeletal movement. The pull exerted by a muscle is called tension, and the amount of force created by this tension can vary. This enables the same muscles to move very light objects and very heavy objects. In individual muscle fibers, the amount of tension produced depends on the cross-sectional area of the muscle fiber and the frequency of neural stimulation.
The number of cross-bridges formed between actin and myosin determine the amount of tension that a muscle fiber can produce. Cross-bridges can only form where thick and thin filaments overlap, allowing myosin to bind to actin. If more cross-bridges are formed, more myosin will pull on actin, and more tension will be produced.
The ideal length of a sarcomere during production of maximal tension occurs when thick and thin filaments overlap to the greatest degree. If a sarcomere at rest is stretched past an ideal resting length, thick and thin filaments do not overlap to the greatest degree, and fewer cross-bridges can form. This results in fewer myosin heads pulling on actin, and less tension is produced. As a sarcomere is shortened, the zone of overlap is reduced as the thin filaments reach the H zone, which is composed of myosin tails. Because it is myosin heads that form cross-bridges, actin will not bind to myosin in this zone, reducing the tension produced by this myofiber. If the sarcomere is shortened even more, thin filaments begin to overlap with each other—reducing cross-bridge formation even further, and producing even less tension. Conversely, if the sarcomere is stretched to the point at which thick and thin filaments do not overlap at all, no cross-bridges are formed and no tension is produced. This amount of stretching does not usually occur because accessory proteins, internal sensory nerves, and connective tissue oppose extreme stretching.
The primary variable determining force production is the number of myofibers within the muscle that receive an action potential from the neuron that controls that fiber. When using the biceps to pick up a pencil, the motor cortex of the brain only signals a few neurons of the biceps, and only a few myofibers respond. In vertebrates, each myofiber responds fully if stimulated. When picking up a piano, the motor cortex signals all of the neurons in the biceps and every myofiber participates. This is close to the maximum force the muscle can produce. As mentioned above, increasing the frequency of action potentials (the number of signals per second) can increase the force a bit more, because the tropomyosin is flooded with calcium.
Summary
The body contains three types of muscle tissue: skeletal muscle, cardiac muscle, and smooth muscle. Skeleton muscle tissue is composed of sarcomeres, the functional units of muscle tissue. Muscle contraction occurs when sarcomeres shorten, as thick and thin filaments slide past each other, which is called the sliding filament model of muscle contraction. ATP provides the energy for cross-bridge formation and filament sliding. Regulatory proteins, such as troponin and tropomyosin, control cross-bridge formation. Excitation–contraction coupling transduces the electrical signal of the neuron, via acetylcholine, to an electrical signal on the muscle membrane, which initiates force production. The number of muscle fibers contracting determines how much force the whole muscle produces.
Art Connections
Figure \(\PageIndex{5}\): Which of the following statements about muscle contraction is true?
- The power stroke occurs when ATP is hydrolyzed to ADP and phosphate.
- The power stroke occurs when ADP and phosphate dissociate from the myosin head.
- The power stroke occurs when ADP and phosphate dissociate from the actin active site.
- The power stroke occurs when Ca 2+ binds the calcium head.
- Answer
-
B
Figure \(\PageIndex{6}\): The deadly nerve gas Sarin irreversibly inhibits acetycholinesterase. What effect would Sarin have on muscle contraction?
- Answer
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In the presence of Sarin, acetycholine is not removed from the synapse, resulting in continuous stimulation of the muscle plasma membrane. At first, muscle activity is intense and uncontrolled, but the ion gradients dissipate, so electrical signals in the T-tubules are no longer possible. The result is paralysis, leading to death by asphyxiation.
Glossary
- actin
- globular contractile protein that interacts with myosin for muscle contraction
- acetylcholinesterase
- (AChE) enzyme that breaks down ACh into acetyl and choline
- cardiac muscle
- tissue muscle tissue found only in the heart; cardiac contractions pump blood throughout the body and maintain blood pressure
- motor end plate
- sarcolemma of the muscle fiber that interacts with the neuron
- myofibril
- long cylindrical structures that lie parallel to the muscle fiber
- myofilament
- small structures that make up myofibrils
- myosin
- contractile protein that interacts with actin for muscle contraction
- sarcolemma
- plasma membrane of a skeletal muscle fiber
- sarcomere
- functional unit of skeletal muscle
- skeletal muscle tissue
- forms skeletal muscles, which attach to bones and control locomotion and any movement that can be consciously controlled
- smooth muscle
- tissue occurs in the walls of hollow organs such as the intestines, stomach, and urinary bladder, and around passages such as the respiratory tract and blood vessels
- thick filament
- a group of myosin molecules
- thin filament
- two polymers of actin wound together along with tropomyosin and troponin
- tropomyosin
- acts to block myosin binding sites on actin molecules, preventing cross-bridge formation and preventing contraction until a muscle receives a neuron signal
- troponin
- binds to tropomyosin and helps to position it on the actin molecule, and also binds calcium ions


17.7: Review Questions
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1.
A nurse assesses a client’s health literacy prior to providing education. The client is able to use interactive websites, identify multiple resources to gather health information, and assist others in health decision-making. Which health literacy level describes the client’s skills?
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Functional health literacy
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Interactive health literacy
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Critical health literacy
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Organizational health literacy
2.
Which secondary data source would the public health nurse utilize to benchmark local health patterns to national (U.S.) patterns?
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Healthy People 2030
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Key informant interview
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U.S. Census data
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Focus group report
3.
Which community nursing diagnosis has the nurse written appropriately?
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Ineffective coping related to multiple stressors as evidenced by the client crying and stating she has no support system
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Risk of ineffective health maintenance among individuals who do not have access to a primary care provider
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Risk of cardiovascular disease related to poor diet, sedentary lifestyle, and lack of access safe places to exercise
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Risk of obesity among school-age children related to lack of access to healthy food choices and opportunities for physical activity
4.
Which of the following would the community health nurse include in an assessment of community systems?
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Age, gender, and race
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Water quality reports
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Health care providers
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Morbidity patterns
5.
A team is using the American Hospital Association Community Health Assessment (CHA) toolkit to conduct a CHA. They have defined the community and gathered data from multiple sources. According to the toolkit, what is the next step?
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Share results with the community
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Prioritize community health problems
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Evaluate progress
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Plan implementation strategies
6.
Which health literacy skill would the nurse anticipate in a client with functional health literacy skills?
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Understanding a prescription label
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Following blood glucose trends
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Recognizing personal health risks for cardiovascular disease
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Following directions to reduce radon levels in the home
7.
The nurse is developing a teaching plan for a client newly diagnosed with diabetes. Which action by the nurse focuses on the psychomotor domain of learning?
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Providing reinforcement for efforts toward self-care
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Teaching about the relationship between insulin and blood glucose levels
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Instructing the client on self-administration of insulin
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Discussing the symptoms of hypoglycemia and hyperglycemia
8.
Which action will the nurse take when utilizing constructivist theory to teach a community health education program about smoking cessation?
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Reward positive behaviors toward smoking cessation
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Encourage reflection about smoking cessation
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Model desired behaviors needed for smoking cessation
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Assess knowledge to build upon what is already known about smoking cessation
9.
Which method will the nurse use when teaching a client who is an auditory learner?
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Pictures and images
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Videos
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Interactive activities
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Discussion
10.
Which activity will the nurse perform during the continuously improve the community phase of the Mobilizing for Action through Planning and Partnership (MAPP) framework to improve the health of a community?
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Complete partner profiles
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Strengthen community engagement
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Perform Community Partners Assessment
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Establish Community Health Improvement (CHI) leadership


3.6: Practice in Identifying Different Types of Tissues
Name 2 major components of Connective Tissue:
1. ____________________________________________________________
2. ____________________________________________________________
Name 2 components of Extracellular Matrix:
1. ____________________________________________________________
2. ____________________________________________________________
Name 3 major connective tissue fibers:
1. ____________________________________________________________
2. ____________________________________________________________
3. ____________________________________________________________
Name the 3 types of muscle tissue:
1. ____________________________________________________________
2. ____________________________________________________________
3. ____________________________________________________________
Name the 2 major cell types in neural tissue:
1. ____________________________________________________________
2. ____________________________________________________________
Epithelia can be divided into two basic groups based on the number of cell layers . These are:
1. ____________________________________________________________
2. ____________________________________________________________
Name two epithelia that can be considered an exception:
1. ____________________________________________________________
2. ____________________________________________________________
LAB 3 EXERCISE \(\PageIndex{1}\)
When storing a microscope, you should always follow this list:
- Remove any slide found on the stage and return it to the slide box.
- Rotate the smallest lens or no lens into place above the stage. Lower the stage a few turns.
- Loosely coil the cord in your hand starting near the microscope and working toward the plug.
- Hang the coiled cord over one ocular lens.
- Look at the number on the back of the microscope, return that scope to its numbered box.
- If there’s already a microscope in that numbered box, check its number and move it. If it is not numbered simply push it to the back of the box and place yours closer to the front. We have a few extra microscopes which we store in this fashion.


3.3G: Electrochemical Gradient
To move substances against the membrane’s electrochemical gradient, the cell utilizes active transport, which requires energy from ATP.
- Define an electrochemical gradient and describe how a cell moves substances against this gradient
Key Points
- The electrical and concentration gradients of a membrane tend to drive sodium into and potassium out of the cell, and active transport works against these gradients.
- To move substances against a concentration or electrochemical gradient, the cell must utilize energy in the form of ATP during active transport.
- Primary active transport, which is directly dependent on ATP, moves ions across a membrane and creates a difference in charge across that membrane.
- Secondary active transport, created by primary active transport, is the transport of a solute in the direction of its electrochemical gradient and does not directly require ATP.
- Carrier proteins such as uniporters, symporters, and antiporters perform primary active transport and facilitate the movement of solutes across the cell’s membrane.
Key Terms
- adenosine triphosphate : a multifunctional nucleoside triphosphate used in cells as a coenzyme, often called the “molecular unit of energy currency” in intracellular energy transfer
- active transport : movement of a substance across a cell membrane against its concentration gradient (from low to high concentration) facilitated by ATP conversion
- electrochemical gradient : The difference in charge and chemical concentration across a membrane.
Electrochemical Gradients
Electrochemical Gradient : Electrochemical gradients arise from the combined effects of concentration gradients and electrical gradients.
Simple concentration gradients are differential concentrations of a substance across a space or a membrane, but in living systems, gradients are more complex. Because ions move into and out of cells and because cells contain proteins that do not move across the membrane and are mostly negatively charged, there is also an electrical gradient, a difference of charge, across the plasma membrane. The interior of living cells is electrically negative with respect to the extracellular fluid in which they are bathed. At the same time, cells have higher concentrations of potassium (K + ) and lower concentrations of sodium (Na + ) than does the extracellular fluid. In a living cell, the concentration gradient of Na + tends to drive it into the cell, and the electrical gradient of Na + (a positive ion) also tends to drive it inward to the negatively-charged interior. The situation is more complex, however, for other elements such as potassium. The electrical gradient of K + , a positive ion, also tends to drive it into the cell, but the concentration gradient of K + tends to drive K + out of the cell. The combined gradient of concentration and electrical charge that affects an ion is called its electrochemical gradient.
Moving Against a Gradient
To move substances against a concentration or electrochemical gradient, the cell must use energy. This energy is harvested from adenosine triphosphate (ATP) generated through the cell’s metabolism. Active transport mechanisms, collectively called pumps, work against electrochemical gradients. Small substances constantly pass through plasma membranes. Active transport maintains concentrations of ions and other substances needed by living cells in the face of these passive movements. Much of a cell’s supply of metabolic energy may be spent maintaining these processes. For example, most of a red blood cell’s metabolic energy is used to maintain the imbalance between exterior and interior sodium and potassium levels required by the cell. Because active transport mechanisms depend on a cell’s metabolism for energy, they are sensitive to many metabolic poisons that interfere with the supply of ATP.
Two mechanisms exist for the transport of small-molecular weight material and small molecules. Primary active transport moves ions across a membrane and creates a difference in charge across that membrane, which is directly dependent on ATP. Secondary active transport describes the movement of material that is due to the electrochemical gradient established by primary active transport that does not directly require ATP.
Carrier Proteins for Active Transport
An important membrane adaption for active transport is the presence of specific carrier proteins or pumps to facilitate movement. There are three types of these proteins or transporters: uniporters, symporters, and antiporters. A uniporter carries one specific ion or molecule. A symporter carries two different ions or molecules, both in the same direction. An antiporter also carries two different ions or molecules, but in different directions. All of these transporters can also transport small, uncharged organic molecules like glucose. These three types of carrier proteins are also found in facilitated diffusion, but they do not require ATP to work in that process. Some examples of pumps for active transport are Na + -K + ATPase, which carries sodium and potassium ions, and H + -K + ATPase, which carries hydrogen and potassium ions. Both of these are antiporter carrier proteins. Two other carrier protein pumps are Ca 2+ ATPase and H + ATPase, which carry only calcium and only hydrogen ions, respectively.
Uniporters, Symporters, and Antiporters : A uniporter carries one molecule or ion. A symporter carries two different molecules or ions, both in the same direction. An antiporter also carries two different molecules or ions, but in different directions.


3.3: Using Sage to Graph Lines and Solve Equations
In this section we will use technology to graph lines and solve for the intersection point. In particular, we will use an open online resource called Sage .
Let's continue to consider the game given in Table \(3.2.1\) by
| Table \(3.3.1\): Example \(3.1.1\) with Named Strategies | ||
|---|---|---|
| C | D | |
| A | \(1\) | \(0\) |
| B | \(-1\) | \(2\) |
Recall, our goal is to determine how often Player 1 should play A and how often she should play B.
We will follow the same steps as in Section 3.2 . Let \(p\) be the probability that Player 1 plays B. Let \(m\) be the payoff to Player 1. Since we are trying to find a mixed strategy for Player 1, we will pick a strategy for Player 2 and try to determine the possible payoffs for Player 1.
Let us determine some pairs \((p, m)\text{.}\)
Step 1: Assume Player 2 plays pure strategy \(C\text{.}\)
Step 1a: If Player 1 always plays \(A\text{,}\) then we are considering the strategy pair \([A, C]\text{.}\) Since Player 1 never plays \(B\text{,}\) \(p=0\text{.}\) The payoff to Player 1 for \([A, C]\) is \(m=1\text{.}\) Thus, for the strategy pair \([A, C]\) we get \((p, m)=(0, 1)\text{.}\)
Step 1b: If Player 1 always plays \(B\text{,}\) then we are considering the strategy pair \([B, C]\text{.}\) Since Player 1 always plays \(B\text{,}\) \(p=1\text{.}\) The payoff to Player 1 for \([A, C]\) is \(m=-1\text{.}\) Thus, for the strategy pair \([B, C]\) we get \((p, m)=(1, -1)\text{.}\)
Step 1c: Now we want to know what Player 1's payoff will be as she varies the probability, \(p\text{,}\) with which she plays \(B\text{.}\) We can draw a graph where the \(x\)-axis represents to probability with which she plays B (\(p\)) and the \(y\)-axis represents the expected payoff (\(m\)). Thus, when Player 1 plays only \(A\text{,}\) she is playing \(B\) with probability \(0\); when Player 1 plays only B, she is playing B with probability \(1\). It might be easier to remember if you label your graph as in Figure \(3.2.1\).
Step 1d: Now we can use Sage to plot the points we determined in Step 1a and Step 1b and the line between them. This line represents Player 2's pure strategy \(C\text{.}\) See Figure \(3.2.2\). Click on the “run” button to plot the line between the points \((0, 1)\) and \((1, -1)\text{.}\)
Before moving on, let's again, make sure we understand what this line represents. Any point on it represents the expected payoff to Player 1 as she varies her strategy, assuming Player 2 only plays \(C\) . In this case, we can see that as she plays \(B\) more often, her expected payoff goes down. You can now use this Sage cell to plot any line for Player 2's pure strategy \(C\text{.}\) Just edit the values for the points \(u\) and \(v\text{.}\)  (Don't worry the original values will reset when you refresh the page.)
Now let's do the same thing, assuming Player 2 plays only \(D\text{.}\)
Step 2: Assume Player 2 plays pure strategy \(D\text{.}\)
Step 2a: If Player 1 always plays \(A\text{,}\) then we are considering the strategy pair \([A, D]\text{.}\) Since Player 1 never plays \(B\text{,}\) \(p=0\text{.}\) The payoff to Player 1 for \([A, D]\) is \(m=0\text{.}\) Thus, for the strategy pair \([A, D]\) we get \((p, m)=(0, 0)\text{.}\)
Step 2b: If Player 1 always plays \(B\text{,}\) then we are considering the strategy pair \([B, D]\text{.}\) Since Player 1 always plays \(B\text{,}\) \(p=1\text{.}\) The payoff to Player 1 for \([B, D]\) is \(m=2\text{.}\) Thus, for the strategy pair \([B, D]\) we get \((p, m)=(1, 2)\text{.}\)
Step 2c: Now, on our same graph from Step 1, we can plot the points we determined in Step 2a and Step 2b. We will connect them with a line representing Player 2's pure strategy \(D\text{.}\) See Figure \(3.2.3\).
Now we can see that if Player 2 plays only \(D\text{,}\) then Player 1 does best by playing only \(B\text{.}\) Again, you can use this Sage cell to plot both Player 2's pure strategies. Points \(u\) and \(v\) are for strategy \(C\text{,}\) while points \(w\) and \(z\) are for strategy \(D\text{.}\)
As we saw in Section 3.2 , for each choice of \(p\text{,}\) the top line represents the highest expected value for Player 1; the bottom line represents the lowest expected value for Player 1; the area between the lines represents the possible expected values for Player 1. Thus, Player 1 wants to maximize the minimum expected value, which means she wants to find the maximin strategy. And, as we saw in Section 3.2 , the maximin strategy occurs at the intersection of the two lines.
Step 3: Find the intersection of the two lines.
Step 3a: Find the equation for Line C. This is the line passing through the points \((0, 1)\) and \((1, -1)\text{.}\) It has slope \(-2\) and \(y\)-intercept \(1\). Thus, it has equation \(m=-2p+1\text{.}\) (Recall the \(x\)-axis represents probability \(p\) and the \(y\)-axis represents expected payoff \(m\text{.}\))
Step 3b: Find the equation for Line D. This is the line passing through the points \((0, 0)\) and \((1, 2)\text{.}\) It has slope \(2\) and \(y\)-intercept \(0\). Thus, it has equation \(m=2p\text{.}\)
Step 3c: Now we can use technology to help us solve the equations.
The solution for Player 1 is \((p, m)\text{.}\) Where \(p\) is the probability Player 1 plays B, and \(m\) is the expected payoff to Player 1.
We can use this Sage cell to solve for \(p\) and \(m\) for any \(2\times 2\) game by editing the equations \(m==-2*p+1, m==2*p\text{.}\)
Step 4: Determine Player 1's maximin mixed strategy. Recalling that \(p\) is the probability that Player 1 plays \(B\text{,}\) we know that Player 1 will play \(B\) with probability \(\dfrac{1}{4}\), and thus, play A with probability \(\dfrac{3}{4}\). The expected payoff for Player 1, \(m\text{,}\) is \(\dfrac{1}{2}\). It is important to check the algebraic solution with where the intersection point appears on the graph. Although we are using technology to help us graph and solve for the intersection point, we need to be able to catch any errors we make entering the information into Sage.
We have seen that we can use the same matrix with Player 1's payoffs to find the strategy for Player 2. Using the same game matrix as above:
and continuing to label Player 1's strategies by \(A\) and \(B\text{,}\) and Player 2's strategies by \(C\) and \(D\text{,}\) we can graph lines for Player 1's pure strategies \(A\) and \(B\text{.}\) We now let the \(x\)--axis represent the probability that Player 2 plays \(D\text{.}\) In the Sage applet below, for \(u\) and \(v\) enter the coordinates of two points that determine the line for when Player 1 plays \(A\text{,}\) then the two points for \(w\) and \(z\) that determine the line for when Player 1 plays \(B\text{.}\) We will then have Sage graph the lines. You can enter new values for \(u, v, w, z\) if you would like to draw the graph for a different matrix.
Now determine and enter the equations of the two lines and have Sage solve for the intersection point.
You can now use these last two Sage cells to solve any \(2 \times 2\) game with a mixed strategy equilibrium. You can also take some time to experiment with what happens if the game has a pure strategy equilibrium.
Now use the Sage cells to help you analyze some more games!
Consider the zero-sum game \(\begin{bmatrix}-4 & 4 \\2 & -2 \end{bmatrix}\).
- Does this game have a pure strategy equilibrium?
- Just by looking at the matrix, do you think this game will be fair? (Would you rather be Player 1 or Player 2?)
- Use Sage to draw the appropriate graph for this game. Which pure strategy does the blue line represent? Which pure strategy does the red line represent?
- Use your graph to determine if there is a mixed strategy equilibrium point. If there is, use Sage to determine how often Player 1 should play each strategy. What is the expected payoff to each player?
- Is this game fair? Explain. Compare your answer to (b).
Consider the zero-sum game \(\begin{bmatrix}0 & 1 \\1 & -10 \end{bmatrix}\).
- Does this game have a pure strategy equilibrium?
- Just by looking at the matrix, do you think this game will be fair? (Would you rather be Player 1 or Player 2?)
- Use Sage to draw the appropriate graph for this game. Which pure strategy does the blue line represent? Which pure strategy does the red line represent?
- Use your graph to determine if there is a mixed strategy equilibrium point. If there is, use Sage to determine how often Player 1 should play each strategy. What is the expected payoff to each player?
- Is this game fair? Explain. Compare your answer to (b).


18.2: The Time-Dependent Schrödinger Equation
Unfortunately, the analogy with the classical wave equation that allowed us to obtain the TISEq in the previous section cannot be extended to the time domain by considering the equation that involves the partial first derivative with respect to time. Schrödinger himself presented his time-independent equation first, and then went back and postulated the more general time-dependent equation. We are following here the same strategy and just give the time-independent variable as a postulate. The single-particle time-dependent Schrödinger equation is:
\[ i\hbar\dfrac{\partial \psi(x,t)}{\partial t}=-\dfrac{\hbar^2}{2m} \dfrac{\partial^2 \psi(x,t)}{\partial x^2}+V(x)\psi(x,t) \label{19.2.1} \]
where \(V \in \mathbb{R}^{n}\) represents the potential energy of the system. Obviously, the time-dependent equation can be used to derive the time-independent equation. If we write the wavefunction as a product of spatial and temporal terms, \(\psi(x, t) = \psi(x) f(t)\), then Equation \ref{19.2.1} becomes:
\[ \psi(x) i \hbar \dfrac{df(t)}{dt} = f(t) \left[-\dfrac{\hbar^2}{2m} \dfrac{\partial^2}{\partial x^2} + V(x) \right] \psi(x), \label{19.2.2} \]
which can be rearranged to:
\[ \dfrac{i \hbar}{f(t)} \dfrac{df(t)}{dt} = \dfrac{1}{\psi(x)} \left[-\dfrac{\hbar^2}{2m} \dfrac{\partial^2}{\partial x^2} + V(x) \right] \psi(x). \label{19.2.3} \]
Since the left-hand side of Equation \ref{19.2.3} is a function of \(t\) only and the right hand side is a function of \(x\) only, the two sides must equal a constant. If we tentatively designate this constant \(E\) (since the right-hand side clearly must have the dimensions of energy), then we extract two ordinary differential equations, namely:
\[ \dfrac{1}{f(t)} \dfrac{df(t)}{dt} = - \dfrac{i E}{\hbar} \label{19.2.4} \]
and: \[ -\dfrac{\hbar^2}{2m} \dfrac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x) = E \psi(x). \label{19.2.5} \]
The latter equation is the TISEq. The former equation is easily solved to yield
\[ f(t) = e^{-iEt / \hbar} \label{19.2.6} \]
The solutions of Equation \ref{19.2.6}, \(f(t)\), are purely oscillatory, since \(f(t)\) never changes in magnitude. Thus if:
\[ \psi(x, t) = \psi(x) \exp\left(\dfrac{-iEt}{\hbar}\right), \label{19.2.7} \]
then the total wave function \(\psi(x, t)\) differs from \(\psi(x)\) only by a phase factor of constant magnitude. There are some interesting consequences of this. First of all, the quantity \(\vert \psi(x, t) \vert^2\) is time independent, as we can easily show:
\[ \vert \psi(x, t) \vert^2 = \psi^{*}(x, t) \psi(x, t)= \psi^{*}(x)\exp\left(\dfrac{iEt}{\hbar}\right)\psi(x)\exp\left(\dfrac{-iEt}{\hbar}\right)= \psi^{*}(x) \psi(x). \label{19.2.8} \]
Wave functions of the form of Equation \ref{19.2.7} are called stationary states. The state \(\psi(x, t)\) is “stationary,” but the particle it describes is not! Of course Equation \ref{19.2.6} represents only a particular solution to the time-dependent Schrödinger equation. The general solution is much more complicated, and the factorization of the temporal part is often not possible:\(^1\)
\[ \psi({\bf r}, t) = \sum_i c_i e^{-iE_it / \hbar} \psi_i({\bf r}) \nonumber \]
- This sections was adapted in part from Prof. C. David Sherrill’s A Brief Review of Elementary Quantum Chemistry Notes available here .


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How many different 3-people committees can be chosen from ten people?
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How many different 5-player teams can be chosen from eight players?
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In how many ways can a person chose to vote for three out of five candidates on a ballot for a school board election?
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Compute the following:
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9C2
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6C4
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8C3
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7C4
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How many 5-card hands can be chosen from a deck of cards?
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How many 13-card bridge hands can be chosen from a deck of cards?
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There are twelve people at a party. If they all shake hands, how many different hand-shakes are there?
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In how many ways can a student choose to do four questions out of five on a test?
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Five points lie on a circle. How many chords can be drawn through them?
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How many diagonals does a hexagon have?
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There are five team in a league. How many games are played if every team plays each other twice?
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A team plays 15 games a season. In how many ways can it have 8 wins and 7 losses?
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In how many different ways can a 4-child family have 2 boys and 2 girls?
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A coin is tossed five times. In how many ways can it fall three heads and two tails?
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The shopping area of a town is a square that is six blocks by six blocks. How many different routes can a taxi driver take to go from one corner of the shopping area to the opposite cater-corner?
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If the shopping area in the previous problem has a rectangular form of 5 blocks by 3 blocks, then how many different routes can a taxi driver take to drive from one end of the shopping area to the opposite kitty corner end?
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A team of 7 workers is assigned to a project. In how many ways can 3 of the 7 workers be selected to make a presentation to the management about their progress on the project?
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A real estate company has 12 houses listed for sale by their clients. In how many ways can 5 of the 12 houses be selected to be featured in an advertising brochures?
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A frozen yogurt store has 9 toppings to choose from. In how many ways can 3 of the 9 toppings be selected ?
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A kindergarten teacher has 14 books about a holiday. In how many ways can she select 4 of the books to read to her class in the week before the holiday?
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 The solvent is treated as a dielectric continuum characterised at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) by its relative permittivity \(\varepsilon_{r}\). The ions are treated as hard non-polarizable spheres, having radius \(r_{j}\). The Born Equation describes the difference in thermodynamic properties of a mole of i-ions forming a perfect gas and a mole of j-ions in an ideal solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). The calculation is not straightforward [3-5]. What emerges is the difference in Helmholtz energies (at constant \(\mathrm{T}\) and \(\mathrm{V}\)) for a mole of \(j\) ions in incompressible liquid phases. The calculated quantities refer to the energies associated with the electric fields over the limits \(r_{j} \leq r \leq \infty\). In these terms the Born Equation describes the electrical part of the change in chemical potential on transferring an ion from the gas phase, permittivity \(\varepsilon_{0}\), to a solvent, relative (electric) permittivity \(\varepsilon_{r}\). In effect the Born Equation yields parameters characterizing the difference between the properties of one mole of \(j\) ions in ideal systems having equal concentrations at fixed \(\mathrm{T}\) and \(\mathrm{p}\) [6].
\[=-N_{A} \,\left(z_{j} \, e\right)^{2} \,\left[1-\left(1 / \varepsilon_{r}\right)\right] / 8 \, \pi \, r_{j} \, \varepsilon_{0} \nonumber \]
Similarly for transfer of one mole of \(j\) ions from an ideal solution in solvent \(\mathrm{s}_{1}\) to an ideal solution in solvent \(\mathrm{s}_{2}\), the transfer chemical potential is given by the Born Equation assuming \(r_{j}\) is independent of solvent.
\[\mathrm{N}_{\mathrm{A}} \,\left(\mathrm{z}_{\mathrm{j}} \, \mathrm{e}\right)^{2} \,\left[\left(1 / \varepsilon_{\mathrm{r}}(\mathrm{s} 2)\right)-\left(1 / \varepsilon_{\mathrm{r}}(\mathrm{s} 1)\right)\right] / 8 \, \pi \, \mathrm{r}_{\mathrm{j}} \, \varepsilon_{0} \nonumber \]
Many attempts have been made to modify the Born Equation in order to attain agreement between theory and measured thermodynamic ionic properties, particularly in the case of aqueous salt solutions. A common concern is the extent to which near-neighbor water molecules form an electrostricted layer [7] around ions in solution, namely a layer of solvent molecules having dielectric properties which differ from those of the pure solvent at the same temperature and pressure [8 - 11]. A common concern in this subject is the definition of the ionic radius for a given ion [12 - 16]. There is no agreement concerning a set of ‘absolute’ ionic radii. As Conway pointed out in 1966, ‘ …theories .. based on the Born equation seem to have reached an asymptotic level of usefulness..’ [17].
Nevertheless correlations involving thermodynamic properties of salt solutions play an important role.
Standard partial molar entropies for alkali metal halides in various solvents are linear functions of the corresponding entropies in aqueous solution [18]. A similar correlation is reported for ionic entropies in mixed aqueous solvents and the corresponding entropies in aqueous solutions [19]. With reference to enthalpies, the analysis also suffers from the fact that the ionic radius is sensitive to temperature [20].
Footnotes
[1] M. Born, Z. Phys., 1920, 1 , 45.
[2] H. S. Frank quoted by H. L. Friedman, J. Electrochem. Soc., 1977, 124 , 421c.
[3] H. S. Frank, J. Chem. Phys., 1955, 23 , 2023.
[4] J. E. Desnoyers and C. Jolicoeur, Modern Aspects of Electrochem, ed. B. Conway and J.O’M. Bockris.1969, 5 ,1.
[5] J. E. Desnoyers, R. E. Verrall and B. E. Conway, J. Chem. Phys., 1965, 43 , 243.
[6]
\[\begin{aligned}
& Delta(\mathrm{pfg} \rightarrow \mathrm{s} \ln ) \mu_{\mathrm{j}}=\left[\mathrm{mol}^{-1}\right] \,[\mathrm{C}]^{2} \,\{[1]-[1]\} /[1] \,[1] \,[\mathrm{m}] \,\left[\mathrm{Fm}^{-1}\right]= \\
&{[\mathrm{mol}]^{-1} \,\left[\mathrm{A}^{2} \mathrm{~s}^{2}\right] /[\mathrm{m}] \,\left[\mathrm{A}^{2} \mathrm{~s}^{4} \mathrm{~kg}^{-1} \mathrm{~m}^{-3}\right]=\left[\mathrm{mol}^{-1}\right] \,\left[\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}\right]=\left[\mathrm{J} \mathrm{mol}^{-1}\right]}
\end{aligned} \nonumber \]
[7] M. H. Abraham, E. Matteoli and J. Liszi, J. Chem. Soc. Faraday Trans.I, 1983, 79 ,2781; and references therein.
[8] D. R. Rosseinsky, Chem. Rev.,1965, 65 ,467; and references therein.
[9] A. A. Rashin and B. Hornig, J. Phys. Chem., 1985, 89 ,5588.
[10] T. Abe, Bull. Chem. Soc. Jpn.,1991, 64 ,3035.
[11] S. Goldman and R.G.Bates, J.Am.Chem.Soc.,1972, 94 ,1476.
[12] L. Pauling, Nature of the Chemical Bond, Cornell University Press, Ithaca, 3rd edn., 1960, chapter 8.
[13] M. Bucher and T. L. Porter, J. Phys. Chem.,1986, 90 ,3406; and references therein.
[14] M. Salomon, J.Phys.Chem.,1970, 74 ,2519.
[15] K. H. Stern and E. S. Amis, Chem. Rev.,1959, 59 ,1.
[16] Y. Marcus, Chem. Rev.,1988, 88 ,1475.
[17] B. E. Conway, Annu. Rev. Phys. Chem.,1966, 17 ,481.
[18] C. M. Criss, R. P. Held and E. Luksha, J.Phys.Chem.,1968, 72 ,2970.
[19] F. Franks and D. S .Reid, J. Phys.Chem.,1969, 73 ,3152.
[20] B. Roux, H.-A. Yu and M. Karplus, J. Phys. Chem.,1990, 94 ,4683.


5.0: Chapter Scenario
Microsoft Word includes some specialized features that can be used in a variety of ways, for both business and personal use. These features are designed to save you time. For your WorldCorp market trends report, you can use these features to gather information from employees in different departments, communicate with vendors and customers, and even create forms to be used for human resources purposes.
The chapters on Creating and Working in Documents and Document Preparation focused on the basic functions of constructing effective documents in Microsoft Word and Google Docs. In this chapter, you will learn enhanced capabilities in Word and Docs that go beyond simply creating a document. You will also spend some time reviewing additional types of documents you might encounter in business, such as invoices and cover letters.


12.11: The Selection Rule for the Rigid Rotor
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A selection rule describes how the probability of transitioning from one level to another cannot be zero . It has two sub-pieces: a gross selection rule and a specific selection rule . A gross selection rule illustrates characteristic requirements for atoms or molecules to display a spectrum of a given kind, such as an IR spectroscopy or a microwave spectroscopy. Once the atom or molecules follow the gross selection rule, the specific selection rule must be applied to the atom or molecules to determine whether a certain transition in quantum number may happen or not.
Selection rules specify the possible transitions among quantum levels due to absorption or emission of electromagnetic radiation. Incident electromagnetic radiation presents an oscillating electric field \(E_0\cos(\omega t)\) that interacts with a transition dipole. The dipole operator is \(\mu = e \cdot r\) where \(r\) is a vector pointing in a direction of space.
A dipole moment of a given state is
\[\mu_z=\int\Psi_1 \,^{*}\mu_z\Psi_1\,d\tau \nonumber \]
A transition dipole moment is a transient dipolar polarization created by an interaction of electromagnetic radiation with a molecule
\[(\mu_z)_{12}=\int\Psi_1 \,^{*}\mu_z\Psi_2\,d\tau \nonumber \]
In an experiment we present an electric field along the z axis (in the laboratory frame) and we may consider specifically the interaction between the transition dipole along the x, y, or z axis of the molecule with this radiation. If \(\mu_z\) is zero then a transition is forbidden. The selection rule is a statement of when \(\mu_z\) is non-zero.
The selection rule is a statement of when \(\mu_z\) is non-zero.
Rotational transitions
We can use the definition of the transition moment and the spherical harmonics to derive selection rules for a rigid rotator. Once again we assume that radiation is along the z axis.
\[(\mu_z)_{J,M,{J}',{M}'}=\int_{0}^{2\pi } \int_{0}^{\pi }Y_{J'}^{M'}(\theta,\phi )\mu_zY_{J}^{M}(\theta,\phi)\sin\theta\,d\phi,d\theta \nonumber \]
Notice that m must be non-zero in order for the transition moment to be non-zero. This proves that a molecule must have a permanent dipole moment in order to have a rotational spectrum. The spherical harmonics can be written as
\[Y_{J}^{M}(\theta,\phi)=N_{\,JM}P_{J}^{|M|}(\cos\theta)e^{iM\phi} \nonumber \]
where \(N_{JM}\) is a normalization constant. Using the standard substitution of \(x = \cos q\) we can express the rotational transition moment as
\[(\mu_z)_{J,M,{J}',{M}'}=\mu\,N_{\,JM}N_{\,J'M'}\int_{0}^{2 \pi }e^{I(M-M')\phi}\,d\phi\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx \nonumber \]
The integral over f is zero unless M = M' so \(\Delta M = \) 0 is part of the rigid rotator selection rule. Integration over \(\phi\) for \(M = M'\) gives \(2\pi \) so we have
\[(\mu_z)_{J,M,{J}',{M}'}=2\pi \mu\,N_{\,JM}N_{\,J'M'}\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx \nonumber \]
We can evaluate this integral using the identity
\[(2J+1)x\,P_{J}^{|M]}(x)=(J-|M|+1)P_{J+1}^{|M|}(x)+(J-|M|)P_{J-1}^{|M|}(x) \nonumber \]
Substituting into the integral one obtains an integral which will vanish unless \(J' = J + 1\) or \(J' = J - 1\).
\[\int_{-1}^{1}P_{J'}^{|M'|}(x)\Biggr(\frac{(J-|M|+1)}{(2J+1)}P_{J+1}^{|M|}(x)+\frac{(J-|M|)}{(2J+1)}P_{J-1}^{|M|}(x)\Biggr)dx \nonumber \]
This leads to the selection rule \(\Delta J = \pm 1\) for absorptive rotational transitions. Keep in mind the physical interpretation of the quantum numbers \(J\) and \(M\) as the total angular momentum and z-component of angular momentum, respectively. As stated above in the section on electronic transitions, these selection rules also apply to the orbital angular momentum (\(\Delta{l} = \pm 1\), \(\Delta{m} = 0\)).


23.2: IV Therapy Basics
Primary IV Fluid Infusion
Primary IV fluid infusions are prescribed by health care providers to restore or maintain hydration and electrolyte status within the body. When administering IV fluids to a patient, the nurse must continually monitor the patient’s fluid and electrolyte status to evaluate the effectiveness of the infusion and to avoid potential complications of fluid overload and electrolyte imbalance.
The most commonly used primary IV fluid bag contains 1,000 mL. There are also 500 mL, 250 mL, 100 mL, and 50 mL bags. The size of the primary fluid bag is based on infusion need, patient condition, and age. Most adult patients receive continuous IV fluids with 1,000 mL bags due to the higher drip (gtt) rate. Many other fluid volume bags are used for intermittent infusions or short-term therapy.
For example, for renal dialysis patients, IV bags smaller than 1,000 mL are used because large amounts of continuous fluids are contraindicated due to their renal impairment. Many institutions will hang smaller volume normal saline continuous infusion bags just to serve as an additional reminder that these patients should not receive large amounts of primary fluids. Another example of patients requiring smaller IV bags are pediatric patients who, due to their smaller anatomical size, do not require large primary fluid infusion volumes.
Primary fluids are typically administered using an IV pump. An IV pump is the safest method of administration to ensure specific amounts of fluid are administered. However, there may be situations when IV pumps are not available and nurses administer primary fluids by gravity using drip tubing. Read more about calculating infusion rates in the “ Math Calculation ” chapter.
Primary fluids are run at consistent infusion rates for a prescribed period of time. For example, a continuous fluid infusion may be ordered at a rate of 125 mL/hour for 24 hours. Continuous fluids may also be ordered to run until the provider gives a follow-up order to discontinue or decrease the fluid rate.
IV primary fluid bags consist of various types of fluid such as 0.9% normal saline, 0.45% (½) normal saline, lactated ringers solution, and dextrose (5%) preparations. They may also contain replacement electrolytes like potassium chloride. The provider will order primary fluids based on the patient’s fluid and electrolyte statuses.
There are three types of intravenous fluid concentrations: isotonic, hypertonic, and hypotonic fluids.
- Isotonic fluids are typically administered for fluid and electrolyte replacement. Isotonic fluids have a similar concentration to the solutes contained in blood, so they do not cause the osmotic movement of fluid into or out of the patient’s individual cells. An example of isotonic fluid is 0.9% normal saline.
- Hypertonic fluids have a higher concentration of solutes than blood. They are typically used in critical care situations to treat hyponatremia and avoid pulmonary edema by relying on osmosis to help remove excess fluid. An example of a hypertonic fluid is dextrose 5% in 0.9% normal saline (D5NS).
- Hypotonic fluids have a lower concentration of solutes than blood. The goal of hypotonic fluid administration is to move fluids into a patient’s cells due to osmosis. Hypotonic solutions are commonly used when a patient has severe intracellular dehydration such as during diabetic ketoacidosis. An example of hypotonic fluid is 0.45% normal saline (1/2NS). See Figure \(\PageIndex{1}\) [1] for an example of the effects of the administration of hypertonic, isotonic, and hypotonic IV fluids on a patient’s red blood cells.
Because a patient’s fluid and electrolyte statuses are constantly changing when receiving IV fluids, it is important for the nurse to monitor for signs of fluid or electrolyte imbalances and appropriately notify the health care provider of any concerns. For example, primary fluids may be started at a higher rate of infusion when a patient is receiving nothing by mouth (NPO), but should be tapered as they resume normal diet and fluid intake. It is important for the nurse to continually monitor a patient’s skin turgor, urinary output, lung sounds, and oxygen requirements and to assess for any new edema to offer important insight into their fluid volume status. A nurse must also evaluate the effects of replacement fluids and discuss their ongoing need with the prescribing provider.
Read more about types of intravenous fluids in the “ Fluids and Electrolytes ” chapter in Open RN Nursing Fundamentals .
Secondary Fluid Infusion
Secondary IV fluid administration is usually an intermittent infusion that infuses at regular intervals (e.g., every 8 hours). This form of IV therapy usually contains medications that are supplied in a smaller infusion bag and mixed with a diluent fluid like saline (e.g., IV antibiotics). Many common preparations come in 25 to 100 mL bags.
Secondary IV therapy is often referred to as “IV piggyback” (IVPB) medication because it is attached to the primary bag of intravenous fluids. In this case, the primary line maintains venous access between drug doses.
It is important to remember that not all IV solutions are compatible with all IV medications. It is vital for the nurse to triple check that the secondary medications/fluids are compatible with primary fluids. If medication and fluids are not compatible, a precipitate may form when the fluids mix within the line, posing a significant health danger for the patient.
IV Administration Equipment
Intravenous (IV) fluids and medications are administered through flexible plastic tubing called an IV administration set. The IV administration set connects the bag of solution to the patient’s IV access site. There are two major types of IV administration sets: primary tubing and secondary tubing. Additionally, IV fluids can be administered by gravity or by infusion pump, and each method requires its own administration set.
Primary and Secondary Administration Sets
Primary IV Administration Sets
Primary IV administration sets are used to infuse continuous or intermittent fluids or medications. Primary IV tubing can be a macro-drip or micro-drip solution set. A macro-drip infusion set delivers 10, 15, or 20 drops per milliliter, whereas a micro-drip infusion set delivers 60 drops per milliliter. The drop factor is located on the packaging of the IV tubing and is important to verify when calculating medication administration rates. Macro-drip sets are used for routine primary infusions for adults. Micro-drip IV tubing is used in pediatric or neonatal care where small amounts of fluids are administered over a long period of time.
Primary IV administration sets consist of the following parts:
Sterile spike: This part of the tubing must be kept sterile as you spike the IV fluid bag.
Drip chamber: The drip chamber allows air to rise out from a fluid so that it is not passed onto the patient. It is also used to calculate the rate at which fluid is administered by gravity (drops per minute). It should be kept ¼ to ½ full of solution.
Backcheck valve: A backcheck valve prevents fluid or medication from travelling up into the primary IV bag.
Access ports: Access ports are used to infuse secondary medications and to administer IV push medications. These may also be referred to as “Y ports.”
Roller clamp: A roller clamp is used to regulate the speed, or stop, an infusion by gravity.
Secondary IV Administration Sets
Secondary IV administration sets are used to intermittently administer a secondary medication, such as an antibiotic, while the primary IV is also running. Secondary IV tubing is shorter in length than primary tubing and is connected to a primary line via an access port or an IV pump. The secondary infusion is hung above the primary infusion and connected at an access port.
Secondary fluids should always be “piggybacked” into primary infusion lines to ensure that the correct amount of medication is infused. By piggybacking a medication, the solution from the primary fluid line is used to prime the secondary tubing. However, if a secondary infusion is run as a primary fluid, there is a risk of losing some of the secondary medication when priming the line, which results in less medication being administered. Loss of medication is considered a medication error because the patient received less active medication than prescribed.
See Figure \(\PageIndex{2}\) [2] for an illustration of the set up of a primary and secondary tubing for administration of fluids and a secondary medication by gravity. See Figure \(\PageIndex{3}\) [3] for an example of an IV infusion pump.
IV Administration
When initiating or changing an IV bag of fluids or medications, it is important to remember these items:
- IV fluids are a medication. Verify physician orders and check that the patient does not have an allergy to this medication. Perform the six rights of medication administration three times as you would when giving any other medication. Check the type of fluid and the expiration date, and verify the fluid is free of discoloration and sediment. Check the expiration date when obtaining a new tubing administration set.
- Examine the bag to ensure that the bag itself is intact and not leaking. There may be moisture on the inside of the plastic IV bag storage container; this is normal.
- Verify the infusion rate of IV fluids is appropriate based on the patient’s age, size, preexisting medical conditions, and prescribed indication. If a manual calculation is needed to set the IV flow rate, calculate the rate and double-check the calculated rate with another registered nurse.
- IV tubing administration sets require routine replacement to prevent infection. Follow agency policy regarding initiating tubing change before initiating a new bag of fluid or medications.
- If administration set tubing is present, trace the tubing from the patient to its point of origin to make sure that you’re accessing the correct port.
- Assess the IV site. Inspect for redness, swelling, or tenderness that can be a sign of irritation, inflammation, or infection.
- Ensure the IV site is patent when initiating new fluid or medication. Aspirate for blood return and flush the IV catheter according to agency policy.
Complications of IV Therapy
While monitoring a patient receiving IV fluids, it is important to assess for potential complications such as infiltration, extravasation, phlebitis, or infection. If these conditions occur, promptly notify the provider for treatment; the IV catheter will need to be removed and replaced at an alternative site, and additional medication may be prescribed.
Infiltration occurs when the tip of the catheter slips out of the vein. The catheter passes through the wall of the vein, or the blood vessel wall allows part of the fluid to infuse into the surrounding tissue, resulting in the leakage of IV fluids into the surrounding tissue. Infiltration may cause pain, swelling, and skin that is cool to the touch. If you are concerned an IV is infiltrated, follow your facility policy and, as a general guideline, discontinue the site and relocate the IV. If the infiltration is severe, you may consider the application of a compress in addition to elevating the affected limb. Check your institution’s policy regarding which type of compress (warm or cold) should be applied. [4] , [5] Additionally, clinical pharmacists can also be helpful resources for determining the appropriate type of infiltration treatment.
Extravasation refers to infiltration of damaging intravenous medications, such as chemotherapy, into the extravascular tissue around the site of infusion. Extravasation causes tissue injury, and depending on the medication, site, and length of exposure, it can cause tissue death, which is also referred to as necrosis. If detected early, extravasation may be treated with medications that help avoid the complication of necrosis.
Phlebitis is inflammation of a vein. Phlebitis of superficial veins can occur due to trauma to the vein during insertion of the IV catheter. It can cause redness and tenderness along the vein and can lead to infection if not treated appropriately. Treatment may include warm compresses and nonsteroidal anti-inflammatory medications.
Infection can occur whenever the skin barrier is broken by the insertion of an IV catheter. Signs of infection include redness, warmth, tenderness, and possible fever. Vascular catheter–associated infection is considered a hospital-acquired condition because it can be prevented using best practices. Be sure to follow evidence-based infection prevention practices, such as performing hand hygiene, performing a vigorous mechanical scrub of needleless connectors, limiting catheter access, and following sterile no-touch technique during intravenous infusion to reduce the risk of vascular catheter–associated infection.
- “ Osmotic pressure on blood cells diagram.svg “ by LadyofHats is in the Public Domain ↵
- “intravenous_equipment_labels-2.png” by British Columbia Institute of Technology is licensed under CC BY 4.0. Access for free at https://opentextbc.ca/clinicalskills/chapter/8-2-types-of-iv-therapy/ ↵
- “DSC_0738-e1443533768679-678x1024.jpg” by British Columbia Institute of Technology is licensed under CC BY 4.0. Access for free at https://opentextbc.ca/clinicalskills/chapter/8-2-types-of-iv-therapy/ ↵
- Drugs.com [Internet]. IV Infiltration . © 2000-2020 [updated 3 February, 2020; cited 7 August, 2020]. https://www.drugs.com/cg/iv-infiltration-aftercare-instructions.html ↵
- AMN Healthcare Education Services. (2015). Know the difference: Infiltration versus extravasation . https://www.rn.com/nursing-news/know-the-difference-infiltration-vs-extravasation/ ↵


16.8: Input and Output
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Overview
Input and output , or I/O is the communication between an information processing system, such as a computer, and the outside world, possibly a human or another information processing system. Discussion
Every task we have the computer do happens inside the central processing unit (CPU) and the associated memory. Once our program is loaded into memory and the operating system directs the CPU to start executing our programming statements the computer looks like this:
Our program now loaded into memory has basically two areas:
- Machine instructions – our instructions for what we want done
- Data storage – our variables that we using in our program
Often our program contains instructions to interact with the input/output devices. We need to move data into (read) and/or out of (write) the memory data area. A device is a piece of equipment that is electronically connected to the memory so that data can be transferred between the memory and the device. Historically this was done with punched cards and printouts. Tape drives were used for electronic storage. With time we migrated to using disk drives for storage with keyboards and monitors (with monitor output called soft copy) replacing punch cards and printouts (called hard copy).
Most computer operating systems and by extension programming languages have identified the keyboard as the standard input device and the monitor as the standard output device . Often the keyboard and monitor are treated as the default device when no other specific device is indicated.
Key Terms
- device
- A piece of equipment that is electronically connected to the memory so that data can be transferred between the memory and the device.
- escape code
- A code directing an output device to do something.
- extraction
- Aka reading or getting data from an input device.
- insertion
- Aka writing or sending data to an output device.
- standard input
- The keyboard.
- standard output
- The monitor.