2.3: The very short history of the S and R
R is the statistical environment. In was created as a freeware analog of commercial S-Plus which is in turn was implementation of the S language concept. The S language was first created in 1976 in Bell Labs, and its name was inspired by famous C language (from same Bell Labs). S-Plus started in the end of 1980s, and as many statistical software, was seriously expensive. In August 1993, two New Zealand scientists, Robert Gentleman and Ross Ihaka, decided to make R (this name was, in turn, inspired by S ). The idea was to make independent realization of S language concept which would differ from S-Plus in some details (for example, in the way it works with local and global variables).
Practically, R is not an imitation of S-Plus but the new “branch” in the family of S software. In 1990s, R was developing slowly, but when users finally realized its truly amazing opportunities (like the system of R extensions— packages , or libraries ) and started to migrate from other statistical systems, R started to grow exponentially. Now, there are thousands of R packages, and R is used almost everywhere! Without any exaggeration, R is now the most important software tool for data analysis .


3.5: Review and Additional Resources
Review
Assess your learning. Start with the "Important Terms and Concepts" to ensure you know the terminology related to the topic of the chapter and concepts discussed. Move on to the "Review Questions" to answer critical thinking questions about concepts and processes discussed in the chapter. Finally, test your overall understanding by taking the "Self-assessment quiz".
-
- Heterosphere
- the outer most sphere where gases are distributed in distinct layers by gravity according to their atomic weight
-
- Homosphere
- lies between the Earth's surface and the heterosphere. Gases are nearly uniformly mixed through this layer even though density decreases with height above the surface
-
- Constant Gases
- Nitrogen, oxygen and argon are called the "constant gases" because their concentration has remained virtually the same for much of recent earth history
-
- Variable Gases
- those present in small and variable amounts. These include carbon dioxide, methane, ozone, water vapor, and particulates among others.
-
- Ozone
- O 3 is both beneficial and harmful to life on Earth
-
- Carbon dioxide
- (CO 2 ) makes up only .036% of the atmosphere by volume. Carbon dioxide is essential to photosynthetic processes of plants.
-
- Methane
- (CH 4 ) is a greenhouse gas contributing to about 18% of global warming and has been on the rise over the last several decades. Methane is a product of the decomposition of organic matter, with major natural sources being that which occurs from wetlands, termites, the oceans, and hydrates.
-
- Particulates (and aerosols)
- very small particles of solid or liquid suspended in the air
-
- Water Vapor
- an extremely important gas found in the atmosphere.
-
- Environmental lapse rate (ELR)
- The rate of change in temperature with altitude
-
- Normal lapse rate of temperature
- the average value of the ELR, .65 o C /100 meters
-
- Inverted lapse rate of temperature (inversion)
- when the air temperature actually increase with an increase in altitude above the Earth
-
- Troposphere
- the layer closest to the Earth's surface
-
- Tropopause
- lies above the troposphere. Here the temperature tends to stay the same with increasing height.
-
- Stratosphere
- Above the tropopause lies the stratosphere. Through most of the stratosphere the air temperature increases with an increase in elevation creating a temperature inversion
-
- Mesosphere
- air temperatures begin to decrease with increasing altitude
-
- Thermosphere
- above the mesosphere. air temperature increasing with increasing altitude
-
- Ionosphere
- not really a layer of the atmosphere, but an electrified field of ions and free electrons
-
- Ozonosphere
- also called the "ozone layer", is the concentrated layer of ozone found in the stratosphere
-
- Greenhouse effect
- longwave radiation emitted from the Earth's surface and directed toward the sky is readily absorbed by greenhouse gases. When absorbed, the temperature of the atmosphere increases. Some of this absorbed energy is emitted to space while some is emitted back towards the Earth.
Explain how the greenhouse effect works.
- Answer
-
Shortwave radiation penetrates through the atmosphere and is absorbed by the surface. The surface radiates longwave radiation toward the atmosphere. The atmosphere absorbs this energy and radiates it back down toward the surface. Increasing "greenhouse gasses" though human activities traps more radiation causing an increase in near-surface temperatures.
Briefly describe how temperature changes from the surface up to the mesosphere.
- Answer
-
Troposphere: temperature decreases with an increase in elevation. Tropopause: no change in temperature with an increase in elevation. Stratosphere: temperature increases with an increase in elevation. Stratopause: no change in temperature with an increase in elevation.
What impact do particulates have on the atmosphere?
- Answer
-
Particulates can reflect incoming shortwave and cause atmospheric cooling. They are also good absorbers of terrestrial earth radiation which would cause atmospheric warming. Particulates serve as condensation nuclei.
What is an inversion?
- Answer
-
An inversion occurs when air temperatures increase with an increase in elevation. Inversions can occur under the right conditions in the troposphere. The stratosphere is noted for its inversion.
What is the environmental lapse rate of temperature?
- Answer
-
The environmental lapse rate of temperature is the decrease in temperature with an increase in elevation through the troposphere. The temperature decreases with elevation because the earth is the immediate source of energy to heat the air.
How do humans contribute to global warming?
- Answer
-
Fossil fuel burning (e.g., emissions from transportation and energy generation), deforestation, cattle farming and rice production to name a few.
What impact does ozone at the surface have on humans?
- Answer
-
Ozone is an eye irritant and causes shortness of breath due to constricting air ways in the throat and lungs.
Explain why temperatures increase with height through the stratosphere.
- Answer
-
Because ozone absorbs incoming solar radiation. The top of the stratosphere is hotter because its closer to the source of shortwave radiation. An ever diminishing amount of shortwave radiation as it passes through the stratosphere causes the temperature to decrease toward the bottom.
Describe how clouds impact climate.
- Answer
-
Clouds can reflect solar radiation off the top of the cloud causing atmospheric cooling. They are good absorbers of terrestrial earth radiation which would cause atmospheric warming. The role of clouds in climate change is one of the most vexing problems to climate scientists.
Where does methane came from and what role does it play in the atmosphere?
- Answer
-
Methane is a product of the decomposition of organic matter, with major natural sources being that which occurs in wetlands, melting permafrost and as a by-product termite actvities. Methane is a greenhouse gas.
-
Free ____ is a product of plant photosynthesis.
- methane
- carbon dioxide
- oxygen
- nitrogen
-
The concentration of carbon dioxide in the atmosphere
- is generally higher during the winter.
- has increased in the last 100 years
- is a greenhouse gas
- all the above
-
Greenhouse gases
- are good absorbers of solar (shortwave) radiation.
- are good absorbers of longwave radiation emitted by the Earth.
- absorb shortwave and longwave radiation equally well.
- absorb neither shortwave or longwave radiation.
-
Temperatures tend to ______ through much of the stratosphere.
- increase with an increase in elevation
- decrease with and increase in elevation
- stay the same with an increase in elevation
- None of the above
-
Clouds are good absorbers of
- insolation
- longwave radiation emitted by the earth
- UV light
- none of the above
-
Termites are thought to be a source of
- carbon dioxide.
- ozone.
- methane.
- none of the above.
-
Which of the following is considered a "permanent gas"?
- Carbon dioxide
- Methane
- Ozone
- None of the above
-
Particulates
- serve as condensation nuclei.
- can cause atmospheric cooling.
- can cause atmospheric warming
- can do all the above.
-
Most of the ozone found in the atmosphere is found in the
- troposphere.
- stratosphere.
- mesosphere.
- thermosphere.
-
An inversion
- is where temperatures increase with increasing altitude.
- is where temperatures decrease with increasing altitude.
- is where temperatures stay the same with increasing altitude.
- is where temperatures increase with decreasing altitude.
- Answer
-
- C
- D
- B
- A
- B
- C
- D
- D
- B
- A
Additional Resources
Use these resources to further explore the world of geography
- Focus on The Physical Environment: " The Carbon Cycle. NASA Earth Observatory " Elizabeth Kolbert, staff writer, The New Yorker WGBH Forum
- Physical Geography Today: AIRNow - real time air quality data (US EPA)
- World of Change: Antarctic Ozone Hole
Multimedia
"The Precious Envelope" ( Annenberg/CPB ) The World of Chemistry video series "The Earth's atmosphere is examined through theories of chemical evolution; ozone depletion and the greenhouse effect are explained. (Material is somewhat dated but still useful.)
Go to the The World of Chemistry site and scroll to "The Precious Envelope". One-time free registration may be required.
The Ozone Layer: Closing the Gap (NASA, 2019) "In the 1980s, scientists began to realize that ozone-depleting chemicals, such as chlorofluorocarbon (CFCs), were creating a thin spot—a hole—in the ozone layer over Antarctica. Through an international effort to decrease the use of CFCs, the ozone layer is starting to mend, and scientists believe it should mostly recover by the middle of the 21st century. This series of satellite images shows the ozone hole on the day of its maximum depth from 1979 through 2018."


9: Modern Applications of Microbial Genetics
-
- 9.1: Microbes and the Tools of Genetic Engineering
- The science of using living systems to benefit humankind is called biotechnology. Technically speaking, the domestication of plants and animals through farming and breeding practices is a type of biotechnology. However, in a contemporary sense, we associate biotechnology with the direct alteration of an organism’s genetics to achieve desirable traits through the process of genetic engineering.
-
- 9.2: Visualizing and Characterizing DNA
- Finding a gene of interest within a sample requires the use of a single-stranded DNA probe labeled with a molecular beacon (typically radioactivity or fluorescence) that can hybridize with a complementary single-stranded nucleic acid in the sample. Agarose gel electrophoresis allows for the separation of DNA molecules based on size. Restriction fragment length polymorphism (RFLP) analysis allows for the visualization by agarose gel electrophoresis of distinct variants of a DNA sequence.
-
- 9.3: Whole Genome Methods and Industrial Applications
- Advances in molecular biology have led to the creation of entirely new fields of science. Among these are fields that study aspects of whole genomes, collectively referred to as whole-genome methods. In this section, we’ll provide a brief overview of the whole-genome fields of genomics, transcriptomics, and proteomics.
-
- 9.4: Genetic Engineering - Risks, Benefits, and Perceptions
- Many types of genetic engineering have yielded clear benefits with few apparent risks. However, many emerging applications of genetic engineering are much more controversial, often because their potential benefits are pitted against significant risks, real or perceived. This is certainly the case for gene therapy, a clinical application of genetic engineering that may one day provide a cure for many diseases but is still largely an experimental approach to treatment.
Thumbnail: A group of Genetically modified GloFish fluorescent fish. (www.glofish.com).


11.1: Introduction
The three letters “DNA” have now become synonymous with crime solving and genetic testing. DNA can be retrieved from hair, blood, or saliva. Each person’s DNA is unique, and it is possible to detect differences between individuals within a species on the basis of these unique features.
DNA analysis has many practical applications beyond forensics. In humans, DNA testing is applied to numerous uses: determining paternity, tracing genealogy, identifying pathogens, archeological research, tracing disease outbreaks, and studying human migration patterns. In the medical field, DNA is used in diagnostics, new vaccine development, and cancer therapy. It is now possible to determine predisposition to diseases by looking at genes.
DNA is the genetic material passed from parent to offspring for all life on Earth. The technology of molecular genetics developed in the last half century has enabled us to see deep into the history of life to deduce the relationships between living things in ways never thought possible. It also allows us to understand the workings of evolution in populations of organisms. Over a thousand species have had their entire genome sequenced, and there have been thousands of individual human genome sequences completed. These sequences will allow us to understand human disease and the relationship of humans to the rest of the tree of life. Finally, molecular genetics techniques have revolutionized plant and animal breeding for human agricultural needs. All of these advances in biotechnology depended on basic research leading to the discovery of the structure of DNA in 1953, and the research since then that has uncovered the details of DNA replication and the complex process leading to the expression of DNA in the form of proteins in the cell.
Contributors and Attributions
Remixed and/or curated from the following works:
Fowler, S., Roush, R., & Wise, J. (2013). 9.0 Introduction . In Concepts of Biology . OpenStax (CC BY 4.0; Access for free at https://openstax.org/books/concepts-biology/pages/1-introduction ).


7.2: Entropy and The Second Law of Thermodynamics
- To understand the relationship between internal energy and entropy.
The first law of thermodynamics governs changes in the state function we have called internal energy (\(U\)). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH 4 NO 3 , NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔH soln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure \(\PageIndex{1}\).
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S) , a thermodynamic property of all substances that is proportional to their degree of "disorder". In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Entropy
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (ΔS > 0) or a decrease in entropy (ΔS < 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: ΔS = S f − S i .
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
We can illustrate the concepts of microstates and entropy using a deck of playing cards, as shown in Figure \(\PageIndex{2}\). In any new deck, the 52 cards are arranged by four suits, with each suit arranged in descending order. If the cards are shuffled, however, there are approximately 10 68 different ways they might be arranged, which corresponds to 10 68 different microscopic states. The entropy of an ordered new deck of cards is therefore low, whereas the entropy of a randomly shuffled deck is high. Card games assign a higher value to a hand that has a low degree of disorder. In games such as five-card poker, only 4 of the 2,598,960 different possible hands, or microstates, contain the highly ordered and valued arrangement of cards called a royal flush, almost 1.1 million hands contain one pair, and more than 1.3 million hands are completely disordered and therefore have no value. Because the last two arrangements are far more probable than the first, the value of a poker hand is inversely proportional to its entropy.
We can see how to calculate these kinds of probabilities for a chemical system by considering the possible arrangements of a sample of four gas molecules in a two-bulb container (Figure \(\PageIndex{3}\)). There are five possible arrangements: all four molecules in the left bulb (I); three molecules in the left bulb and one in the right bulb (II); two molecules in each bulb (III); one molecule in the left bulb and three molecules in the right bulb (IV); and four molecules in the right bulb (V). If we assign a different color to each molecule to keep track of it for this discussion (remember, however, that in reality the molecules are indistinguishable from one another), we can see that there are 16 different ways the four molecules can be distributed in the bulbs, each corresponding to a particular microstate. As shown in Figure \(\PageIndex{3}\), arrangement I is associated with a single microstate, as is arrangement V, so each arrangement has a probability of 1/16. Arrangements II and IV each have a probability of 4/16 because each can exist in four microstates. Similarly, six different microstates can occur as arrangement III , making the probability of this arrangement 6/16. Thus the arrangement that we would expect to encounter, with half the gas molecules in each bulb, is the most probable arrangement. The others are not impossible but simply less likely.
There are 16 different ways to distribute four gas molecules between the bulbs, with each distribution corresponding to a particular microstate. Arrangements I and V each produce a single microstate with a probability of 1/16. This particular arrangement is so improbable that it is likely not observed. Arrangements II and IV each produce four microstates, with a probability of 4/16. Arrangement III, with half the gas molecules in each bulb, has a probability of 6/16. It is the one encompassing the most microstates, so it is the most probable.
Instead of four molecules of gas, let’s now consider 1 L of an ideal gas at standard temperature and pressure (STP), which contains 2.69 × 10 22 molecules (6.022 × 10 23 molecules/22.4 L). If we allow the sample of gas to expand into a second 1 L container, the probability of finding all 2.69 × 10 22 molecules in one container and none in the other at any given time is extremely small, approximately \(\frac{2}{2.69 \times 10^{22}}\). The probability of such an occurrence is effectively zero. Although nothing prevents the molecules in the gas sample from occupying only one of the two bulbs, that particular arrangement is so improbable that it is never actually observed. The probability of arrangements with essentially equal numbers of molecules in each bulb is quite high, however, because there are many equivalent microstates in which the molecules are distributed equally. Hence a macroscopic sample of a gas occupies all of the space available to it, simply because this is the most probable arrangement.
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔS fus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔS vap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔS vap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔS vap . For instance, ΔS vap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure \(\PageIndex{4}\), the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na + ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl − ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔS soln > 0.
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Predict which substance in each pair has the higher entropy and justify your answer.
- 1 mol of NH 3 (g) or 1 mol of He(g), both at 25°C
- 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given : amounts of substances and temperature
Asked for : higher entropy
Strategy :
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution :
- Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH 3 molecules. With four atoms instead of one, the NH 3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH 3 sample will have the higher entropy.
- The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Predict which substance in each pair has the higher entropy and justify your answer.
- 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
- a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm or a sample of 2 mol of NH 3 (g) at 25°C and 1 atm
- Answer a
-
1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
- Answer a
-
a mixture of 3 mol of H 2 (g) and 1 mol of N 2 (g) at 25°C and 1 atm (more molecules of gas are present)
Video Solution
Reversible and Irreversible Changes
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (P ext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = - P ext ΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: w rev ≥ w irrev . Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So
\[ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1} \]
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: w rev ≥ w irrev .
In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (q rev ) to define entropy quantitatively.
The Relationship between Internal Energy and Entropy
Because the quantity of heat transferred (q rev ) is directly proportional to the absolute temperature of an object (T) (q rev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ q rev ). Combining these relationships for any reversible process,
\[q_{\textrm{rev}}=T\Delta S\;\textrm{ and }\;\Delta S=\dfrac{q_{\textrm{rev}}}{T} \label{Eq2} \]
Because the numerator (q rev ) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is w rev = −PΔV, we can express Equation \(\ref{Eq1}\) as follows:
\[ \begin{align} ΔU &= q_{rev} + w_{rev} \\[4pt] &= TΔS − PΔV \label{Eq3} \end{align} \]
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the \(PV\) work done.
To illustrate the use of Equation \(\ref{Eq2}\) and Equation \(\ref{Eq3}\), we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its \(T\) constant (Figure \(\PageIndex{5}\)). The internal energy of the gas does not change because the temperature of the gas does not change; that is, \(ΔU = 0\) and \(q_{rev} = −w_{rev}\). During expansion, ΔV > 0, so the gas performs work on its surroundings:
\[w_{rev} = −PΔV < 0. \nonumber \]
According to Equation \(\ref{Eq3}\), this means that q rev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore ΔS sys = +q rev /T, and the entropy change of the surroundings is
\[ΔS_{surr} = −\dfrac{q_{rev}}{T}. \nonumber \]
The corresponding change in entropy of the universe is then as follows:
\[ \begin{align*} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \\[4pt] &= \dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \\[4pt] &= 0 \label{Eq4} \end{align*} \]
Thus no change in ΔS univ has occurred.
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure \(\PageIndex{6}\). The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation \(\ref{Eq2}\), we see that the entropy of fusion of ice can be written as follows:
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case, ΔS fus = (6.01 kJ/mol)/(273 K) = 22.0 J/(mol•K) = ΔS sys . The amount of heat lost by the surroundings is the same as the amount gained by the ice, so ΔS surr = q rev /T = −(6.01 kJ/mol)/(273 K) = −22.0 J/(mol•K). Once again, we see that the entropy of the universe does not change:
ΔS univ = ΔS sys + ΔS surr = 22.0 J/(mol•K) − 22.0 J/(mol•K) = 0
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔS cold = q/T cold . Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔS hot = −q/T hot , where T cold and T hot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
\[\Delta S_{\textrm{univ}}=\Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}}=\dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6} \]
The numerators on the right side of Equation \(\ref{Eq6}\) are the same in magnitude but opposite in sign. Whether ΔS univ is positive or negative depends on the relative magnitudes of the denominators. By definition, T hot > T cold , so −q/T hot must be less than q/T cold , and ΔS univ must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔS univ is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔS univ is negative will not occur as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure \(\PageIndex{7}\)).
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to have plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and may have disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
- What is ΔS for this process?
- Which is the more highly ordered form of tin—white or gray?
Given : ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy :
Use Equation \(\ref{Eq2}\) to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution
- We know from Equation \(\ref{Eq2}\) that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where q rev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
- The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C)
Elemental sulfur exists in two forms: an orthorhombic form (S α ), which is stable below 95.3°C, and a monoclinic form (S β ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.
- What is ΔS for this process?
- Which is the more highly ordered form of sulfur—S α or S β ?
- Answer a
-
1.09 J/(mol•K)
- Answer b
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S α
Entropy: Entropy(opens in new window) [youtu.be]
Summary
For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases. \[\Delta S=\frac{q_{\textrm{rev}}}{T} \nonumber \]
A measure of the disorder of a system is its entropy (S), a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.


25.E: Diseases of the Immune System (Exercises)
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- Ying Liu
- City College of San Francisco
19.1: Hypersensitivities
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
Multiple Choice
Which of the following is the type of cell largely responsible for type I hypersensitivity responses?
- erythrocyte
- mast cell
- T lymphocyte
- antibody
- Answer
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B
Type I hypersensitivities require which of the following initial priming events to occur?
- sensitization
- secondary immune response
- cellular trauma
- degranulation
- Answer
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A
Which of the following are the main mediators/initiators of type II hypersensitivity reactions?
- antibodies
- mast cells
- erythrocytes
- histamines
- Answer
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A
Inflammatory molecules are released by mast cells in type I hypersensitivities; type II hypersensitivities, however, are characterized by which of the following?
- cell lysis (cytotoxicity)
- strong antibody reactions against antigens
- leukotriene release upon stimulation
- localized tissue reactions, such as hives
- Answer
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A
An immune complex is an aggregate of which of the following?
- antibody molecules
- antigen molecules
- antibody and antigen molecules
- histamine molecules
- Answer
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C
Which of the following is a common treatment for type III hypersensitivity reactions?
- anti-inflammatory steroid treatments
- antihistamine treatments
- hyposensitization injections of allergens
- RhoGAM injections
- Answer
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A
Which of the following induces a type III hypersensitivity?
- release of inflammatory molecules from mast cells
- accumulation of immune complexes in tissues and small blood vessels
- destruction of cells bound by antigens
- destruction of cells bound by antibodies
- Answer
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B
Which one of the following is not an example of a type IV hypersensitivity?
- latex allergy
- Contact dermatitis (e.g., contact with poison ivy)
- a positive tuberculin skin test
- hemolytic disease of the newborn
- Answer
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D
Fill in the Blank
Antibodies involved in type I hypersensitivities are of the ________ class.
- Answer
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IgE
Allergy shots work by shifting antibody responses to produce ________ antibodies.
- Answer
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IgG
A person who is blood type A would have IgM hemagglutinin antibodies against type ________ red blood cells in their plasma.
- Answer
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B
The itchy and blistering rash that develops with contact to poison ivy is caused by a type ________ hypersensitivity reaction.
- Answer
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IV
Short Answer
Although both type I and type II hypersensitivities involve antibodies as immune effectors, different mechanisms are involved with these different hypersensitivities. Differentiate the two.
What types of antibodies are most common in type III hypersensitivities, and why?
Critical Thinking
Patients are frequently given instructions to avoid allergy medications for a period of time prior to allergy testing. Why would this be important?
In some areas of the world, a tuberculosis vaccine known as bacillus Calmette-Guérin (BCG) is used. It is not used in the United States. Every person who has received this vaccine and mounted a protective response will have a positive reaction in a tuberculin skin test. Why? What does this mean for the usefulness of this skin test in those countries where this vaccine is used?
19.2: Autoimmune Disorders
Autoimmune diseases result from a breakdown in immunological tolerance. The actual induction event(s) for autoimmune states are largely unknown. Some autoimmune diseases attack specific organs, whereas others are more systemic. Organ-specific autoimmune diseases include celiac disease, Graves disease, Hashimoto thyroiditis, type I diabetes mellitus, and Addison disease.
Multiple Choice
Which of the following is an example of an organ-specific autoimmune disease?
- rheumatoid arthritis
- psoriasis
- Addison disease
- myasthenia gravis
- Answer
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C
Which of the following is an example of a systemic autoimmune disease?
- Hashimoto thyroiditis
- type I diabetes mellitus
- Graves disease
- myasthenia gravis
- Answer
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D
Fill in the Blank
The thyroid-stimulating immunoglobulin that acts like thyroid-stimulating hormone and causes Graves disease is an antibody to the ________.
- Answer
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thyroid-stimulating hormone receptor
19.3: Organ Transplantation and Rejection
Grafts and transplants can be classified as autografts, isografts, allografts, or xenografts based on the genetic differences between the donor’s and recipient’s tissues. Genetic differences, especially among the MHC (HLA) genes, will dictate the likelihood that rejection of the transplanted tissue will occur. Transplant recipients usually require immunosuppressive therapy to avoid rejection, even with good genetic matching.
Matching
Match the graft with its description.
| ___autograft | A. donor is a different species than the recipient |
| ___allograft | B. donor and recipient are the same individual |
| ___xenograft | C. donor is an identical twin of the recipient |
| ___isograft | D. donor is the same species as the recipient, but genetically different |
- Answer
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B, D, A, C
Fill in the Blank
For a transplant to have the best chances of avoiding rejection, the genes coding for the ________ molecules should be closely matched between donor and recipient.
- Answer
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MHC
Because it is a “transplant” that can include APCs and T cells from the donor, a bone marrow transplant may induce a very specific type of rejection known as ________ disease.
- Answer
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graft-versus-host
Short Answer
Why is a parent usually a better match for transplanted tissue to a donor than a random individual of the same species?
19.4: Immunodeficiency
Primary immunodeficiencies are caused by genetic abnormalities; secondary immunodeficiencies are acquired through disease, diet, or environmental exposures. Primary immunodeficiencies may result from flaws in phagocyte killing of innate immunity, or impairment of T cells and B cells. Primary immunodeficiencies include chronic granulomatous disease, X-linked agammaglobulinemia, selective IgA deficiency, and severe combined immunodeficiency disease.
Multiple Choice
Which of the following is a genetic disease that results in lack of production of antibodies?
- agammaglobulinemia
- myasthenia gravis
- HIV/AIDS
- chronic granulomatous disease
- Answer
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A
Which of the following is a genetic disease that results in almost no adaptive immunity due to lack of B and/ or T cells?
- agammaglobulinemia
- severe combined immunodeficiency
- HIV/AIDS
- chronic granulomatous disease
- Answer
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B
All but which one of the following are examples of secondary immunodeficiencies?
- HIV/AIDS
- malnutrition
- chronic granulomatous disease
- immunosuppression due to measles infection
- Answer
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C
Fill in the Blank
Diseases due to ________ abnormalities are termed primary immunodeficiencies.
- Answer
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genetic
A secondary immunodeficiency is ________, rather than genetic.
- Answer
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acquired
Short Answer
Compare the treatments for primary and secondary immunodeficiencies.
19.5: Cancer Immunobiology and Immunotherapy
When control of the cell cycle is lost, the affected cells rapidly divide and often lose the ability to differentiate into the cell type appropriate for their location in the body. In addition, they lose contact inhibition and can start to grow on top of each other. This can result in formation of a tumor. It is important to make a distinction here: The term “cancer” is used to describe the diseases resulting from loss of cell-cycle regulation and subsequent cell proliferation.
Multiple Choice
Cancer results when a mutation leads to which of the following?
- cell death
- apoptosis
- loss of cell-cycle control
- shutdown of the cell cycle
- Answer
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C
Tumor antigens are ________ that are inappropriately expressed and found on abnormal cells.
- self antigens
- foreign antigens
- antibodies
- T-cell receptors
- Answer
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A
Fill in the Blank
A ________ cancer vaccine is one that stops the disease from occurring in the first place.
- Answer
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preventive
A ________ cancer vaccine is one that will help to treat the disease after it has occurred.
- Answer
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therapeutic
Short Answer
How can tumor antigens be effectively targeted without inducing an autoimmune (anti-self) response?


13.4: The Female Reproductive Organs
The female reproductive system consists of a pair of ovaries that produce egg cells or ova and fallopian tubes where fertilisation occurs and which carry the fertilised ovum to the uterus . Growth of the foetus takes place here. The cervix separates the uterus from the vagina or birth canal, where the sperm are deposited (see diagram 13.6).
Diagram 13.6. - The reproductive system of a female rabbit
Note that primates like humans have a uterus with a single compartment but in most mammals the uterus is divided into two separate parts or horns as shown in diagram 13.6.
The Ovaries
Ovaries are small oval organs situated in the abdominal cavity just ventral to the kidneys. Most animals have a pair of ovaries but in birds only the left one is functional to reduce weight (see below).
The ovary consists of an inner region ( medulla ) and an outer region ( cortex ) containing egg cells or ova. These are formed in large numbers around the time of birth and start to develop after the animal becomes sexually mature. A cluster of cells called the follicle surrounds and nourishes each ovum.
The Ovarian Cycle
The ovarian cycle refers to the series of changes in the ovary during which the follicle matures, the ovum is shed and the corpus luteum develops (see diagram 13.7).
Numerous undeveloped ovarian follicles are present at birth but they start to mature after sexual maturity. In animals that normally have only one baby at a time only one ovum will mature at once but in litter animals several will. The mature follicle consists of outer cells that provide nourishment. Inside this is a fluid-filled space that contains the ovum.
A mature follicle can be quite large, ranging from a few millimetres in small mammals to the size of a golf ball in large animals. It bulges out from the surface of the ovary before eventually rupturing to release the ovum into the abdominal cavity. Once the ovum has been shed, a blood clot forms in the empty follicle. This develops into a tissue called the corpus luteum that produces the hormone progesterone (see diagram 13.9). If the animal becomes pregnant the corpus luteum persists, but if there is no pregnancy it degenerates and a new ovarian cycle usually.
Diagram 13.7 - The ovarian cycle showing from the top left clockwise: the maturation of the ovum over time, followed by ovulation and the development of the corpus luteum in the empty follicle
The Ovum
When the ovum is shed the nucleus is in the final stages of meiosis (cell division). It is surrounded by few layers of follicle cells and a tough membrane called the zona pelluc ida (see diagram 13.8).
Diagram 13.8 - An ovum
The Oestrous Cycle
The oestrous cycle is the sequence of hormonal changes that occurs through the ovarian cycle . These changes influence the behaviour and body changes of the female (see diagram 13.9).
Diagram 13.9 - The oestrous cycle
The first hormone involved in the oestrous cycle is follicle stimulating hormone (F.S.H.), secreted by the anterior pituitary gland (see chapter 16). It stimulates the follicle to develop. As the follicle matures the outer cells begin to secrete the hormone oestrogen and this stimulates the mammary glands to develop. It also prepares the lining of the uterus to receive a fertilised egg. Ovulation is initiated by a surge of another hormone from the anterior pituitary, luteinising hormone (L.H.). This hormone also influences the development of the corpus luteum, which produces progesterone , a hormone that prepares the lining of the uterus for the fertilised ovum and readies the mammary glands for milk production. If no pregnancy takes place the corpus luteum shrinks and the production of progesterone decreases. This causes FSH to be produced again and a new oestrous cycle begins.
For fertilisation of the ovum by the sperm to occur, the female must be receptive to the male at around the time of ovulation. This is when the hormones turn on the signs of “ heat ”, and she is “ in season ” or “ in oestrous ”. These signs are turned off again at the end of the oestrous cycle.
During the oestrous cycle the lining of the uterus ( endometrium ) thickens ready for the fertilised ovum to be implanted. If no pregnancy occurs this thickened tissue is absorbed and the next cycle starts. In humans and other higher primates, however, the endometrium is shed as a flow of blood and instead of an oestrous cycle there is a menstrual cycle .
The length of the oestrous cycle varies from species to species. In rats the cycle only lasts 4–5 days and they are sexually receptive for about 14 hours. Dogs have a cycle that lasts 60–70 days and heat lasts 7–9 days and horses have a 21-day cycle and heat lasts an average of 6 days.
Ovulation is spontaneous in most animals but in some, e.g. the cat, and the rabbit, ovulation is stimulated by mating. This is called induced ovulation .

- Female cats “call” at night, roll and tread the carpet and are generally restless but will “stand” firm when pressure is placed on the pelvic region (this is the lordosis response).
- A female rat shows the lordosis response when on heat. It will “mount” other females and be more active than normal.
- A cow mounts other cows (bulling), bellows, is restless and has a discharge from the vulva.
Breeding Seasons And Breeding Cycles
Only a few animals breed throughout the year. This includes the higher primates (humans, gorillas and chimpanzees etc.), pigs, mice and rabbits. These are known as continuous breeders .
Most other animals restrict reproduction to one or two seasons in the year- seasonal breeders (see diagram 13.10). There are several reasons for this. It means the young can be born at the time (usually spring) when feed is most abundant and temperatures are favourable. It is also sensible to restrict the breeding season because courtship, mating, gestation and the rearing of young can exhaust the energy resources of an animal as well as make them more vulnerable to predators.
Diagram 13.10 - Breeding cycles
The timing of the breeding cycle is often determined by day length. For example the shortening day length in autumn will bring sheep and cows into season so the foetus can gestate through the winter and be born in spring. In cats the increasing day length after the winter solstice (shortest day) stimulates breeding. The number of times an animal comes into season during the year varies, as does the number of oestrous cycles during each season. For example a dog usually has 2-3 seasons per year, each usually consisting of just one oestrous cycle. In contrast ewes usually restrict breeding to one season and can continue to cycle as many as 20 times if they fail to become pregnant.


9.2A: Viral Morphology
- Describe the relationship between the viral genome, capsid, and envelope
Viral Morphology
Viruses are acellular, meaning they are biological entities that do not have a cellular structure. Therefore, they lack most of the components of cells, such as organelles, ribosomes, and the plasma membrane. A virion consists of a nucleic acid core, an outer protein coating or capsid, and sometimes an outer envelope made of protein and phospholipid membranes derived from the host cell. The capsid is made up of protein subunits called capsomeres. Viruses may also contain additional proteins, such as enzymes. The most obvious difference between members of viral families is their morphology, which is quite diverse. An interesting feature of viral complexity is that host and virion complexity are uncorrelated. Some of the most intricate virion structures are observed in bacteriophages, viruses that infect the simplest living organisms: bacteria.
Morphology
Viruses come in many shapes and sizes, but these are consistent and distinct for each viral family. In general, the shapes of viruses are classified into four groups: filamentous, isometric (or icosahedral), enveloped, and head and tail. Filamentous viruses are long and cylindrical. Many plant viruses are filamentous, including TMV (tobacco mosaic virus). Isometric viruses have shapes that are roughly spherical, such as poliovirus or herpesviruses. Enveloped viruses have membranes surrounding capsids. Animal viruses, such as HIV, are frequently enveloped. Head and tail viruses infect bacteria. They have a head that is similar to icosahedral viruses and a tail shape like filamentous viruses.
Many viruses use some sort of glycoprotein to attach to their host cells via molecules on the cell called viral receptors. For these viruses, attachment is a requirement for later penetration of the cell membrane, allowing them to complete their replication inside the cell. The receptors that viruses use are molecules that are normally found on cell surfaces and have their own physiological functions. Viruses have simply evolved to make use of these molecules for their own replication.
Overall, the shape of the virion and the presence or absence of an envelope tell us little about what disease the virus may cause or what species it might infect, but they are still useful means to begin viral classification. Among the most complex virions known, the T4 bacteriophage, which infects the Escherichia coli bacterium, has a tail structure that the virus uses to attach to host cells and a head structure that houses its DNA. Adenovirus, a non-enveloped animal virus that causes respiratory illnesses in humans, uses glycoprotein spikes protruding from its capsomeres to attach to host cells. Non-enveloped viruses also include those that cause polio (poliovirus), plantar warts (papillomavirus), and hepatitis A (hepatitis A virus).
Enveloped virions like HIV consist of nucleic acid and capsid proteins surrounded by a phospholipid bilayer envelope and its associated proteins. Glycoproteins embedded in the viral envelope are used to attach to host cells. Other envelope proteins include the matrix proteins that stabilize the envelope and often play a role in the assembly of progeny virions. Chicken pox, influenza, and mumps are examples of diseases caused by viruses with envelopes. Because of the fragility of the envelope, non-enveloped viruses are more resistant to changes in temperature, pH, and some disinfectants than are enveloped viruses.
Types of Nucleic Acid
Unlike nearly all living organisms that use DNA as their genetic material, viruses may use either DNA or RNA. The virus core contains the genome or total genetic content of the virus. Viral genomes tend to be small, containing only those genes that encode proteins that the virus cannot obtain from the host cell. This genetic material may be single- or double-stranded. It may also be linear or circular. While most viruses contain a single nucleic acid, others have genomes that have several, called segments.
In DNA viruses, the viral DNA directs the host cell’s replication proteins to synthesize new copies of the viral genome and to transcribe and translate that genome into viral proteins. DNA viruses cause human diseases, such as chickenpox, hepatitis B, and some venereal diseases, like herpes and genital warts.
RNA viruses contain only RNA as their genetic material. To replicate their genomes in the host cell, the RNA viruses encode enzymes that can replicate RNA into DNA, which cannot be done by the host cell. These RNA polymerase enzymes are more likely to make copying errors than DNA polymerases and, therefore, often make mistakes during transcription. For this reason, mutations in RNA viruses occur more frequently than in DNA viruses. This causes them to change and adapt more rapidly to their host. Human diseases caused by RNA viruses include hepatitis C, measles, and rabies.
Key Points
- Viruses are classified into four groups based on shape: filamentous, isometric (or icosahedral), enveloped, and head and tail.
- Many viruses attach to their host cells to facilitate penetration of the cell membrane, allowing their replication inside the cell.
- Non-enveloped viruses can be more resistant to changes in temperature, pH, and some disinfectants than are enveloped viruses.
- The virus core contains the small single- or double-stranded genome that encodes the proteins that the virus cannot get from the host cell.
Key Terms
- capsid : the outer protein shell of a virus
- envelope : an enclosing structure or cover, such as a membrane
- filamentous : Having the form of threads or filaments
- isometric : of, or being a geometric system of three equal axes lying at right angles to each other (especially in crystallography)


17.7: Sample problems and solutions
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Exercise \(\PageIndex{1}\)
Consider a charged sphere of radius, \(R\) , which has a non-uniform charge density, that varies with radius, as \(\rho(r) = ar^2\) .
- What is the total charge on the sphere?
- What is the electric field as a function of distance from the center of the sphere outside the sphere, \(r>R\) ?
- What is the electric field as a function of distance from the center of the sphere inside the sphere, \(r\leq R\) ?
- Answer
-
a. In order to determine the total charge of the sphere, we divide the sphere into infinitesimal shells of radius \(r\) , and thickness, \(dr\) . The volume, \(dV\) , of one of these infinitesimal shells is their area (given by the area of the surface of a sphere of radius \(r\) ), multiplied by their thickness, \(dr\) :
\[\begin{aligned} dV = 4\pi r^2 dr\end{aligned}\]
The charge, \(dQ\) , of one of those shells is given by the charge per unit volume, \(\rho(r)\) :
\[\begin{aligned} dQ = \rho(r) dV = ar^24\pi r^2 dr = 4a\pi r^4 dr\end{aligned}\]
The total charge of the sphere is found by summing the charge from each shell:
\[\begin{aligned} Q=\int dQ =\int_0^R4a\pi r^4 dr=\frac{4}{5}a\pi R^5\end{aligned}\]
b. Outside of the sphere, we can use a spherical gaussian surface of radius \(r\) , so that the flux is given by:
\[\begin{aligned} \oint \vec E\cdot d\vec A=4\pi r^2 E\end{aligned}\]
The entire charge of the sphere is enclosed. Applying Gauss’ Law, we can determine the electric field outside the sphere:
\[\begin{aligned} \oint \vec E\cdot d\vec A&= \frac{Q^{enc}}{\epsilon_0}\\[4pt] 4\pi r^2 E&= \frac{4a\pi R^5}{5\epsilon_0}\\[4pt] \therefore E(r)&=\frac{aR^5}{5\epsilon_0r^2}\end{aligned}\]
and we see that the electric field decreases as the radius squared, which makes sense, since from outside the sphere, we do not know how the charge is distributed within.
c. Inside the volume of the sphere, we still use a gaussian spherical surface of radius, \(r\) , so that the flux is given by:
\[\begin{aligned} \oint \vec E\cdot d\vec A=4\pi r^2 E\end{aligned}\]
However, inside the sphere, the gaussian surface only encloses the charge up to a radius of \(r\) , which we find by integration, similar to part a):
\[\begin{aligned} Q^{enc}=\int dQ =\int_0^r 4a\pi r^4 dr=\frac{4}{5}a\pi r^5\end{aligned}\]
Applying Gauss’ Law:
\[\begin{aligned} \oint \vec E\cdot d\vec A&= \frac{Q^{enc}}{\epsilon_0}\\[4pt] 4\pi r^2 E&= \frac{4a\pi r^5}{5\epsilon_0}\\[4pt] \therefore E(r)&=\frac{ar^3}{5\epsilon_0r^2}\end{aligned}\]
and we find that the electric field is zero at the center of the sphere and increases with radius cube inside the sphere.
Exercise \(\PageIndex{2}\)
C onsider two conducting plates which are illustrated in Figure \(\PageIndex{1}\). Both plates have a hollow circle of radius, \(R\) , at their center. One plate is a square on the outside and the other is a triangle on the outside, both of the outside shapes have a side length of \(L\) . A point charge of charge \(+Q\) is placed at the center of the hollowed out circle of both plates.
- What is the electric field outside of the shells?
- What is the average linear charge density on the inner and outer surfaces of the shells?
- Which sections of the two plates would have the largest charge density?
- Answer
-
a. The conducting shells have no net charge, so the only charge in the system is the point charge \(Q\) . If we draw a spherical gaussian surface, the only enclosed charge will be \(Q\) , and we can ignore the charges on the plates. The electric field is thus the field from a point charge:
\[\begin{aligned} E= \frac{kQ}{r^2} \end{aligned}\]
b. Let’s begin with the shell that has a triangle on the outside. We will use Gauss’ law to determine the charge density of the inner and outer shells. To do this, we will draw a circle within the shell, \(S_1\) and a triangle outside of the outer shell, \(S_2\) as shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\) : A solution to the triangular conducting shell.
When considering \(S_1\) , we know that the electric field inside of the (conducting) shell is \(0\) , so that the flux out of \(S_1\) will be zero. This means that the point charge on the inside of the shell will be equal and opposite to the sum of the surface charges on the inner shell. From here, we divide the net charge by the circumference of the inner shell to determine the linear charge density:
\[\begin{aligned} \lambda_{circle} = \frac{-Q}{2\pi R} \end{aligned}\]
When considering \(S_2\) , we know that the \(Q_{enc} = +Q\) , which means that the total linear charge on the outer triangle will be \(+Q\) such that it cancels the \(-Q\) along the inner circle. The sum of charges would be \(Q_{enc} = Q_{point}+Q_{triangle}-Q_{circle}\) . Knowing this, we must divide the total charge on the outer surface by the sum of the length of each of the triangle’s sides in order to find the average linear charge density:
\[\begin{aligned} \lambda_{circle} = \frac{Q}{3L} \end{aligned}\]
Now, we consider the inner and outer linear charge densities of the square conducting shell. We choose two gaussian surfaces, \(S_1\) , and \(S_2\) , as shown in Figure \(\PageIndex{3}\):
Figure \(\PageIndex{3}\) : A solution to the square conducting shell.
For \(S_1\) , the circle is treated as it was while solving the triangular shell. The electric field is also \(0\) within the square conducting shell, so we know that the average linear charge density is \(\frac{-Q}{2\pi R}\) .
When considering \(S_2\) , we know that \(Q_{enc}\) is \(+Q\) , so we know that the total charge on the square surface of the shell will be \(+Q\) . This leave us with the following average linear charge density:
\[\begin{aligned} \lambda_{square} = \frac{Q}{4L} \end{aligned}\]
c. These plates are charged by the electric field generated by the point charge held within them, which means that the linear charge density of the two plates will be highest at the points along the outer sides which are the shortest distance from the point charge. These points occur in the triangular and square plates at the point which is the closest to the charge, \(Q\) , as shown in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\) : A solution to the square conducting shell.


5.7: Weekend assignment 5
There is a whole world of Arabic-dubbed movies especially Disney movies. Experiment searching for these on YouTube. Type ديزني بالعربي and you will find a lot even full movies. Many are dubbed in Egyptian Arabic like this song.
Additional resources for keeping up with Arabic :
There are channels for Arabic fairy tales like Cinderella and lion king and many others:
There is a collection here for these in children books
Also check out some collected resources that we put together on my page here:
I encourage you to keep Arabic in your ears even if you are not paying attention :
1)You can check Aljazeera news channel or BBC Arabic or CNN arabic websites for streaming or video news
2)Search Youtube for “Egyptian Movies with English subtitles” or “Arabic movies with subtitles ” . Even if you are not paying attention, you will be able to pick up a lot.
Writing prompt:
ماذا تفعلون في عيد الشكر؟ ماذا تفعلون في الكريسماس؟
What do you like to do in Thanksgiving, Christmas? What do your family members and friends like to do?
Talk in some details about types of possible activities that you, your family members and friends like to do in vacation , holidays, thanksgiving, and Christmas. Where do you like to go, play, watch, eat etc.
Use as many verbs and a variety of vocabulary as you can. You can use some helping vocabulary:
| Vacation/holiday | أجازة |
| weekend | نهِاية الأسبوع |
| I go out | أخرُج |
| I rest | أستريح |
| sometimes | أحياناً |
| Always | دائماً |
| A lot | كثيراً |
| turkey | ديك رومي |
| potatoes | بطاطِس |
| beans | فاصوليا |
| I cook | أطبُخ |
| I stay up late | أسهَر |
| After that | بعد ذَلِك |
| I celebrate | أحتَفِل |
| I buy | أشتري |
| gift/s | هَدِيّة – هدايا |
| thanksgiving | عِيد الشُّكر |
Target : 60-70 words or more.


9.8: 20th Century Podcasts
Here are some podcasts to help you in your studies of our final period of music history, the 20th Century.
Listening Help 1
An audio element has been excluded from this version of the text. You can listen to it online here: http://pb.libretexts.org/map/?p=348
Listening Help 2
An audio element has been excluded from this version of the text. You can listen to it online here: http://pb.libretexts.org/map/?p=348
Listening Help 3
An audio element has been excluded from this version of the text. You can listen to it online here: http://pb.libretexts.org/map/?p=348
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution


2.2.16: Olive Schreiner (1855-1920)
The Story of an African Farm South African Realism Olive Schreiner, a South African writer and activist, is best known for her work to advance women's rights and political independence for South Africa. As a writer, she is best known for her 1883 novel, The Story of an African Farm , and for a later work, Woman and Labour (1911), which became a seminal work of the Woman's Movement, asserting the need for women to be engaged in meaningful work. Schreiner was the daughter of English missionary parents in South Africa. Her father, an unsuccessful missionary, was also an ineffectual businessman, so the family lived in abject poverty for most of Schreiner's life. Always an intellectually curious child and a religious skeptic, the turning point in Schreiner's life was her reading of Herbert Spencer's First Principles , which reinforced her rejection of organized religion. After working as a governess for a number of employers, Schreiner travelled to England in 1881 to study medicine. However, her ill health—she suffered all of her life from acute asthma—prevented her working as a nurse, so she reconciled herself to working as a full-time writer, a profession she pursued for the rest of her life. Upon its publication in 1883, The Story of an African Farm (1833) became an immediate sensation with readers in both England and America. Readers were fascinated by its experimental form, which included allegorical and dream sequences, a sexually active feminist protagonist, and an account of life in an exotic colonial setting. The novel's themes range from bold feminist statements about the oppression of women, gender ambivalence on the part of a male character who finds fulfillment in dressing as a woman, frank discussions of sex, and an examination of religious agnosticism. Consider while reading:
- What is the effect of the opening descriptions? Which features of the environment seem to be emphasized?
- Contrast Waldo, Lyndall, and Em. How would you characterize each of the children?
- How are the aboriginal peoples (the Bushmen) and colonized peoples ("Kaffirs" and "Hottentots") represented? Are these representations consistent with Schreiner's later advocacy for the equality and independence of African peoples?
Written by Anita Turlington


2.5: Family Life
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Parenting Styles
Relationships between parents and children continue to play a significant role in children’s development during early childhood. We will explore two models of parenting styles. Keep in mind that these most parents do not follow any model completely. Real people tend to fall somewhere in between these styles. And sometimes parenting styles change from one child to the next or in times when the parent has more or less time and energy for parenting. Parenting styles can also be affected by concerns the parent has in other areas of his or her life. For example, parenting styles tend to become more authoritarian when parents are tired and perhaps more authoritative when they are more energetic. Sometimes parents seem to change their parenting approach when others are around, maybe because they become more self-conscious as parents or are concerned with giving others the impression that they are a “tough” parent or an “easy-going” parent. And of course, parenting styles may reflect the type of parenting someone saw modeled while growing up.
Baumrind (1971) offers a model of parenting that includes three styles. The first, authoritarian , is the traditional model of parenting in which parents make the rules and children are expected to be obedient. Baumrind suggests that authoritarian parents tend to place maturity demands on their children that are unreasonably high and tend to be aloof and distant. Consequently, children reared in this way may fear rather than respect their parents and, because their parents do not allow discussion, may take out their frustrations on safer targets-perhaps as bullies toward peers.
Permissive parenting involves holding expectations of children that are below what could be reasonably expected from them. Children are allowed to make their own rules and determine their own activities. Parents are warm and communicative, but provide little structure for their children. Children fail to learn self-discipline and may feel somewhat insecure because they do not know the limits.
Authoritative parenting involves being appropriately strict, reasonable, and affectionate. Parents allow negotiation where appropriate and discipline matches the severity of the offense. A popular parenting program that is offered in many school districts is called “Love and Logic” and reflects the authoritative or democratic style of parenting just described. Uninvolved parents are disengaged from their children. They do not make demands on their children and are non-responsive. These children can suffer in school and in their relationships with their peers (Gecas & Self, 1991).
Lemasters and Defrain (1989) offer another model of parenting. This model is interesting because it looks more closely at the motivations of the parent and suggests that parenting styles are often designed to meet the psychological needs of the parent rather than the developmental needs of the child. The martyr is a parent who will do anything for the child; even tasks that the child should do for himself or herself. All of the good deeds performed for the child, in the name of being a “good parent”, may be used later should the parent want to gain compliance from the child. If a child goes against the parent’s wishes, the parent can remind the child of all of the times the parent helped the child and evoke a feeling of guilt so that the child will do what the parent wants. The child learns to be dependent and manipulative as a result. (Beware! A parent busy whipping up cookies may really be thinking “control”!)
The pal is like the permissive parent described in Baumrind’s model above. The pal wants to be the child’s friend. Perhaps the parent is lonely or perhaps the parent is trying to win a popularity contest against an ex-spouse. Pals let children do what they want and focus most on being entertaining and fun and set few limitations. Consequently, the child may have little self-discipline and may try to test limits with others.
The police officer/drill sergeant style of parenting is similar to the authoritarian parent described above. The parent focuses primarily making sure that the child is obedient and that the parent has full control of the child. Sometimes this can be taken to extreme by giving the child tasks that are really designed to check on their level of obedience. For example, the parent may require that the child fold the clothes and place items back in the drawer in a particular way. If not, the child might be scolded or punished for not doing things “right”. This type of parent has a very difficult time allowing the child to grow and learn to make decisions independently. And the child may have a lot of resentment toward the parent that is displaced on others.
The teacher-counselor parent is one who pays a lot of attention to expert advice on parenting and who believes that as long as all of the steps are followed, the parent can rear a perfect child. “What’s wrong with that?” you might ask. There are two major problems with this approach. First, the parent is taking all of the responsibility for the child’s behavior-at least indirectly. If the child has difficulty, the parent feels responsible and thinks that the solution lies in reading more advice and trying more diligently to follow that advice. Parents can certainly influence children, but thinking that the parent is fully responsible for the child’s outcome if faulty. A parent can only do so much and can never have full control over the child. Another problem with this approach is that the child may get an unrealistic sense of the world and what can be expected from others. For example, if a teacher-counselor parent decides to help the child build self-esteem and has read that telling the child how special he or she is or how important it is to compliment the child on a job well done, the parent may convey the message that everything the child does is exceptional or extraordinary. A child may come to expect that all of his efforts warrant praise and in the real world, this is not something one can expect. Perhaps children get more of a sense of pride from assessing their own performance than from having others praise their efforts.
So what is left? Lemasters and Defrain (1989) suggest that the athletic coach style of parenting is best. Before you draw conclusions here, set aside any negative experiences you may have had with coaches in the past. The principles of coaching are what are important to Lemasters and Defrain. A coach helps players form strategies, supports their efforts, gives feedback on what went right and what went wrong, and stands at the sideline while the players perform. Coaches and referees make sure that the rules of the game are followed and that all players adhere to those rules. Similarly, the athletic coach as parent helps the child understand what needs to happen in certain situations whether in friendships, school, or home life, and encourages and advises the child about how to manage these situations. The parent does not intervene or do things for the child. Rather, the parent’s role is to provide guidance while the child learns first hand how to handle these situations. And the rules for behavior are consistent and objective and presented in that way. So, a child who is late for dinner might hear the parent respond in this way, “Dinner was at six o’clock.” Rather than, “You know good and well that we always eat at six. If you expect me to get up and make something for you now, you have got another thing coming! Just who do you think you are showing up late and looking for food? You’re grounded until further notice!”
The most important thing to remember about parenting is that you can be a better, more objective parent when you are directing your actions toward the child’s needs and while considering what they can reasonably be expected to do at their stage of development. Parenting is more difficult when you are tired and have psychological needs that interfere with the relationship. Some of the best advice for parents is to try not to take the child’s actions personally and be as objective as possible.
The impact of class and culture cannot be ignored when examining parenting styles. The two models of parenting described above assume that authoritative and athletic coaching styles are best because they are designed to help the parent raise a child who is independent, self-reliant and responsible. These are qualities favored in “individualistic” cultures such as the United States, particularly by the middle class. African-American, Hispanic and Asian parents tend to be more authoritarian than non-Hispanic whites. However, in “collectivistic” cultures such as China or Korea, being obedient and compliant are favored behaviors. Authoritarian parenting has been used historically and reflects cultural need for children to do as they are told. In societies where family members’ cooperation is necessary for survival, as in the case of raising crops, rearing children who are independent and who strive to be on their own makes no sense. But in an economy based on being mobile in order to find jobs and where one’s earnings are based on education, raising a child to be independent is very important.
Working class parents are more likely than middle class parents to focus on obedience and honesty when raising their children. In a classic study on social class and parenting styles called Class and Conformity, Kohn (1977) explains that parents tend to emphasize qualities that are needed for their own survival when parenting their children. Working class parents are rewarded for being obedient, reliable, and honest in their jobs. They are not paid to be independent or to question the management; rather, they move up and are considered good employees if they show up on time, do their work as they are told, and can be counted on by their employers. Consequently, these parents reward honesty and obedience in their children. Middle class parents who work as professionals are rewarded for taking initiative, being self-directed, and assertive in their jobs. They are required to get the job done without being told exactly what to do. They are asked to be innovative and to work independently. These parents encourage their children to have those qualities as well by rewarding independence and self-reliance. Parenting styles can reflect many elements of culture.
Child Care Concerns
About 77.3 percent of mothers of school-aged and 64.2 percent of mothers of preschool aged children in the United States work outside the home (Cohen and Bianchi, 1999; Bureau of Labor Statistics, 2010). Since more women have been entering the workplace, there has been a concern that families do not spend as much time with their children. This, however, is not true. Between 1981 and 1997, the amount of time that parents spent with children has increased overall (Sandberg and Hofferth, 2001).
Seventy-five percent of children under age 5 are in scheduled child care programs. Others are cared for by family members, friends, or are in Head Start Programs. Older children are often in after school programs, before school programs, or stay at home alone after school once they are older. Quality childcare programs can enhance a child’s social skills and can provide rich learning experiences. But long hours in poor quality care can have negative consequences for young children in particular. What determines the quality of child care? One consideration is the teacher/child ratio . States specify the maximum number of children that can be supervised by one teacher. In general, the younger the children, the more teachers required for a given number of children. The higher the teacher to child ratio, the more time the teacher has for involvement with the children and the less stressed the teacher may be so that the interactions can be more relaxed, stimulating and positive. The more children there are in a program, the less desirable the program as well. This is because the center may be more rigid in rules and structure to accommodate the large number of children in the facility.
The physical environment should be colorful, stimulating, clean, and safe. The philosophy of the organization and the curriculum available should be child-centered, positive, and stimulating. Providers should be trained in early childhood education as well. A majority of states do not require training for their child care providers. And while formal education is not required for a person to provide a warm, loving relationship to a child, knowledge of a child’s development is useful for addressing their social, emotional, and cognitive needs in an effective way. By working toward improving the quality of childcare and increasing family-friendly workplace policies such as more flexible scheduling and perhaps childcare facilities at places of employment, we can accommodate families with smaller children and relieve parents of the stress sometimes associated with managing work and family life.
Global Concerns: The Market Women of Liberia
Work and mothering go hand in hand in many parts of the world. Consider the market women of Liberia. These are women who work as street vendors and are primary providers for their families.  Recently, they have been supported by President Sirleaf whose grandmother was a market woman. President Sirleaf has worked to raise funds to improve the marketplaces and conditions for mothers and children. The hope has been to make these marketplaces more safe, to provide childcare, and social services to improve the lives of mothers and children (Nance-Nash, 2009).
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- Content Attribution
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Faculty Resources
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Unit 1: Introduction and Prenatal Development
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Module 1: Lifespan Psychology
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- Introduction to Life Span, Growth and Development
- The Cohort Effect
- Culture
- Periods of Development
- Research Methods
- Video: 49 Up
- Video: Meet Neil
- Lecture: Introduction to Life Span, Growth and Development
- PowerPoint: Introduction to Life Span Development
- Discussion: Life Stages
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Module 2: Developmental Theories
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- Introduction to Developmental Theories
- Psychodynamic Theory
- Psychosocial Theory
- Exploring Behavior
- Exploring Cognition
- Research Designs
- Lecture Transcript: Developmental Theories
- Slideshow: Developmental Theories
- Essay: Lifespan Psychology
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Module 3: Prenatal Development
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- Introduction to Heredity, Prenatal Development, and Birth
- Prenatal Development
- Environmental Risks
- Pregnancy
- Childbirth
- Maternal Mortality
- Video: "Life's Greatest Miracle"
- Lecture: Prenatal Development
- Slideshow: Prenatal Development
- Study Guide: Introduction and Prenatal Development
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Unit 2: Childhood
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Module 4: Infancy
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- Introduction to Infancy
- Physical Development
- Nutrition
- Cognitive Development
- Language Development
- Psychosocial Development and Attachment
- Temperament
- Psychosocial Development
- Lecture: Infancy
- Slideshow: Infancy
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Module 5: Early Childhood
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- Introduction to Early Childhood
- Physical Development
- Cognitive Development
- Psychosocial Development
- Family Life
- Childhood Stress and Development
- Lecture: Early Childhood
- Slideshow: Early Childhood
- Discussion: Labeling and Children
- Essay: Childhood
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Module 6: Middle Childhood
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- Introduction to Middle Childhood
- Physical Development
- Cognitive Development
- Kohlberg’s Stages of Moral Development
- Developmental Problems
- Learning and Intelligence
- Psychosocial Development
- Lecture: Middle Childhood
- Slideshow: Middle Childhood
- Study Guide: Childhood
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Unit 3: Adolescence and Early Adulthood
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Module 7: Adolescence
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- Introduction to Adolescense
- Physical Development
- Cognitive Development
- Social Development
- Lecture: Adolescence
- Slideshow: Adolescence
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Module 8: Early Adulthood
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- Introduction to Early Adulthood
- Physical Development
- Cognitive Development
- Psychosocial Development
- Types of Love
- Activity: Love Attitude Scale
- Assignment: Love Styles
- Lecture: Early Adulthood
- Slideshow: Early Adulthood
- Essay Assignment: Adolescence and Early Adulthood
- Study Guide: Adolescence and Early Adulthood
- Practice Test: Adolescence and Early Adulthood
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Unit 4: Middle and Late Adulthood
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Module 9: Middle Adulthood
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- Introduction to Middle Adulthood
- Physical Development
- Cognitive Development
- Psychosocial Development
- Relationships
- Work and Personality
- Lecture: Middle Adulthood
- Slideshow: Middle Adulthood
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Module 10: Late Adulthood
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- Introduction to Late Adulthood
- Physical Development
- Cognitive Development
- Psychosocial Development
- Relationships
- Listen: Treating Delirium
- Lecture: Late Adulthood
- Slideshare: Late Adulthood
- Additional Links
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Module 11: Death and Dying
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- Introduction to Death and Dying
- Most Common Causes of Death
- The Process of Dying
- Five Stages of Loss
- Palliative Care and Hospice
- Euthanasia
- Bereavement and Grief
- Lecture: Death and Dying
- Slideshow: Death and Dying
- Additional Links
- Essay: Middle and Late Adulthood
- Study Guide: Middle and Late Adulthood
- Practice Test: Middle and Late Adulthood
- EXAM: Unit 4
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