8.5: Educational Access and Quality
On the whole, higher levels of education are linked with better paying jobs and better health over the lifespan. Graduating from college decreases the risk of future unemployment (and thus many of the health factors that come with economic instability), and college graduates report better health than those who only complete high school. College education is more often required for white-collar jobs, which tend to pay better wages and are more likely to provide health insurance benefits. This also means college graduates may be able to afford better quality housing, and have more access to healthy dietary patterns and leisure-time physical activity. Unhealthy behaviors like smoking and drinking excessively are lower in more educated populations ( Enrollment in Higher Education - Healthy People 2030 , n.d.).
Education levels of parents are also correlated with the health of the rest of the family. Childhood obesity is negatively correlated with the education level of the head of the household (HHS reports). If parents have higher education, children are less likely to experience adverse childhood experiences (ACEs). These types of experiences can include trauma from physical or sexual abuse, neglect, witnessing a family member use drugs or have mental health problems - including attempting or committing suicide - or having a family member become incarcerated. They can also include witnessing violence in the home or community, becoming homeless, or experiencing stress from housing and food insecurity (U.S. Department of Health and Human Services, Health Resources and Services Administration, Office of Health Equity., 2020).
As mentioned earlier, racial disparities in health and life expectancy persist across educational strata. However, one increasingly common cause of death primarily affects white, non-college educated, middle-aged men, who live in mostly rural areas and small towns. Correlated with the decline in blue-collar job opportunities in middle America, so-called “deaths of despair” have increased over the last decade. These include deaths from suicide, drug (mostly opioid) overdoses, and liver disease, all associated with physical and psychological pain (Scutchfield & Keck, 2017). This is a tragic example of a combination of social determinants of health converging: unemployment, poverty, and a lack of access to healthcare (mental health care in particular), all contributing to higher rates of depression and substance abuse.


6: Column chromatography
Summary
Column chromatography is a method for separating a mixture of substances based on their chemical and physical characteristics. In biological research, this is usually a mixture of proteins.
Also known as
Variants: Affinity chromatography, ion exchange chromatography, size exclusion chromatography
Samples needed
In biology, most column chromatography is performed on a solution containing a mixture of proteins
Method
Although column chromatography is used widely for many applications in chemistry, in biology and biochemistry, it is most often used to separate an aqueous mixture of proteins. The column is a tube filled with a stationary phase, also known as a resin. A solution called the mobile phase is then run through the column, either by gravity, or under pressure. The separation of the mixture into component proteins is based on the difference in how each protein interacts with the stationary phase vs. the mobile phase. For instance, a protein that interacts strongly with the stationary phase will take longer to move from the beginning to the end of the column, and subsequently, it will be eluted from the column later than another protein that interacts only weakly with the stationary phase. The liquid phase that comes off the end of the column can be collected in a series of fractions, which can then be used for further applications.
There are many variants of column chromatography that differ by the characteristic of the protein that is used to separate the mixture.
Ion exchange chromatography
Ion exchange chromatography separates proteins based on their overall charge. In this method, the stationary phase is either positively or negatively charged. The mobile phase is a solution that has a low salt concentration initially, but over the course of the run, the concentration of salt in the mobile phase is increased so that proteins that were initially strongly attracted to the stationary phase are eluted. For instance, if the stationary phase is negatively charged, proteins that have an overall negative charge will run through the column very quickly. However, proteins with an overall positive charge will interact with the stationary phase and therefore have a longer elution time. As the salt concentration increases, proteins that are weakly positively charged will elute first, followed by those that are strongly positively charged.
Size exclusion chromatography
As the name implies, size exclusion chromatography separates proteins based on their size and shape, and it is also called gel filtration. The resin in size exclusion chromatography is made of beads that have pores or channels running through them. The size of the beads and the channels can be used to fine-tune the separation achieved by this method. Large proteins will not be able to fit into the channels in the beads, and will therefore have the shortest elution times. However, smaller proteins will get caught up in the winding channels, and therefore, they will have a longer path through the column and longer elution times.
The outcome of size exclusion chromatography is the opposite of gel electrophoresis , which can be confusing! In size exclusion chromatography, large proteins move the fastest through the column, whereas in gel electrophoresis, they move through the gel the slowest , since the gel acts as a simple sieve.
Affinity chromatography
Affinity chromatography achieves separation based on the affinity of one or a small number of proteins for a particular substance affixed to the stationary phase. For instance, if ATP is affixed to the stationary phase, ATP-binding proteins would interact with it, and non-ATP-binding proteins would be quickly eluted. The ATP-binding proteins could be eluted by changing the mobile phase to a buffer containing “free” (not resin-bound) ATP. Another extremely common example is purification of proteins with a 6xHis epitope tag with a Ni2+ or Co2+ column. This method is known as immobilized metal affinity chromatography, or IMAC. Some of the N atoms in the histidine ring coordinate the metal ion, so that tagged proteins are bound to the column, and untagged proteins flow through. The tagged proteins can be eluted using a buffer containing imidazole, which has the same chemical structure as the ring in the amino acid histidine. Finally, if a column is derivatized with an antibody that binds a specific protein, that can also be considered affinity chromatography. However, when purifying a protein using an antibody, immunoprecipitation is often performed instead of chromatography.
Interpretation
In this experiment, researchers showed that they successfully developed a relatively simple protocol to separate extracellular vesicles (EVs) from soluble proteins in human serum. Obviously, extracellular vesicles are far larger than individual proteins, so the use of size exclusion chromatography was a good choice. To make the protocol as straightforward as possible, all the mobile phase eluted from the column was divided into only two “fractions,” eluate 1 and eluate 2. Through extensive protocol optimization, the researchers were able to establish a cutoff between eluates 1 and 2 that resulted in good separation between the EVs and the proteins from the serum. In panel a, they measured the amount of particles, i.e. EVs, in each eluate using a special instrument. The instrument determined particle concentration through a mixture of light scattering and Brownian motion. In panel b, they measured protein concentration of each eluate using a Bradford assay. As expected, EVs were far more abundant in eluate 1, since they are larger and would not get caught up in the channels of the resin. Eluate 2 contained most of the proteins not associated with the EVs, since they were much smaller.
Image Descriptions
Figure 1 image description:
Two column graphs. Graph a shows particle concentration. Eluate 1 contains ~3.4 x 10 12 ± 0.8 x 10 12 particles/ml and Eluate 2 contains ~0.25 x 10 12 ± 0.05 x 10 12 particles/ml. p = 0.04. Graph b shows protein concentration in μg/μl. Eluate 1 contains ~5 ± 1 μg/μl and Eluate 2 contains ~240 ± 20 μg/μl. p < 0.01. ↵
Thumbnail
"Chromatography column.PNG" ↗ by Klaas1978 is in the Public Domain ↗.
Description: Chromatography column with sample in three different stages.
Author
Katherine Mattaini, Tufts University
-
Guo, J., C. Wu, X. Lin, J. Zhou, J. Zhang, W. Zheng, T. Wang, and Y. Cui. 2021. Establishment of a simplified dichotomic size-exclusion chromatography for isolating extracellular vesicles toward clinical applications. Journal of Extracellular Vesicles 10:e12145. ↵


2.3: Assignment- History of Management
Preparation
Our text introduced us to many famous leaders and theorists in the subject of management practices. Each was notable for the different studies and practices or methods they analyzed and professed. Additionally, we examined current developments in management practices. In this assignment, your mission is to match key definitions, works, or concepts with the leaders or theorists who promoted them.
Your Task
Click the link to download one of the following formats:
Read the answer choices in the left column and the definition or associated terms, famous work, or key concepts in the center column. Complete the table by inserting the correct answer choice in the far right column. Hint: Each definition has only one correct method or person. Submit your completed table for grading.
Grading
Your assignment will be graded on the basis of number of correct matches out of a total of 22.
Contributors and Attributions
- Assignment: History of Management. Authored by : Betty Fitte and Lumen Learning. License : CC BY: Attribution


8.4: Mole-Mass Conversions
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Learning Objectives
- To convert quantities between mass units and mole units.
A previous Example stated that the mass of 2 mol of U is twice the molar mass of uranium. Such a straightforward exercise does not require any formal mathematical treatment. Many questions concerning mass are not so straightforward, however, and require some mathematical manipulations.
The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).
We also established that 1 mol of Al has a mass of 26.98 g (Example \(\PageIndex{1}\)). Stated mathematically,
1 mol Al = 26.98 g Al
We can divide both sides of this expression by either side to get one of two possible conversion factors:
\[\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber \]
The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”
Example \(\PageIndex{1}\)
What is the mass of 3.987 mol of Al?
Solution
The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the \(\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) conversion factor. We start with the given quantity and multiply by the conversion factor:
\(\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\)
Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives
\(\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}\)
Our final answer is expressed to four significant figures.
Exercise \(\PageIndex{1}\)
How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.)
- Answer
- \(\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}\)
Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure \(\PageIndex{1}\) is a chart for determining what conversion factor is needed, and Figure \(\PageIndex{2}\) is a flow diagram for the steps needed to perform a conversion.
Example \(\PageIndex{2}\)
A biochemist needs 0.00655 mol of bilirubin (C 33 H 36 N 4 O 6 ) for an experiment. How many grams of bilirubin will that be?
Solution
To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula:
| 33 C molar mass: | 33 × 12.01 g = | 396.33 g |
|---|---|---|
| 36 H molar mass: | 36 × 1.01 g = | 36.36 g |
| 4 N molar mass: | 4 × 14.01 g = | 56.04 g |
| 6 O molar mass: | 6 × 16.00 g = | 96.00 g |
| Total: | 584.73 g |
The molar mass of bilirubin is 584.73 g. Using the relationship
1 mol bilirubin = 584.73 g bilirubin
we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure \(\PageIndex{2}\):
\(\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}\)
The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin.
Exercise \(\PageIndex{2}\)
A chemist needs 457.8 g of KMnO 4 to make a solution. How many moles of KMnO 4 is that?
- Answer
- \(\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}\)
To Your Health: Minerals
For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDI s of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.
| Mineral | Male (age 19–30 y) | Female (age 19–30 y) | ||
|---|---|---|---|---|
| Ca | 1,000 mg | 0.025 mol | 1,000 mg | 0.025 mol |
| Cr | 35 µg | 6.7 × 10 −7 mol | 25 µg | 4.8 × 10 −7 mol |
| Cu | 900 µg | 1.4 × 10 −5 mol | 900 µg | 1.4 × 10 −5 mol |
| F | 4 mg | 2.1 × 10 −4 mol | 3 mg | 1.5 × 10 −4 mol |
| I | 150 µg | 1.2 × 10 −6 m ol | 150 µg | 1.2 × 10 −6 mol |
| Fe | 8 mg | 1.4 × 10 −4 mol | 18 mg | 3.2 × 10 −4 mol |
| K | 3,500 mg | 9.0 × 10 −2 mol | 3,500 mg | 9.0 × 10 −2 mol |
| Mg | 400 mg | 1.6 × 10 −2 mol | 310 mg | 1.3 × 10 −2 mol |
| Mn | 2.3 mg | 4.2 × 10 −5 mol | 1.8 mg | 3.3 × 10 −5 mol |
| Mo | 45 mg | 4.7 × 10 −7 mol | 45 mg | 4.7 × 10 −7 mol |
| Na | 2,400 mg | 1.0 × 10 −1 mol | 2,400 mg | 1.0 × 10 −1 mol |
| P | 700 mg | 2.3 × 10 −2 mol | 700 mg | 2.3 × 10 −2 mol |
| Se | 55 µg | 7.0 × 10 −7 mol | 55 µg | 7.0 × 10 −7 mol |
| Zn | 11 mg | 1.7 × 10 −4 mol | 8 mg | 1.2 × 10 −4 mol |
Table \(\PageIndex{1}\) illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women.
Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly.
Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich ).
Key Takeaway
- It is possible to convert between moles of material and mass of material.


24.3: Conclusion, and Further Resources
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This chapter on one-dimensional waves has barely scratched the surface of the extremely rich world of wave phenomena. I have only given you a passing glance at interference, and I have not said anything at all about diffraction, the Doppler effect, polarization, refraction.... Many of these things you will learn about in later courses, most likely when you encounter electromagnetic waves (which are non-mechanical, but described by the same mathematical equation).
Waves are such an intrinsically kinetic phenomenon that they are best appreciated by watching them in action, or, as a second-best alternative, through animations. A wonderful repository of such movies and animations is PHYSCLIPS at the University of New South Wales:
http://www.animations.physics.unsw.edu.au/waves-sound/oscillations/index.html
They also have a set of pages on the “physics of music” that I have already mentioned a couple of times. If you are interested in this topic, you should go spend some time there!
http://newt.phys.unsw.edu.au/music
Finally, closer to home, the fellows at PhET (University of Colorado), have this great interactive app to explore waves on a string:


8.2.1: Simulating Multi-step Experiments
Lesson
Let's simulate more complicated events.
Exercise \(\PageIndex{1}\): Notice and Wonder: Ski Business
What do you notice? What do you wonder?
Exercise \(\PageIndex{2}\): Alpine Zoom
Alpine Zoom is a ski business. To make money over spring break, they need it to snow at least 4 out of the 10 days. The weather forecast says there is a \(\frac{1}{3}\) chance it will snow each day during the break.
Use the applet to simulate the weather for 10 days of break to see if Alpine Zoom will make money.
- Describe a chance experiment that you could use to simulate whether it will snow on the first day of break.
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How could this chance experiment be used to determine whether Alpine Zoom will make money?
- In each trial, spin the spinner 10 times, and then record the number of 1’s that appeared in the row.
- The applet reports if the Alpine Zoom will make money or not in the last column.
- Click Next to get the spin button back to start the next simulation.
- Based on your simulations, estimate the probability that Alpine Zoom will make money.
 A playing piece starts on a white square and must advance 4 squares to the other side of the board within 5 turns to win the game.
 If the block color matches the color of the next square on the board, the playing piece moves onto it. If it does not match, the playing piece stays on its current square.
- Take turns playing the game until each person in your group has played the game twice.
- Use the results from all the games your group played to estimate the probability of winning Kiran’s game.
- Do you think your estimate of the probability of winning is a good estimate? How could it be improved?
Are you ready for more?
How would each of these changes, on its own, affect the probability of winning the game?
- Change the rules so that the playing piece must move 7 spaces within 8 moves.
- Change the board so that all the spaces are black.

Exercise \(\PageIndex{4}\): Simulation Nation
Match each situation to a simulation .
Situations:
- In a small lake, 25% of the fish are female. You capture a fish, record whether it is male or female, and toss the fish back into the lake. If you repeat this process 5 times, what is the probability that at least 3 of the 5 fish are female?
- Elena makes about 80% of her free throws. Based on her past successes with free throws, what is the probability that she will make exactly 4 out of 5 free throws in her next basketball game?
- On a game show, a contestant must pick one of three doors. In the first round, the winning door has a vacation. In the second round, the winning door has a car. What is the probability of winning a vacation and a car?
- Your choir is singing in 4 concerts. You and one of your classmates both learned the solo. Before each concert, there is an equal chance the choir director will select you or the other student to sing the solo. What is the probability that you will be selected to sing the solo in exactly 3 of the 4 concerts?
Simulations:
- Toss a standard number cube 2 times and record the outcomes. Repeat this process many times and find the proportion of the simulations in which a 1 or 2 appeared both times to estimate the probability.
- Make a spinner with four equal sections labeled 1, 2, 3, and 4. Spin the spinner 5 times and record the outcomes. Repeat this process many times and find the proportion of the simulations in which a 4 appears 3 or more times to estimate the probability.
- Toss a fair coin 4 times and record the outcomes. Repeat this process many times, and find the proportion of the simulations in which exactly 3 heads appear to estimate the probability.
- Place 8 blue chips and 2 red chips in a bag. Shake the bag, select a chip, record its color, and then return the chip to the bag. Repeat the process 4 more times to obtain a simulated outcome. Then repeat this process many times and find the proportion of the simulations in which exactly 4 blues are selected to estimate the probability.
Summary
The more complex a situation is, the harder it can be to estimate the probability of a particular event happening. Well-designed simulations are a way to estimate a probability in a complex situation, especially when it would be difficult or impossible to determine the probability from reasoning alone.
To design a good simulation, we need to know something about the situation. For example, if we want to estimate the probability that it will rain every day for the next three days, we could look up the weather forecast for the next three days. Here is a table showing a weather forecast:
| today (Tuesday) | Wednesday | Thursday | Friday | |
|---|---|---|---|---|
| probability of rain | \(0.2\) | \(0.4\) | \(0.5\) | \(0.9\) |
We can set up a simulation to estimate the probability of rain each day with three bags.
- In the first bag, we put 4 slips of paper that say “rain” and 6 that say “no rain.”
- In the second bag, we put 5 slips of paper that say “rain” and 5 that say “no rain.”
- In the third bag, we put 9 slips of paper that say “rain” and 1 that says “no rain.”
Then we can select one slip of paper from each bag and record whether or not there was rain on all three days. If we repeat this experiment many times, we can estimate the probability that there will be rain on all three days by dividing the number of times all three slips said “rain” by the total number of times we performed the simulation.
Practice
Exercise \(\PageIndex{5}\)
Priya’s cat is pregnant with a litter of 5 kittens. Each kitten has a 30% chance of being chocolate brown. Priya wants to know the probability that at least two of the kittens will be chocolate brown.
To simulate this, Priya put 3 white cubes and 7 green cubes in a bag. For each trial, Priya pulled out and returned a cube 5 times. Priya conducted 12 trials.
Here is a table with the results.
| trial number | outcome |
|---|---|
| 1 | ggggg |
| 2 | gggwg |
| 3 | wgwgw |
| 4 | gwggg |
| 5 | gggwg |
| 6 | wwggg |
| 7 | gwggg |
| 8 | ggwgw |
| 9 | wwwgg |
| 10 | ggggw |
| 11 | wggwg |
| 12 | gggwg |
- How many successful trials were there? Describe how you determined if a trial was a success.
- Based on this simulation, estimate the probability that exactly two kittens will be chocolate brown.
- Based on this simulation, estimate the probability that at least two kittens will be chocolate brown.
- Write and answer another question Priya could answer using this simulation.
- How could Priya increase the accuracy of the simulation?
Exercise \(\PageIndex{6}\)
A team has a 75% chance to win each of the 3 games they will play this week. Clare simulates the week of games by putting 4 pieces of paper in a bag, 3 labeled “win” and 1 labeled “lose.” She draws a paper, writes down the result, then replaces the paper and repeats the process two more times. Clare gets the result: win, win, lose. What can Clare do to estimate the probability the team will win at least 2 games?
Exercise \(\PageIndex{7}\)
- List the sample space for selecting a letter a random from the word “PINEAPPLE.”
- A letter is randomly selected from the word “PINEAPPLE.” Which is more likely, selecting “E” or selecting “P?” Explain your reasoning.
(From Unit 8.1.5)
Exercise \(\PageIndex{8}\)
On a graph of side length of a square vs. its perimeter, a few points are plotted.
1. Add at least two more ordered pairs to the graph.
2. Is there a proportional relationship between the perimeter and side length? Explain how you know.
(From Unit 2.4.2)


15.5E: Other Fungi Involved in Respiratory Disease
- Compare and contrast the various forms of sporotrichosis: cutaneous/skin, pulmonary and disseminated sporotrichosis
Sporotrichosis (also known as “Rose gardener’s disease”) is caused by the infection of the fungus Sporothrix schenckii . This fungal disease usually affects the skin, although other rare forms can affect the lungs, joints, bones, and even the brain. Because roses can spread the disease, it is one of a few diseases referred to as rose-thorn or rose-gardeners’ disease.
Because S. schenckii is naturally found in soil, hay, sphagnum moss, and plants, it usually affects farmers, gardeners, and agricultural workers. It enters through small cuts and abrasions in the skin to cause the infection. In cases of sporotrichosis affecting the lungs, the fungal spores enter through the respiratory pathways. Sporotrichosis can also be acquired from handling cats with the disease; it is an occupational hazard for veterinarians.
Sporotrichosis progresses slowly – the first symptoms may appear from one to 12 weeks (average three weeks) after the initial exposure to the fungus. Serious complications can also develop in patients who have a compromised immune system.
Forms and symptoms of sporotrichosis include: cutaneous or skin sporotrichosis; pulmonary sporotrichosis; and disseminated sporotrichosis.
Cutaneous or skin sporotrichosis: This is the most common form of the disease. Symptoms include nodular lesions or bumps in the skin at the point of entry and also along lymph nodes and vessels. The lesion starts off small and painless, and ranges in color from pink to purple. Left untreated, the lesion becomes larger and looks similar to a boil. More lesions will appear, until a chronic ulcer develops. Usually, cutaneous sporotrichosis lesions occur in the finger, hand, and arm.
Pulmonary sporotrichosis: This rare form of the disease occurs when S. schenckii spores are inhaled. Symptoms include productive coughing, nodules and cavitations of the lungs, fibrosis, and swollen hilar lymph nodes. Patients with this form of sporotrichosis are susceptible to developing tuberculosis and pneumonia.
Disseminated sporotrichosis: this occurs when the infection spreads from the primary site to secondary sites in the body and develops into a rare and critical form. The infection can spread to joints and bones (called osteoarticular sporotrichosis) as well as the central nervous system and the brain (called sporotrichosis meningitis). The symptoms include weight loss, anorexia, and the appearance of bony lesions.
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- fungal. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/fungal . License : CC BY-SA: Attribution-ShareAlike
- Boundless. Provided by : Boundless Learning. Located at : www.boundless.com//microbiolo.../blastomycosis . License : CC BY-SA: Attribution-ShareAlike
- inhalation. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/inhalation . License : CC BY-SA: Attribution-ShareAlike
- Chest%20X-ray%20acute%20pulmonary%20histoplasmosis%20PHIL%203954. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Chest_X-ray_acute_pulmonary_histoplasmosis_PHIL_3954.jpg . License : Public Domain: No Known Copyright
- Histoplasmosis%20capsulatum. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Histoplasmosis_capsulatum.jpg . License : Public Domain: No Known Copyright
- Coccidioidomycosis 01. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...mycosis_01.jpg . License : Public Domain: No Known Copyright
- Pneumocystis. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Pneumocystis.jpg . License : Public Domain: No Known Copyright
- Blastomycosis cropped. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Bl...is_cropped.JPG . License : CC BY-SA: Attribution-ShareAlike
- Sporotrichosis. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sporotrichosis . License : CC BY-SA: Attribution-ShareAlike
- fungus. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/fungus . License : CC BY-SA: Attribution-ShareAlike
- Sporotrichosis. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/Sporotrichosis . License : CC BY-SA: Attribution-ShareAlike
- Chest%20X-ray%20acute%20pulmonary%20histoplasmosis%20PHIL%203954. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Chest_X-ray_acute_pulmonary_histoplasmosis_PHIL_3954.jpg . License : Public Domain: No Known Copyright
- Histoplasmosis%20capsulatum. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Histoplasmosis_capsulatum.jpg . License : Public Domain: No Known Copyright
- Coccidioidomycosis 01. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...mycosis_01.jpg . License : Public Domain: No Known Copyright
- Pneumocystis. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Pneumocystis.jpg . License : Public Domain: No Known Copyright
- Blastomycosis cropped. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Bl...is_cropped.JPG . License : CC BY-SA: Attribution-ShareAlike
- Feline sporotrichosis 4. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Fe...richosis_4.jpg . License : CC BY-SA: Attribution-ShareAlike
Key Points
- In cases of sporotrichosis affecting the lungs, the fungal spores enter through the respiratory pathways.
- Sporotrichosis progresses slowly – the first symptoms may appear from one to 12 weeks (average three weeks) after the initial exposure to the fungus.
- Forms and symptoms of sporotrichosis include: cutaneous or skin sporotrichosis; pulmonary sporotrichosis; and disseminated sporotrichosis.
Key Terms
- fungus : Any member of the kingdom Fungi; a eukaryotic organism typically having chitin cell walls but no chlorophyll or plastids. Fungi may be unicellular or multicellular.
- Sporotrichosis : A disease caused by the infection of the fungus Sporothrix schenckii, usually affecting the skin, although other rare forms can affect the lungs, joints, bones, and even the brain. Because roses can spread the disease, it is often referred to as rose-thorn or rose-gardeners’ disease.


3.1: Groups of Small Order
Let's start exploring groups in order of increasing, um, order. Before we do this, it will be helpful to introduce the notion of a group table (also known as a Cayley table 1 ) for a finite group. Given a finite group \(G\text{,}\) list its elements in some fixed order, say, \(a_1, a_2, \ldots, a_n\text{,}\) and then construct its group table by creating an array with exactly one row and exactly one column corresponding to each group element. We then put in the row \(i\) and column \(j\) the element \(a_ia_j\) of \(G\text{.}\) Note that a single group can have group tables that look different from one another, since reordering a group's elements will change its table.
Example \(\PageIndex{1}\)
Consider the group \(\mathbb{Z}_4\) under addition modulo \(4\). Ordering the elements of \(\mathbb{Z}_4\) as \(1,2,3,4\text{,}\) we have the following group table for \(\langle \mathbb{Z}_4,+\rangle\text{:}\)
| \(+\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(0\) | \(0\) | \(1\) | \(2\) | \(3\) |
| \(1\) | \(1\) | \(2\) | \(3\) | \(0\) |
| \(2\) | \(2\) | \(3\) | \(0\) | \(1\) |
| \(3\) | \(3\) | \(0\) | \(1\) | \(2\) |
Now, clearly, there is no group of order \(0\) (do you see why?). Is there a group of order \(1\)? Well, suppose \(\langle G,*\rangle\) is such a group. Since \(G\) must contain an identity element \(e\text{,}\) we must have \(G=\{e\}\text{,}\) and since \(e\) is \(G\)'s identity element, we must have \(e*e=e\text{.}\) Clearly, in this case, the three group axioms hold. So \(G\) is a valid group, without much going on in it.
Definition: Trivial Group
If \(G\) is a group with \(|G|=1\text{,}\) then \(G\) is called the trivial group .
Remark
A good question to ask here is why it's called “the” trivial group, rather than “a” trivial group. Indeed, there are infinitely many groups of order 1! (Do you see why?) But it turns out that all of these groups are structurally the same. Hence mathematicians end up thinking of them as various instantiations of one group, rather than separate groups. We will discuss this in more depth shortly, when we introduce the idea of isomorphism .
Next suppose that group \(\langle G, \;\cdot \rangle\) has order 2. Then \(G\) must contain an identity element, \(e\text{,}\) and a non-identity element, \(a\text{.}\) Since \(e\) is its own inverse, and inverses are unique, \(a\) must be its own inverse as well. So \(G\) must have the following table.
| \(*\) | \(e\) | \(a\) |
|---|---|---|
| \(e\) | \(e\) | \(a\) |
| \(a\) | \(a\) | \(e\) |
It is straightforward to show that such a structure does satisfy all the group axioms (the only one we really need to check is associativity).
Now, what if group \(\langle G, \;\cdot \rangle\) has order \(3\)? Note that you can't have any entry appear more than once in the same row or same column (excluding of course the labels outside the grid we're filling in), given Theorem \(2.5.3\). Is there only one way of filling in the table for a group of order \(3\)? (Hint: consider what element must be the second row, third column entry.)
| \(*\) | \(e\) | \(a\) | \(b\) |
|---|---|---|---|
| \(e\) | \(\) | \(\) | \(\) |
| \(a\) | \(\) | \(\) | \(\) |
| \(b\) | \(\) | \(\) | \(\) |
Finally, what if group \(\langle G, \;\cdot \rangle\) has order \(4\)? It turns out in this case there are two valid ways of filling in a group table!
What we have been doing here is really getting into the idea of the structure of groups, and when we can consider groups to be essentially “the same” or fundamentally “different.” We approach this more formally via the concepts of homomorphism and isomorphism.


4.10: Reactions with cyclic transition state
- Learn examples of reactions that involve five- or six-member transition state, including cyclonic hemiacetal formation of monosaccharides, Diels-Alter reactions producing six-membered cyclic products, and decarboxylation of \(\beta\)-keto acids.
Intramolecular reactions happen if the two reacting groups are on the same molecule and can come to a bonding distance through a five- or six-member cyclic transition state. Some examples of it are described in the next sections.
Cyclic hemiacetal formation of monosaccharides
Monosaccharides, like glucose, fructose, galactose, etc., have a \(\ce{C=O}\)-group on one \(\ce{C}\) and \(\ce{-OH}\)-group on every other \(\ce{C}\). \(\ce{-OH}\)-group can add to \(\ce{C=O}\)-group forming a hemiacetal. Monosaccharides exist primarily in a five- or six-membered hemiacetal form because one of their \(\ce{-OH}\)-group can form a five- or six-membered transition state for the reaction, as shown in Figure \(\PageIndex{1}\) for the case of D-glucose and D-fructose.
Diels–Alder reaction
A conjugated diene, e.g., butadiene, and an alkene, e.g., ethene, make a cyclic six-member transition state. They react by the Diels-Alder reaction mechanism and produce a six-member cyclic product. This reaction can be intermolecular, e.g., between butadiene and then, or intramolecular, e.g., in the biosynthesis of antibiotic lovastatin, illustrated in Figure \(\PageIndex{2}\).

This reaction requires high temperatures, such as in the thermal decomposition process. However, if there is a second carbonyl (\(\ce{C=O}\))) group \(\beta\}\) to the \(\ce{-COOH}\) group, it can easily acquire a six-member transition state and decarboxylate at moderate temperatures, as illustrated in Figure \(\PageIndex{3}\).
Acetoacetic acid and its reduced product \(\beta\)-hydroxybutyric acid, shown below, are produced in the liver as a result of the metabolism of fatty acids and some amino acids.
Acetoacetic acid and \(\beta\)-hydroxybutyric acid are called ketone bodies . Their concentration in the blood of healthy persons is about 0.01 mmol/L but in persons suffering from starvation or diabetes mellitus may be up to 500 times higher.
Carboxylic acids exist as carboxylate anions under physiological conditions. Decarboxylation of the \(\beta\)-keto carboxylates happens spontaneously under physiological conditions. For example, acetoacetate decarboxylates and produces carbon dioxide and acetone, as illustrated in Figure \(\PageIndex{4}\).
Carbon dioxide leaves under moderate conditions in this case because the anion left behind is in resonance with the \(\beta\)-\(\ce{C=O}\) group. The body does not metabolize acetone but exhales through the lungs. Acetone is responsible for its characteristic sweet smell in the breath of people with swear diabetes.


8: Cosmic Topology
Cosmic topology can be described as the effort to determine the shape of our universe through observational techniques. In this chapter we discuss two programs of research in cosmic topology: the cosmic crystallography method and the circles-in-the-sky method. Both programs search for topology by assuming the universe is finite in volume without boundary. The chapter begins with a discussion of three-dimensional geometry and some \(3\)-manifolds that have been given consideration as models for the shape of our universe.
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- 8.1: Three-Dimensional Geometry and 3-Manifolds
- Euclidean geometry is the geometry of our experience in three dimensions. Planes look like infinite tabletops, lines in space are Euclidean straight lines. Any planar slice of 3-space inherits two-dimensional Euclidean geometry. The Poincaré disk model of hyperbolic geometry may also be extended to three dimensions. Three-dimensional elliptic geometry is derived from the fact that the 3-sphere consists of all points in 4-dimensional space one unit from the origin.
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- 8.2: Cosmic Crystallography
- Imagine once again that we are two-dimensional beings living in a two-dimensional universe. In fact, suppose we are living in the torus in Figure 8.2.1 at point E (for Earth). Our world is homogeneous and isotropic, and adheres to Euclidean geometry. Our lines of sight follow Euclidean lines. If we can see far enough, we ought to be able to see an object, say G (for galaxy), in different directions. Three different lines of sight are given in the figure.
-
- 8.3: Circles in the Sky
- Immediately after the big bang, the universe was so hot that the usual constituents of matter could not form. Photons could not move freely in space, as they were constantly bumping into free electrons. Eventually, about 350,000 years after the big bang, the universe had expanded and cooled to the point that light could travel unimpeded. This free radiation is called the cosmic microwave background (CMB) radiation, and much of it is still travelling today.
-
- 8.4: Our Universe
- Our universe appears to be homogeneous and isotropic. The presence of cosmic microwave background radiation is evidence of this: it is coming to us from every direction with more or less constant temperature. The assumptions of isotropy and homogeneity are remarkably fruitful when one approaches the geometry and topology of the universe from a mathematical point of view. The mathematical point of view gives us our candidate geometries.


Chapter 27: The Reproductive System
- Describe the anatomy of the male and female reproductive systems, including their accessory structures
- Explain the role of hypothalamic and pituitary hormones in male and female reproductive function
- Trace the path of a sperm cell from its initial production through fertilization of an oocyte
- Explain the events in the ovary prior to ovulation
- Describe the development and maturation of the sex organs and the emergence of secondary sex characteristics during puberty
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- 27.2: Anatomy and Physiology of the Male Reproductive System
- The penis is the male organ of copulation.  Testosterone regulates and maintains the sex organs and sex drive, and induces the physical changes of puberty. Interplay between the testes and the endocrine system precisely control the production of testosterone with a negative feedback loop.
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- 27.3: Anatomy and Physiology of the Female Reproductive System
- The female reproductive system functions to produce gametes and reproductive hormones, just like the male reproductive system; however, it also has the additional task of supporting the developing fetus and delivering it to the outside world. Unlike its male counterpart, the female reproductive system is located primarily inside the pelvic cavity (Figure 27.2.1). Recall that the ovaries are the female gonads. The gamete they produce is called an oocyte.
-
- 27.4: Development of the Male and Female Reproductive Systems
- The development of the reproductive systems begins soon after fertilization of the egg, with primordial gonads beginning to develop approximately one month after conception. Reproductive development continues in utero, but there is little change in the reproductive system between infancy and puberty.
Thumbnail: Monoamniotic twins. Häggström, Mikael (2014). " Medical gallery of Mikael Häggström 2014 ". WikiJournal of Medicine 1 (2). DOI:10.15347/wjm/2014.008. ISSN 2002-4436. Medicine 1 (2). DOI : 10.15347/wjm/2014.008 . ISSN 2002-4436 . (Public Domain; Mikael Häggström via Wikipedia )


2: Knots and Numbers
- Determine and contrast several ways to notate (tabulate) a knot using its crossings.
- Use numerical invariants for knots including the unknotting, bridge, and crossing numbers, to investigate relationships among classes of knots.
Studying rational tangles was a way to focus in a limited fashion on how crossings interact with one another to build intricate local structures that define a knot. But as an invariant for knots, the tangle number isn't perfect: it's most useful for rational knots, and even then, it can be challenging to rearrange a knot diagram into a twist-form rational tangle.
What we'd like instead are more global invariants that work for knots, invariants that capture the whole structure of the topology without relying upon making a specific set of choices along the way. This will come at the cost of needing invariants capable of conveying more algebraic information than a single rational number does: polynomials on one hand, and algebraic groups on the other.
References
- Adams, C. C. (2004). The Knot Book: an elementary introduction to the mathematical theory of knots. American Mathematical Society, ISBN 0-8218-3678-1. Chapters 2 and 3.
- Austin, D. (2016). Knot quandaries quelled by quandles. American Mathematical Society Feature Column, accessed at http://www.ams.org/publicoutreach/feature-column/fc-2016-03 .
- Portnoy, N. and Mattman, T. (Undated). Knot Theory for Preservice and Practicing Secondary Mathematics Teachers. Accessed at http://www.csuchico.edu/math/mattman/NSF.html .
- Rolfsen, D. (1990). Knots and Li nks. Corrected reprint of the 1976 original. Mathematics Lecture Series (7). American Mathematical Society. Chapter 3


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3.S: Chemical Bonding and Nomenclature (Summary)
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Last updated
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Save as PDF
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In
covalent bonding
, electrons are
shared
between atoms. In
ionic bonding
, electrons are transferred from one atom to another. Compounds that are formed using only covalent bonds are termed
molecular compounds
.
-
The
outermost
electron level in any atom is referred to as the
valence shell
. The electron configuration of the valence shell of an atom can be shown graphically using a
Lewis diagram
(or
electron-dot structure
). The arrangement of “dots” around the chemical symbol for the element are shown singly up through four electrons, and then paired until eight electrons are present.
-
To form ions from individual elements, electrons are added or subtracted from the valence shell in order to completely fill the shell with eight electrons (the
octet rule
). The charge on the ion reflects the electrons added or removed. Representative elements through Group
IIIA
will lose electrons to form cations, while those in groups
IVA
–
VIIA
will gain electrons and form anions.
-
A
covalent bond
is constructed in a Lewis diagram by pairing a set of unpaired electrons from two different atoms. For the purposes of the “octet rule”, a pair of shared electrons is counted as two electrons for each atom. Multiple covalent bonds (
double bonds
and
triple bonds
) are used, if necessary, to give each bonded atom a full octet (except, of course, for helium and hydrogen). When two or more atoms are bonded together utilizing covalent bonds, the compound is referred to as a
molecule
.
-
As a rule of thumb,
ionic
bonds will be formed whenever the compound contains a
metal
.
Covalent
bonding will be observed in compounds containing only
semimetals
or
nonmetals
.
-
Groups of covalently bonded semimetals or nonmetals which are charged are called
polyatomic ions
. Common examples include sulfate dianion, nitrate anion, phosphate trianion, etc. These polyatomic ions are commonly paired with metals forming ionic compounds.
-
Many (but not all) polyatomic ions can be drawn in two or more equivalent Lewis representations. These are called
resonance forms
of the ion. The actual electronic structure of the ion is a combination of these Lewis structures and is called the
resonance hybrid
.
-
The
electronegativity
of an element is a measure of the tendency of that element to attract electrons towards itself. Electronegativities range from 0.6 to 4.0, with fluorine as the most electronegative element (a value of 4.0). The general trend in the periodic table is for electronegativity to increase from the lower left-hand corner (Fr) to the upper right-hand corner (F).
-
Covalent bonds formed between atoms with different electronegativity will be
polarized
with the greatest electron density localized around the most electronegative atom. The effect of electronegativity on electron distribution within a molecule can be shown using a computer-calculated
electrostatic potential map
where colors are used to represent electron density.
-
Elements in periods 3 – 7 can accommodate more that eight electrons in there valence shells. This phenomena is called
valence shell expansion
and molecules involving these elements may have 10 – 14 valence electrons in properly drawn Lewis diagrams. Exceptions to the “octet rule” also exist where the valence shell contains less than eight electrons, or contains unpaired electrons.
-
When naming simple, binary
ionic
compounds, the cation is named first using the name of the element, followed by the anion, where the suffix
ide
is added to the root name of the element. Multipliers are
not used
. For transition metals in which the metal can assume a variety of oxidation states (different positive charges), the charge of the metal ion is shown in the name using Roman numerals, in parenthesis, following the name of the element (i.e., iron (III) chloride).
-
When naming simple binary
molecular
compounds (compounds containing only covalent bonds) the
least electronegative
element is (generally) named first, followed by the second element, where the suffix
ide
is again added to the root name of the element. In molecular compounds multipliers are used to indicate the number of each atom present (mono-, di-, tri-, tetra-, etc.) with the exception that
mono
is not used for the first element in the compound.


10.6.6: Class
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- Miliann Kang, Donovan Lessard, Laura Heston, and Sonny Nordmarken
- University of Massachusetts Amherst via UMass Amherst Libraries
- Miliann Kang, Donovan Lessard, Laura Heston, and Sonny Nordmarken
- University of Massachusett Amherst
Socio-economic class differences are particularly hidden in the US context. Part of this can be explained by the ideology of the American Dream. According to a popular belief in meritocracy , anyone who works hard enough will succeed, and those who do not succeed must not have worked hard enough. There is a logical error in this form of reasoning, which does not explain the following two scenarios: What about people who do not work very hard at all and still succeed? What about those who work exceptionally hard and never succeed? Part of this, of course, is about how we define success. Succeeding at the American Dream means something akin to having a great job, making a lot of money, and owning a car, a house, and all the most-recent gadgets. These are markers of material, that is, economic, wealth. Wealth is not only captured in personal income, but other assets as well (house, car, stocks, inheritances), not all of which are necessarily earned by hard work alone, but can come from inheritance, marriage, or luck.
Though rich/poor may be the binary associated with class, most people in the US context (no matter how much wealth they have) consider themselves “middle-class.” (Pew Research Center, 2010). The label “middle-class” represents more than what people have in their bank accounts—it reflects a political ideology. When politicians run for election or argue over legislation they often employ the term “middle-class” to stand in for “average,” “tax-paying,” “morally upstanding” constituents and argue for their collective voice and prosperity. Rhetorically, the “middle class” is not compared to the super rich (since, in the US, you can never be too rich or too thin), but rather the poor. So, when people talk about the middle class they are also often implying that they are NOT those “deviant,” “tax-swindling,” “immoral,” poor people. This may seem harsh, but this is truly how the poor are represented in news media (Mantsios, 2007). If this still seems far-fetched, just replace with phrase “the poor” with “welfare recipients.” Welfare recipients are often faceless but framed as undeserving of assistance since they are assumed to be cheating the system, addicted to alcohol or drugs, and have only themselves to blame for their poverty (Mantsios, 2007). Welfare recipients are the implied counterparts to the middle-class everymen that populate political speeches and radio rants. Thus, in the United States, socioeconomic class has been constructed as a binary between the middle-class and the poor.
Furthermore, these class-based categories also carry racial and sexual meanings, as the “welfare queen” stereotype conjures images of poor, black, sexually-promiscuous women, contrary to the fact that white women as a group are the largest recipients of welfare. Fred Block and colleagues (2006) discuss how these stereotypes about the poor are written into American poverty policies. For instance, in 1996, President Bill Clinton passed the Personal Responsibility/Work Opportunities Reconciliation Act (PRWORA), which fundamentally rewrote prior US welfare policy. This act limits lifetime receipt of welfare to a maximum of 60 months, or 5 years, and requires that able-bodied recipients work or job-train for low-skill jobs while receiving checks. Under PRWORA, recent immigrants cannot receive welfare for their first five years of legal residence, and undocumented immigrants can never receive welfare benefits (Block et al. 2006). These restrictions are based on the assumption that welfare recipients are ultimately cheating the American taxpayer and looking for a free ride. In spite of these changes, most people still believe that being on government assistance means a lifetime of free money. Media contempt for welfare recipients is accomplished by not humanizing the experience of poverty. People experiencing poverty can face tough choices; for instance, working more hours or getting a slightly better paying job can cause one to fail the “means test” (an income level above which people are ineligible for welfare benefits) for food stamps or Medicaid. The poor are increasingly forced to decide between paying for rent versus food and other bills, as the cost of living has risen dramatically in the past few decades while working-class wages have not risen comparably.
The SPENT game captures and humanizes this process of making tough decisions on a tight budget. Try it out and see how you fare: http://playspent.org/ .
However, class issues are not only about income differences. Cultural capital is a term coined by the late sociologist Pierre Bourdieu (1984) to address non-monetary class differences such as tastes in food and music or knowledge of high culture. Bourdieu explained that even when a formerly poor individual experiences economic mobility and becomes middle-class, there are still markers of her former status in the way she carries herself and the things she knows. We see many examples of this in popular films. When someone goes from rags to riches, they often use the wrong utensils at a dinner party, call something by the wrong name, cannot tell the difference between a Chardonnay and a Merlot (wines ), or spend their money in a showy way. Thus, someone can have high cultural capital and not be wealthy, or have low cultural capital and be a millionaire. For instance, in the popular (and very campy) movie Showgirls (Verhoeven, 1995), the main character, Nomi Malone, goes from homeless and unemployed to a well paid Las Vegas showgirl at record speed. Along the way, she buys an expensive Versace dress and brags about it. Unfortunately, she reveals her lack of cultural capital, and thus her former status as poor, by mispronouncing the brand (saying ‘Verse-ACE’ instead of ‘Vers-a-Chee’) and is humiliated by some rather mean bystanders. In sum, the concept of cultural capital highlights the ways in which social class is not just about wealth and income, but that social classes develop class cultures.


1.1: Cellular Foundations of Biochemistry
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Understand the Role of Water and Ions in Biochemistry
- Importance of water’s unique properties in biochemical systems.
- pH and ion gradients influence cellular function and metabolism.
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Characterize the Structure and Function of Biomolecules
- The four major classes of biomolecules: proteins, nucleic acids, lipids, and carbohydrates.
- Macromolecular structure relates to function in metabolism and enzyme activity.
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Explore Metabolism and Enzyme Function
- Catabolic and anabolic reactions in metabolism.
- Role of enzymes as biological catalysts and how they regulate metabolic pathways.
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Examine the Biochemical Environment of the Cell
- Crowded nature of the cytoplasm and its effects on protein stability and enzyme function.
By achieving these learning goals, students will develop a solid understanding of the cellular foundations that underlie all biochemical processes, preparing them for more advanced topics in biochemistry and related disciplines.
prompt: Write a series of learning goals for the following web page. The page is designed for junior and senior biochemistry majors.
Introduction
Biochemistry explores the chemical processes that sustain life. This text focuses on four key biomolecule classes: lipids, proteins, nucleic acids, and carbohydrates. It examines their roles in biochemical reactions, metabolism, and cellular function.
Cells function as chemical factories. They import, synthesize, use, and degrade biomolecules to sustain life. Metabolism includes both the energy-releasing breakdown of molecules ( catabolism ) and the energy-consuming synthesis of new compounds ( anabolism ). These processes support growth, reproduction, and adaptation to environmental changes.
Enzymes are essential to metabolism. They act as catalysts that speed up chemical reactions. Enzymes also help regulate metabolic pathways by responding to conditions inside the cell and external signals. Their structure determines specificity, either through a lock-and-key model or an induced-fit model of substrate binding.
Metabolic reactions occur in pathways , where one molecule is gradually converted into another. Each step is often controlled by a specific enzyme. This organization allows efficient energy production, biosynthesis, and waste elimination.
Biological membranes play a critical role in biochemistry. The lipid bilayer and membrane proteins regulate molecular transport. Passive and active transport mechanisms control the movement of nutrients, ions, and waste.
pH and Ion Regulation in Cells
Cellular function depends on the regulation of pH and ion concentrations. The cytosol typically has a pH between 7.0 and 7.4, but some organelles maintain distinct pH environments. For example, lysosomes have a pH of about 4.8, which is necessary for enzymatic degradation. The mitochondria create a pH gradient across the inner membrane, which drives ATP synthesis.
Ion gradients across membranes are critical for transport, energy production, and cell signaling. The concentration of potassium ions is higher inside the cell, while sodium, chloride, and calcium ions are more concentrated outside. These gradients are maintained by ion transporters and require ATP hydrolysis.
| Ion | Inside (mM) | Outside (mM) |
|---|---|---|
| Na + | 140 | 5 |
| K + | 12 | 140 |
| Cl- | 4 | 15 |
| Ca 2 + | 1 uM | 2 |
The cell is an amazingly crowded place
In chemistry labs, we typically work with dilute solutions of solutes in a solvent. However, the cellular environment is very different. While the body is about 68% water, the concentration of water varies within cells due to the presence of densely packed biomolecules. Proteins, carbohydrates, and other solutes fill much of the available space, often leaving less room between molecules than the size of a single protein.
This crowded environment has several important effects. Studies show that high intracellular concentrations help stabilize proteins, keeping them in their correctly folded, native state. Crowding also slows the diffusion of molecules, which affects reaction rates and the localization of cellular functions. As a result, biochemical processes are often restricted to specific regions within the cytoplasm, creating specialized microenvironments that enhance efficiency.


4.5: Getting Ready for Chapter 4
Chapter 3 gave us an introduction to the structure and composition of the atmosphere, the gaseous envelope that supports life as we know it. We've found that some gases act to absorb heat released from the surface, creating a greenhouse environment and making earth habitable. But the concentration of greenhouse gases is on the rise, and human activities are responsible for much of it. Other gases shield the earth from receiving too much ultra-violet light. Their concentrations have been affected by human activities as well.
Ahead in Chapter 4, Energy and Radiation , we'll examine the sources of heat that not only warm the atmosphere, but drive most of the environmental processes acting at the surface of the earth. We'll gain insight into how exogenic sources of energy ultimately heat the air, change the state of water, and warm the surface of the earth. Using these new insights, we'll be able to explain the geographical variation of radiation and energy. A peek into future changes in the earth's heat balance as a result of human-induced climate change will conclude the chapter.
What you should already know...
The content of Chapter 4 Energy and Radiation is some of the most important material for understanding the functioning of the earth system. Chapter 4 delves into the nature of energy and heat. Much of the chapter discusses energy received from the Sun, the ultimate source of energy to drive most environmental processes acting at the surface of the Earth. The knowledge gained in this chapter will serves as the foundation for understanding the geography of temperature, phase changes of water, circulation of the air, and much more.
Before beginning you should be comfortable with the material covered in Chapter 2 The Earth System . Have a good understanding of the Earth in space, and how the orientation, rotation, and revolution of the Earth around the Sun affects sunlight reaching the earth system.
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The Earth is closest to the Sun in
- January
- March
- September
- July
-
The Earth's axis of rotation is tilted
- 23.5 degrees from perpendicular to the plane of the ecliptic
- 66.5 degrees from perpendicular to the plane of the ecliptic
- 0 degrees from perpendicular to the plane of the ecliptic
- 90 from perpendicular to the plane of the ecliptic
-
The subsolar point on June 21st is
- 23.5 degrees north latitude
- 0 degrees (the equator)
- 23.5 degrees south latitude
- 66.5 degrees north latitude.
-
There is 12 hours of day and 12 hours of night
- on June 21
- on December 22
- every day of the year at the equator
- none of the above.
-
Tangent rays
- strike the north and south poles on the solstices
- strike the equator only on the solstice.
- strike the north and south poles on the equinox
- only strike the earth on the solstice
-
24 hours
- of daylight occurs at the South Pole on June 21
- of daylight occurs at the South Pole on March 21
- of darkness occurs at the South Pole on June 21
- of darkness does not occur anywhere
-
The subtropical latitude zone is located
- 10 degrees N - 10 degrees S
- 10 degrees - 25 degrees N and S latitude
- 25 degrees - 35 degrees N and S latitude
- 35 degrees - 55 degrees N and S latitude
-
Most greenhouse gases
- are good absorbers of solar radiation
- are good absorbers of radiation emitted by the Earth
- are known as "uniform absorbers"
- are all the above
-
The largest sources of particulates entering the atmosphere is from
- salt from sea spray and bursting bubbles.
- windblown dust
- volcanoes
- biomass burning
-
Solar radiation is considered
- an exogenic source of energy
- an endogenic source of energy
- an adiabatic source of energy
- a diabatic source of energy
- Answer
-
- A
- A
- A
- C
- C
- C
- C
- B
- A
- A
About your score ....
If you scored 80% or above, Great! ... start reading the chapter.
If you scored 70% to 80% you should consider reviewing the previous material.
If you scored less than 70% you should consider reviewing the previous material and seeking help from your instructor.


15.17B: Trichomoniasis
- Outline the causes and symptoms associated with trichomoniasis
Trichomoniasis, sometimes referred to as “trich”, is a common cause of vaginitis. It is a sexually transmitted disease, and is caused by the single-celled protozoan parasite Trichomonas vaginalis producing mechanical stress on host cells and then ingesting cell fragments after cell deat. Trichomoniasis is primarily an infection of the urogenital tract; the most common site of infection is the urethra and the vagina in women. Typically, only women experience symptoms associated with Trichomonas infection.
Symptoms include inflammation of the cervix (cervicitis), urethra (urethritis), and vagina (vaginitis) which produces an itching or burning sensation. Discomfort may increase during intercourse and urination. There may also be a yellow-green, itchy, frothy, foul-smelling (“fishy” smell) vaginal discharge. In rare cases, lower abdominal pain can occur. Symptoms usually appear in women within 5 to 28 days of exposure. In many cases, men may hold the parasite for some years without any signs (dormant). Some sexual health specialists have stated that the condition can probably be carried in the vagina for years, despite standard tests being negative. While symptoms are most common in women, some men may temporarily exhibit symptoms such as an irritation inside the penis, mild discharge or slight burning after urination or ejaculation. Trichomoniasis is diagnosed by visually observing the trichomonads via a microscope. In women, the examiner collects the specimen during a pelvic examination by inserting a speculum into the vagina and then using a cotton-tipped applicator to collect the sample. The sample is then placed onto a microscopic slide and sent to a laboratory to be analyzed.
Key Points
- Trichomoniasis is a sexually transmitted disease, and is caused by the single-celled protozoan parasite Trichomonas vaginalis producing mechanical stress on host cells and then ingesting cell fragments after cell death.
- Symptoms usually appear in women within 5 to 28 days of exposure. In many cases, men may hold the parasite for some years without any signs (dormant).
- Trichomoniasis is diagnosed by visually observing the trichomonads via a microscope.
Key Terms
- trichomoniasis : A common sexually transmitted disease caused by the parasite Trichomonas vaginalis and infecting the urinary tract or vagina.
- vaginitis : Inflammation of the vagina.
- trichomonads : flagellate protozoa of the genus Trichomonas.


1.3: Determining the Equation of a Line
Learning Objectives
In this section, you will learn to:
- Find an equation of a line if a point and the slope are given.
- Find an equation of a line if two points are given.
Prerequisite Skills
Before you get started, take this prerequisite quiz.
1. Simplify each expression:
a. \(3(2x-5)\)
b. \(-7(4x-2)\)
- Click here to check your answer
-
a. \(6x-15\)
b. \(-28x+14\)
If you missed any part of this problem, review here . (Note that this will open a different textbook in a new window.)
2. Find the slope of the line containing the points (1, 4) and (5, -2).
- Click here to check your answer
-
\(-\dfrac{3}{2}\)
If you missed this problem, review Section 1.2 . (Note that this will open in a new window.)
3. Find the slope of the line containing the points (-2, 5) and (10, -4).
- Click here to check your answer
-
\(-\dfrac{3}{4}\)
If you missed this problem, review Section 1.2 . (Note that this will open in a new window.)
So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on the line, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.
An equation of a line can be written in three forms, the slope-intercept form , the point-slope form, or the standard form . We will discuss each of them in this section.
A line is completely determined by two points, or by a point and slope. The information we are given about a particular line will influence which form of the equation is most convenient to use. Once we know any form of the equation of a line, it is easy to re-express the equation in the other forms if needed.
THE SLOPE-INTERCEPT FORM OF A LINE: \(y = mx + b\)
In the last section we learned that the equation of a line whose slope = \(m\) and \(y\)-intercept = \(b\) is \[y=mx+b.\] This is called the slope-intercept form of the line and is the most commonly used form.
Example \(\PageIndex{1}\)
Find an equation of a line whose slope is 5, and \(y\)-intercept is 3.
Solution
Since the slope is \(m = 5\), and the \(y\)- intercept is \(b = 3\), the equation is \(y = 5x + 3\).
Example \(\PageIndex{2}\)
Find the equation of the line that passes through the point (\(2, 7\)) and has slope \(3\).
Solution
Since \(m = 3\), the partial equation is \(y = 3x + b\).
Now \(b\) can be determined by substituting the point (\(2, 7\)) in the equation \(y = 3x + b\).
\begin{aligned}
&7=3(2)+b \nonumber \\
&b=1 \nonumber
\end{aligned}
Therefore, the equation is \(y = 3x + 1\).
Example \(\PageIndex{3}\)
Find an equation of the line that passes through the points (-1, 2), and (1, 8).
Solution
We need to find the slope of the line containing these points first. Remember that if (\(x_1\), \(y_1\)) and (\(x_2\), \(y_2\)) are two different points on a line, the slope of the line is
\[\text{slope}=m=\frac{y_2-y_1}{x_2-x_1} \label{slope}\nonumber\]
\(m=\frac{8-2}{1-(-1)}=\frac{6}{2}=3\). So the partial equation is \(y = 3x + b\).
We can use either of the two points (-1, 2) or (1, 8), to find \(b\). Substituting (-1, 2) gives
\begin{aligned}
&2=3(-1)+b \nonumber \\
&5=b \nonumber
\end{aligned}
So the equation is \(y = 3x +5\).
Example \(\PageIndex{4}\)
Find an equation of the line that has \(x\)-intercept 3, and \(y\)-intercept 4.
Solution
\(x\)-intercept = 3, and \(y\)-intercept = 4 correspond to the points (3, 0), and (0, 4), respectively.
\[ m=\frac{4-0}{0-3} = -\frac{4}{3} \nonumber \]
We are told the \(y\)-intercept is 4; thus \(b\) = 4
Therefore, the equation is \(y = -\frac{4}{3} x + 4\).
THE POINT-SLOPE FORM OF A LINE: \(y - y_1 = m(x - x_1)\)
The point-slope form is useful when we know two points on the line and want to find the equation of the line.
The definition of a slope leads us to the point-slope formula. Using two points (\(x_1\), \(y_1\)) and any other (\(x,y\)), the slope is \( \frac{y-y_1}{x-x_1}= m\).
Multiplying both sides by (\(x-x_1\)) gives the point-slope form: \(y - y_1 = m(x - x_1)\).
A line with slope which contains a specific point (\(x_1, y_1)\) can be expressed in the form \[y - y_1 = m(x - x_1).\]
Example \(\PageIndex{5}\)
Find the point-slope form of the equation of the line given in Example \(\PageIndex{2}\). ( Find the equation of the line that passes through the point (\(2, 7\)) and has slope \(3\). ) Show that the two forms of the equations are equivalent.
Solution
Substituting the point \((x_1,y_1) = (2,7)\) and \(m= 3\) in the point-slope formula, we get
\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]
\[y - 7 = 3(x - 2) \nonumber \]
We can show that the forms are equivalent by solving this equation for \(y\).
\[y - 7 = 3(x - 2) \nonumber \]
\[y - 7 = 3x - 6 \nonumber \]
\[y = 3x + 1 \nonumber \]
This is the slope-intercept form of the equation that we found in Example \(\PageIndex{2}\).
Example \(\PageIndex{6}\)
Find the point-slope form of the equation of a line that has slope \(\frac{1}{2}\) and passes through the point (12,4). Then write the equation in slope-intercept form.
Solution
Substituting the point \((x_1,y_1) = (12,4)\) and \(m= \frac{1}{2}\) in the point-slope formula, we get
\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]
\[y - 4 = \frac{1}{2}(x - 12) \nonumber \]
\[y - 4 = \frac{1}{2}x - 6 \nonumber \]
\[y = \frac{1}{2}x - 2 \nonumber \]
THE STANDARD FORM OF A LINE: \(Ax + By = C\)
Another useful form of the equation of a line is the standard form.
If we know the equation of a line in point-slope form, \(y - y_1 = m(x - x_1)\), or if we know the equation of the line in slope-intercept form \(y = mx + b\), we can simplify the formula to have all terms for the \(x\) and \(y\) variables on one side of the equation, and the constant on the other side of the equation.
The result is referred to as the standard form of the line: \[\mathbf{Ax + By = C}.\]
We should always be able to convert from one form of an equation to another. For example, if we are given a line in the slope-intercept form, we should be able to express it in the standard form, and vice versa.
Example \(\PageIndex{7}\)
Write the equation \(y = -\frac{2}{3}x + 3\) in the standard form.
Solution
Multiplying both sides of the equation by 3, we get
\begin{aligned}
&3y = -2x + 9 \\
&2x + 3y = 9 \quad \text { Standard Form }
\end{aligned}
Example \(\PageIndex{8}\)
Write the equation \(3x - 4y = 10\) in the slope-intercept form.
Solution
Solving for \(y\), we get
\begin{aligned}
&-4y = -3x + 10 \\
&y = \frac{3}{4}x - \frac{5}{2} \quad \text { Standard Form }
\end{aligned}
Example \(\PageIndex{9}\)
Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope \(-3/5\).
Solution: Substituting the point (2, 3) and \(m= - 3/5\) in the point-slope formula, we get
\[y - 3 = - 3/5(x - 2) \nonumber \]
Multiplying both sides by 5 gives us
\begin{aligned}
&5(y-3)=-3(x-2)\\
&5 y-15=-3 x+6\\
&3 x+5 y=21 \quad \text { Standard Form }
\end{aligned}
Example \(\PageIndex{10}\)
Find the standard form of the line that passes through the points (1, -2), and (4, 0).
Solution
First we find the slope: \(m = \frac{0-(-2)}{4-1} = \frac{2}{3}\)
Then, the point-slope form is: \(y - (-2) = \frac{2}{3}(x -1)\)
Multiplying both sides by 3 gives us
\begin{aligned}&3(y+2)=2(x-1) \\
&3 y+6=2 x-2\\
&-2 x+3 y=-8\\
&2 x-3 y=8 \quad \text{ Standard Form } \end{aligned}
Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.
By solving for \(y\), it can easily be shown that the slope of the line \(Ax + By = C\) is \(-A/B\).
The reader should verify this.
Example \(\PageIndex{11}\)
Find the slope of the following lines, by inspection.
- \(3x-5y=10\)
- \(2x+7y=20\)
- \(4x-3y=8\)
Solution
- \(A=3\), \(B=-5\), therefore, \(m=\frac{3}{-5}=\frac{3}{5}\)
- \(A=2\), \(B=7\), therefore, \(m=-\frac{2}{7}\)
- \(m=\frac{4}{-3}=\frac{4}{3}\)
Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be easy.
Example \(\PageIndex{12}\)
Find an equation of the line that passes through (2, 3) and has slope - 4/5.
Solution
Since the slope of the line is - 4/5, we know that the left side of the equation is \(4x + 5y\), and the partial equation is going to be
\[4x + 5y = c \nonumber\]
Of course, \(c\) can easily be found by substituting for \(x\) and \(y\).
\begin{aligned}
&4(2)+5(3)=c\\
&23=c
\end{aligned}
The desired equation is
\[4x + 5y = 23. \nonumber\]
If you use this method often enough, you can do these problems very quickly.
We summarize the forms for equations of a line below:
Slope Intercept form:
\(\mathbf{y = mx + b}\),
where \(m\) = slope, \(b\) = \(y\)-intercept
Point Slope form:
\(\mathbf{y - y_1 = m(x - x_1)}\),
where \(m\) = slope, \((x_1,y_1)\) is a point on the line
Standard form: \(\mathbf{Ax + By = C}\)
Horizontal Line:
\(\mathbf{y = b}\)
where \(b\) = \(y\)-intercept
Vertical Line:
\(\mathbf{x = a}\)
where \(a\) = \(x\)-intercept


4.4: Exceptions to autosomal inheritance
Learning Objectives
- Compare different modes of sex determination, such as mammalian XY, fruit flies, insects, or environmental.
- Explain why dosage compensation is needed and the role of the non-coding RNA Xist in mammalian dosage compensation.
- Predict the number of Barr bodies in a particular cell.
- Predict offspring phenotypes for X-linked genes.
For loci on autosomes, in which all individuals have two copies of the chromosome and either copy is equally likely to be passed to offspring, the alleles follow the normal Mendelian pattern of inheritance. However, for loci on the allosomes (also called sex chromosomes) individuals may have zero, one, or two copies of a chromosome. Because homologous chromosomes contain the same genes, individuals may have zero, one, or two copies of some genes. In humans, the allosomes are the X and Y chromosomes, which contain different genes, even though they act as a homologous pair during meiosis. Instead, genes on the X chromosome will follow an X-linked pattern of inheritance, while genes on the Y chromosome will follow a Y-linked pattern of inheritance.
X-Linked inheritance example: the white gene in Drosophila melanogaster
A well-studied sex-linked gene is the white gene on the X chromosome of Drosophila melanogaster . Normally flies have red eyes but flies with a mutant allele of this gene called white - ( w - ) have white eyes because the red pigments are absent. Because this mutation is recessive to the wild type w + allele females that are heterozygous have normal red eyes. Female flies that are homozygous for the mutant allele have white eyes. Because there is no white gene on the Y chromosome, male flies can only be hemizygous for the wild type allele or the mutant allele.
Note about nomenclature
In genetics, genes are often named after their mutant (frequently, but not always, loss-of-function) phenotypes. Remember, genetics has its origin in observing phenotypes. The white allele described here was identified in 1910 (Morgan, T.H.) long before gene sequences could be determined, in fact, even before DNA had been identified as the unit of heredity. The DNA sequence of this locus was not published until 1984 (O'Hare et al). We know now that this gene encodes a transporter protein that deposits pigment and that the mutant allele does not produce a functional transporter, but the gene name white continues to be used.
The process of identifying mutant phenotypes and then later determining which gene is involved is called forward genetics .
The alternative, to identify a candidate gene and manipulate it to observe the phenotype, is called reverse genetics , and was only made possible by the advent of modern molecular biology tools.
A researcher may not know beforehand whether a novel mutation is sex-linked. The definitive method to test for sex-linkage is reciprocal crosses , in which a male and a female that have different phenotypes are crossed. In a second cross, the phenotypes are reversed relative to the sex of the parents in the first cross. Whenever reciprocal crosses give different results in the F 1 and F 2 , and whenever the male and female offspring have different phenotypes, the usual explanation is sex-linkage. Remember, if the locus were autosomal, the F 1 and F 2 progeny are expected to show the same genotypes and phenotypes from either of these crosses.
A similar pattern of sex-linked inheritance is seen for X-chromosome loci in other species with an XX-XY sex chromosome system, including mammals and humans. The ZZ-ZW system is similar, but reversed (see below).
Diversity in Sex Determination
Sex determination refers to the process by which reproductive organs are specified to become ovaries or testes.
Some organisms used chromosomal sex determination. For example in humans and other mammals XY embryos commonly develop testes while XX embryos develop ovaries. This difference in development is usually initiated by the presence of a single gene, the SRY gene ( s ex-determining r egion on Y ), on the Y-chromosome. SRY encodes is a transcription factor that activates expression of genes that promote testes differentiation. Without SRY, these genes are not activated and genes that promote ovary development are transcribed. Notably, SRY initiates the process of testes development, but a series of genes are required to complete testes and reproductive organ development. If the downstream genes are not fully functional, development of these structures may not occur despite the presence of the Y chromosome and SRY.
Query \(\PageIndex{1}\)
Although Drosophila melanogaster also has a Y chromosome, there is not an SRY gene and the X to Autosome (X:A) ratio determines male or female differentiation. Individuals with two autosome sets and two X chromosomes (2A:2X) will develop ovaries, while those with only one X chromosome (2A:1X) will develop testes. The presence/absence of the Y-chromosome and its genes are not significant. For example, both an XO and XY fruit fly will develop testes and other male phenotypes, if they both have two sets of autosomes.
In other animal species, the number of chromosome sets can determine sex. For example the haploid-diploid system is used in bees, ants, and wasps. Typically haploid individuals, which develop from unfertilized eggs, develop testes, while diploid individuals, which develop from fertilized eggs, form ovaries. For this system to work, meiosis does not occur during sperm production in haploids; instead sperm are identical copies produced by mitosis because chromosome number does not need to be reduced by half.
In other species, the environment can determine an individuals sex. In alligators (and some other reptiles) the temperature of development dictates the sex, while in many reef fish, the population sex ratio can produce environmental signals that cause some individuals to change sex.
Dosage Compensation
Mammals and Drosophila both have XX - XY sex determination systems. However, because these systems evolved independently they work differently with regard to compensating for the difference in gene dosage. Remember, in most cases the sex chromosomes act as a homologous pair even though the Y chromosome has lost most of the loci when compared to the X chromosome. Comparison of X and Y chromosomes had led to the view that Y chromosomes have evolved from autosomes that degenerated, slowly mutating and losing loci. In modern day mammals, the Y chromosomes have very few genes, while the X chromosomes remain as they were. This is a general feature of all organisms that use chromosome based sex determination systems. Chromosomes found in both sexes (the X or the Z) have retained their genes while the chromosome found in only one sex (the Y or the W) have lost most of their genes.
Human X and Y Compared
| X chromosome | Y chromosome | |
| Number of base pairs | 156,040,895 | 57,227,415 |
| Number of coding genes | 857 | 64 |
| Contains essential genes for viability? | Yes | No |
Ensembl 2021. (Release 104) Nucleic Acids Res. 2021, vol. 49(1):884–891 PubMed PMID: 33137190. doi:10.1093/nar/gkaa942
Chromosomal sex determination requires dealing with the gene dosage difference between individuals with different chromosome complements: e.g. XX females have two copies ("doses") of X-chromosome genes while XY males only have one copy of these genes. For all the autosomes, XX and XY individuals have the same number of copies of genes. Producing too much of some proteins can be problematic, even lethal, for organisms. This difference in gene dosage needs to be compensated in a process called dosage compensation . In Drosophila and many other insects, individuals with single X chromosome express the genes on it at twice the normal rate. This mechanism of dosage compensation restores a balance of proteins encoded by X-linked genes and those made by autosomal genes.
X-chromosome Inactivation in Mammals
In mammals the dosage compensation system operates in any cell with more than one X chromosome. In XX embryos, one X in each cell is randomly chosen and marked for inactivation by expressing a long non-coding RNA Xist ( X i nactive s pecific t ranscript). The Xist RNA stays in the nucleus and "coats" the X chromosome, recruiting proteins that affect chromosome packing. This X chromosome becomes highly compact heterochromatin. The expression of Xist is repressed on the other chromosome, so it remains more loosely packed euchromatin. The X i is replicated during S phase and transmitted during mitosis the same as any other chromosome, but most of its genes are never allowed to turn on and the chemical modifications that regulate chromatin are passed on following each round of cell division.
Make a prediction
What would happen to XY or XX mouse embryos that lack the Xist gene?
- Answer
-
XY mouse embryos with a deletion of the Xist gene were viable and fully fertile. The XX mouse embryos died during embryonic development.
(Marahrens Y, Panning B, Dausman J, Strauss W, Jaenisch R. Xist-deficient mice are defective in dosage compensation but not spermatogenesis. Genes Dev. 1997 Jan 15;11(2):156-66. doi: 10.1101/gad.11.2.156. PMID: 9009199.)
From this point forward this chromosome will be inactive, abbreviated X inactive (X i ). The chromosome appears as a condensed mass within interphase nuclei called the Barr body . With the inactivation of genes on one X chromosome, XX individuals have the same number of functioning X-linked genes as XY individuals. The other X chromosome, the X active (X a ), is unaffected. The mechanisms that determine X inactivation can also be used in other situations, such as XXX, in which cells inactivate two X chromosomes, or XXY individuals, in which cells inactivate one X chromosome. This remarkable system always leaves one X chromosome active, such that every mammalian cell expresses the correct dose of X chromosome genes.
Counting chromosomes
The exact mechanism cells use to "count" their X chromosomes is not entirely clear. Essentially all mammalian cells leave only one X chromosome active, even when abnormal numbers of X chromosomes are present.
Examples:
- Females with an extra X (XXX genotype) have 2 Barr bodies and 1 active X.
- Males with an extra X (XXY) have 1 Barr body and 1 active X.
This random inactivation of one X-chromosome leads to a commonly observe phenomenon in cats. A familiar X-linked gene is the Orange gene ( O ) in cats. The O O allele encodes an enzyme that results in orange pigment for the hair. The O B allele causes the hairs to be black. Heterozygous females have an orange and black mottled phenotype known as tortoiseshell. This is due to patches of skin cells having different X-chromosomes inactivated. In each orange hair the X i chromosome carrying the O B allele is inactivated. The O O allele on the X a is functional and orange pigments are made.  Because the inactivation decision happens early during embryogenesis, the cells continue to divide to make large patches on the adult cat skin where one or the other X is inactivated. The Orange gene in cats is a good demonstration of how the mammalian dosage compensation system affects gene expression. However, most X-linked genes do not produce such dramatic mosaic phenotypes in heterozygous females.
Another example is the F8 gene in humans that encodes Factor VIII, a protein needed for blood clotting that is produced in liver cells. If an XY individual is hemizygous for a mutant allele ( F8 – /F8 – ) the result is hemophilia type A. An XX individual homozygous for mutant alleles will also have hemophilia. Heterozygous F8 + / F8 – individuals do not have hemophilia because even though half of their liver cells do not make Factor VIII (because the X with the F8 + allele is inactive) the other 50% can. Because some of their liver cells are exporting Factor VIII proteins into the blood stream, they have the ability to form blood clots throughout their bodies. The genetic mosaicism in the cells of their bodies does not produce a visible mosaic phenotype.
Example \(\PageIndex{1}\)
Predicting genetic crosses with X-linked genes follows that of autosomal genes, remembering that each gamete contains only one sex chromosome, unless otherwise specified.
 How many are expected to be tortoiseshell?
Solution
Using a Punnett square reveals that 25% (or two) offspring are predicted to be tortoiseshell.
Exercise \(\PageIndex{1}\)
If you were able to clone a tortoiseshell cat, would the pattern of orange and black markings be identical in the clone? Why or why not?
- Answer
-
The pattern is not likely to be identical because X chromosome inactivation happens during embryogenesis. When the cloned cat is in the early stages of development, the X chromosomes with the orange and black alleles will be randomly inactivated and passed on during each mitosis as more cells are formed.
Exercise \(\PageIndex{2}\)
A heterozygous individual (genotype F8 + / F8 — ) is a carrier for the hemophilia allele. Her partner is an XY male with normal blood clotting. What is the probability their son will have hemophilia? What about a daughter?
- Answer
-
A son will be either F8 + Y or F8 — Y (50% with hemophilia). A daughter will be either F8 + / F8 — or F8 + / F8 + (0% with hemophilia).
Z-linked genes
The ZW system is distinguished from the XY system based on which sex has two different sex chromosomes, also called the heterogametic sex. In this case, male birds commonly have ZZ chromosomes, whereas female birds commonly have ZW chromosomes. Thus, female birds may have only one copy of Z-linked genes. For example, a Z-linked gene influences feather color in turkeys. Turkeys are birds, which use the ZZ-ZW sex chromosome system. The E allele makes the feathers bronze and the e allele makes the feathers brown (Figure \(\PageIndex{5}\)). Only turkeys with two Z chromosomes, commonly males, can be heterozygous for this locus. Heterozygous turkeys ( Ee ) are uniformly bronze because the E allele is completely dominant to the e allele and birds use a dosage compensation system similar to Drosophila and not mammals. Reciprocal crosses between turkeys from pure-breeding bronze and brown breeds would reveal that this gene is in fact Z-linked.
Contributors and Attributions
-
Dr. Todd Nickle and Isabelle Barrette-Ng (Mount Royal University) The content on this page is licensed under CC SA 3.0 licensing guidelines.
References
Bachtrog D, Mank JE, Peichel CL, Kirkpatrick M, Otto SP, Ashman T-L, et al. (2014) Sex Determination: Why So Many Ways of Doing It? PLoS Biol 12(7): e1001899. https://doi.org/10.1371/journal.pbio.1001899


10.5: Unintentional Injuries
Distracted Driving
Each day in the United States, over 8 people are killed and 1,161 injured in crashes that are reported to involve a distracted driver.
Distracted driving is driving while doing another activity that takes your attention away from driving. Distracted driving can increase the chance of a motor vehicle crash.
Public domain content
- Distracted Driving. Authored by : National Center for Injury Prevention and Control, Division of Unintentional Injury Prevention. Provided by : Centers for Disease Control and Prevention. Located at : https://www.cdc.gov/motorvehiclesafety/Distracted_Driving/index.html . License : Public Domain: No Known Copyright
- Unintentional Injuries Infographic. Authored by : Centers for Disease Control and Prevention, National Center for Injury Prevention and Control. Provided by : U.S. Department of Health and Human Services. Located at : http://www.cdc.gov/injury/wisqars/overview/key_data.html . License : Public Domain: No Known Copyright


Chapter 7: Teachers’ Purposes and Beliefs
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- SUNY Oneonta via LumenLearning
One of the main charges to teachers is to convey content to their students. Teachers need to express why the content they teach is important to learn. For elementary teachers, the necessity for students to learn how to calculate, read, and write is a given, but the answer is not as clear for other subjects: music, science, physical education, history, and art. Many parents do not see the need for their children to study a subject past a certain point. In order to ground why content is important, reflection and creation of a personal philosophy is essential.
In conveying content, teachers need to use the most effective strategies available. Different content might require a variety of approaches. Another factor to consider is if technology will enhance student understanding. Teachers with a good grasp of strategies along with a wide variety of ways to instruct students during a lesson will be more successful.
Students’ mindsets indicate how well and how much they are able to learn. Psychologist Carol Dweck, (2008) defines a growth mindset as the increase in ability to learn when a learner accepts that they may improve, and this improvement will lead to increased ability to learn more. Effort is valued because effort and self-efficacy lead to knowing more and therefore having more ability to learn. Individuals with a growth mindset also ask for help when needed and respond well to constructive feedback. In contrast, individuals with a fixed mindset assume that some people naturally have more ability than others and nothing can be done to change that. Individuals with a fixed mindset often view effort in opposition to ability (“Smart people don’t have to study”) and so do not try as hard and are less likely to ask for help since they believe that asking questions indicates that they are not smart. There are individual differences in students’ beliefs about their views of intelligence. However, teachers’ beliefs and classroom practices influence these students’ perceptions, behaviors, and willingness to adopt a growth mindset.
Teachers with a growth mindset believe that the goal of learning is mastering the material and figuring things out. Assessment is used by these teachers to understand what students know so they can decide whether to move to the next topic, re-teach the entire class, or provide remediation for a few students. Assessment also helps students understand their own learning and demonstrate their competence. Teachers with these views say things like, “We are going to practice over and over again. That’s how you get good. And you’re going to make mistakes. That’s how you learn” (Patrick, Anderman, Ryan, Edelin, & Midgley, 2001, p. 45).
In contrast, teachers with a fixed mindset are more likely to believe that the goal of learning is doing well on tests – and especially outperforming others. These teachers are more likely to say things that imply fixed abilities such as, “This test will determine what your math abilities are,” or stress the importance of interpersonal competition, “We will have speech competition and the top person will compete against all the other district schools and last year the winner got a big award and their photo in the paper.” When teachers stress competition some students will be motivated; however, there can only a few winners so there are many more students who believe they have no chance of winning. Another problem with competition as an assessment is that the focus can become winning rather than understanding the material.
Teachers who view assessment as promoting and developing learning rather than as a means of ranking students, or awarding prizes to those who did very well, or catching those who did not pay attention, are likely to enhance student willingness to identify and correct gaps in learning and understanding.
LICENSES AND ATTRIBUTIONS
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- Foundations of Education. Authored by : SUNY Oneonta Education Department. License : CC BY: Attribution
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- Educational Psychology. Authored by : Kevin Seifert and Rosemary Sutton. Located at : open.umn.edu/opentextbooks/BookDetail.aspx?bookId=153 . License : CC BY: Attribution