4.7C: Comparing Plant and Animal Cells
- Differentiate between the structures found in animal and plant cells
Animal Cells versus Plant Cells
Each eukaryotic cell has a plasma membrane, cytoplasm, a nucleus, ribosomes, mitochondria, peroxisomes, and in some, vacuoles; however, there are some striking differences between animal and plant cells. While both animal and plant cells have microtubule organizing centers (MTOCs), animal cells also have centrioles associated with the MTOC: a complex called the centrosome. Animal cells each have a centrosome and lysosomes, whereas plant cells do not. Plant cells have a cell wall, chloroplasts and other specialized plastids, and a large central vacuole, whereas animal cells do not.
The Centrosome
The centrosome is a microtubule-organizing center found near the nuclei of animal cells. It contains a pair of centrioles, two structures that lie perpendicular to each other. Each centriole is a cylinder of nine triplets of microtubules. The centrosome (the organelle where all microtubules originate) replicates itself before a cell divides, and the centrioles appear to have some role in pulling the duplicated chromosomes to opposite ends of the dividing cell. However, the exact function of the centrioles in cell division isn’t clear, because cells that have had the centrosome removed can still divide; and plant cells, which lack centrosomes, are capable of cell division.
The Centrosome Structure : The centrosome consists of two centrioles that lie at right angles to each other. Each centriole is a cylinder made up of nine triplets of microtubules. Nontubulin proteins (indicated by the green lines) hold the microtubule triplets together.
Lysosomes
Animal cells have another set of organelles not found in plant cells: lysosomes. The lysosomes are the cell’s “garbage disposal.” In plant cells, the digestive processes take place in vacuoles. Enzymes within the lysosomes aid the breakdown of proteins, polysaccharides, lipids, nucleic acids, and even worn-out organelles. These enzymes are active at a much lower pH than that of the cytoplasm. Therefore, the pH within lysosomes is more acidic than the pH of the cytoplasm. Many reactions that take place in the cytoplasm could not occur at a low pH, so the advantage of compartmentalizing the eukaryotic cell into organelles is apparent.
The Cell Wall
The cell wall is a rigid covering that protects the cell, provides structural support, and gives shape to the cell. Fungal and protistan cells also have cell walls. While the chief component of prokaryotic cell walls is peptidoglycan, the major organic molecule in the plant cell wall is cellulose, a polysaccharide comprised of glucose units. When you bite into a raw vegetable, like celery, it crunches. That’s because you are tearing the rigid cell walls of the celery cells with your teeth.
Chloroplasts
Like mitochondria, chloroplasts have their own DNA and ribosomes, but chloroplasts have an entirely different function. Chloroplasts are plant cell organelles that carry out photosynthesis. Photosynthesis is the series of reactions that use carbon dioxide, water, and light energy to make glucose and oxygen. This is a major difference between plants and animals; plants (autotrophs) are able to make their own food, like sugars, while animals (heterotrophs) must ingest their food.
Like mitochondria, chloroplasts have outer and inner membranes, but within the space enclosed by a chloroplast’s inner membrane is a set of interconnected and stacked fluid-filled membrane sacs called thylakoids. Each stack of thylakoids is called a granum (plural = grana). The fluid enclosed by the inner membrane that surrounds the grana is called the stroma.
The chloroplasts contain a green pigment called chlorophyll, which captures the light energy that drives the reactions of photosynthesis. Like plant cells, photosynthetic protists also have chloroplasts. Some bacteria perform photosynthesis, but their chlorophyll is not relegated to an organelle.
The Central Vacuole
The central vacuole plays a key role in regulating the cell’s concentration of water in changing environmental conditions. When you forget to water a plant for a few days, it wilts. That’s because as the water concentration in the soil becomes lower than the water concentration in the plant, water moves out of the central vacuoles and cytoplasm. As the central vacuole shrinks, it leaves the cell wall unsupported. This loss of support to the cell walls of plant cells results in the wilted appearance of the plant. The central vacuole also supports the expansion of the cell. When the central vacuole holds more water, the cell gets larger without having to invest a lot of energy in synthesizing new cytoplasm.
Key Points
- Centrosomes and lysosomes are found in animal cells, but do not exist within plant cells.
- The lysosomes are the animal cell’s “garbage disposal”, while in plant cells the same function takes place in vacuoles.
- Plant cells have a cell wall, chloroplasts and other specialized plastids, and a large central vacuole, which are not found within animal cells.
- The cell wall is a rigid covering that protects the cell, provides structural support, and gives shape to the cell.
- The chloroplasts, found in plant cells, contain a green pigment called chlorophyll, which captures the light energy that drives the reactions of plant photosynthesis.
- The central vacuole plays a key role in regulating a plant cell’s concentration of water in changing environmental conditions.
Key Terms
- protist : Any of the eukaryotic unicellular organisms including protozoans, slime molds and some algae; historically grouped into the kingdom Protoctista.
- autotroph : Any organism that can synthesize its food from inorganic substances, using heat or light as a source of energy
- heterotroph : an organism that requires an external supply of energy in the form of food, as it cannot synthesize its own


3.1.3: Conversación
Preparación comunicativa 1: La familia de Mariana
Paso 1.
The following is Mariana’s family tree. Describe each person’s relationship to Mariana. Follow the example provided .
Modelo:
Carlos → Carlos es el abuelo de Mariana.
Paso 2.
Ahora observa el árbol familiar de Mariana y decide si las oraciones a continuación son Ciertas (C) o Falsas (F). If the statements are false, explain why .
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
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C |
F |
Preparación comunicativa 2: Unas familias colombianas y argentinas
Mira las familias y responde las preguntas de tu compañero.
Estudiante 1
La familia colombiana
- Click here to see Student 1´s information:
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Add texts here. Do not delete this text first.
Estudiante 2
La familia argentina
- Click here to view Student 2´s information:
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Attribution: La familia by Community College of Baltimore County is licensed under CC BY-NC 4.0
Preparación comunicativa 3: Tu árbol genealógico
Paso 1.
Work with a classmate to construct each other’s family trees (real or imagined). Each partner creates the other’s tree based on the information provided by the partner. Make sure to include names and each person’s relation to your partner. While working together, try to use as much Spanish as possible.
Modelo:
Emily → Emily es mi hermana.
Paso 2.
How does your partner’s family tree compare to Mariana’s? How does it compare to your family tree? Are there any similarities? Are there any differences? If so, explain. Be ready to present your work to the rest of the class.
Preparación comunicativa 4: ¿Hay una persona en tu familia?
When we describe our families in Spanish, we often use the verb ser accompanied by one or more adjectives. Below you will find a list of adjectives appropriate for describing people. Use this list to ask your partner if these adjectives describe anyone in their family. Use as much Spanish as possible and take turns asking each other about your families.
Modelo:
E1: ¿Hay una persona simpática en tu familia?
E2: Sí, mis primos son simpáticos . Se llaman Alfredo y Gloria.
Cognados útiles: admirable, elegante, importante, inteligente, materialista, pesimista, responsable, arrogante, flexible, independiente, interesante, optimista, realista, sentimental, extravagante, idealista, informal, liberal, paciente, rebelde, tolerante


5.1: Introduction
Language is central to daily human existence. It is the principal means by which we conduct our social lives at home, neighborhood, school, work place and recreation area. It is the tool we use to plan our lives, remember our past, and express our cultural identity. We create meaning when we talk on the cell phone, send an e-mail message, read a newspaper and interpret a graph or chart. Many persons conduct their social lives using only one language. Many others, however, rely on two languages in order to participate effectively in the community, get a job, obtain a college degree and enjoy loving relationships. We live in a discourse world that incorporates ways of speaking, reading and writing, but also integrates ways of behaving, interacting, thinking and valuing. Language is embedded in cultural practices and, at the same time, symbolizes cultural reality itself.


3.9: Scientific Notation
- Write numbers in standard or scientific notation.
- Convert numbers between standard and scientific notation.
- Add and subtract numbers in scientific notation.
- Multiply and divide numbers in scientific notation.
- Use scientific notation in computing real-world applications.
The amount of information available on the Internet is simply incomprehensible. One estimate for the amount of data that will be on the Internet by 2025 is 175 Zettabytes. A single zettabyte is one billion trillion. Written out, it is 1,000,000,000,000,000,000,000. One estimate is that we’re producing 2.5 quintillion bytes of data per day. A quintillion is a trillion trillion, or, written out, 1,000,000,000,000,000,000. To determine how many days it takes to increase the amount of information that is on the Internet by 1 zettabyte, divide these two numbers, a zettabyte being 1,000,000,000,000,000,000,000, and 2.5 quintillion, being 2,500,000,000,000,000,000, shows it takes 400 days to generate 1 zettabyte of information. But doing that calculation is awkward with a calculator. Keeping track of the zeros can be tedious, and a mistake can easily be made.
On the other end of the scale, a human red blood cell has a diameter of 7.8 micrometers. One micrometer is one millionth of a meter. Written out, 7.8 micrometers is 0.0000078 meters. Smaller still is the diameter of a virus, which is about 100 nanometers in diameter, where a nanometer is a billionth of a meter. Written out, 100 nanometers is 0.0000001 meters. To compare that to engineered items, a single transistor in a computer chip can be 14 nanometers in size (0.000000014 meters). Smaller yet is the diameter of an atom, at between 0.1 and 0.5 nanometers.
Sometimes we have numbers that are incredibly big, and so have an incredibly large number of digits, or sometimes numbers are incredibly small, where they have a large number of digits after the decimal. But using those representations of the names of the sizes makes comparing and computing with these numbers problematic. That’s where scientific notation comes in.
Writing Numbers in Standard or Scientific Notation Form
When we say that a number is in scientific notation , we are specifying the form in which that number is written. That form begins with an integer with an absolute value between 1 and 9, then perhaps followed the decimal point and then some more digits. This is then multiplied by 10 raised to some power. When the number only has one non-zero digit, the scientific notation form is the digit multiplied by 10 raised to an exponent. When the number has more than one non-zero digit, the scientific notation form is a single digit, followed by a decimal, which is then followed by the remaining digits, which is then multiplied by 10 to a power.
The following numbers are written in scientific notation:
- \(1.45 \times 10^3 \)
- \(-8.345 \times 10^{-4} \)
- \(3 \times 10^2 \)
- \(3.14159 \times 10^0\)
The following numbers are not written in scientific notation:
- because it isn't multiplied by 10 raised to a power
- because the absolute value of −50.053 is not at least 1 and less than 10
- because 41.7 is not at least 1 and less than 10
- because 0.036 is not at least one and less than 10
Which of the following numbers are in scientific notation? If the number is not in scientific notation, explain why it is not.
- Answer
-
- The number is in scientific notation because the absolute value of −9.67 is at least 1 and less than 10.
- The number is not in scientific notation because 145 is not at least 1 and less than 10.
- The number is not in scientific notation form. Even though it is at least 1 but less than 10, it is not multiplied by 10 raised to a power.
Which of the following numbers are in scientific notation? If the number is not in scientific notation, explain why it is not.
- \(42.67 \times 10^{13}\)
- \(7.113 \times 10^{-2}\)
- -80.91
Some numbers are so large or so small that it is impractical to write them out fully. Avogadro’s number is important in chemistry. It represents the number of units in 1 mole of any substance. The substance many be electrons, atoms, molecules, or something else. Written out, the number is: 602,214,076,000,000,000,000,000. Another example of a number that is impractical to write out fully is the length of a light wave. The wavelength of the color blue is about 0.000000450 to 0.000000495 meters. Such numbers are awkward to work with, and so scientific notation is often used. We need to discuss how to convert numbers into scientific notation, and also out of scientific notation.
Recall that multiplying a number by 10 adds a 0 to the end of the number or moves the decimal one place to the right, as in
or
. And if you multiply by 100, it adds two zeros to the end of the number or moves the decimal two places to the right, and so on. For example,
and
. Multiplying a number by 1 followed by some number of zeros just adds that many zeros to the end of the number or moves the decimal place that many places to the right. Numbers written as 1 followed by some zeros are just powers of 10, as in
,
,
, etc. Generally,
We can use this to write very large numbers. For instance, Avogadro’s number is 602,214,076,000,000,000,000,000, which can be written as . The multiplication moves the decimal 23 places to the right.
Similarly, when we divide by 10, we move the decimal one place to the left, as in . If we divide by 100, we move the decimal two places to the left, as in . In general, when you divide a number by a 1 followed by zeros, you move the decimal places to the left, as in . This denominator could be written as . If we use that in the expression and allow for negative exponents, rewrite the number as . With this, we can write division by a 1 followed by zeros as multiplication by 10 raised to .
Using that information, we can demonstrate how to convert from a number in standard form into scientific notation form.
Case 1: The number is a single-digit integer.
In this case, the scientific notation form of the number is
Case 2: The absolute value of the number is less than 1.
Follow the process below.
- Step 1: Count the number of zeros between the decimal and the first non-zero digit. Label this .
- Step 2: Starting with the first non-zero digit of the number, write the digits. If the number was negative, include the negative sign.
- Step 3: If there is more than one digit, place the decimal after the first digit from Step 2.
-
Step 4:
Multiply the number from Step 3 by
10 n + 1 10 n .+ 1
Case 3: The absolute value of the number is 10 or larger.
Follow the process below.
- Step 1: Count the number of digits that are to the left of the decimal point. Label this .
- Step 2: Write the digits of the number without the decimal place, if one was present. If the number was negative, include the negative sign.
- Step 3: If there is more than one digit, place the decimal point after the first digit.
-
Step 4:
Multiply the number from Step 3 by
10 n − 1 10 n .− 1
Write the following numbers in scientific notation form:
- 428.9
- −0.00000981
- 8
- Answer
-
-
Since the absolute value of 428.9 is 10 or larger, so we use the process from Case 3, above.
Step 1: There are three digits to the left of the decimal point, so .
Step 2: Write the digits of the number without the decimal place, which is 4289.
Step 3: Since there is more than one digit, place the decimal point after the first digit. We now have 4.289.
Step 4: Since , we multiply 4.289 by 10 raised to the second power, .
The scientific notation form of 428.9 is .
-
Since the absolute value of −0.00000981 is less than 1, we use the process from Case 2.
Step 1: The number of zeros between the decimal and the first non-zero digit is 5, so .
Step 2: We write the non-zero digits, including the negative sign, yielding −981.
Step 3: The decimal gets placed to the right of the first digit, resulting in −9.81.
Step 4: Since , we multiply −9.81 by 10 raised to the fourth power, .
The scientific notation form of −0.00000981 is .
- Since 8 is a single-digit integer, apply Case 1. The scientific notation form of 8 is .
-
Since the absolute value of 428.9 is 10 or larger, so we use the process from Case 3, above.
Write the following numbers in scientific notation form:
- -38300
- 0.0045
- 1
When we write numbers in scientific notation form, we can manipulate the representation of the number by moving the decimal around, and making an appropriate change to the exponent of the 10. For instance, let’s look at
. If we wanted to move the decimal one place to the left, we’d have to increase the power of 10, as shown here:
. Since we moved the decimal one to the left, we balance that with moving the exponent up by one. Similarly, if we move the decimal one place to the right, we have to balance that by moving the exponent one to the left, or subtracting one from the exponent, as shown here:
. Generally, for a number in the form
- If you move the decimal to the left by digits, you increase the exponent by .
- If you move the decimal to the right by digits, you decrease the exponent by digits.
Change by moving the decimal two places to the left.
- Answer
-
Since we are moving the decimal to the left by two places, we increase the exponent of 10 by 2, so that the exponent is now 7. This gives us .
Change \(46.113 \times 10^8\) by moving the decimal four places to the left.
Change by moving the decimal five places to the right.
- Answer
-
Since we are moving the decimal to the right by five places, we decrease the exponent of 10 by 5, so that exponent is now −3. This give us .
Change \(149.11 \times 10^{-4}\) by moving the decimal two places to the right
Converting Numbers from Scientific Notation to Standard Form
In the previous section, converting a number from standard form to scientific notation was explored. Now, we explore converting from scientific notation back into standard form. Doing so involves moving the decimal according to the power of the 10. The decimal is moved a number of steps equal to the exponent of the 10. As demonstrated previously, when the exponent of the 10 is negative, the decimal is moved to the left and when the exponent of the 10 is positive, the decimal is moved to the right.
Convert the following into standard form:
- Answer
-
- Since the exponent is positive, the decimal moves nine places to the right, so is .
- Since the exponent is negative, the decimal moves eight places to the left, so is .
Convert the following into standard form:
- \(1.02 \times 10^6\)
- \(4.09 \times 10^{-5}\)
Converting from Standard Form to Scientific Notation Form
Converting from Scientific Notation Form to Standard Form
Most scientific and graphing calculators come with the ability to directly convert from standard form to scientific notation. On the TI-83, it is accessed through the MODE menus. For a commonly used, free phone scientific calculator, the calculator can be forced to work in scientific notation mode through its settings.
Some calculators, such as the Desmos online calculator, display scientific notation as a number times 10 to a power as you’ve seen in this section. However, some calculators indicate scientific notation by replacing the
Adding and Subtracting Numbers in Scientific Notation
To add or subtract numbers in scientific notation, the numbers first need to have the same exponent for the 10s. It is possible to add the following since the powers of 10 match:
Notice that the number parts were added, but the exponent part remained the same. This is due to the distributive property of the real numbers. The is factored from the two terms, as shown:
Numbers in scientific notation can be added or subtracted directly using a calculator. Simply enter the values in scientific form and set your calculator to display scientific notation.
Calculate the following:
- Answer
-
- Since the powers of 10 match, we use the distributive property of real numbers to factor 10 −3 from the numbers. We then add the number parts separately to get 4.806.
- Since the powers of 10 match, we use the distributive property of real numbers to factor 10 8 from the numbers. We then subtract the number parts separately to get 5.76.
Calculate the following:
- \(7.57 \times {10^{13}} + 2.031 \times {10^{13}}\)
- \(3.03 \times {10^{ - 6}} - 1.5 \times {10^{ - 6}}\)\(7.57 \times {10^{13}} + 2.031 \times {10^{13}}\)
Adding and subtracting in scientific notation is straightforward when the exponents are the same. There are two issues that can arise. The first issue is what to do if after adding or subtracting the result is not in scientific notation.
Calculate the following:
- Answer
-
-
Since the powers of 10 match, we add the number parts and multiply that by
:
.
However, is not in scientific notation because the absolute value of 15.53 is more than 10. To put this number in scientific notation, the decimal needs to move one to the left. To balance that move, the power of 10 must be increased by 1. So, the answer in scientific notation is .
- Since the powers of 10 match, we add the number parts:
-
Since the powers of 10 match, we add the number parts and multiply that by
:
.
Calculate the following:
- \(5.08 \times {10^3} + 6.9 \times {10^3}\(
- \(8.968 \times {10^{ - 38}} - 8.761 \times {10^{ - 38}}\(
The second issue that might be encountered when adding or subtracting is that the powers of 10 do not match. In that case, one of the numbers must be changed so that the powers of 10 match. It is easiest to make the smaller power of 10 larger to match the other power of 10.
For example, to perform the following, , we’d change the so that the power of 10 is 5. To do so, we need to increase the power of 10 and move the decimal in the number part two places to the left. That would alter into . We would use in the addition problem, so that the exponents match, allowing the addition to occur.
The steps to take when the exponents of the 10s are not equal are:
Step 1: Increase the smaller exponent to equal the larger exponent. Label the amount increased as .
Step 2: For the number with the smaller power of 10, move the decimal point of the number part to the left places.
Step 3: Perform the addition or subtraction.
Step 4: If the result is not in scientific notation, adjust the number to be in scientific notation.
Calculate the following:
\[6.1 \times 10^4+4.8 \times 10^5 \nonumber\]
- Answer
-
Step 1: The lower exponent is 4. To make this equal to the larger exponent, we increased it by 1.
Step 2: Since the smaller exponent was increased by 1, move the decimal one to the left, so the addition become .
Step 3: Now add the numbers,
Step 4: The result is in scientific notation, so no additional adjustment is necessary.
Calculate the following:
\[1.14 \times {10^{ - 43}} + 2.56\,{ \times\, 10^{ - 46}}\]
Calculate the following:
\[7.9 \times 10^{-15}-6.8 \times 10^{-13}\]
- Answer
-
Step 1: The lower exponent is −15 and the larger is −13. To make −15 equal to the larger exponent, we increased it by 2.
Step 2: Since the smaller exponent increased by 2, move the decimal two to the left. The subtraction changes to .
Step 3: Subtract the numbers, .
Step 4: The result is in scientific notation, so no additional adjustment is necessary.
Calculate the following:
\[9.15 \times {10^{28}} - 7.23 \times {10^{26}}\]
Multiplying and Dividing Numbers in Scientific Notation
Multiplying and dividing numbers in scientific notation is somewhat easier than adding or subtracting, because the exponents of the 10s do not have to match. However, it is much more likely that the result will not be in scientific notation, and so that will have to be adjusted at the end. Generally, we multiply or divide the number parts of the two values, and then apply exponent rules to the 10 raised to the powers.
To multiply two numbers in scientific notation:
Step 1: Multiply the number parts.
Step 2: Add the exponents of the 10s.
Step 3: The result is the answer from Step 1 times 10 raised to the answer from Step 2.
Step 4: If the number is not in scientific notation, adjust it appropriately.
Calculate the following:
- Answer
-
-
Step 1:
Multiply the number parts to get
.
Step 2: Add the exponents of the 10s to get .
Step 3: The result is then .
Step 4: This number is already in scientific notation, so no additional adjustment is necessary, .
-
Step 1:
Multiply the number parts to get
.
Step 2: Add the exponents of the 10s to get .
Step 3: The result then is .
Step 4: Since the number is not in scientific notation, it must be adjusted. To put into scientific notation, the decimal moves one to the left, so the exponent would be increased by 1, giving .
-
Step 1:
Multiply the number parts to get
.
Calculate the following:
- \((2.29 \times {10^3}) \times (3 \times {10^4})\)
- \((6.91 \times {10^{ - 3}}) \times (9.1 \times {10^5})\)
Multiplying Numbers in Scientific Notation
Dividing Numbers in Scientific Notation
To divide two numbers that are in scientific notation:
Step 1: Divide the number parts.
Step 2: Subtract the exponent of the denominator from the exponent of the numerator.
Step 3: The answer is the result from Step 1 times 10 raised to the result from Step 2.
Step 4: If the number is not in scientific notation, adjust it appropriately.
Calculate the following:
- Answer
-
-
Step 2:
Subtract the exponent of the denominator from the exponent of the numerator to get
.
Step 3: The result is then .
Step 4: This number is already in scientific notation, so no adjustment is necessary.
-
Step 2:
Subtract the exponent of the denominator from the exponent of the numerator to get
.
Step 3: The result then is .
Step 4: Since this number is not in scientific notation, it must be adjusted. To put into scientific notation, the decimal needs to move one to the right, so the exponent is decreased by 1, giving .
-
Step 2:
Subtract the exponent of the denominator from the exponent of the numerator to get
.
Calculate the following:
- \(\left(3.6 \times 10^{-2}\right) /\left(1.5 \times 10^3\right)\)
- \(\left(1.8 \times 10^4\right) /\left(4.8 \times 10^3\right)\)
Dividing Numbers in Scientific Notation
Using Scientific Notation in Computing Real-World Applications
As noted at the start of this section, scientific notation is useful when the standard representation of a number is awkward or impractical, which occurs when the numbers being used are extremely large or extremely small. For example, Venus is 67,667,000 miles from the sun. In scientific notation, this is . Planetary and galaxy distances is one set of numbers that is easier to express using scientific notation.
How much farther from the sun is Earth compared to Venus if Venus is miles from the sun and Earth is miles from the sun?
- Answer
-
To determine how much farther Earth is compared to Venus, we’d subtract the distances.
.So, Earth is miles farther from the sun than Venus.
Earlier we saw that a single transistor in a computer chip 0.000000014 meters, or \(1.4 \times 10^{-8} \mathrm{~m}\), in size, and that the diameter of an atom could be 0.2 nanometers, or \(2 \times 10^{-10} \mathrm{~m}\) in size. How much larger is the transistor than the atom?
The probability of winning the Mega Millions lottery is published as . The probability of being hit by lightning is approximated to be . How many times more likely are you to be hit by lightning than win the Mega Millions?
- Answer
-
To find out how many times more likely you are to be hit by lightning, divide the probability of being hit by lightning by the probability of winning the Mega Millions.
Step 1: Divide the number parts to get = (rounded to the fourth digit).
Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get .
Step 3: The result then is .
Step 4: Since this number is not in scientific notation, it must be adjusted. To put into scientific notation, the decimal needs to move one place to the right, so the exponent is decreased by 1, giving .
You are , or 605.2, times more likely to be hit by lightning than you are to win the Mega Millions.
Mercury is about \(3.114 \times 10^7\) miles from the sun. Neptune is about \(2.781 \times 10^9\) miles from the sun. How many times further is Neptune from the sun than Mercury?
Sometimes it is entertaining to determine the time it takes for something to happen. Fingernails grow about km per minute. How many kilometers long would fingernails be after minutes?
- Answer
-
To find the length of the fingernails after the specified time, we multiply their rate of growth and the time they’ve grown.
So, after minutes, the fingernails would be km long. To put this in perspective, km is a millimeter, and minutes is about 4.16 days. So, after about 4.16 days, fingernails have grown about 4.8 millimeters.
There are approximately \(1 \times 10^{12}\) grains of sand in a cubic meter. If the number of grains of sand on the Australian coastline is roughly \(7.5 \times 10^{21}\) grains, roughly how many cubic meters of sand is there on the Australian coastline? of sand on the Australian coastline is roughly \(7.5 \times {10^{21}}\) grains, roughly how many cubic meters of sand is there on the Australian coastline?
As mentioned in the opening to this section, it is estimated that we’re producing 2.5 quintillion bytes of data per day. A good estimate is that there are 7.674 billion people on the planet. Convert both of those numbers to scientific notation, and then determine how much data is being generated per person each day.
- Answer
-
Written in standard form, 2.5 quintillion is 2,500,000,000,000,000,000. Changing that to scientific notation, move the decimal 18 places, so 2.5 quintillion bytes = bytes. Writing 7.647 billion in scientific notation would be because a billion is 1,000,000,000 = . So, to find out how much data is being produced daily per person, we would divide these two numbers.
In standard form, that’s 327,000,000 bytes per person, so 327 million bytes of data daily are being produced per person.
Humans collectively exhale approximately \(6.4235 \times 10^{12}\) pounds of carbon dioxide per year. There are approximately \(7.647 \times 10^9\) humans currently living on Earth. How many pounds of carbon dioxide does a single human, on average, exhale per year?
Application of Scientific Notation
What Numbers Could Be Considered “Too Big” or “Too Small”?
One wonders when the numbers we represent become too large or small for consideration. Perhaps the following examples put limits on what is meaningful. The number of particles in the known universe has been estimated at
particles. The smallest distance that has been measured is
Check Your Understanding
- Write 0.00456 in scientific notation.
- Write \(5.67 \times {10^8}\) in standard form.
- Calculate \(4.5 \times {10^3} + 9.8 \times {10^2}\).
- Calculate \(2.5 \times {10^5} - 9.8 \times {10^6}\).
- Calculate \((7.4 \times {10^4}) \times (4.8 \times {10^3})\).
-
Calculate \(\frac
(click for details)
Callstack: at (Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/03:_Real_Number_Systems_and_Number_Theory/3.09:__Scientific_Notation), /content/body/div[7]/section/div/ol/li[6]/span[1], line 1, column 5
(click for details)\).Callstack: at (Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/03:_Real_Number_Systems_and_Number_Theory/3.09:__Scientific_Notation), /content/body/div[7]/section/div/ol/li[6]/span[2], line 1, column 3
- The distance from Earth to the moon is \(1.514 \times {10^{10}}\) inches. The thickness of a dollar bill is \(4.3 \times {10^{ - 3}}\) inches. How many dollar bills must be stacked so the pile reaches the moon?


0: Unidad Introductoria Last updated Save as PDF Page ID 282937 0.0: Unidad Introductoria - Objetivos 0.1: What is Proficiency? 0.1.0: Three Modes of Communication 0.1.1: Path to Proficiency 0.1.2: Goal Setting 0.2: Why Study Spanish? 0.2.0: Benefits of Language Learning 0.2.1: Spanish in the Workforce 0.2.2: Spanish Certificates at Butler 0.3: Tips to Studying Spanish 0.3.0: Cognates 0.3.1: Context Clues 0.3.2: Auditory Studying 0.3.3: Study Schedule 0.4: Class Expectations 0.4.0: Course Load Recommendations 0.4.1: Growth Mindset 0.4.2: Required Class Materials 0.4.3: Academic Honesty Policy (Use of Online Translators and Artificial Intelligence) 0.5: Class Resources 0.5.0: Using Accents Online and the Spanish Keyboard 0.5.1: Viewing Instructor Feedback 0.5.2: "What If" Score using Canvas Grade Book


13.1: Reading- Access Training
Required Trainings
Read, view, and complete the following tutorials at GCF LearnFree.org , ( http://www.gcflearnfree.org/access2013 ).
Access Basics
- #3 – INTRODUCTION TO DATABASES
- #4 – INTRODUCTION TO OBJECTS
- #5 – GETTING STARTED WITH ACCESS
- #6 – MANAGING DATABASES AND OBJECTS
Working with Data
- #7 – WORKING WITH TABLES
- #8 – WORKING WITH FORMS
- #9 – SORTING AND FILTERING RECORDS
Running Queries and reports
- #10- DESIGNING A SIMPLE QUERY
- #11- DESIGNING A MULTI-TABLE QUERY
- #12- MORE QUERY DESIGN OPTIONS
- #13- CREATING REPORTS
- #14- ADVANCED REPORT OPTIONS
Database Design Tips
- #15- MODIFYING TABLES
- #16- CREATING FORMS
- #17- FORMATTING FORMS
- #18 – DESIGNING YOUR OWN DATABASE
More Access Tasks
- #19 – CREATING CALCULATED FIELDS AND TOTALS ROWS
- #20 – CREATING A PARAMETER QUERY
- #21 – CREATING A FIND DUPLICATES QUERY
Optional Additional Tutorials are available from GCF Learnfree.org under EXTRAS.
Supplemental Trainings
You can find the following supplemental training by clicking the MS Office Tutorials link in the course menu.
- Atomic Learning – Use these videos to supplement your GCF Learn Free Tutorials as necessary.
- MS Office Tutorials – Watch the Access tutorials in MS Office 2013 Video Tutorials link or the MS Office 2010 Tutorials link .
- Access Help Menu – Additional learning materials are also available from the Access HELP menu. Click on the BLUE QUESTION MARK in the upper right hand corner of your application HOME page. This will take you to the Access Help page where you will find tutorials and a searchable index of topics.
- Introduction to Computer Applications and Concepts. Authored by : Jim Shannon. Provided by : Extended Learning Institute of Northern Virginia Community College. Located at : http://www.nvcc.edu/eli/index.html . License : CC BY: Attribution


17.11: Other Useful Data Structures
-
- Last updated
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Up to this point we have encountered several different kinds of variables. At the simplest level, we’ve seen numeric data, logical data and character data. However, we’ve also encountered some more complicated kinds of variables, namely factors, formulas, data frames and lists. We’ll see a few more specialised data structures later on in this book, but there’s a few more generic ones that I want to talk about in passing. None of them are central to the rest of the book (and in fact, the only one we’ll even see anywhere else is the matrix), but they do crop up a fair bit in real life.
Matrices
In various different places in this chapter I’ve made reference to an R data structure called a
matrix
, and mentioned that I’d talk a bit more about matrices later on. That time has come. Much like a data frame, a matrix is basically a big rectangular table of data, and in fact there are quite a few similarities between the two. However, there are also some key differences, so it’s important to talk about matrices in a little detail. Let’s start by using
rbind()
to create a small matrix:
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row.1 <- c( 2,3,1 ) # create data for row 1
row.2 <- c( 5,6,7 ) # create data for row 2
M <- rbind( row.1, row.2 ) # row bind them into a matrix
print( M ) # and print it out...
## [,1] [,2] [,3]
## row.1 2 3 1
## row.2 5 6 7
The variable
M
is a matrix, which we can confirm by using the
class()
function. Notice that, when we bound the two vectors together, R retained the names of the original variables as row names. We could delete these if we wanted by typing
rownames(M)<-NULL
, but I generally prefer having meaningful names attached to my variables, so I’ll keep them. In fact, let’s also add some highly unimaginative column names as well:
colnames(M) <- c( "col.1", "col.2", "col.3" )
print(M)
## col.1 col.2 col.3
## row.1 2 3 1
## row.2 5 6 7
You can use square brackets to subset a matrix in much the same way that you can for data frames, again specifying a row index and then a column index. For instance,
M[2,3]
pulls out the entry in the 2nd row and 3rd column of the matrix (i.e.,
7
), whereas
M[2,]
pulls out the entire 2nd row, and
M[,3]
pulls out the entire 3rd column. However, it’s worth noting that when you pull out a column, R will print the results horizontally, not vertically. The reason for this relates to how matrices (and arrays generally) are implemented. The original matrix
M
is treated as a two-dimensional objects, containing 2 rows and 3 columns. However, whenever you pull out a single row or a single column, the result is considered to be one-dimensional. As far as R is concerned there’s no real reason to distinguish between a one-dimensional object printed vertically (a column) and a one-dimensional object printed horizontally (a row), and it prints them all out horizontally.
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There is also a way of using only a single index, but due to the internal structure to how R defines a matrix, it works very differently to what we saw previously with data frames.
The single-index approach is illustrated in Table 7.5 but I don’t really want to focus on it since we’ll never really need it for this book, and matrices don’t play anywhere near as large a role in this book as data frames do. The reason for these differences is that for this is that, for both data frames and matrices, the “row and column” version exists to allow the human user to interact with the object in the psychologically meaningful way: since both data frames and matrices are basically just tables of data, it’s the same in each case. However, the single-index version is really a method for you to interact with the object in terms of its internal structure, and the internals for data frames and matrices are quite different.
Table 7.5: The row and column version, which is identical to the corresponding indexing scheme for a data frame of the same size.
| Row | Col.1 | Col.2 | Col.3 |
|---|---|---|---|
| Row 1 | [1,1] | [1,2] | [1,3] |
| Row 2 | [2,1] | [2,2] | [2,3] |
Table 7.5: The single-index version, which is quite different to what we would get with a data frame.
| Row | Col.1 | Col.2 | Col.3 |
|---|---|---|---|
| Row 1 | 1 | 3 | 5 |
| Row 2 | 2 | 4 | 6 |
The critical difference between a data frame and a matrix is that, in a data frame, we have this notion that each of the columns corresponds to a different variable: as a consequence, the columns in a data frame can be of different data types. The first column could be numeric, and the second column could contain character strings, and the third column could be logical data. In that sense, there is a fundamental asymmetry build into a data frame, because of the fact that columns represent variables (which can be qualitatively different to each other) and rows represent cases (which cannot). Matrices are intended to be thought of in a different way. At a fundamental level, a matrix really is just one variable: it just happens that this one variable is formatted into rows and columns. If you want a matrix of numeric data, every single element in the matrix must be a number. If you want a matrix of character strings, every single element in the matrix must be a character string. If you try to mix data of different types together, then R will either spit out an error, or quietly coerce the underlying data into a list. If you want to find out what class R secretly thinks the data within the matrix is, you need to do something like this:
class( M[1] )
## [1] "numeric"
You can’t type
class(M)
, because all that will happen is R will tell you that
M
is a matrix: we’re not interested in the class of the matrix itself, we want to know what class the underlying data is assumed to be. Anyway, to give you a sense of how R enforces this, let’s try to change one of the elements of our numeric matrix into a character string:
M[1,2] <- "text"
M
## col.1 col.2 col.3
## row.1 "2" "text" "1"
## row.2 "5" "6" "7"
It looks as if R has coerced all of the data in our matrix into character strings. And in fact, if we now typed in
class(M[1])
we’d see that this is exactly what has happened. If you alter the contents of one element in a matrix, R will change the underlying data type as necessary.
There’s only one more thing I want to talk about regarding matrices. The concept behind a matrix is very much a mathematical one, and in mathematics a matrix is a most definitely a two-dimensional object. However, when doing data analysis, we often have reasons to want to use higher dimensional tables (e.g., sometimes you need to cross-tabulate three variables against each other). You can’t do this with matrices, but you can do it with
arrays
. An array is just like a matrix, except it can have more than two dimensions if you need it to. In fact, as far as R is concerned a matrix is just a special kind of array, in much the same way that a data frame is a special kind of list. I don’t want to talk about arrays too much, but I will very briefly show you an example of what a 3D array looks like. To that end, let’s cross tabulate the
speaker
and
utterance
variables from the
nightgarden.Rdata
data file, but we’ll add a third variable to the cross-tabs this time, a logical variable which indicates whether or not I was still awake at this point in the show:
dan.awake <- c( TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,FALSE )
Now that we’ve got all three variables in the workspace (assuming you loaded the
nightgarden.Rdata
data earlier in the chapter) we can construct our three way cross-tabulation, using the
table()
function.
xtab.3d <- table( speaker, utterance, dan.awake )
xtab.3d
## , , dan.awake = FALSE
##
## utterance
## speaker ee onk oo pip
## makka-pakka 0 2 0 2
## tombliboo 0 0 1 0
## upsy-daisy 0 0 0 0
##
## , , dan.awake = TRUE
##
## utterance
## speaker ee onk oo pip
## makka-pakka 0 0 0 0
## tombliboo 1 0 0 0
## upsy-daisy 0 2 0 2
Hopefully this output is fairly straightforward: because R can’t print out text in three dimensions, what it does is show a sequence of 2D slices through the 3D table. That is, the
, , dan.awake = FALSE
part indicates that the 2D table that follows below shows the 2D cross-tabulation of
speaker
against
utterance
only for the
dan.awake = FALSE
instances, and so on.
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Ordered factors
One topic that I neglected to mention when discussing factors previously (Section 4.7 is that there are actually two different types of factor in R, unordered factors and ordered factors. An unordered factor corresponds to a nominal scale variable, and all of the factors we’ve discussed so far in this book have been unordered (as will all the factors used anywhere else except in this section). However, it’s often very useful to explicitly tell R that your variable is
ordinal scale
, and if so you need to declare it to be an
ordered factor
. For instance, earlier in this chapter we made use of a variable consisting of Likert scale data, which we represented as the
likert.raw
variable:
likert.raw
## [1] 1 7 3 4 4 4 2 6 5 5
We can declare this to be an ordered factor in by using the
factor()
function, and setting
ordered = TRUE
. To illustrate how this works, let’s create an ordered factor called
likert.ordinal
and have a look at it:
likert.ordinal <- factor( x = likert.raw, # the raw data
levels = seq(7,1,-1), # strongest agreement is 1, weakest is 7
ordered = TRUE ) # and it's ordered
print( likert.ordinal )
## [1] 1 7 3 4 4 4 2 6 5 5
## Levels: 7 < 6 < 5 < 4 < 3 < 2 < 1
Notice that when we print out the ordered factor, R explicitly tells us what order the levels come in. Because I wanted to order my levels in terms of
increasing
strength of agreement, and because a response of 1 corresponded to the strongest agreement and 7 to the strongest disagreement, it was important that I tell R to encode 7 as the lowest value and 1 as the largest. Always check this when creating an ordered factor: it’s very easy to accidentally encode your data “upside down” if you’re not paying attention. In any case, note that we can (and should) attach meaningful names to these factor levels by using the
levels()
function, like this:
levels( likert.ordinal ) <- c( "strong.disagree", "disagree", "weak.disagree",
"neutral", "weak.agree", "agree", "strong.agree" )
print( likert.ordinal )
## [1] strong.agree strong.disagree weak.agree neutral
## [5] neutral neutral agree disagree
## [9] weak.disagree weak.disagree
## 7 Levels: strong.disagree < disagree < weak.disagree < ... < strong.agree
One nice thing about using ordered factors is that there are a lot of analyses for which R automatically treats ordered factors differently from unordered factors, and generally in a way that is more appropriate for ordinal data. However, since I don’t discuss that in this book, I won’t go into details. Like so many things in this chapter, my main goal here is to make you aware that R has this capability built into it; so if you ever need to start thinking about ordinal scale variables in more detail, you have at least some idea where to start looking!
Dates and times
Times and dates are very annoying types of data. To a first approximation we can say that there are 365 days in a year, 24 hours in a day, 60 minutes in an hour and 60 seconds in a minute, but that’s not quite correct. The length of the solar day is not exactly 24 hours, and the length of solar year is not exactly 365 days, so we have a complicated system of corrections that have to be made to keep the time and date system working. On top of that, the measurement of time is usually taken relative to a local time zone, and most (but not all) time zones have both a standard time and a daylight savings time, though the date at which the switch occurs is not at all standardised.  Unfortunately, they’re also important. Sometimes it’s possible to avoid having to use any complicated system for dealing with times and dates. Often you just want to know what year something happened in, so you can just use numeric data: in quite a lot of situations something as simple as
this.year <- 2011
works just fine. If you can get away with that for your application, this is probably the best thing to do. However, sometimes you really do need to know the actual date. Or, even worse, the actual time. In this section, I’ll very briefly introduce you to the basics of how R deals with date and time data. As with a lot of things in this chapter, I won’t go into details because I don’t use this kind of data anywhere else in the book. The goal here is to show you the basics of what you need to do if you ever encounter this kind of data in real life. And then we’ll all agree never to speak of it again.
To start with, let’s talk about the date. As it happens, modern operating systems are very good at keeping track of the time and date, and can even handle all those annoying timezone issues and daylight savings pretty well. So R takes the quite sensible view that it can just ask the operating system what the date is. We can pull the date using the
Sys.Date()
function:
today <- Sys.Date() # ask the operating system for the date
print(today) # display the date
## [1] "2018-12-30"
Okay, that seems straightforward. But, it does rather look like
today
is just a character string, doesn’t it? That would be a problem, because dates really do have a numeric character to them, and it would be nice to be able to do basic addition and subtraction to them. Well, fear not. If you type in
class(today)
, R will tell you that the class of the
today
variable is
"Date"
. What this means is that, hidden underneath this text string that prints out an actual date, R actually has a numeric representation.
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What that means is that you actually can add and subtract days. For instance, if we add 1 to
today
, R will print out the date for tomorrow:
today + 1
## [1] "2018-12-31"
Let’s see what happens when we add 365 days:
today + 365
## [1] "2019-12-30"
This is particularly handy if you forget that a year is a leap year since in that case you’d probably get it wrong is doing this in your head. R provides a number of functions for working with dates, but I don’t want to talk about them in any detail. I will, however, make passing mention of the
weekdays()
function which will tell you what day of the week a particular date corresponded to, which is extremely convenient in some situations:
weekdays( today )
## [1] "Sunday"
I’ll also point out that you can use the
as.Date()
to convert various different kinds of data into dates. If the data happen to be strings formatted exactly according to the international standard notation (i.e.,
yyyy-mm-dd
) then the conversion is straightforward, because that’s the format that R expects to see by default. You can convert dates from other formats too, but it’s slightly trickier, and beyond the scope of this book.
What about times? Well, times are even more annoying, so much so that I don’t intend to talk about them at all in this book, other than to point you in the direction of some vaguely useful things. R itself does provide you with some tools for handling time data, and in fact there are two separate classes of data that are used to represent times, known by the odd names
POSIXct
and
POSIXlt
. You can use these to work with times if you want to, but for most applications you would probably be better off downloading the
chron
package, which provides some much more user friendly tools for working with times and dates.


14.13: Introduction to Source Analysis
What you’ll learn to do: examine methods used to assess the quality and reliability of sources
You have been asked to write an article for your university newspaper about what we’ve learned in the 30+ years since the Chernobyl Disaster. In 1986, an explosion at the Chernobyl nuclear power plant in the Soviet Union released radioactive particles into the air over much of the USSR and Western Europe. You’ve already done some basic research on Wikipedia, then used Google Scholar to investigate some more of the health effects. Finally, you searched in your university health, science, and medicine databases to learn specifics about the impact of the disaster.
How do you know which sources are worth using in your article? How will you know if the sources are even good? Journalists famously cover the 5 Ws (who, what, where, when, why…and how) in their articles, and these similar questions can be used to evaluate your search results:
- Who : Who is the author and what are his/her credentials in this topic?
- What: Is the material primary or secondary in nature?
- Where: Is the publisher or organization behind the source considered reputable? Does the website appear legitimate?
- When: Is the source current or does it cover the right time period for your topic?
- Why: Is the opinion or bias of the author apparent and can it be taken into account?
- How: Is the source written at the right level for your needs? Is the research well-documented?
If you can answer all of these questions, you’ll understand more about the quality and usefulness of a source for your article.
In this section, you’ll learn more about tools like this that help you examine the usefulness and appropriateness of information for your research. You’ll use the C.R.A.A.P. test to evaluate a source, four moves to evaluate online information, and consider techniques to help you synthesize pieces from multiple sources in your writing.
Contributors and Attributions
- Outcome: Source Analysis. Provided by : Lumen Learning. License : CC BY-SA: Attribution-ShareAlike
- Research strategy guide for finding quality, credible sources. Provided by : Journalist's Resource . Located at : http://journalistsresource.org/tip-sheets/research/research-strategy-guide . Project : Project of the Harvard Kennedy School's Shorenstein Center and the Carnegie-Knight Initiative. License : CC BY-SA: Attribution-ShareAlike
- Radiation warning sign. Authored by : Diana Markosian. Provided by : Wikipedia. Located at : https://en.Wikipedia.org/wiki/Chernobyl_disaster#/media/File:VOA_Markosian_-_Chernobyl02.jpg . License : Public Domain: No Known Copyright


6.2: Orthogonal Complements and the Matrix Tranpose
\(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \)
\(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \)
\(\newcommand{\fourvec}[4]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \end{pmatrix} } \)
\(\newcommand{\fivevec}[5]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{pmatrix} } \)
We've now seen how the dot product enables us to determine the angle between two vectors and, more specifically, when two vectors are orthogonal. Moving forward, we will explore how the orthogonality condition simplifies many common tasks, such as expressing a vector as a linear combination of a given set of vectors.
This section introduces the notion of an orthogonal complement, the set of vectors each of which is orthogonal to a prescribed subspace. We'll also find a way to describe dot products using matrix products, which allows us to study orthogonality using many of the tools for understanding linear systems that we developed earlier.
Preview Activity 6.2.1.
- Sketch the vector \(\mathbf v=\twovec{-1}2\) on Figure 6.2.1 and one vector that is orthogonal to it.
- If a vector \(\mathbf x\) is orthogonal to \(\mathbf v\text{,}\) what do we know about the dot product \(\mathbf v\cdot\mathbf x\text{?}\)
- If we write \(\mathbf x=\twovec xy\text{,}\) use the dot product to write an equation for the vectors orthogonal to \(\mathbf v\) in terms of \(x\) and \(y\text{.}\)
- Use this equation to sketch the set of all vectors orthogonal to \(\mathbf v\) in Figure 6.2.1.
- Section 3.5 introduced the column space \(Col(A)\) and null space \(Nul(A)\) of a matrix \(A\text{.}\) Suppose that \(A\) is a matrix and \(\mathbf x\) is a vector satisfying \(A\mathbf x=\mathbf 0\text{.}\) Does \(\mathbf x\) belong to \(Nul(A)\) or \(Col(A)\text{?}\)
- Suppose that the equation \(A\mathbf x=\mathbf b\) is consistent. Does \(\mathbf b\) belong to \(Nul(A)\) or \(Col(A)\text{?}\)
Orthogonal complements
The preview activity presented us with a vector \(\mathbf v\) and led us through the process of describing all the vectors orthogonal to \(\mathbf v\text{.}\) Notice that the set of scalar multiples of \(\mathbf v\) describes a line \(L\text{,}\) a 1-dimensional subspace of \(\mathbb R^2\text{.}\) We then described a second line consisting of all the vectors orthogonal to \(\mathbf v\text{.}\) Notice that every vector on this line is orthogonal to every vector on the line \(L\text{.}\) We call this new line the orthogonal complement of \(L\) and denote it by \(L^\perp\text{.}\) The lines \(L\) and \(L^\perp\) are illustrated on the left of Figure 6.2.2.
The next definition places this example into a more general context.
Given a subspace \(W\) of \(\mathbb R^m\text{,}\) the orthogonal complement of \(W\) is the set of vectors in \(\mathbb R^m\) each of which is orthogonal to every vector in \(W\text{.}\) We denote the orthogonal complement by \(W^\perp\text{.}\)
A typical example appears on the right of Figure 6.2.2. Here we see a plane \(W\text{,}\) a two-dimensional subspace of \(\mathbb R^3\text{,}\) and its orthogonal complement \(W^\perp\text{,}\) which is a line in \(\mathbb R^3\text{.}\)
As we'll soon see, the orthogonal complement of a subspace \(W\) is itself a subspace of \(\mathbb R^m\text{.}\)
Activity 6.2.2.
Suppose that \(\mathbf w_1=\threevec10{-2}\) and \(\mathbf w_2=\threevec11{-1}\) form a basis for \(W\text{,}\) a two-dimensional subspace of \(\mathbb R^3\text{.}\) We will find a description of the orthogonal complement \(W^\perp\text{.}\)
- Suppose that the vector \(\mathbf x\) is orthogonal to \(\mathbf w_1\text{.}\) If we write \(\mathbf x=\threevec{x_1}{x_2}{x_3}\text{,}\) use the fact that \(\mathbf w_1\cdot\mathbf x = 0\) to write a linear equation for \(x_1\text{,}\) \(x_2\text{,}\) and \(x_3\text{.}\)
- Suppose that \(\mathbf x\) is also orthogonal to \(\mathbf w_2\text{.}\) In the same way, write a linear equation for \(x_1\text{,}\) \(x_2\text{,}\) and \(x_3\) that arises from the fact that \(\mathbf w_2\cdot\mathbf x = 0\text{.}\)
- If \(\mathbf x\) is orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\text{,}\) these two equations give us a linear system \(B\mathbf x=\mathbf 0\) for some matrix \(B\text{.}\) Identify the matrix \(B\) and write a parametric description of the solution space to the equation \(B\mathbf x = \mathbf 0\text{.}\)
-
Since \(\mathbf w_1\) and \(\mathbf w_2\) form a basis for the two-dimensional subspace \(W\text{,}\) any vector in \(\mathbf w\) in \(W\) can be written as a linear combination
\begin{equation*} \mathbf w = c_1\mathbf w_1 + c_2\mathbf w_2\text{.} \end{equation*}
If \(\mathbf x\) is orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\text{,}\) use the distributive property of dot products to explain why \(\mathbf x\) is orthogonal to \(\mathbf w\text{.}\)
- Give a basis for the orthogonal complement \(W^\perp\) and state the dimension \(DimW^\perp\text{.}\)
- Describe \((W^\perp)^\perp\text{,}\) the orthogonal complement of \(W^\perp\text{.}\)
If \(L\) is the line defined by \(\mathbf v=\threevec1{-2}3\) in \(\mathbb R^3\text{,}\) we will describe the orthogonal complement \(L^\perp\text{,}\) the set of vectors orthogonal to \(L\text{.}\)
If \(\mathbf x\) is orthogonal to \(L\text{,}\) it must be orthogonal to \(\mathbf v\) so we have
We can describe the solutions to this equation parametrically as
Therefore, the orthogonal complement \(L^\perp\) is a plane, a two-dimensional subspace of \(\mathbb R^3\text{,}\) spanned by the vectors \(\threevec210\) and \(\threevec{-3}01\text{.}\)
Suppose that \(W\) is the \(2\)-dimensional subspace of \(\mathbb R^5\) with basis
We will give a description of the orthogonal complement \(W^\perp\text{.}\)
If \(\mathbf x\) is in \(W^\perp\text{,}\) we know that \(\mathbf x\) is orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\text{.}\) Therefore,
In other words, \(B\mathbf x=\mathbf 0\) where
The solutions may be described parametrically as
The distributive property of dot products implies that any vector that is orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\) is also orthogonal to any linear combination of \(\mathbf w_1\) and \(\mathbf w_2\) since
Therefore, \(W^\perp\) is a \(3\)-dimensional subspace of \(\mathbb R^5\) with basis
One may easily check that the vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\) are orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\text{.}\)
The matrix transpose
The previous activity and examples show how we can describe the orthogonal complement of a subspace as the solution set of a particular linear system. We will make this connection more explicit by defining a new matrix operation called the transpose .
The transpose of the \(m\times n\) matrix \(A\) is the \(n\times m\) matrix \(A^T\) whose rows are the columns of \(A\text{.}\)
If \(A=\begin{bmatrix} 4 & -3 & 0 & 5 \\ -1 & 2 & 1 & 3 \\ \end{bmatrix} \text{,}\) then \(A^T=\begin{bmatrix} 4 & -1 \\ -3 & 2 \\ 0 & 1 \\ 5 & 3 \\ \end{bmatrix}\)
Activity 6.2.3.
This activity illustrates how multiplying a vector by \(A^T\) is related to computing dot products with the columns of \(A\text{.}\) You'll develop a better understanding of this relationship if you compute the dot products and matrix products in this activity without using technology.
- If \(B = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & -2 \\ \end{bmatrix} \text{,}\) write the matrix \(B^T\text{.}\)
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Suppose that
\begin{equation*} \mathbf v_1=\threevec20{-2},\hspace{24pt} \mathbf v_2=\threevec112,\hspace{24pt} \mathbf w=\threevec{-2}23\text{.} \end{equation*}
Find the dot products \(\mathbf v_1\cdot\mathbf w\) and \(\mathbf v_2\cdot\mathbf w\text{.}\)
- Now write the matrix \(A = \begin{bmatrix} \mathbf v_1 & \mathbf v_2 \end{bmatrix}\) and its transpose \(A^T\text{.}\) Find the product \(A^T\mathbf w\) and describe how this product computes both dot products \(\mathbf v_1\cdot\mathbf w\) and \(\mathbf v_2\cdot\mathbf w\text{.}\)
- Suppose that \(\mathbf x\) is a vector that is orthogonal to both \(\mathbf v_1\) and \(\mathbf v_2\text{.}\) What does this say about the dot products \(\mathbf v_1\cdot\mathbf x\) and \(\mathbf v_2\cdot\mathbf x\text{?}\) What does this say about the product \(A^T\mathbf x\text{?}\)
- Use the matrix \(A^T\) to give a parametric description of all the vectors \(\mathbf x\) that are orthogonal to \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
- Remember that \(Nul(A^T)\text{,}\) the null space of \(A^T\text{,}\) is the solution set of the equation \(A^T\mathbf x=\mathbf 0\text{.}\) If \(\mathbf x\) is a vector in \(Nul(A^T)\text{,}\) explain why \(\mathbf x\) must be orthogonal to both \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
- Remember that \(Col(A)\text{,}\) the column space of \(A\text{,}\) is the set of linear combinations of the columns of \(A\text{.}\) Therefore, any vector in \(Col(A)\) can be written as \(c_1\mathbf v_1+c_2\mathbf v_2\text{.}\) If \(\mathbf x\) is a vector in \(Nul(A^T)\text{,}\) explain why \(\mathbf x\) is orthogonal to every vector in \(Col(A)\text{.}\)
The previous activity demonstrates an important connection between the matrix transpose and dot products. More specifically, the components of the product \(A^T\mathbf x\) are simply the dot products of the columns of \(A\) with \(\mathbf x\text{.}\) We will put this observation to use quite often so let's record it as a proposition.
If \(A\) is the matrix whose columns are \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\text{,}\) then
Suppose that \(W\) is a subspace of \(\mathbb R^4\) having basis
and we wish to describe the orthogonal complement \(W^\perp\text{.}\)
If \(A\) is the matrix \(A = \begin{bmatrix}\mathbf w_1 & \mathbf w_2\end{bmatrix}\) and \(\mathbf x\) is in \(W^\perp\text{,}\) we have
Describing vectors \(\mathbf x\) that are orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\) is therefore equivalent to the more familiar task of describing the solution set \(A^T\mathbf x = \mathbf 0\text{.}\) To do so, we find the reduced row echelon form of \(A^T\) and write the solution set parametrically as
Once again, the distributive property of dot products tells us that such a vector is also orthogonal to any linear combination of \(\mathbf w_1\) and \(\mathbf w_2\) so this solution set is, in fact, the orthogonal complement \(W^\perp\text{.}\) Indeed, we see that the vectors
form a basis for \(W^\perp\text{,}\) which is a two-dimensional subspace of \(\mathbb R^4\text{.}\)
To place this example in a slightly more general context, note that \(\mathbf w_1\) and \(\mathbf w_2\text{,}\) the columns of \(A\text{,}\) form a basis of \(W\text{.}\) Since \(Col(A)\text{,}\) the column space of \(A\) is the subspace of linear combinations of the columns of \(A\text{,}\) we have \(W=Col(A)\text{.}\)
This example also shows that the orthogonal complement \(W^\perp = Col(A)^\perp\) is described by the solution set of \(A^T\mathbf x = \mathbf 0\text{.}\) This solution set is what we have called \(Nul(A^T)\text{,}\) the null space of \(A^T\text{.}\) In this way, we see the following proposition, which is illustrated in Figure 6.2.11.
For any matrix \(A\text{,}\) the orthogonal complement of \(Col(A)\) is \(Nul(A^T)\text{;}\) that is,
Properties of the matrix transpose
The transpose is a simple algebraic operation performed on a matrix. The next activity explores some of its properties.
Activity 6.2.4.
In Sage, the transpose of a matrix
A
is given by
A.T
. Define the matrices
- Evaluate \((A+B)^T\) and \(A^T+B^T\text{.}\) What do you notice about the relationship between these two matrices?
- What happens if you transpose a matrix twice; that is, what is \((A^T)^T\text{?}\)
- Find \(\det(C)\) and \(\det(C^T)\text{.}\) What do you notice about the relationship between these determinants?
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- Find the product \(AC\) and its transpose \((AC)^T\text{.}\)
- Is it possible to compute the product \(A^TC^T\text{?}\) Explain why or why not.
- Find the product \(C^TA^T\) and compare it to \((AC)^T\text{.}\) What do you notice about the relationship between these two matrices?
- What is the transpose of the identity matrix \(I\text{?}\)
- If a square matrix \(D\) is invertible, explain why you can guarantee that \(D^T\) is invertible and why \((D^T)^{-1} = (D^{-1})^T\text{.}\)
In spite of the fact that we are looking at some specific examples, this activity demonstrates the following general properties of the tranpose, which may be verified with a little effort.
Properties of the transpose.
Here are some properties of the matrix transpose, expressed in terms of general matrices \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\) We assume that \(C\) is a square matrix.
- If \(A+B\) is defined, then \((A+B)^T = A^T+B^T\text{.}\)
- \((sA)^T = sA^T\text{.}\)
- \((A^T)^T = A\text{.}\)
- \(\det(C) = \det(C^T)\text{.}\)
- If \(AB\) is defined, then \((AB)^T = B^TA^T\text{.}\) Notice that the order of the multiplication is reversed.
- \((C^T)^{-1} = (C^{-1})^T\text{.}\)
There is one final property we wish to record though we will wait until Section 7.4 to explain why it is true.
For any matrix \(A\text{,}\) we have
This proposition is important because it implies a relationship between the dimensions of a subspace and its orthogonal complement. For instance, if \(A\) is an \(m\times n\) matrix, we saw in Section 3.5 that \(Dim\;Col(A) = Rank(A)\) and \(Dim\;Nul(A) = n-Rank(A)\text{.}\)
Now suppose that \(W\) is an \(n\)-dimensional subspace of \(\mathbb R^m\) with basis \(\mathbf w_1,\mathbf w_2,\ldots,\mathbf w_n\text{.}\) If we form the \(m\times n\) matrix \(A=\begin{bmatrix}\mathbf w_1 & \mathbf w_2 & \ldots & \mathbf w_n\end{bmatrix}\text{,}\) then \(Col(A) = W\) so that
The transpose \(A^T\) is an \(n\times m\) matrix having \(Rank(A^T) = Rank(A)= n\text{.}\) Since \(W^\perp = Nul(A^T)\text{,}\) we have
This explains the following proposition.
If \(W\) is a subspace of \(\mathbb R^m\text{,}\) then
In Example 6.2.4, we constructed the orthogonal complement of a line in \(\mathbb R^3\text{.}\) The dimension of the orthogonal complement should be \(3 - 1 = 2\text{,}\) which explains why we found the orthogonal complement to be a plane.
In Example 6.2.5, we looked at \(W\text{,}\) a \(2\)-dimensional subspace of \(\mathbb R^5\) and found its orthogonal complement \(W^\perp\) to be a \(5-2=3\)-dimensional subspace of \(\mathbb R^5\text{.}\)
Activity 6.2.5.
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Suppose that \(W\) is a \(5\)-dimensional subspace of \(\mathbb R^9\) and that \(A\) is a matrix whose columns form a basis for \(W\text{;}\) that is, \(Col(A) = W\text{.}\)
- What are the dimensions of \(A\text{?}\)
- What is the rank of \(A\text{?}\)
- What are the dimensions of \(A^T\text{?}\)
- What is the rank of \(A^T\text{?}\)
- What is \(Dim\;Nul(A^T)\text{?}\)
- What is \(DimW^\perp\text{?}\)
- How are the dimensions of \(W\) and \(W^\perp\) related?
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Suppose that \(W\) is a subspace of \(\mathbb R^4\) having basis
\begin{equation*} \mathbf w_1 = \fourvec102{-1},\hspace{24pt} \mathbf w_2 = \fourvec{-1}2{-6}3. \end{equation*}
- Find the dimensions \(DimW\) and \(DimW^\perp\text{.}\)
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Find a basis for \(W^\perp\text{.}\) It may be helpful to know that the Sage command
A.right_kernel()
produces a basis for \(Nul(A)\text{.}\) - Verify that each of the basis vectors you found for \(W^\perp\) are orthogonal to the basis vectors for \(W\text{.}\)
Summary
This section introduced the matrix tranpose, its connection to dot products, and its use in describing the orthogonal complement of a subspace.
- The columns of the matrix \(A\) are the rows of the matrix transpose \(A^T\text{.}\)
- The components of the product \(A^T\mathbf x\) are the dot products of \(\mathbf x\) with the columns of \(A\text{.}\)
- The orthogonal complement of the column space of \(A\) equals the null space of \(A^T\text{;}\) that is, \(Col(A)^\perp = Nul(A^T)\text{.}\)
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If \(W\) is a subspace of \(\mathbb R^p\text{,}\) then
\begin{equation*} Dim\;W+ DimW^\perp = p. \end{equation*}
Exercises 6.2.5Exercises
Suppose that \(W\) is a subspace of \(\mathbb R^4\) with basis
- What are the dimensions \(DimW\) and \(DimW^\perp\text{?}\)
- Find a basis for \(W^\perp\text{.}\)
- Verify that each of the basis vectors for \(W^\perp\) are orthogonal to \(\mathbf w_1\) and \(\mathbf w_2\text{.}\)
Consider the matrix \(A = \begin{bmatrix} -1 & -2 & -2 \\ 1 & 3 & 4 \\ 2 & 1 & -2 \end{bmatrix}\text{.}\)
- Find \(Rank(A)\) and a basis for \(Col(A)\text{.}\)
- Determine the dimension of \(Col(A)^\perp\) and find a basis for it.
Suppose that \(W\) is the subspace of \(\mathbb R^4\) defined as the solution set of the equation
- What are the dimensions \(DimW\) and \(DimW^\perp\text{?}\)
- Find a basis for \(W\text{.}\)
- Find a basis for \(W^\perp\text{.}\)
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In general, how can you easily find a basis for \(W^\perp\) when \(W\) is defined by
\begin{equation*} Ax_1+Bx_2+Cx_3+Dx_4 = 0? \end{equation*}
Determine whether the following statements are true or false and explain your reasoning.
- If \(A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \\ -3 & 1 \end{bmatrix}\text{,}\) then \(\mathbf x=\threevec{4}{-5}1\) is in \(Col(A)^\perp\text{.}\)
- If \(A\) is a \(2\times3\) matrix and \(B\) is a \(3\times4\) matrix, then \((AB)^T = A^TB^T\) is a \(4\times2\) matrix.
- If the columns of \(A\) are \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\) and \(A^T\mathbf x = \threevec{2}01\text{,}\) then \(\mathbf x\) is orthogonal to \(\mathbf v_2\text{.}\)
- If \(A\) is a \(4\times 4\) matrix with \(Rank(A) = 3\text{,}\) then \(Col(A)^\perp\) is a line in \(\mathbb R^4\text{.}\)
- If \(A\) is a \(5\times 7\) matrix with \(Rank(A) = 5\text{,}\) then \(Rank(A^T) = 7\text{.}\)
Apply properties of matrix operations to simplify the following expressions.
- \(\displaystyle A^T(BA^T)^{-1} \)
- \(\displaystyle (A+B)^T(A+B) \)
- \(\displaystyle [A(A+B)^T]^T \)
- \(\displaystyle (A + 2I)^T \)
A symmetric matrix \(A\) is one for which \(A=A^T\text{.}\)
- Explain why a symmetric matrix must be square.
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If \(A\) and \(B\) are general matrices and \(D\) is a square diagonal matrix, which of the following matrices can you guarantee are symmetric?
- \(\displaystyle D\)
- \(\displaystyle BAB^{-1} \)
- \(AA^T\text{.}\)
- \(\displaystyle BDB^T\)
If \(A\) is a square matrix, remember that the characteristic polynomial of \(A\) is \(\det(A-\lambda I)\) and that the roots of the characteristic polynomial are the eigenvalues of \(A\text{.}\)
- Explain why \(A\) and \(A^T\) have the same characteristic polynomial.
- Explain why \(A\) and \(A^T\) have the same set of eigenvalues.
- Suppose that \(A\) is diagonalizable with diagonalization \(A=PDP^{-1}\text{.}\) Explain why \(A^T\) is diagonalizable and find a diagonlization.
This exercise introduces a version of the Pythagorean theorem that we'll use later.
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Suppose that \(\mathbf v\) and \(\mathbf w\) are orthogonal to one another. Use the dot product to explain why
\begin{equation*} |{\mathbf v+\mathbf w}^2| = |{\mathbf v}^2 + |{\mathbf w}^2|. \end{equation*}
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Suppose that \(W\) is a subspace of \(\mathbb R^m\) and that \(\mathbf z\) is a vector in \(\mathbb R^m\) for which
\begin{equation*} \mathbf z = \mathbf x + \mathbf y, \end{equation*}
where \(\mathbf x\) is in \(W\) and \(\mathbf y\) is in \(W^\perp\text{.}\) Explain why
\begin{equation*} |{\mathbf z}^2| = |{\mathbf x}^2| + |{\mathbf y}^2|, \end{equation*}which is an expression of the Pythagorean theorem.
In the next chapter, symmetric matrices---that is, matrices for which \(A=A^T\)---play an important role. It turns out that eigenvectors of a symmetric matrix that are associated to different eigenvalues are orthogonal. We will explain this fact in this exercise.
- Viewing a vector as a matrix having one column, we may write \(\mathbf x\cdot\mathbf y = \mathbf x^T\mathbf y\text{.}\) If \(A\) is a matrix, explain why \(\mathbf x\cdot (A\mathbf y) = (A^T\mathbf x) \cdot \mathbf y\text{.}\)
- We have seen that the matrix \(A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) has eigenvectors \(\mathbf v_1=\twovec11\text{,}\) with associated eigenvalue \(\lambda_1=3\text{,}\) and \(\mathbf v_2 = \twovec{1}{-1}\text{,}\) with associated eigenvalue \(\lambda_2 = -1\text{.}\) Verify that \(A\) is symmetric and that \(\mathbf v_1\) and \(\mathbf v_2\) are orthogonal.
- Suppose that \(A\) is a general symmetric matrix and that \(\mathbf v_1\) is an eigenvector associated to eigenvalue \(\lambda_1\) and that \(\mathbf v_2\) is an eigenvector associated to a different eigenvalue \(\lambda_2\text{.}\) Beginning with \(\mathbf v_1\cdot (A\mathbf v_2)\text{,}\) apply the identity from the first part of this exercise to explain why \(\mathbf v_1\) and \(\mathbf v_2\) are orthogonal.
Given an \(m\times n\) matrix \(A\text{,}\) the row space of \(A\) is the column space of \(A^T\text{;}\) that is, \(Row(A) = Col(A^T)\text{.}\)
- Suppose that \(A\) is a \(7\times 15\) matrix. For what \(p\) is \(Row(A)\) a subspace of \(\mathbb R^p\text{?}\)
- How can Proposition 6.2.10 help us describe \(Row(A)^\perp\text{?}\)
- Suppose that \(A = \begin{bmatrix} -1 & -2 & 2 & 1 \\ 2 & 4 & -1 & 5 \\ 1 & 2 & 0 & 3 \end{bmatrix}\text{.}\) Find bases for \(Row(A)\) and \(Row(A)^\perp\text{.}\)


2.1: Introduction
As we begin this fingerspelling journey, we can take to heart Socrates words, “To know thyself is the beginning of wisdom.” What is your motivation for wanting to improve your fingerspelling skills? Is it that you tire of missing the lightning fast words flying off your Deaf friend’s hands while they giggle and poke fun?  Maybe you’re an awful speller and letters do a little dance in your head. There’s a common theme in almost all these examples—tension! We’ll get to talking about tension a bit later. But for now, let’s get to knowing thyself and spend some time with your motivation to improve your fingerspelling. There are two helpful perspectives that might drive your goals for this class:
1) Clarity – you may wish to improve your readability, smooth out the finger tangles. How do you do this?
Mindful Tip 1: Consciously (i.e. mindfully) move your palm orientation to the side to naturally emphasize the shape of your letters.
o Letters in which this can be done and still maintain conformity with ASL are: C, D, F, G, H, K, O, P, Q, X
- These 10 letters are almost 40% of the manual alphabet letters so...being mindful of your palm orientation is paramount to pull this off with success!
Why would you want to sign with the purpose of clarity or readability? Perhaps your goal is to be in the field of education or you work with children and they are just becoming familiar with the English language, or...perhaps you like to take it slow and chill when you’re chatting with your ASL using friends
2) Speed – you may wish to improve your ability to go fast. Now, before we go any further, here is a rule of thumb that brings us to another tip...
Mindful Tip 2: Never, ever, ever, never compromise clarity for speed.
Stop, please and say the italicized aloud to yourself three times. Never compromise clarity for speed can become a kind of mantra that each time you say it, waters the seeds of clarity in your beautiful brain!
Mindful Tip 3: For speed, consciously (i.e. mindfully) keep your palm orientation always pointed outward.
Palm facing out eliminates all movement and allows the hand to stay still, steady, and unmoving with no wasted effort. This translates to speed and ease of letters transitioning one to the other.
o Letters in which this can be done and still maintain conformity with ASL are same as the above: C, D, F, G, H, K, O, P, Q, X
- Again, much of the ASL manual alphabet can be signed with the palm orientation unmoving, however, clarity of each letter will go down.


8.20: Assignment- My Development Journal
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Last updated
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Save as PDF
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STEP 1
: Consider the following questions as you reflect on your developmental experiences.
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Think back to your elementary and junior high years. How would you describe your social competence and friendships?
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Body image is an important part of your self-esteem and is especially true during adolescence. Because of the rapid changes taking place, many adolescents are dissatisfied with their bodies. Think back to high school. How did you feel about your body? Try to imagine yourself on a typical day in high school. Compared to others, how did you feel about your: height, weight, breasts, muscles, ears, hips, legs, nose, hair, eyes, face, clothes, posture, nails, athletic ability, skin, teeth, smile, sexuality, and overall body? How do you think these early adolescent feelings have affected how you feel about yourself now?
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Now reflect on your adulthood experience. How do your feelings now contrast with your experiences in high school? How are your attitudes the same or different? How are your relationships the same or different? Do you plan to get married or are you already married or in a relationship? Do you have any children?
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What are your goals? When did you decide to go to college and why did you choose this college? Is this your first time in college? What difficulties are you experiencing? Is college like what you expected it to be; how is it different?
STEP 2
: Submit your assignment as
either
detailed responses to the five questions, OR as a reflection essay (between 400-600 words) that describes your experiences through life.
Sample Grading Rubric
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Criteria
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Proficient
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Developing
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Not Evident
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Points
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Complete response
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Writes in clear, descriptive sentences with no or few grammatical errors. Answers responses correctly and appropriately.
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Does not provide enough detail in the responses, or responds incorrectly.
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Incomplete responses or difficult to understand or follow.
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__/5
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Personal reflection
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Shows personal reflection and includes examples from own experiences.
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Demonstrates some personal reflection but is not thorough.
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Does not include personal reflection or demonstrate understanding of the material.
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__/5
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Total:
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__/10
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Contributors and Attributions
CC licensed content, Original
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Assignment: My Development Journal.
Authored by
: Nancee Ott.
License
:
CC BY: Attribution
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Modification, adaptation, and original content.
Provided by
: Lumen Learning.
License
:
CC BY: Attribution


8.3: Unit I- The Chemistry of Life
Our opening unit introduces students to the sciences, including the scientific method and the fundamental concepts of chemistry and physics that provide a framework within which learners comprehend biological processes.
Thumbnail: A cluster of Escherichia coli bacteria magnified 10,000 times. (Public Domain; Eric Erbe, digital colorization by Christopher Pooley, both of USDA, ARS, EMU).
- Contributors
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Connie Rye (East Mississippi Community College), Robert Wise (University of Wisconsin, Oshkosh), Vladimir Jurukovski (Suffolk County Community College), Jean DeSaix (University of North Carolina at Chapel Hill), Jung Choi (Georgia Institute of Technology), Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax (CC BY 4.0; Download for free at<EMAIL_ADDRESS>).


7.3.6D: Natural History of a Typical HIV Infection
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- Last updated
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- Ying Liu
- City College of San Francisco
State the median incubation period for AIDS and, in terms of viral load, exhaustion of the lymphopoietic system, and immune responses, briefly describe what marks the progression to AIDS.
Briefly describe the following:
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early or acute HIV infection
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chronic HIV infection
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AIDS
- early or acute HIV infection
- chronic HIV infection
- AIDS
According to WHO estimates from 2004, HIV has now infected 50 to 60 million people worldwide. The virus has killed over 22 million children adults and has left 14 million children orphaned. Worldwide, over 42 million people are currently living with HIV infection/AIDS - approximately 70% of these live in Africa, 20% in Asia. Around 3 million people die each year of AIDS and it is estimated that each day 14,000 people in the world become newly infected with HIV.
The median incubation period for AIDS is around 10 years. During early or acute HIV infection the virus primarily infects and destroys memory T4-lymphocytes which express the chemokine receptor CCR5 and are very abundant in mucosal lymphoid tissues. Here HIV also encounters the dendritic cells located throughout the epithelium of the skin and the mucous membranes where in their immature form called Langerhans cells they are attached by long cytoplasmic processes. The envelope glycoproteins gp41 and gp120 of HIV contain mannose-rich glycans that bind to mannan-binding proteins (pattern recognition receptors; also called lectin receptors) on the dendritic cells.
Upon capturing antigens through pinocytosis and phagocytosis and becoming activated by pro-inflammatory cytokines, the dendritic cells detach from the epithelium, enter lymph vessels, and are carried to regional lymph nodes. By the time they enter the lymph nodes, the dendritic cells have matured and are now able to present antigens of HIV to naive T-lymphocytes located in the the lymph nodes in order to induce adaptive immune responses.
At this point the infection has transitioned from the acute phase to the chronic phase. The chronic phase of HIV infection is characterized by viral dissemination, viremia, and induction of adaptive immune responses. The viremia allows the viruses to spread and infect T4-helper lymphocytes, macrophages, and dendritic cells found in peripheral lymphoid tissues.
During the chronic phase of HIV infection, the lymph nodes and the spleen become sites for continuous viral replication and host cell destruction. During most of this phase, the immune system remains active and competent and there are few clinical symptoms. A steady state-infection generally persists where T4-lymphocyte death and T4-lymphocyte replacement by the body are in equilibrium. In a person infected with HIV, somewhere between one and two billion of these T4-cells die each day as a result of HIV infection and must be replaced by the body's lymphopoietic system in the bone marrow. It is estimated that 10 billion virions are produced and cleared in an infected individual each day. However, the enormous turnover of T4-lymphocytes eventually exhausts the lymphopoietic system and it becomes unable to replace the T4-cells being destroyed. A variety of mechanisms then eventually lead to immunodeficiency.
Mechanisms of HIV-induced immunodeficiency include:
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Direct HIV-induced cytopathic effect on infected T4-lymphocytes. This can occur through:
- Increased cell permeability as a result of gp41 expression in the host cell membrane and viral release by budding;
- Inhibition of host cell protein synthesis as a result of viral replication within the infected cell; and
- Fusion of infected T4-cells with numerous uninfected T4-cells resulting in syncytia formation.
- Killing of HIV-infected T4-cells by cytotoxic T-lymphocytes or CTLs.
- Killing of HIV-infected T4-cells by antibody-dependent cytotoxicity or ADCC.
- Apoptosis of T4-cells as a result of chronic activation by HIV and by cytokines.
- Shedding of gp120 molecules by HIV. This subsequently triggers a series of events that cause the adaptive immune system to become less and less effective, primarily by altering the normal balance of immunoregulatory T H 1 and T H 2 cells in the body.
- Impaired function of HIV infected macrophages and dendritic cells.
These mechanisms will be discussed in greater detail in Unit 5 under secondary immunodeficiency.
Progression to AIDS is marked by a viral load that progressively increases in number while the immune system weakens as a result of the destruction of increasing numbers of T4-lymphocytes and the inability of the body to continually replace these destroyed cells. As will be seen in Unit 5, the loss of T4-helper lymphocytes leads to a marked decline in cells called cytotoxic T-lymphocytes ( CTLs ), the primary cells the body's immune responses use to destroy virus-infected cells. Once a person progresses to full-blown AIDS he or she becomes susceptible to a variety of opportunistic infections by:
- bacteria such as Mycobacterium avium complex (MAC), Salmonella , and Nocardia ;
- protozoa such as Cryptosporidium and Toxoplasma ;
- viruses such as cytomegalovirus (CMV), herpes simplex viruses types 1 and 2 (HSV-1, HSV-2), and varicella zoster virus (VZV);
- Candida, Cryptococcus, Coccidioides, Histoplasma , and Pneumocystis .
There is also an increased incidence of tumors, such Epstein-Barr virus-associated B-cell lymphomas, other lymphomas, cervical cancer, and Kaposi’s sarcoma. Wasting syndrome and encephalopathy are also common.
Why do you think the incubation period between HIV infection and AIDS has typically been 10 years or more?
Highly active anti-retroviral therapy (HAART) with a combination of reverse transcriptase inhibitors and protease inhibitors, as will be discussed later in Unit 4 under "Control of Viruses," has had relatively good success in both improving T4-lymphocyte levels and reducing the levels of HIV in the body - sometimes to undetectable levels. However, even with undetected levels of HIV, most infected persons continue to harbor relatively small amounts of replication-competent HIV, most likely in the resting T4-memory cells produced as a normal part of the immune responses. These infected T4-memory cells probably persist for years after antiretroviral therapy has reduced viral load below the limit of laboratory detection and could represent a pool that can keep HIV infection going or reactivate the infection. Macrophages and dendritic cells may also serve as a reservoir for HIV.
Summary
- The median incubation period for AIDS is around 10 years.
- During early or acute HIV infection the virus primarily infects and destroys memory T4-lymphocytes which express the chemokine receptor CCR5 and are very abundant in mucosal lymphoid tissues. Here HIV also encounters the dendritic cellslocated throughout the epithelium of the skin and the mucous membranes.
- The dendritic cells detach from the epithelium, enter lymph vessels, and are carried to regional lymph nodes where they are now able to present antigens of HIV to naive T-lymphocytes in order toinduce adaptive immune responses.
- The virus transitions from the acute phase to the chronic phase characterized by viral dissemination, viremia, and induction of adaptive immune responses.
- The viremia allows the viruses to spread and infect T4-helper lymphocytes, macrophages, and dendritic cells found in peripheral lymphoid tissues.
- During the chronic phase of HIV infection, the lymph nodes and the spleen become sites for continuous viral replication and host cell destruction whereby a steady state-infection generally persists where T4-lymphocyte death and T4-lymphocyte replacement by the body are in equilibrium.
- The enormous turnover of T4-lymphocytes eventually exhausts the lymphopoietic system and it becomes unable to replace the T4-cells being destroyed eventually leading to immunodeficiency.
- Progression to AIDS is marked by a viral load that progressively increases in number while the immune system weakens as a result of the destruction of increasing numbers of T4-lymphocytes and the inability of the body to continually replace these destroyed cells.
- As a result of immunosuppression, the person becomes susceptible to a variety of opportunistic infections and secondary cancers.
Questions
Study the material in this section and then write out the answers to these questions. Do not just click on the answers and write them out. This will not test your understanding of this tutorial.
- State the median incubation period for AIDS. (ans)
- In terms of viral load, exhaustion of the lymphopoietic system, and immune responses, briefly describe what marks the progression to AIDS. (ans)
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Briefly describe the following:
- early or acute HIV infection (ans)
- chronic HIV infection (ans)
- Multiple Choice (ans)