5.14: Case Study- Using Stents to Prevent Strokes
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Section 1.1 introduces a classic challenge in statistics: evaluating the efficacy of a medical treatment. Terms in this section, and indeed much of this chapter, will all be revisited later in the text. The plan for now is simply to get a sense of the role statistics can play in practice.
In this section we will consider an experiment that studies effectiveness of stents in treating patients at risk of stroke 1 . Stents are devices put inside blood vessels that assist in patient recovery after cardiac events and reduce the risk of an additional heart attack or death. Many doctors have hoped that there would be similar bene ts for patients at risk of stroke. We start by writing the principle question the researchers hope to answer:
1 Chimowitz MI, Lynn MJ, Derdeyn CP, et al. 2011. Stenting versus Aggressive Medical Therapy for Intracranial Arterial Stenosis. New England Journal of Medicine 365:993-1003. http://www.nejm.org/doi/full/10.1056/NEJMoa1105335 . NY Times article reporting on the study: http://www.nytimes.com/2011/09/08/health/research/08stent.html .
Does the use of stents reduce the risk of stroke?
The researchers who asked this question collected data on 451 at-risk patients. Each volunteer patient was randomly assigned to one of two groups:
- Treatment group . Patients in the treatment group received a stent and medical management. The medical management included medications, management of risk factors, and help in lifestyle modi cation.
- Control group . Patients in the control group received the same medical manage-ment as the treatment group, but they did not receive stents.
Researchers randomly assigned 224 patients to the treatment group and 227 to the control group. In this study, the control group provides a reference point against which we can measure the medical impact of stents in the treatment group.
Researchers studied the effect of stents at two time points: 30 days after enrollment and 365 days after enrollment. The results of 5 patients are summarized in Table 1.1. Patient outcomes are recorded as "stroke" or "no event", representing whether or not the patient had a stroke at the end of a time period.
|
Patient |
group |
0-30 days |
0-365 days |
|---|---|---|---|
| 1 |
treatment |
no event |
no event |
| 2 |
treatment |
stroke |
stroke |
| 3 |
treatment |
no event |
no event |
|
.. |
.. |
.. |
.. |
| 450 |
control |
no event |
no event |
| 451 |
control |
no event |
no event |
Considering data from each patient individually would be a long, cumbersome path towards answering the original research question. Instead, performing a statistical data analysis allows us to consider all of the data at once. Table 1.2 summarizes the raw data in a more helpful way. In this table, we can quickly see what happened over the entire study. For instance, to identify the number of patients in the treatment group who had a stroke within 30 days, we look on the left-side of the table at the intersection of the treatment and stroke: 33.
| 0-30 days | 0-365 days | |||
|---|---|---|---|---|
| stroke | no event | stroke | no event | |
|
treatment |
33 | 191 | 45 | 179 |
|
control |
13 | 214 | 28 | 199 |
| Total | 46 | 405 | 73 | 378 |
Exercise
Exercise 1.1 Of the 224 patients in the treatment group, 45 had a stroke by the end of the first year. Using these two numbers, compute the proportion of patients in the treatment group who had a stroke by the end of their rst year. (Please note: answers to all in-text exercises are provided using footnotes.) 2
Answer
2 The proportion of the 224 patients who had a stroke within 365 days: \(\frac {45}{224}\) = 0.20.
We can compute summary statistics from the table. A summary statistic is a single number summarizing a large amount of data (formally, a summary statistic is a value computed from the data. Some summary statistics are more useful than others). For instance, the primary results of the study after 1 year could be described by two summary statistics: the proportion of people who had a stroke in the treatment and control groups.
- Proportion who had a stroke in the treatment (stent) group: \(\frac {45}{224}\) = 0.20 = 20%.
- Proportion who had a stroke in the control group: \(\frac {28}{227}\) = 0.12 = 12%.
These two summary statistics are useful in looking for differences in the groups, and we are in for a surprise: an additional 8% of patients in the treatment group had a stroke! This is important for two reasons. First, it is contrary to what doctors expected, which was that stents would reduce the rate of strokes. Second, it leads to a statistical question: do the data show a "real" difference between the groups?
This second question is subtle. Suppose you flip a coin 100 times. While the chance a coin lands heads in any given coin flip is 50%, we probably won't observe exactly 50% heads. This type of fluctuation is part of almost any type of data generating process. It is possible that the 8% difference in the stent study is due to this natural variation. However, the larger the difference we observe (for a particular sample size), the less believable it is that the difference is due to chance. So what we are really asking is the following: is the difference so large that we should reject the notion that it was due to chance?
While we don't yet have our statistical tools to fully address this question on our own, we can comprehend the conclusions of the published analysis: there was compelling evidence of harm by stents in this study of stroke patients.
Be careful
Do not generalize the results of this study to all patients and all stents. This study looked at patients with very speci c characteristics who volunteered to be a part of this study and who may not be representative of all stroke patients. In addition, there are many types of stents and this study only considered the self-expanding Wingspan stent (Boston Scientific). However, this study does leave us with an important lesson: we should keep our eyes open for surprises.


11.5: Evaluate Nutritional Strategies to Impact Cardiovascular Wellness
By the end of this section, you should be able to:
- 11.4.1 Evaluate a nutritional plan for its effect on cardiovascular wellness.
- 11.4.2 Modify a nutritional plan to promote cardiovascular wellness.
Evaluating Client Readiness to Follow a Nutritional Plan
Client readiness is an important metric in their desire to make necessary life and nutritional changes. The nurse needs to work with the client to evaluate their desire to change. Most changes involve significant lifestyle adjustments, and if the client isn’t ready, diet changes alone may not be enough. If the client is willing to follow the nutritional plan, the nurse should educate the client on diet modifications. This will help the client understand how much needs to change and ensure they feel that they can follow the plan.
The nurse also needs to evaluate ableness. The nurse should evaluate the challenges and provide recommendations on how to remove those challenges. The nurse must ensure the client is capable of comprehending the information given to them and that they can make decisions on how best to follow the plan. One step to aid in this evaluation is to ask that the client return with a documented meal plan and include appropriate substitutions for daily meals.
Evaluating the Effectiveness of a Nutritional Plan
Continual monitoring of lab values and blood pressure is the most effective way to evaluate a nutritional plan. The client’s LDL levels should trend downward along with total cholesterol levels, if they were not in normal or optimal ranges. Their blood pressure reading should trend downward to a normal level if it was increased. Levels of prealbumin (a protein produced by the liver) should increase to the normal range if they had been low. BMI should trend downward if it was high. Overall, all values should slowly trend toward standard levels if the client adheres to the nutritional plan and the plan is effective.
Read the following clinical scenario and then answer the questions that follow. This case study is a follow-up to Case Study Parts A and B.
Mr. Thompson has returned for his 3-month follow-up. He claims to be compliant with all the required changes he and the provider and nurse previously discussed, and he said he is feeling great. The nurse documents that his blood pressure is 122/78 mm Hg, his BMI is 20 kg/m 2 (a slight change), and his waist measurement is 38 inches.
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They appear ineffective.
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They are not effective enough.
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This is not enough information to determine.
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The client appears to be compliant.
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The client appears to be partially compliant.
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The client is not compliant.
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There is not enough information to determine.


26.0: Introduction to Vision and Optical Instruments
Explore how the image on the computer screen is formed. How is the image formation on the computer screen different from the image formation in your eye as you look down the microscope? How can videos of living cell processes be taken for viewing later on, and by many different people?
Seeing faces and objects we love and cherish is a delight--one’s favorite teddy bear, a picture on the wall, or the sun rising over the mountains. Intricate images help us understand nature and are invaluable for developing techniques and technologies in order to improve the quality of life. The image of a red blood cell that almost fills the cross-sectional area of a tiny capillary makes us wonder how blood makes it through and not get stuck. We are able to see bacteria and viruses and understand their structure. It is the knowledge of physics that provides fundamental understanding and models required to develop new techniques and instruments. Therefore, physics is called an enabling science --a science that enables development and advancement in other areas. It is through optics and imaging that physics enables advancement in major areas of biosciences. This chapter illustrates the enabling nature of physics through an understanding of how a human eye is able to see and how we are able to use optical instruments to see beyond what is possible with the naked eye. It is convenient to categorize these instruments on the basis of geometric optics (see "Geometric Optics") and wave Optics (see "Wave Optics").


2.6: Substitution
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Just as when solving integrals, one method to try is to change variables to end up with a simpler equation to solve.
Substitution
The equation
\[ y' = (x-y + 1)^2 \label{1.5.1} \]
is neither separable nor linear. What can we do? How about trying to change variables, so that in the new variables the equation is simpler. We use another variable \(v\), which we treat as a function of \(x\). Let us try
\[v=x-y+1. \label{1.5.2} \]
We need to figure out \(y'\) in terms of \(v'\), \(v\) and \(x\). We differentiate (in \(x\)) to obtain \(v' = 1 -y'\). So \( y' = 1 - v' \). We plug this into the equation to get
\[ 1 - v' = v^2 \label{1.5.3} \]
In other words, \( v' = 1 - v^2\). Such an equation we know how to solve by separating variables:
\[ \dfrac {1}{1 -v^2} dv = dx \label{1.5.5} \]
So
\[\frac{1}{2}\ln\left|\frac{v+1}{v-1}\right| =x+C,\quad\text{or}\quad\left|\frac{v+1}{v-1}\right|=e^{2x+2C},\quad\text{or}\quad\frac{v+1}{v-1}=De^{2x}, \nonumber \]
for some constant \(D\). Note that \( v = 1\) and \( v = -1\) are also solutions.
Now we need to “unsubstitute” to obtain
\[ \dfrac {x - y + 2}{x -y} = De^{2x} \label{1.5.8} \]
and also the two solutions \( x - y + 1 = 1\) or \(y = x\), and \( x - y + 1 = -1\) or \( y = x + 2\). We solve the first equation for \(y\).
\[\begin{align}\begin{aligned} x - y + 2 &= \left ( x - y \right ) De^{2x}, \\ x - y + 2 &= Dxe^{2x} - yDe^{2x}, \\ -y + yDe^{2x} &= Dxe^{2x} - x - 2, \\ y( -1 + De^{2x}) &= Dxe^{2x} - x - 2, \\ y &= \dfrac {Dxe^{2x} - x - 2}{De^{2x} - 1}.\end{aligned}\end{align} \nonumber \]
Note that \(D = 0\) gives \( y = x + 2\), but no value of \(D\) gives the solution \(y = x\).
Substitution in differential equations is applied in much the same way that it is applied in calculus. You guess. Several different substitutions might work. There are some general things to look for. We summarize a few of these in a table.
| When you see | Try substituting |
|---|---|
| \(yy'\) | \(v = y^2\) |
| \(y^2y'\) | \(v = y^3\) |
| \((\cos y )y'\) | \( v = \sin y \) |
| \((\sin y )y'\) | \( v = \cos y \) |
| \( y'e^y\) | \( v = e^y \) |
Usually you try to substitute in the “most complicated” part of the equation with the hopes of simplifying it. The above table is just a rule of thumb. You might have to modify your guesses. If a substitution does not work (it does not make the equation any simpler), try a different one.
Bernoulli Equations
There are some forms of equations where there is a general rule for substitution that always works. One such example is the so-called Bernoulli equation. \(^{1}\)
\[ y' + p(x)\,y = q(x)\, y^n \label{1.5.15} \]
This equation looks a lot like a linear equation except for the \(y^n\). If \(n = 0\) or \(n =1\), then the equation is linear and we can solve it. Otherwise, the substitution \(v = y^{1 - n}\) transforms the Bernoulli equation into a linear equation. Note that \(n\) need not be an integer.
Example 1.5.1: Bernoulli Equation
Solve
\[xy' + y(x +1) + xy^5 = 0, ~~ y(1) = 1 \nonumber \]
Solution
First, the equation is Bernoulli \(p(x) = \dfrac{x + 1}{x}\) ( and \(q(x) = -1\) ). We substitute
\[v = y^{1 - 5} = y^{-4}, \quad v' = -4y^{-5}y' \nonumber \]
In other words, \( \left ( \dfrac {-1}{4} \right ) y^5 v' = y' \). So
\[\begin{align}\begin{aligned} xy' + y(x + 1) + xy^5 &= 0, \\ \dfrac {-xy^5}{4} v' + y(x + 1) + xy^5 &= 0, \\ \dfrac {-x}{4} v' + y^{-4} (x + 1) + x &= 0, \\ \dfrac {-x}{4} v' + v(x + 1) + x &= 0,\end{aligned}\end{align} \nonumber \]
\[ v' - \dfrac {4(x + 1)}{x} v = 4 \nonumber \]
Now the equation is linear. We can use the integrating factor method. In particular, we use formula (1.4.17). Let us assume that \(x > 0 \) so \( \left \vert x \right \vert = x \). This assumption is OK, as our initial condition is \( x = 1\). Let us compute the integrating factor. Here \( p(s)\) from formula (1.4.17) is \( \dfrac {-4(s + 1)}{s}\).
\[\begin{align}\begin{aligned} e^{\int _1^x p(s) ds} &= \text {exp} \left ( \int ^x_1 \dfrac {-4(s + 1)}{s} ds \right ) = e^{-4x - 4ln(x) + 4} = e^{-4x + 4} x^{-4} = \dfrac {e^{-4x + 4}}{x^4}, \\ e^{- \int_1^x p(s) ds} &= e^{4x + 4ln(x) - 4} = e^{4x - 4} x^4\end{aligned}\end{align} \nonumber \]
We now plug in to (1.4.17)
\[\begin{align}\begin{aligned} v(x) &= e^{- \int_1^x p(s)ds} \left ( \int_1^x e^{\int_1^t p(s)ds} 4 dt + 1\right ), \\ &= e^{4x - 4} x^4 \left ( \int_1^x 4 \dfrac {e^{-4t+4}}{t^4} dt + 1 \right ) \end{aligned}\end{align} \nonumber \]
Note that the integral in this expression is not possible to find in closed form. As we said before, it is perfectly fine to have a definite integral in our solution. Now “unsubstitute”
\[\begin{align}\begin{aligned} y^{-4} &= e^{4x - 4} x^{4} \left ( 4 \int_1^x \dfrac {e^{-4t + 4}}{t^4} dt + 1 \right ), \\ y &= \dfrac {e^{-x+1}}{x \left (4 \int_1^x \dfrac {e^{-4t+4}}{t^4} dt + 1\right )^{1/4}} \end{aligned}\end{align} \nonumber \]
Homogeneous Equations
Another type of equations we can solve by substitution are the so-called homogeneous equations . Suppose that we can write the differential equation as
\[ y' = F \left ( \dfrac {y}{x} \right ) \label{1.5.16} \]
Here we try the substitutions
\[ v = \dfrac {y}{x} \quad\text{and therefore}\quad y' = v + xv' \label{1.5.17} \]
We note that the equation is transformed into
\[ v + xv' = F(v) \quad\text{or}\quad xv' = F(v) - v \quad\text{or}\quad \dfrac {v'}{F(v) - v} = \dfrac {1}{x} \label{1.5.18} \]
Hence an implicit solution is
\[ \int \dfrac {1}{F(v) - v} dv = \ln \left \vert x \right \vert + C \label{1.5.19} \]
Example 1.5.2
Solve
\[ x^2y' = y^2 + xy, \quad y(1) = 1 \nonumber \]
Solution
We put the equation into the form \(y' = \left(\frac{y}{x}\right)^{2}+\frac{y}{x} \). We substitute \( v = \dfrac {y}{x}\) to get the separable equation
\[ xv' = v^2 + v - v = v^2 \nonumber \]
\[ v^{-2}dv = \dfrac{1}{x} dx \nonumber \]
which has a solution
\[\begin{align}\begin{aligned} \int \dfrac {1}{v^2} dv &= ln \left \vert x \right \vert + C, \\ \dfrac{-1}{v} &= ln \left \vert x \right \vert + C, \\ v &= \dfrac {-1}{ ln \left \vert x \right \vert + C}. \end{aligned}\end{align} \nonumber \]
We unsubstitute
\[\begin{align}\begin{aligned} \dfrac {y}{x} &= \dfrac {-1}{ ln \left \vert x \right \vert + C }, \\ y &= \dfrac {-x}{ln \left \vert x \right \vert + C }\end{aligned}\end{align} \nonumber \]
We want \( y(1) = 1\), so
\[ 1 = y(1) = \dfrac {-1}{ln \left \vert 1 \right \vert + C } = \dfrac {-1}{C} \nonumber \]
Thus \( C = -1 \) and the solution we are looking for is
\[ y = \dfrac {-x}{ ln \left \vert x \right \vert - 1} \nonumber \]
Footnotes
[1] There are several things called Bernoulli equations, this is just one of them. The Bernoullis were a prominent Swiss family of mathematicians. These particular equations are named for Jacob Bernoulli (1654–1705).
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- Jiří Lebl ( Oklahoma State University ).These pages were supported by NSF grants DMS-0900885 and DMS-1362337.


6.5: Limiting Reactants
Gloves will typically come is right- and left-handed models. In order to make a pair of gloves, you need one that is designed to fit each hand. If you had a box containing 50 left-handed gloves and another box containing 40 right-handed gloves, you could make 40 proper pairs and you would have ten left-handed gloves left over. The number of pairs of gloves that you could assemble is limited by the glove in the smallest number (the right-handed glove). The other glove in this example, the left, is present in excess.
The same sort of logic applies to chemical reactions in which there are two or more reactants. In Example 6.4, we carefully weighed out 0.010 mole of solid zinc and solid sulfur in order to react them to form 0.010 mole of the product, ZnS. If instead, we had reacted 0.010 mole of Zn with 0.020 moles of sulfur, how much ZnS would have (theoretically) formed? The answer is still 0.010 mole of ZnS. What happens to the leftover sulfur? It just sits there! When the reaction is complete, there is (theoretically) 0.010 mole of ZnS mixed with the remaining 0.010 mole of sulfur. 10 atoms of Zn reacting with 20 atoms of S yield 10 molecules of ZnS with 10 atoms of S remaining.
You may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is determined by the limiting reactant in the reaction, and some of the excess reactant is left over. If aluminum and chlorine gas, a diatomic gas, react to form aluminum chloride according to the equation shown below,
2 Al (s) + 3 Cl 2 (g) → 2 AlCl 3 (s)
and there are 2.0 moles of aluminum and 14 moles of chlorine present, two moles of aluminum chloride are formed and 11 moles of chlorine gas remain in excess. Our stoichiometry is:
\[\left ( \frac{3\: mol\: Cl_{2}}{2\: mol\: AlCl_{3}} \right ) \nonumber \]
If three moles of chlorine gas are used in the reaction, (14 – 3) = 11 moles of chlorine must remain. When a problem is presented and one reactant is labeled as excess, the theoretical yield of product is equal to the moles of the limiting reagent, adjusted for the stoichiometry of the reaction.
Although it would be easier if reactants were routinely labeled as “limiting” or “excess”, more commonly problems are written in such a way that it is not always trivial to identify the limiting reactant. For example, if you were told that 6.0 grams of aluminum was reacted with 3.8 grams of chlorine gas and you were asked to calculate the mass of AlCl 3 that would be formed, there would be no simple way to identify the limiting- and excess reactants. In a case like this what you want to do is to simply solve the problem twice . First you would calculate the number of moles of aluminum in 6.0 grams, and then calculate how many moles of AlCl 3 could be formed. Next, you calculate how many moles of chlorine are present in 3.8 grams of chlorine gas and again, calculate how many moles of AlCl 3 could be formed. Whichever reagent produces the smallest number of moles of product must be limiting and the other reagent must be in excess .
Find : x moles of AlCl 3
We set up the problem to solve for mol product for each reactant. The general equation is:
\[(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber \]
The solutions for both reactants are:
\[x\: mol\: AlCl_{3}=(6.0\: g\: Al)\left ( \frac{1\: mol\: Al}{26.98\: g\: Al} \right )\left ( \frac{1\: mol\: AlCl}{1\: mol\: Al} \right ) \nonumber \]
\[x\: mol\: AlCl_{3}=(3.8\: g\: Cl_{2})\left ( \frac{1\: mol\: Cl_{2}}{70.90\: g\: Cl_{2}} \right )\left ( \frac{2\: mol\: AlCl}{3\: mol\: Cl_{2}} \right ) \nonumber \]
Solving these equations, we see that, beginning with 6.0 grams of aluminum, 0.22 moles of AlCl 3 can be formed, and that, beginning with 3.8 grams of chlorine, 0.036 moles of AlCl 3 can be formed.
Keeping score, 6.0 grams of Al yields 0.22 moles of AlCl 3 , and 3.8 grams of chlorine gas yields 0.036 moles of AlCl 3 . The lowest yield comes from the chlorine gas, therefore it must be limiting and aluminum must be in excess. The reaction in the problem will therefore produce 0.036 moles of product, which is equivalent to:
\[0.036\: mol\: AlCl_{3}(133.33\: g/mol)=4.8\: g\: AlCl_{3} \nonumber \]
Lead (IV) chloride reacts with fluorine gas to give lead (IV) fluoride and Cl 2 . If 0.023 moles of fluorine gas reacts with 5.3 grams of lead (IV) chloride, what mass of lead (IV) fluoride will be formed?
Although “limiting reactant problems” may be tedious , they are not difficult. When you are faced with a limiting reactant problem, just remember, you do the simple molar yield calculations twice , one for each reactant. The reactant that yields the lowest molar quantity is your limiting reagent and the molar value you calculate determines the theoretical yield in the problem.
Ammonia, which is the active ingredient in “smelling salts”, is prepared from nitrogen and hydrogen according to the equation shown below.
N 2 + 3 H 2 → 2 NH 3
- If you mix 5.0 mol of nitrogen and 10.0 moles of hydrogen how many moles of ammonia would you produce? Which reactant is in excess?
- If you have 6.2 grams of nitrogen and you react it with 6.2 grams of hydrogen how many grams of ammonia would you produce? Which reactant is the limiting reactant?


9.2: DNA Replication
As we’ve seen, DNA strands have directionality, with a 5’ nucleotide-phosphate and a 3’ deoxyribose hydroxyl end. This is even true for circular bacterial chromosomes…, if the circle is broken! Because the strands of the double helix are antiparallel , the 5’ end of one strand aligns with the 3’end of the other at both ends of the double helix. The complementary pairing of bases in DNA means that the base sequence of one strand can be used as a template to make a new complementary strand. As we’ll see, this structure of DNA created some interesting dilemmas for understanding the biochemistry of replication. The puzzlement surrounding how replication proceeds begins with experiments that visualize replicating DNA.
A. Visualizing Replication and Replication Forks
Recall the phenomenon of bacterial conjugation allowed a demonstration bacterial chromosomes were circular. In 1963, John Cairns confirmed this fact by direct visualization of bacterial DNA. He cultured E. coli cells for long periods on 3Hthymidine (3H-T) to make all of their cellular DNA radioactive. He then disrupted the cells gently to minimize damage to the DNA. The DNA released was allowed to settle and adhere to membranes. A sensitive film was placed over the membrane and time was allowed for the radiation to expose the film. After Cairns developed the autoradiographs, he examined the results in the electron microscope. He saw tracks of silver grains in the autoradiographs (the same kind of silver grains that create an image on film in old-fashioned photography). Look at the two drawings of his autoradiographs on the next page.
Cairns measured the length of the “silver” tracks, which usually consisted of three possible closed loops, or circles. The circumferences of two of these circles were always equal, their length closely predicted by the DNA content of a single, nondividing cell. Cairns therefore interpreted these images to be bacterial DNA in the process of replication. Cairns’ autoradiographs and the measurements that led him to conclude that he was looking at images of bacterial circular chromosomes are illustrated below.
He arranged his autoradiograph images in a sequence (below) to make his point.
Because the replicating chromosomes looked (vaguely!) like the Greek letter \(\theta \) , Cairns called them theta images . He inferred that replication starts at a single origin of replication on the bacterial chromosome, proceeding around the circle to completion.
Subsequent experiments by David Prescott demonstrated bidirectional replication …, that replication did indeed begin at an origin of replication, after which the double helix was unwound and replicated in both directions, away from the origins, forming two replication forks (illustrated below).
176 Semiconservative Bidrectional Replication From Two RFs
Bacterial cells can divide every hour (or even less); the rate of bacterial DNA synthesis is about 2 X 106 base pairs per hour. A typical eukaryotic cell nucleus contains thousands of times as much DNA as a bacterium, and typical eukaryotic cells double every 15-20 hours. Even a small chromosome can contain hundreds or thousands of times as much DNA as a bacterium. It appeared that eukaryotic cells could not afford to double their DNA at a bacterial rate of replication! Eukaryotes solved this problem not by evolving a faster biochemistry of replication, but by using multiple origins of replication from which DNA synthesis proceeds in both directions. This results in the creation of multiple replicons .
Each replicon enlarges, eventually meeting other growing replicons on either side to replicate most of each linear chromosome, suggested in the illustration below.
Before we consider the biochemical events at replication forks in detail, let's look at the role of DNA polymerase enzymes in the process.
B. DNA Polymerases Catalyze Replication
The first of these enzymes was discovered in E. coli by Arthur Kornberg, for which he received the 1959 Nobel Prize in Chemistry. Thomas Kornberg, one of Arthur’s sons later found two more of DNA polymerases! All DNA polymerases require a template strand against which to synthesize a new complementary strand. They all grow new DNA by adding to the 3’ end of the growing DNA chain in successive condensation reactions. And finally, all DNA polymerases also have the odd property that they can only add to a pre-existing strand of nucleic acid, raising the question of where the ‘preexisting’ strand comes from! DNA polymerases catalyze the formation of a phosphodiester linkage between the end of a growing strand and the incoming nucleotide complementary to the template strand. The energy for the formation of the phosphodiester linkage comes in part from the hydrolysis of two phosphates (pyrophosphate) from the incoming nucleotide during the reaction. While replication requires the participation of many nuclear proteins in both prokaryotes and eukaryotes, DNA polymerases perform the basic steps of replication, as shown in the illustration below.
178 DNA Polymerases & Their Activities
Although DNA polymerases replicate DNA with high fidelity with as few as one error per 107 nucleotides, mistakes do occur. The proofreading ability of some DNA polymerases corrects many of these mistakes. The polymerase can sense a mismatched base pair, slow down and then catalyze repeated hydrolyses of nucleotides until it reaches the mismatched base pair. This basic proofreading by DNA polymerase is shown below.
After mismatch repair, DNA polymerase resumes forward movement. Of course, not all mistakes are caught by this or other repair mechanisms (see DNA Repair , below). Mutations in the eukaryotic germ line cells that elude correction can cause genetic diseases. However, most are the mutations that fuel evolution. Without mutations in germ line cells (egg and sperm), there would be no mutations and no evolution, and without evolution, life itself would have reached a quick dead end! Other replication mistakes can generate mutations somatic cells. If these somatic mutations escape correction, they can have serious consequences, including the generation of tumors and cancers.
C. The Process of Replication
DNA replication is a sequence of repeated condensation (dehydration synthesis) reactions linking nucleotide monomers into a DNA polymer. Like all biological polymerizations, replication proceeds in three enzymatically catalyzed and coordinated steps: initiation , elongation and termination .
1. Initiation
As we have seen, DNA synthesis starts at one or more origins or replication. These are DNA sequences targeted by initiator proteins in E. coli (below).
After breaking hydrogen bonds at the origin of replication, the DNA double helix is progressively unzipped in both directions (i.e., by bidirectional replication ). The separated DNA strands serve as templates for new DNA synthesis. Sequences at replication origins that bind to initiation proteins tend to be rich in adenine and thymine bases. This is because A-T base pairs have two hydrogen (H-) bonds that require less energy to break than the three H-bonds holding G-C pairs together. Once initiation proteins loosen H-bonds at a replication origin, DNA helicase uses the energy of ATP hydrolysis to unwind the double helix. DNA polymerase III is the main enzyme that then elongates new DNA. Once initiated, a replication bubble (replicon) forms as repeated cycles of elongation proceed at opposite replication forks.
179 Replication Initiation in E. coli
Recalling that new nucleotides can only be added to the free 3' hydroxyl group of a pre-existing nucleic acid strand. Since no known DNA polymerase can start synthesizing new DNA strands from scratch, this is a problem! The action of DNA polymerases therefore requires a primer , a nucleic acid strand to which to add nucleotides. The questions were…, what is the primer and where does it come from? Since RNA polymerases (enzymes that catalyze RNA synthesis) are the only nucleotide polymerase that can grow a new nucleic acid strand against a DNA template from scratch (i.e., from the first base), it was suggested that RNA might be the primer, After synthesis of a short RNA primer, new deoxynucleotides would be added to its 3’ end by DNA polymerase. The discovery of short stretches of RNA nucleotides at the 5’ end of Okazaki fragments confirmed the notion of RNA primers. We now know that cells use primase , a special RNA polymerase active during replication, to make those RNA primers against DNA templates before a DNA polymerase can grow the DNA strands at replication forks. As we will see now, the requirement for RNA primers is nowhere more in evidence in events at a replication fork.
2. Elongation
Looking at elongation at one replication fork (below), we see another problem:
One of the two new DNA strands can grow continuously towards the replication fork as the double helix unwinds. But what about the other strand? Either this other strand must grow in pieces in the opposite direction, or it must wait to begin synthesis until the double helix is fully unwound. If one strand of DNA must be replicated in fragments, then those fragments would have to be stitched (i.e., ligated) together. The problem is illustrated below.
According to this hypothesis, a new leading strand of DNA is lengthened continuously by sequential addition of nucleotides to its 3’ end against its leading strand template . The other strand however, would be made in pieces that would be joined in phosphodiester linkages in a subsequent reaction. Because of the extra step and presumably extra time it takes to make and join these new DNA fragments, this new DNA is called the lagging strand , making its template the lagging strand template .
Reiji Okazaki and his colleagues were studying mutants of T4 phage that grew slowly in their E. coli host cells. They graphed the growth rates of wild-type and mutant T4 phage and demonstrated that slow growth was due to a deficient DNA ligase enzyme, already known to catalyze the circularization of linear phage DNA molecules being replicated in infected host cells. The graph below summarizes their results.
Okazaki’s hypothesis was that the deficient DNA ligase in the mutant phage not only slowed down circularization of replicating T4 phage DNA, but would also be slow at joining phage DNA fragments replicated against at least one of the two template DNA strands. When the hypothesis was tested, the Okazakis found that short DNA fragments did indeed accumulate in E. coli cells infected with ligasedeficient mutants, but not in cells infected with wild type phage. The lagging strand fragments are now called Okazaki fragments .
180 Okazaki Experiments & Fragments - Solving a Problem at an RF
181 Okazaki Fragments are Made Beginning with RNA Primers
You can check out Okazaki’s original research at this link.
Each Okazaki fragment would have to begin with a 5’ RNA primer, creating yet another dilemma! The RNA primer must be replaced with deoxynucleotides before stitching the fragments together. This in fact happens, and the process illustrated below
Removal of RNA primer nucleotides from Okazali fragments requires the action of DNA polymerase I, an enzyme that can also catalyze hydrolysis of the phosphodiester bonds between the RNA (or DNA) nucleotides from the 5’-end of a nucleic acid strand. Flap Endonuclease 1 ( FEN 1 ) also plays a role in removing ‘flaps’ of nucleic acid from the 5’ ends of the fragments often displaced by polymerase as it replaces the replication primer. At the same time as the RNA nucleotides are removed, DNA polymerase I catalyzes their replacement by the appropriate deoxynucleotides. Finally, when a fragment is entirely DNA, DNA ligase links it to the rest of the already assembled lagging strand DNA. Because of its 5’ exonuclease activity (not found in other DNA polymerases), DNA polymerase 1 also plays unique roles in DNA repair (discussed further below). As Cairn’s suggested and others demonstrated, replication proceeds in two directions from the origin to form a replicon with its two replication forks (RFs). Each RF has a primase associated with replication of Okazaki fragments along lagging strand templates.
As Cairn’s suggested and others demonstrated, replication proceeds in two directions from the origin to form a replicon with its two replication forks (RFs). Each RF has a primase associated with replication of Okazaki fragments along lagging strand templates.
The requirement for primases at replication forks is shown below.
Now we can ask what happens when replicons reach the ends of linear chromosomes in eukaryotes.
3. Termination
In prokaryotes, replication is complete when two replication forks meet after replicating their portion of the circular DNA molecule. In eukaryotes, many replicons fuse to become larger replicons, eventually reaching the ends of the chromosomes…, where there is yet another problem (below)!
When a replicon nears the end of a chromosome (i.e. a double-stranded DNA molecule), the strand synthesized continuously stops when it reaches the 5’ end of its template DNA. In theory, synthesis of a last Okazaki fragment can be primed from the 3’ end of the lagging template strand. The illustration above implies removal of a primer from the penultimate Okazaki fragment and DNA polymerase catalyzed replacement with DNA nucleotides. But what about the last Okazaki fragment? Would its primer be hydrolyzed? Moreover, without a free 3’ end to add to, how are those RNA nucleotides replaced with DNA nucleotides? The problem here is that every time a cell replicates, one strand of new DNA (likely both) would get shorter and shorter. Of course, this would not do…, and does not happen! Eukaryotic replication undergoes a termination process involving extending the length of one of the two strands by the enzyme telomerase . The action of telomerase is summarized in the illustration below
Telomerase consists of several proteins and an RNA. From the drawing,, the RNA component serves as a template for 5’-> 3’ extension of the problematic DNA strand. The protein with the requisite reverse transcriptase activity is called Te lomerase R everse T ranscriptase , or TERT . The Te lomerase R NA C omponent is called TERC . Carol Greider, Jack Szostak and Elizabeth Blackburn shared the 2009 Nobel Prize in Physiology or Medicine for discovering telomerase.
183 Telomerase Replication Prevents Chromosome Shortening https ://youtu.be/M4dmfrxGKKU
One of the more interesting recent observations was that differentiated, nondividing cells no longer produce the telomerase enzyme. On the other hand, the telomerase genes are active in normal dividing cells (e.g., stem cells) and cancer cells, which contain abundant telomerase.
4. Is Replication Processive?
Drawings of replicons and replication forks suggest separate events on each DNA strand. Yet events at replication forks seem to be coordinated. Replication may be processive , meaning both new DNA strands are replicated in the same direction at the same time, smoothing out the process. How might this be possible? The drawing below shows lagging strand template DNA bending, so that it faces in the same direction as the leading strand at the replication fork.
The replisome structure cartooned at the replication fork consists of clamp proteins , primase, helicase, DNA polymerase and single-stranded binding proteins among others.
Newer techniques of visualizing replication by real-time fluorescence videography call the processive model into question, suggesting that the replication process is anything but smooth! Are lagging and leading strand replication not in fact coordinated? Alternatively, is the jerky movement of DNA elongation in the video an artifact, so that the model of smooth, coordinated replication integrated at a replisome still valid? Or is coordination defined and achieved in some other way? Check out the video yourself in the article here.
5. One More Problem with Replication
Cairns recorded many images of E.coli of the sort shown below.
The coiled, twisted appearance of the replicating circles were interpreted to be a natural consequence of trying to pull apart helically intertwined strands of DNA… or intertwined strands of any material! As the strands continued to unwind, the DNA should twist into a supercoil of DNA. Increased DNA unwinding would cause the phosphodiester bonds in the DNA to rupture, fragmenting the DNA. Obviously, this does not happen. Experiments were devised to demonstrate supercoiling, and to test hypotheses explaining how cells relax the supercoils during replication. Testing these hypotheses revealed the topoisomerase enzymes. These enzymes bind and hold on to DNA, catalyze hydrolysis of phosphodiester bonds, control unwinding of the double helix, and finally catalyze the re-formation of the phosphodiester linkages. It is important to note that the topoisomerases are not part of a replisome, but can act far from a replication fork, probably responding to the tensions in overwound DNA. Recall that topoisomerases comprise much of the protein lying along eukaryotic chromatin.
185 Topoisomerases Relieve Supercoiling During Replication
We have considered most of the molecular players in replication. Below is a list of the key replication proteins and their functions (from here).
| Enzyme | Function in DNA Replication |
| DNA Helicase | Also known as helix destabilizing enzyme. Unwinds the DNA double helix at the Replication Fork. |
| DNA Polymerase |
Builds a new duplex DNA strand by adding nucleotides in the 5' to 3' direction. Also performs proof-reading and error correction. |
| DNA clamp | A protein which prevents DNA polymerase III from dissociating from the DNA parent strand. |
| Single-Strand Binding (SSB) Proteins | Bind to ssDNA and preven the DNA double helix from re-annealing after DNA helicase unwinds it thus maintaining the strand separation. |
| Topoisomerase | Relaxes the DNA from its super-coiled nature. |
| DNA Gyrase | Relieves strain of unwinding by DNA helicase; this is a specific type of topoisomerase |
| DNA Ligase | Re-anneals the semi-conservative strands and joins Okazaki Fragments of the lagging strand |
| Primase | Provides a starting point of RNA (or DNA) for DNA polymerase to begin synthesis of the new DNA strand. |
| Telomerase | Lengthens telomeric DNA by adding repetitive nucleotide sequences to the ends of eukaryotic chromosomes. |


5.5: Chapter Resources
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Summary
Populations are individuals of a species that live in a particular habitat. Ecologists measure characteristics of populations: size, density, and distribution pattern. Life tables are useful to calculate life expectancies of individual population members. Survivorship curves show the number of individuals surviving at each age interval plotted versus time. Populations with unlimited resources grow exponentially—with an accelerating growth rate. When resources become limiting, populations follow a logistic growth curve in which population size will level off at the carrying capacity. Humans have increased their carrying capacity through technology, urbanization, and harnessing the energy of fossil fuels. Unchecked human population growth could have dire long-term effects on human welfare and Earth’s ecosystems. Communities include all the different species living in a given area. The variety of these species is referred to as biodiversity. Species may form symbiotic relationships such as commensalism, mutualism, or parasitism. Community structure is described by its foundation and keystone species. Communities respond to environmental disturbances by succession: the predictable appearance of different types of plant species, until a stable community structure is established.
Review Questions
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You are working as a biologist and perform the “mark and recapture” technique to estimate the number of endangered lemurs living within a particular habitat. You initially capture 37 lemurs, marking them all before releasing them. Two months later, you capture 49 lemurs, of which 11 are those originally captured and marked. What is the estimated size of the lemur population, rounded to the nearest whole number?
- 60
- 86
- 97
- 165
- 407
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Which one of the following measuring techniques would best enable you to determine the distribution pattern of a population of zebra?
- Track how many and which individuals use a central watering hole
- Use a drone to capture aerial photographs of their habitat range
- Employ a camera trap in the middle of their habitat
- Use the mark-recapture method
- Collect and analyze DNA from hair samples collected at 2 locations
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A species that you are studying has a type III survivorship curve. Which one of the following describes this type of curve?
- Survivorship rates are lowest during early parts of its lifecycle
- Survivorship rates are lowest during the late parts of its lifecycle
- Survivorship rates are highest during early parts of its lifecycle
- Survivorship rates are consistent through the lifecycle
- Survivorship rates are highest in the early and middle parts of its lifecycle
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A population has unlimited resources and exhibits rapid and sustained population growth. This type of growth would be best described by which one of the following?
- Exponential
- Logistic
- Sigmoidal
- Parabolic
- Inverse
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What single factor has most strongly contributed to the rapid population growth in the human population witnessed over the last 150 years?
- Increased fertility rates
- Reduced death rates
- Longer life spans
- Economic growth
- Increased morbidity
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Which one of the following age groups would most likely to lead to rapid population growth in the future if it contained the greatest relative abundance within that population?
- 0-15 years old
- 16-30 years old
- 31-45 years old
- 46-60 years old
- 61 years old and greater
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Two species have the same ecological niche. If they lived in the same habitat, both would compete until one species became predominant and the other became locally extinct. This process is summarized by which one of the following?
- Niche warfare
- Competitive exclusion principle
- Species selection principle
- Exclusion through competition theorem
- Exclusive ecological fractioning
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Which form of symbiosis benefits one member of the interaction, but neither benefits nor harms the other member?
- Parasitism
- Commensalism
- Sequentialism
- Mutualism
- Natural selection
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Biologists examined the effects of reintroducing wolves into Yellowstone National Park of the United States. They found that by preying on elk, wolves altered the foraging behavior of the elk; the elk spent less time browsing near streambanks. This allowed the regrowth of important vegetation, which had large positive impacts on the ecosystem at large. When a relatively small number of individuals, like wolves, have disproportionate impacts on the ecosystem, they are referred to as a…
- Foundation species
- Portal species
- Keystone species
- Cornerstone species
- Pivotal species
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The 1980 volcanic explosion of Mt. St. Helens in the United States devasted the north side of the mountain and it’s forests. The forests were demolished and replaced with volcanic debris that formed the new soil, free of any remnants of the previous ecosystem (such as seeds stored in the soil). Such an event would lead to which one of the following processes?
- Primary succession
- Secondary succession
- Tertiary succession
- Quaternary succession
- Pioneering succession
See Appendix for answers
Attributions
OpenStax College. (2013). Concepts of biology . Retrieved from<EMAIL_ADDRESS>. OpenStax CNX. Available under Creative Commons Attribution License 3.0 ( CC BY 3.0 ). Modified from Original.
Page attribution: Essentials of Environmental Science by Kamala Doršner is licensed under CC BY 4.0. “Review Questions” is licensed under CC BY 4.0 by Matthew R. Fisher.


16.3D: Ocean Floor
- Explain the importance of microbes and hydrothermal vents to underwater ecosystems
Microorganisms, by their omnipresence, impact the entire biosphere. Microbial life plays a primary role in regulating biogeochemical systems in virtually all of our planet ‘s environments, including some of the most extreme, from frozen environments and acidic lakes, to hydrothermal vents at the bottom of deepest oceans, and some of the most familiar, such as the human small intestine.
Microbes, especially bacteria, often engage in symbiotic relationships (either positive or negative) with other organisms, and these relationships affect the ecosystem. One example of these fundamental symbioses are chloroplasts, which allow eukaryotes to conduct photosynthesis. Chloroplasts are considered to be endosymbiotic cyanobacteria, a group of bacteria that are thought to be the origins of aerobic photosynthesis.
They are the backbone of all ecosystems, but even more so in the zones where light cannot approach and therefore photosynthesis cannot be the basic means to collect energy. In such zones, chemosynthetic microbes provide energy and carbon to the other organisms. Other microbes are decomposers, with the ability to recycle nutrients from other organisms’ waste poducts. These microbes play a vital role in biogeochemical cycles. The nitrogen cycle, the phosphorus cycle and the carbon cycle all depend on microorganisms in one way or another. For example, nitrogen which makes up 78% of the planet’s atmosphere is “indigestible” for most organisms, and the flow of nitrogen into the biosphere depends on a microbial process called fixation.
Recently there has been the discovery of abundant marine life in the deep sea, especially around hydrothermal vents. Large deep sea communities of marine life have been discovered around black and white smokers – hydrothermal vents emitting typical chemicals toxic to humans and most of the vertebrates. This marine life receives its energy from both the extreme temperature difference (typically a drop of 150 degrees) and from chemosynthesis by bacteria.
Brine pools are another seabed feature, usually connected to cold seeps. Hydrothermal vents along the mid-ocean ridge spreading centers act as oases, as do their opposites, cold seeps. Such places support unique biomes and many new microbes and other lifeforms have been discovered at these locations. The deepest recorded oceanic trench measured to date is the Mariana Trench, near the Philippines, in the Pacific Ocean at 10,924 m (35,838 ft). At such depths, water pressure is extreme. There is no sunlight, but some life still exists. A white flatfish, a shrimp, and a jellyfish were seen by the American crew of the bathyscaphe Trieste when it dove to the bottom in 1960. Marine life also flourishes around seamounts that rise from the depths, where fish and other sea life congregate to spawn and feed.
Key Points
- Recently, there has been the discovery of abundant marine life in the deep sea, especially around hydrothermal vents.
- Hydrothermal vents along the mid-ocean ridge spreading centers act as oases and support unique biomes and many new microbes.
- Each area of the seabed has typical features such as common soil composition, typical topography, salinity of water layers above it, marine life, magnetic direction of rocks, and sedimenting.
Key Terms
- plankton : Plankton (singular plankter) are any organisms that live in the water column and are incapable of swimming against a current. They provide a crucial source of food to many large aquatic organisms, such as fish and whales.
- ecosystems : Communities of living organisms (plants, animals and microbes) in conjunction with the nonliving components of their environment (things like air, water, and mineral soil), interacting as a system; linked together through nutrient cycles and energy flows.
- biogeochemical cycles : A biogeochemical cycle or substance turnover or cycling of substances is a pathway by which a chemical element or molecule moves through both biotic (biosphere) and abiotic (lithosphere, atmosphere, and hydrosphere) compartments of Earth. A cycle is a series of change which comes back to the starting point and which can be repeated.


Skip to main content
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Crowdfunding platforms, such as Kickstarter or GoFundMe, allow startups to access many different investors who may be able to contribute only small amounts in return for relatively small rewards. In the case of the iBackPack, those providing funds through Kickstarter were largely unaware of the company’s postfunding operations.
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How did the lack of oversight harm investors?
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How might this situation have been different if a single angel investor had been involved?
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Put yourself in the shoes of the manager of a nonprofit organization, such as a homeless shelter. The ultimate goal is to help people who have suffered misfortune get back on their feet. They likely cannot afford to pay or otherwise contribute to the costs of running the shelter. Thus, you must look to other sources to help pay the bills. Visit the website
https://blog.hubspot.com/marketing/e...undations-list
. What are some ways you might be able to attract donors for the shelter?
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While thinking about or visiting the coffee shop in your area, look around (or visualize) and identify items or activities that are the expenses of the coffee shop. Remember, expenses for the coffee shop are related to resources consumed while generating revenue from selling coffee and related items. Do not forget about any expenses that might not be so obvious—as a general rule, every activity in a business has an associated cost.


8.1: Introduction to Set Estimation
Basic Theory
The Basic Statistical Model
As usual, our starting point is a random experiment with an underlying sample space and a probability measure \(\P\). In the basic statistical model, we have an observable random variable \(\bs{X}\) taking values in a set \(S\). In general, \(\bs{X}\) can have quite a complicated structure. For example, if the experiment is to sample \(n\) objects from a population and record various measurements of interest, then \[ \bs{X} = (X_1, X_2, \ldots, X_n) \] where \(X_i\) is the vector of measurements for the \(i\)th object. The most important special case occurs when \((X_1, X_2, \ldots, X_n)\) are independent and identically distributed. In this case, we have a random sample of size \(n\) from the common distribution.
Suppose also that the distribution of \(\bs{X}\) depends on a parameter \(\theta\) taking values in a parameter space \(\Theta\). The parameter may also be vector-valued, in which case \(\Theta \subseteq \R^k\) for some \(k \in \N_+\) and the parameter vector has the form \(\bs{\theta} = (\theta_1, \theta_2, \ldots, \theta_k)\).
Confidence Sets
A confidence set is a subset \(C(\bs{X})\) of the parameter space \(\Theta\) that depends only on the data variable \(\bs{X}\), and no unknown parameters. the confidence level is the smallest probability that \(\theta \in C(\bs{X})\): \[ \min\left\{\P[\theta \in C(\bs{X})]: \theta \in \Theta\right\} \]
Thus, in a sense, a confidence set is a set-valued statistic . A confidence set is an estimator of \(\theta\) in the sense that we hope that \(\theta \in C(\bs{X})\) with high probability, so that the confidence level is high. Note that since the distribution of \( \bs{X} \) depends on \( \theta \), there is a dependence on \( \theta \) in the probability measure \( \P \) in the definition of confidence level. However, we usually suppress this, just to keep the notation simple. Usually, we try to construct a confidence set for \(\theta\) with a prescribed confidence level \(1 - \alpha\) where \(0 \lt \alpha \lt 1\). Typical confidence levels are 0.9, 0.95, and 0.99. Sometimes the best we can do is to construct a confidence set whose confidence level is at least \(1 - \alpha\); this is called a conservative \(1 - \alpha\) confidence set for \(\theta\).
Suppose that \(C(\bs{X})\) is \(1 - \alpha\) level confidence set for a parameter \(\theta\). Note that when we run the experiment and observe the data \(\bs{x}\), the computed confidence set is \(C(\bs{x})\). The true value of \(\theta\) is either in this set, or is not, and we will usually never know. However, by the law of large numbers, if we were to repeat the confidence experiment over and over, the proportion of sets that contain \(\theta\) would converge to \(\P[\theta \in C(\bs{X})] = 1 - \alpha\). This is the precise meaning of the term confidence . In the usual terminology of statistics, the random set \(C(\bs{X})\) is the estimator ; the deterministic set \(C(\bs{x})\) based on an observed value \(\bs{x}\) is the estimate .
Next, note that the quality of a confidence set, as an estimator of \(\theta\), is based on two factors: the confidence level and the
precision
as measured by the
size
of the set. A good estimator has small size (and hence gives a precise estimate of \(\theta\)) and large confidence. However, for a given \(\bs{X}\), there is usually a tradeoff between confidence level and precision—increasing the confidence level comes only at the expense of increasing the size of the set, and decreasing the size of the set comes only at the expense of decreasing the confidence level. How we measure the
size
of the confidence set depends on the dimension of the parameter space and the nature of the confidence set. Moreover, the size of the set is usually random, although in some special cases it may be deterministic.
Considering the extreme cases may give us some insight. First, suppose that \(C(\bs{X}) = \Theta\). This set estimator has maximum confidence 1, but no precision and hence it is worthless (we already knew that \(\theta \in \Theta\)). At the other extreme, suppose that \(C(\bs{X})\) is a singleton set. This set estimator has the best possible precision, but typically for continuous distributions, would have confidence 0. In between these extremes, hopefully, are set estimators that have high confidence and high precision.
Suppose that \(C_i(\bs{X})\) is a \(1 - \alpha_i\) level confidence set for \(\theta\) for \(i \in \{1, 2, \ldots, k\}\). If \(\alpha = \alpha_1 + \alpha_2 + \cdots + \alpha_k \lt 1\) then \(C_1(\bs{X}) \cap C_2(\bs{X}) \cap \cdots \cap C_k(\bs{X})\) is a conservative \(1 - \alpha \) level confidence set for \(\theta\).
Proof
This follows from Bonferroni's inequality.
Real-Valued Parameters
In many cases, we are interested in estimating a real-valued parameter \(\lambda = \lambda(\theta)\) taking values in an interval parameter space \((a, b)\), where \(a, \, b \in \R\) with \(a \lt b\). Of course, it's possible that \(a = -\infty\) or \(b = \infty\). In this context our confidence set frequently has the form \[ C(\bs{X}) = \left\{\theta \in \Theta: L(\bs{X}) \lt \lambda(\theta) \lt U(\bs{X})\right\} \] where \(L(\bs{X})\) and \(U(\bs{X})\) are real-valued statistics. In this case \((L(\bs{X}), U(\bs{X}))\) is called a confidence interval for \(\lambda\). If \(L(\bs{X})\) and \(U(\bs{X})\) are both random, then the confidence interval is often said to be two-sided . In the special case that \(U(\bs{X}) = b\), \(L(\bs{X})\) is called a confidence lower bound for \(\lambda\). In the special case that \(L(\bs{X}) = a\), \(U(\bs{X})\) is called a confidence upper bound for \(\lambda\).
Suppose that \(L(\bs{X})\) is a \(1 - \alpha\) level confidence lower bound for \(\lambda\) and that \(U(\bs{X})\) is a \(1 - \beta\) level confidence upper bound for \(\lambda\). If \(\alpha + \beta \lt 1\) then \((L(\bs{X}), U(\bs{X}))\) is a conservative \(1 - (\alpha + \beta)\) level confidence interval for \(\lambda\).
Proof
This follows immediately from (2) .
Pivot Variables
You might think that it should be very difficult to construct confidence sets for a parameter \(\theta\). However, in many important special cases, confidence sets can be constructed easily from certain random variables known as pivot variables .
Suppose that \(V\) is a function from \(S \times \Theta\) into a set \(T\). The random variable \(V(\bs{X}, \theta)\) is a pivot variable for \(\theta\) if its distribution does not depend on \(\theta\). Specifically, \(\P[V(\bs{X}, \theta) \in B]\) is constant in \(\theta \in \Theta\) for each \(B \subseteq T\).
The basic idea is that we try to combine \(\bs{X}\) and \(\theta\) algebraically in such a way that we factor out the dependence on \(\theta\) in the distribution of the resulting random variable \(V(\bs{X}, \theta)\). If we know the distribution of the pivot variable, then for a given \(\alpha\), we can try to find \(B \subseteq T\) (that does not depend on \(\theta\)) such that \( \P_\theta\left[V(\bs{X}, \theta) \in B\right] = 1 - \alpha \). It then follows that a \(1 - \alpha\) confidence set for the parameter is given by \( C(\bs{X}) = \{ \theta \in \Theta: V(\bs{X}, \theta) \in B \} \).
Suppose now that our pivot variable \(V(\bs{X}, \theta)\) is real-valued, which for simplicity, we will assume has a continuous distribution. For \(p \in (0, 1)\), let \(v(p)\) denote the quantile of order \(p\) for the pivot variable \(V(\bs{X}, \theta)\). By the very meaning of pivot variable, \(v(p)\) does not depend on \(\theta\).
For any \(p \in (0, 1)\), a \(1 - \alpha\) level confidence set for \(\theta\) is \[ \left\{\theta \in \Theta: v(\alpha - p \alpha) \lt V(\bs{X}, \theta) \lt v(1 - p \alpha)\right\} \]
Proof
By definition, the probability of the event is \((1 - p \alpha) - (\alpha - p \alpha) = 1 - \alpha\).
The confidence set above corresponds to \((1 - p) \alpha\) in the left tail and \(p \alpha\) in the right tail, in terms of the distribution of the pivot variable \(V(\bs{X}, \lambda)\). The special case \(p = \frac{1}{2}\) is the equal-tailed case , the most common case.
The confidence set (5) is decreasing in \(\alpha\) and hence increasing in \(1 - \alpha\) (in the sense of the subset relation) for fixed \(p\).
For the confidence set (5) , we would naturally like to choose \(p\) that minimizes the size of the set in some sense. However this is often a difficult problem. The equal-tailed interval, corresponding to \(p = \frac{1}{2}\), is the most commonly used case, and is sometimes (but not always) an optimal choice. Pivot variables are far from unique; the challenge is to find a pivot variable whose distribution is known and which gives tight bounds on the parameter (high precision).
Suppose that \(V(\bs{X}, \theta)\) is a pivot variable for \(\theta\). If \(g\) is a function defined on the range of \(V\) and \(g\) involves no unknown parameters, then \(U = g[V(\bs{X}, \theta)]\) is also a pivot variable for \(\theta\).
Examples and Special Cases
Location-Scale Families
In the case of location-scale families of distributions, we can easily find pivot variables. Suppose that \(Z\) is a real-valued random variable with a continuous distribution that has probability density function \(g\), and no unknown parameters. Let \(X = \mu + \sigma Z\) where \(\mu \in \R\) and \(\sigma \in (0, \infty)\) are parameters. Recall that the probability density function of \(X\) is given by \[ f_{\mu, \sigma}(x) = \frac{1}{\sigma} g\left( \frac{x - \mu}{\sigma} \right), \quad x \in \R\] and the corresponding family of distributions is called the location-scale family associated with the distribution of \(Z\); \(\mu\) is the location parameter and \(\sigma\) is the scale parameter . Generally, we are assuming that these parameters are unknown.
Now suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the distribution of \(X\); this is our observable outcome vector. For each \(i\), let \[ Z_i = \frac{X_i - \mu}{\sigma} \]
The random vector \(\bs{Z} = (Z_1, Z_2, \ldots, Z_n)\) is a random sample of size \(n\) from the distribution of \(Z\).
In particular, note that \(\bs{Z}\) is a pivot variable for \((\mu, \sigma)\), since \(\bs{Z}\) is a function of \(\bs{X}\), \(\mu\), and \(\sigma\), but the distribution of \(\bs{Z}\) does not depend on \(\mu\) or \(\sigma\). Hence, any function of \(\bs{Z}\) will also be a pivot variable for \((\mu, \sigma)\), (if the function does not involve the parameters). Of course, some of these pivot variables will be much more useful than others in estimating \(\mu\) and \(\sigma\). In the following exercises, we will explore two common and important pivot variables.
Let \(M(\bs{X})\) and \(M(\bs{Z})\) denote the sample means of \(\bs{X}\) and \(\bs{Z}\), respectively. Then \(M(\bs{Z})\) is a pivot variable for \((\mu, \sigma)\) since \[ M(\bs{Z}) = \frac{M(\bs{X}) - \mu}{\sigma} \]
Let \(m\) denote the quantile function of the pivot variable \(M(\bs{Z})\). For any \(p \in (0, 1)\), a \(1 - \alpha\) confidence set for \((\mu, \sigma)\) is \[ Z_{\alpha, p}(\bs{X}) = \{(\mu, \sigma): M(\bs{X}) - m(1 - p \alpha) \sigma \lt \mu \lt M(\bs{X}) - m(\alpha - p \alpha) \sigma \} \]
The confidence set constructed
above
is a
cone
in the \((\mu, \sigma)\) parameter space, with vertex at \( (M(\bs{X}), 0) \) and boundary lines of slopes \(-1 / m(1 - p \alpha)\) and \(-1 / m(\alpha - p \alpha)\), as shown in the graph below. (Note, however, that both slopes might be negative or both positive.)
The fact that the confidence set is unbounded is clearly not good, but is perhaps not surprising; we are estimating two real parameters with a single real-valued pivot variable. However, if \(\sigma\) is known, the confidence set defines a confidence interval for \(\mu\). Geometrically, the confidence interval simply corresponds to the horizontal cross section at \(\sigma\).
\(1 - \alpha\) confidence sets for \((\mu, \sigma)\) are
- \(Z_{\alpha, 1}(\bs{X}) = \{(\mu, \sigma): M(\bs{X}) - m(1 - \alpha) \sigma \lt \mu \lt \infty\}\)
- \(Z_{\alpha, 0}(\bs{X}) = \{(\mu, \sigma): - \infty \lt \mu \lt M(\bs{X}) - m(\alpha) \sigma\}\)
Proof
In the confidence set constructed above , let \(p \uparrow 1\) and \(p \downarrow 0\), respectively.
If \(\sigma\) is known, then (a) gives a \(1 - \alpha\) confidence lower bound for \(\mu\) and (b) gives a \(1 - \alpha\) confidence upper bound for \(\mu\).
Let \(S(\bs{X})\) and \(S(\bs{Z})\) denote the sample standard deviations of \(\bs{X}\) and \(\bs{Z}\), respectively. Then \(S(\bs{Z})\) is a pivot variable for \((\mu, \sigma)\) and a pivot variable for \(\sigma\) since \[ S(\bs{Z}) = \frac{S(\bs{X})}{\sigma} \]
Let \(s\) denote the quantile function of \(S(\bs{Z})\). For any \(\alpha \in (0, 1)\) and \(p \in (0, 1)\), a \(1 - \alpha\) confidence set for \((\mu, \sigma)\) is \[ V_{\alpha, p}(\bs{X}) = \left\{(\mu, \sigma): \frac{S(\bs{X})}{s(1 - p \alpha)} \lt \sigma \lt \frac{S(\bs{X})}{s(\alpha - p \alpha)} \right\} \]
Note that the confidence set gives no information about \(\mu\) since the random variable above is a pivot variable for \(\sigma\) alone. The confidence set can also be viewed as a bounded confidence interval for \(\sigma\).
\(1 - \alpha\) confidence sets for \((\mu, \sigma)\) are
- \(V_{\alpha, 1}(\bs{X}) = \left \{ (\mu, \sigma): S(\bs{X}) / s(1 - \alpha) \lt \sigma \lt \infty \right \}\)
- \(V_{\alpha, 0}(\bs{X}) = \left \{ (\mu, \sigma): 0 \lt \sigma \lt S(\bs{X}) / s(\alpha) \right \}\)
Proof
In the confidence set constructed above, let \(p \uparrow 1\) and \(p \downarrow 0\), respectively.
The set in part (a) gives a \(1 - \alpha\) confidence lower bound for \(\sigma\) and the set in part (b) gives a \(1 - \alpha\) confidence upper bound for \(\sigma\).
We can intersect the confidence sets corresponding to the two pivot variables to produce conservative, bounded confidence sets.
If \(\alpha, \; \beta, \; p, \; q \in (0, 1)\) with \(\alpha + \beta \lt 1\) then \(Z_{\alpha, p} \cap V_{\beta, q}\) is a conservative \(1 - (\alpha + \beta)\) confidence set for \((\mu, \sigma)\).
Proof
The most important location-scale family is the family of normal distributions. The problem of estimation in the normal model is considered in the next section. In the remainder of this section, we will explore another important scale family.
The Exponential Distribution
Recall that the exponential distribution with scale parameter \(\sigma \in (0, \infty)\) has probability density function \(f(x) = \frac{1}{\sigma} e^{-x / \sigma}, \; x \in [0, \infty)\). It is the scale family associated with the standard exponential distribution, which has probability density function \(g(x) = e^{-x}, \; x \in [0, \infty)\). The exponential distribution is widely used to model random times (such as lifetimes and
arrival
times), particularly in the context of the Poisson model. Now suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the exponential distribution with unknown scale parameter \(\sigma\). Let \[ Y = \sum_{i=1}^n X_i \]
The random variable \(\frac{2}{\sigma} Y\) has the chi-square distribution with \(2 n\) degrees of freedom, and hence is a pivot variable for \(\sigma\).
Note that this pivot variable is a multiple of the variable \( M \) constructed above for general location-scale families (with \(\mu = 0\)). For \(p \in (0, 1)\) and \(k \in (0, \infty)\), let \(\chi_k^2(p)\) denote the quantile of order \(p\) for the chi-square distribution with \(k\) degrees of freedom. For selected values of \(k\) and \(p\), \(\chi_k^2(p)\) can be obtained from the special distribution calculator or from most statistical software packages.
Recall that
- \(\chi_k^2(p) \to 0\) as \(p \downarrow 0\)
- \(\chi_k^2(p) \to \infty\) as \(p \uparrow 1\)
For any \(\alpha \in (0, 1)\) and any \(p \in (0, 1)\), a \(1 - \alpha\) confidence interval for \(\sigma\) is \[ \left( \frac{2\,Y}{\chi_{2n}^2(1 - p \alpha)}, \frac{2\,Y}{\chi_{2n}^2(\alpha - p \alpha)} \right) \]
Note that
- \(2 Y \big/ \chi_{2n}^2(1 - \alpha)\) is a \(1 - \alpha\) confidence lower bound for \(\sigma\).
- \(2 Y \big/ \chi_{2n}^2(\alpha)\) is a \(1 - \alpha\) confidence lower bound for \(\sigma\).
Of the two-sided confidence intervals constructed above, we would naturally prefer the one with the smallest length, because this interval gives the most information about the parameter \(\sigma\). However, minimizing the length as a function of \(p\) is computationally difficult. The two-sided confidence interval that is typically used is the equal tailed interval obtained by letting \(p = \frac{1}{2}\): \[ \left( \frac{2\,Y}{\chi_{2n}^2(1 - \alpha/2)}, \frac{2\,Y}{\chi_{2n}^2(\alpha/2)} \right) \]
The lifetime of a certain type of component (in hours) has an exponential distribution with unknown scale parameter \(\sigma\). Ten devices are operated until failure; the lifetimes are 592, 861, 1470, 2412, 335, 3485, 736, 758, 530, 1961.
- Construct the 95% two-sided confidence interval for \(\sigma\).
- Construct the 95% confidence lower bound for \(\sigma\).
- Construct the 95% confidence upper bound for \(\sigma\).
Answer
- \((769.1, 2740.1)\)
- 836.7
- 2421.9


4.9: Key Takeaways
The key takeaways of this chapter are:
- Vaccine safety is a highly regulated, scrutinous process in Canada.
- Vaccines undergo various stages of research before receiving license issuance by Health Canada.
- Vaccine evaluation is ongoing, even after licensure. Post-marketing surveillance is important to ensure vaccines continue to be safe and to detect rare or serious adverse events.
- Health professionals are required by law to report adverse events following immunization.
- Adverse events following immunization should be reported when the event is serious or unexpected, and there is no other clear cause for the event. A causal relationship is not required to report the adverse event.


2.4: The Euler and De Moivre Identities
The Euler and De Moivre identities are the fundamental identities for deriving trigonometric formulas. From the identity \(e^{jθ}=\cosθ+j\sinθ\) and the conjugate identity \(e^{−jθ}=(e^{jθ})^∗=\cosθ−j\sinθ\), we have the Euler identities for \(\cosθ\) and \(\sinθ\):
\[\cosθ=\frac {e^{jθ}+e^{−jθ}} {2} \nonumber \]
\[\sinθ=\frac {e^{jθ}−e^{−jθ}} {2j} \nonumber \]
These identities are illustrated in the Figure.
The identity \(e^{jθ}=\cosθ+j\sinθ\) also produces the De Moivre identity:
\[(\cosθ+j\sinθ)n=(e^{jθ})^n=e^{jnθ}=\cos nθ+j\sin nθ \nonumber \]
When the left-hand side of this equation is expanded with the binomial expansion, we obtain the identity
\[∑_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix} (cosθ)n−k(jsinθ)k=cosnθ+jsinnθ \nonumber \]
Binomial Coefficients and Pascal's Triangle
The binomial coefficients (nk) in Equation are shorthand for the number
\[\begin{pmatrix}n\\k\end{pmatrix}=n!(n−k)!k!'k=0,1,...,n \nonumber \]
This number gives the coefficient of \(x^{n−k}y^k\) in the expansion of \((x+y)^n\). How do we know that there are \(\begin {pmatrix} n\\k\end {pmatrix}\) terms of the form \(x^{n−k}y^k\)? One way to answer this question is to use Pascal's triangle, illustrated in Figure. Each node on Pascal's triangle shows the number of routes that terminate at that node. This number is always the sum of the number of routes that terminate at the nodes just above the node in question. If we think of a left-hand path as an occurrence of an x and a right-hand path as an occurrence of a y, then we see that Pascal's triangle keeps track of the number of occurrences of \(x^{n−k}y^k\).
Prove \(\begin{pmatrix}n\\k\end{pmatrix}=(n^{n−k})\).
Find an identity for \(\begin{pmatrix}n−1\\k−1\end{pmatrix}+\begin{pmatrix}k\\n−1)\end{pmatrix}\).
Find “half-angle” formulas for \(\cos2θ\) and \(\sin2θ\).
Show that
- \(\cos3θ=\cos^2θ−3 \cosθ\sin2θ\)
- \(\sin3θ=34\cos^2θ \sinθ−\sin3θ\)


17.7: Beats
- Determine the beat frequency produced by two sound waves that differ in frequency
- Describe how beats are produced by musical instruments
The study of music provides many examples of the superposition of waves and the constructive and destructive interference that occurs. Very few examples of music being performed consist of a single source playing a single frequency for an extended period of time. You will probably agree that a single frequency of sound for an extended period might be boring to the point of irritation, similar to the unwanted drone of an aircraft engine or a loud fan. Music is pleasant and interesting due to mixing the changing frequencies of various instruments and voices.
An interesting phenomenon that occurs due to the constructive and destructive interference of two or more frequencies of sound is the phenomenon of beats . If two sounds differ in frequencies, the sound waves can be modeled as
\[y_{1} = A \cos (k_{1} x - 2 \pi f_{1} t)\]
and
\[\; y_{2} = A \cos (k_{2} x - 2 \pi f_{2} t) \ldotp \nonumber\]
Using the trigonometric identity
\[\cos u + \cos v = 2 \cos \left(\dfrac{u + v}{2}\right) \cos \left(\dfrac{u − v}{2}\right)\]
and considering the point in space as \(x = 0.0\, m\), we find the resulting sound at a point in space, from the superposition of the two sound waves, is equal to Figure \(\PageIndex{1}\):
\[y(t; x=0\,m) = 2A \cos (2 \pi f_{avg} t) \cos \left[ 2 \pi \left(\dfrac{|f_{2} - f_{1}|}{2}\right) t \right], \nonumber\]
where the beat frequency is
\[f_{beat} = |f_{2} - f_{1}| \ldotp \label{17.17}\]
These beats can be used by piano tuners to tune a piano. A tuning fork is struck and a note is played on the piano. As the piano tuner tunes the string, the beats have a lower frequency as the frequency of the note played approaches the frequency of the tuning fork.
What is the beat frequency produced when a tuning fork of a frequency of 256 Hz and a tuning fork of a frequency of 512 Hz are struck simultaneously? Strategy The beat frequency is the difference of the two frequencies.
Solution
We use Equation \ref{17.17} to get the beat frequency
\[f_{beat} = |f_2 − f_1| = (512 - 256)\; Hz = 256\; Hz \ldotp \nonumber \]
Significance
The beat frequency is the absolute value of the difference between the two frequencies. A negative frequency would not make sense.
What would happen if more than two frequencies interacted? Consider three frequencies.
The study of the superposition of various waves has many interesting applications beyond the study of sound. In later sections, we will discuss the wave properties of particles. The particles can be modeled as a “wave packet” that results from the superposition of various waves, where the particle moves at the “group velocity” of the wave packet.


15.5: Chapter Summary
This chapter focused on defining the various aspects of substance use disorders and ways to manage three of the most common substance use disorders in the world today. Substance use and abuse were described in terms of the short-term effects of intoxication as well as more long-term effects such as physical dependence, tolerance, and withdrawal. Emphasis was placed on how the presence of things like tolerance and physical dependence does not always mean a client has a substance use disorder.
Drug classifications covered in this chapter focused on medications to treat opioid, alcohol, and nicotine use disorders. Medications used to treat opioid use disorders included agents to minimize withdrawal effects or to prevent opioid effects should a client relapse. Naloxone, the major opioid reversal agent, was also discussed, along with how important it is for clients to have ready access to this in case of overdose. Medications for alcohol use disorders included agents to minimize the withdrawal effects of alcohol. Also addressed was how alcohol withdrawal can be potentially fatal and should be treated as a medical emergency. Agents to aid success in smoking cessation were covered, including bupropion, which blunts cravings by affecting the reward center in the brain, and nicotine replacement therapies such as gums, patches, and inhalers.


9.10: Text Boxes
Learning Outcomes
- Insert text boxes.
Another type of edit you may wish to perform on a presentation slide is to add a text box. Text boxes are used for emphasis or as supplements to the standard layout space. For example, perhaps you want to draw attention to a specific feature on an image or in data. An easy way to do this is with a text box.
- Select the slide you wish to edit with a text box.
- Click on the Insert Tab
- Select Text Box from the menu in the Text group of the ribbon
- Click on the slide to place your text box.
- Enter your text.
After placing your text box, you can move it by clicking on the text box and dragging it to a new location. On the screenshot below the text box has been moved to the left side of the table, from the right.
Contributors and Attributions
CC licensed content, Original
- Text Boxes. Authored by : Shelli Carter. Provided by : Lumen Learning. License : CC BY: Attribution


2.6.S: Chromosomes, Mitosis, and Meiosis (Summary)
-
- Last updated
- Save as PDF
- Ying Liu
- City College of San Francisco
- Chromosomes are complex and dynamic structures consisting of DNA and proteins (chromatin).
- The degree of chromatin compaction involves proteins and varies between heterochromatic and euchromatic regions and among stages of the cell cycle.
- Chromosomes can be distinguished cytologicaly based on their length, centromere position, and banding patterns when stained dyes or labeled with sequence-specific probes.
- Homologous chromosomes contain the same series of genes along their length, but not necessarily the same alleles.Sister chromatids initially contain the same alleles.
- Chromosomes are replicated by DNA polymerases and begin at an origin. Replication is bi-directional. Eukaryotes have multiple origins along each chromosome and have telomerase to replicate the ends.
- Mitosis reduces the c-number, but not the n-number.Meiosis reduces both c and n.
- Homologous chromosomes pair (sysnapse) with each other during meiosis, but not mitosis.
- Several types of structural defects in chromosomes occur naturally, and can affect cellular function and even evolution.
- Aneuploidy results from the addition or subtraction of one or more chromosomes from a group of homologs, and is usually deleterious to the cell.
- Polyploidy is the presence of more than two complete sets of chromosomes in a genome.Even-numbered multiple sets of chromosomes can be stably inherited in some species, especially plants.
- Endopolyploidy is tissue-specific type of polyploidy observed in some species, including diploids.
- Both aneuploidy and structural defects such as duplications can affect gene balance.
- Organelles also contain chromosomes, but these are much more like prokaryotic chromosomes than the nuclear chromosomes of eukaryotes.
Key Terms
|
chromosome core histones nucleosome 30nm fiber histone H1 scaffold proteins heterochromatin euchromatin satellite DNA chromatid centromere metacentric acrocentric telocentric holocentric telomere homologous non-homologous chromatid sister chromatid non-sister chromatid interphase mitosis prophase metaphase anaphase telophase DNA polymerase origin of replication telomerase riboprotein Hayflick limit HeLa cells cytokinesis |
meiosis gametes prophase (I, II) metaphase (I, II) anaphase (I, II) telophase (I, II) cytokinesis meiocyte bivalent syanapse, pair up synaptonemal complex reductional division equational division leptotene zygotene pachytene diplotene diakinesis crossing over chiasma (chiasmata) polar bodies G 1 G 2 S M G 0 interphase n c replicated chromosome karyotype/karyogram autosome sex-chromosome homogametic |
heterogametic aneuploidy monsomic trisomic Down syndrome deletion duplication insertion inversion translocation non-disjunction chromosome breakage polyploidy x monoploid sterile tetravalent octoploid hexaploid triploid endoreduplication endopolyploidy salivary gland chromosome polytene gene balance cellular network chloroplast mitochondria endosymbiont endosymbiont theory organellar chromosome mtDNA |


1: Overview
This is Part I, Overview of a ten-part series of technical articles on Distillation Science, as is currently practiced on an industrial level. It is organized as it would be used for the industrial process design of distillation columns, which may differ from the way it is introduced to students academically. As a part of distillation column design, these articles also involve the vapor-liquid equilibria (VLE) of binary systems, with potential extension to multi-component systems.
It is assumed that the reader has a basic understanding of distillation, as a means of separating two or more volatile fluids by the difference between their boiling points; and also the purpose of a distillation column's various components. A variety of texts exist that explain these concepts on a basic level. From such pre-requisites, the purpose of this series of articles is to expand these basics to the degree necessary for commercial applications with real fluids - which typically do not follow ideal behavior. It is also assumed that the reader is familiar with the concepts of molar units, mole fractions, the bonding of chemical elements, vapor pressure, latent heat of vaporization, and a compound's critical point - all of which are found in college freshman-level textbooks. In discussing mathematical relationships, it is assumed that the concepts of differentiation and integration are known to the reader.
While chlorosilanes and their electronic impurities are used as recurring examples, the technology developed here has great application in other mildly polar and hydrogenated compounds, which are normally excluded from theoretical treatment in many texts. The chlorosilane homologue starts with silane (SiH 4 ), and concludes with silicon tetrachloride (SiCl 4 ), as the Si-H bonds are incrementally swapped for Si-Cl bonds.
Chlorosilanes (and many of the electronic impurities) are non-naturally occurring compounds, but are key to the manufacture of high-purity silicon (solar photovoltaics and modern electronic integrated circuits, aka “computer chips”); as well as silicon-based chemicals such as silicones and organic/inorganic coupling agents. Applications for distillation science are found in both bulk separations of these fluids, as well as the purification by high-reflux processing, to parts-per-billion levels.
Furthermore, this technology is applicable to a wide variety of other non-organic fluids such as refrigerants, biologic pre-cursors, and pharmaceutical pre-cursors (i.e., not the final biological or pharmaceutical product, but rather the compounds used as catalysts and building-block agents for assembling these specialized chemical products).
To make the discussion of Distillation Science more generic, temperature, pressure, and molar volume ( T, P, and V) are often expressed in dimensionless reduced units, designated respectively as T r , P r , and V r . T r = T/T c ; P r =P/P c ; V r = V/V c . By definition, T r =1, P r =1, and V r =1 simultaneously at the critical point. When temperature and pressure are given as T and P, these are in absolute units of Kelvins and atmospheres. For those more familiar with pressures in other absolute units such as PSIA, bar absolute, and Pascals absolute, converting those units to absolute atmospheres is easily done using internet-accessible conversion tables and calculators.
- Part II, Existing Vapor Pressure Equations deals with the pure component VP relationships commonly found in textbooks, as well as their limitations. This article lays the basis for subsequent articles.
- Part III, Critical Properties and Acentric Factor deals with the tabulation of these properties for selected fluids based on globally collected data. Data analysis and validation are discussed, as well as estimation techniques for those fluids for which data is either poor or non-existent. Critical properties are used to convert temperature, pressure, and specific volume from conventional units used in both chemistry and chemical engineering, to the reduced form of Part IV . In Parts V and VII , the value of the acentric factor is important.
- Part IV, New Vapor Pressure Equation expands on the basics of Part II and the results of Part III , to show how a new vapor pressure equation allows the practice of distillation applications at the elevated pressures more common to industry. The culmination of this article is a thermodynamically consistent equation that is valid between the atmospheric boiling point and the critical point, and which allows evaluation of other required distillation properties such as saturated phase densities and latent heat of vaporization.
- Part V, Equation of State deals with the recommendations for best EOS, as well as the mathematical techniques for solving such non-intrinsic EOS equation forms.
- Part VI, Fugacity deals with the departure of apparent pure-component vapor pressure in binary mixtures, as is commonly found in practical application of distillation science. The equations for evaluating fugacity coefficents are given.
- Part VII, Liquid Activity Coefficient s deals with the application and estimation of Liquid Activity Coefficients, as commonly found in practical application of distillation science. Various Liquid Activity Coefficient models are reviewed along with some limited data and a recommended estimation technique given where data is lacking.
- Part VIII, VLE Analysis Methods discusses the recommended methodology used when data-collecting binary systems, to assure that systemic errors are minimized. This topic also deals with validation whenever data collection is done on reactive fluids, which can disproportionate or dimerize during study.
- Part IX, Putting It All Together shows how to combine the components of the above distillation science articles in a practical application.
- Part X, Convergence Strategy illustrates how solutions are best obtained for additional fluids using non-intrinsic or nested-loop equation methods, such as the recommended vapor pressure equations. While this topic is more mathematical or computer-science oriented than chemistry-centered, it is a necessary technique to understand when dealing with modern technology. No complex mathematics are required.
| Notation | Usage | Units | Notation | Usage | Units |
|---|---|---|---|---|---|
| NBP | Normal Boiling Point | °K | q | Watson exponent | - |
| P, VP | Pressure, Vapor Pressure | atmospheres | ΔH vap | Latent heat of Vaporization | Cal/g-mol |
|
P c |
Critical Pressure | atmospheres | ΔH vb | Latent heat @ T b | Cal/g-mol |
|
P r |
Reduced Pressure | - | T-S ΔH vb | Thek-Stiel VP equation parameter | Cal/g-mol |
|
R |
Gas constant | Cal/g-mol°K | Z | compressibility | - |
| T | Temperature | °K | Z v & Z L | compressibility of saturated vapor & liquid | - |
|
T b |
Temperature @ Normal Boiling Point | °K | ΔZ vap | (Z v -Z L ) @ vaporization | - |
|
T c |
Critical Temp | °K | α | Riedel derivative | - |
|
T r |
Reduced Temp | - | α c | α @ critical | - |
|
T br |
Reduced T b | - | \(\psi \nonumber\) | Vapor Pressure derivative | - |
|
V c |
Critical Volume | cc/g-mol | ø | fugacity coefficient | - |
|
k |
H bonding parameter | - | \(\gamma \nonumber\) | liquid activity coefficient | - |
| k ij | Binary Interaction Coefficient | - | ω | acentric factor | - |
Units
The units used in these articles are temperatures in °K, pressures in atmospheres (absolute), and molar volumes in cc/gram-mole. Molecular weight (MW), and compressibility (Z) are dimensionless. For the above units, the value of the gas constant (R) is 82.057 atm-cc/mole-°K.


7.7: Franchises
What you’ll learn to do: discuss the advantages and disadvantages of franchises
For aspiring business owners who lack the time, vision, or resources to start from scratch, franchising is a viable alternative. Everyone is familiar with franchises—many industries such as fast food are almost wholly comprised of franchises. As appealing as it may seem, there are still risks to franchising for both the franchisor and franchisee.
Learning Objectives
- Discuss the advantages and disadvantages of franchising for the franchisee
- Discuss the advantages and disadvantages of franchising for the franchisor
Franchises
A franchise is a business model that involves one business owner (the franchisor ) licensing trademarks and methods to an independent entrepreneur (the franchisee ) for a prescribed period of time. For the franchisor, the franchise is an alternative to expanding through the establishment of a new location, which avoids the financial investment and liability of a chain of stores. Ultimately, the franchisor’s success depends on the success of the franchisees. If the franchisees are successful then the franchisor can grow its brand and market presence while the franchisee, in effect, does all of the work.
The United States is a leader in franchising, a position it has held since the 1930s when it used the approach for fast-food restaurants, food inns, and, slightly later, motels during the time of the Great Depression. Today, the world’s largest franchise chains are U.S. companies:
- Subway : start-up costs $84,300–$258,300; 41,916 locations worldwide in 2015
- McDonald’s : start-up costs in 2010: $995,900–$1,842,700; 36,368 Locations in 2015
- 7-Eleven Inc. : start-up costs in 2010: $40,500–$775,300; 56,439 locations in 2015
- Hampton Inns & Suites: start-up costs in 2010: $3,716,000–$15,148,800
- Great Clips: start-up costs in 2010: $109,000 – $203,000; 3,694 locations in 2015
Buying a Franchise
The decision to purchase a franchise involves many factors, including how much you can afford to invest, what abilities you have, and what your goals are.   By law, franchisors must disclose certain information about their business to potential buyers. Make sure you get all the information you need first before entering into this form of business.
The following strategies can help you gain a solid understanding of what to expect as well as the risks that could be involved:
- Be a Detective. In addition to the routine investigation that should be conducted prior to any business purchase, you should be able to contact other franchisees before deciding to invest. You can obtain a Uniform Franchise Offering Circular (UFOC), which contains vital details about the franchise’s legal, financial, and personnel history, before you sign a contract.
- Know What You Are Getting Into. Before entering into any contract as a franchisee, you should make sure that you would have the right to use the franchise name and trademark, receive training and management assistance from the franchisor, use the franchisor’s expertise in marketing, advertising, facility design, layouts, displays and fixtures, and do business in an area protected from other competing franchisees.
- Watch Out for Possible Pitfalls. The contract between the two parties usually benefits the franchisor far more than the franchisee. The franchisee is generally subject to meeting sales quotas and is required to purchase equipment, supplies, and inventory exclusively from the franchisor.
- Seek Professional Help. The tax rules surrounding franchises are often complex, and an attorney, preferably a specialist in franchise law, should assist you to evaluate the franchise package and tax considerations. An accountant may be needed to determine the full costs of purchasing and operating the business as well as to assess the potential profit to the franchisee.
Franchise Taxes
The taxation of a franchise depends on the underlying form of ownership. Generally franchises are required by the franchisor to be established as a corporation or LLC. Ultimately, the franchise agreement governs this, and individuals looking to purchase a franchise should scrutinize any agreements with regard to prescribed legal ownership structure.
Advantages for the Franchisor
- Access to Capital for Growth and Expansion. After the brand and formula are carefully designed and properly executed, franchisors are able to sell franchises and expand rapidly across countries and continents using the capital and resources of their franchisees.
- Cash Flow for Operations. In addition to initial franchise fees that can range from $50,000 to $5 million, franchisors receive payments in the form of royalties from each franchisee. These royalties typically range from 4 percent to 8 percent of gross revenues. In addition, franchisees are also assessed for marketing and advertising.
- Economies of Scale . Once a franchise is established with multiple locations, the company may be able to leverage its buying power to realize economies of scale with suppliers, advertisers, and vendors. If purchasing and distribution for the franchise locations can be centralized, then the cost savings will increase the franchisor’s bottom line, particularly if the franchise agreement provides for a percent-of-sales payment to the franchisor.
Disadvantages for the Franchisor
- Lack of Control. Despite the language of the franchise agreement, once the franchisee has established their location, the franchisor may have difficulty ensuring that quality standards are met and the franchise is operating in a manner that benefits the brand. A Dunkin’ Donuts franchise in Russia had to be closed after it was discovered that instead of serving donuts and coffee, the franchisee was serving vodka and meat pies.
- Trade Secrets. If the success of a business is based on a trade secret, special process, or innovative technology, establishing a franchise may make the business vulnerable to knock-offs or imitation. Although the franchise agreement specifically prohibits the disclosure of trade secrets, the fact that the franchisee may see opportunities to improve upon the process and become a competitor is not expressly prohibited.
- Overexposure, Brand Dilution . One or two locations of a business is unique and may generate enough demand that the business can charge top dollar for goods or services. When franchises appear on almost every street corner, the allure of the business may fade and the brand or business may suffer.
Advantages for the Franchisee
- Less Risk. In certain industries, when compared with starting one’s own business from scratch, buying a franchise enables the franchisee to own a business with a proven track record and an established market presence, thereby reducing the risk of failure. However, purchasing a franchise still doesn’t guarantee success, and many franchisees go out of business, losing their initial investment and start-up capital.
- Name/Brand Recognition. The franchise has an established image and identity already, which can reduce or simplify marketing efforts. Many franchises are nationally advertised brands, shortening the time it takes for the franchisee to establish a market presence.
- Access to Expertise, Ongoing Support. Franchisee often receives help with site selection, training materials, product supply, and marketing plans. The franchisee gets to take advantage of a business model whose strategies and processes have already been tested and streamlined.
- Relative Autonomy. Franchisee must comply with the terms and standards of the franchisor, but otherwise has a fair amount control over the day-to-day operations of the franchise.
Disadvantages for the Franchisee
- Cost . Buying and running a franchise can be very expensive. Jimmy John’s Subs was listed as one of the top franchises in 2016, but the initial investment to open a location was $325,000–$555,000. Franchise fees generally run in the $20,000-to-$30,000 range, though they can top $100,000 for higher-end, more established brands. Once open, there are ongoing royalties to pay, which typically range from 4 percent to 8 percent of gross revenues and include an ongoing assessment for marketing and advertising.
- Unequal Partnership . The franchisor sets the rules, and the franchisee must follow them. The franchisee doesn’t have much leverage if the franchisor falls short on promises or makes unreasonable demands.
- Rules and Enforcement. Franchisor rules imposed by the franchising authority are becoming increasingly strict. Some franchisors are using minor rule violations to terminate contracts and seize the franchise without any reimbursement. Often this happens when a franchise location becomes very profitable or the franchisor sees an opportunity to profit by seizing and liquidating the location.