7.4: Distribution Needed for Hypothesis Testing
Earlier, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. We will perform hypotheses tests of a population mean using either a normal distribution or a Student's \(t\)-distribution. (Remember, use a Student's \(t\)-distribution when the population standard deviation is unknown and the sample size is small, where small is considered to be less than 100 observations.) We perform tests of a population proportion using a normal distribution when we can assume that the distribution is normally distributed. We consider this to be true if the sample size is 100 or more. This is the same rule of thumb we used when developing the formula for the confidence interval for a population proportion.
Hypothesis Test for the Mean
Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means.
\[z_{obs}=\frac{\overline{x}-\mu_{0}}{\sigma / \sqrt{n}}\nonumber\]
The standardizing formula can not be solved as it is because we do not have \(\mu\), the population mean. However, if we substitute in the hypothesized value of the mean, \(\mu_0\) in the formula as above, we can compute a \(z\) value. This is the test statistic for a test of hypothesis for a mean and is presented in Figure \(\PageIndex{1}\). We interpret this \(z\) value as the associated probability that a sample with a sample mean of \(\overline x\) could have come from a distribution with a population mean of \(H_0\) and we call this \(z\) value \(z_{obs}\) for “observed”. At times notation \(z_c\) for “calculated” is used (see figures below). Figure \(\PageIndex{1}\) and Figure \(\PageIndex{2}\) show this process.
In Figure \(\PageIndex{1}\) two of the three possible outcomes are presented. \(\overline x_1\) and \(\overline x_3\) are in the tails of the hypothesized distribution of \(H_0\). Notice that the horizontal axis in the top panel is labeled \(\overline x\)'s. This is the same theoretical distribution of \(\overline x\)'s, the sampling distribution, that the Central Limit Theorem tells us is normally distributed. This is why we can draw it with this shape. The horizontal axis of the bottom panel is labeled \(z\) and is the standard normal distribution. \(z_{\frac{\alpha}{2}}\) and \(-z_{\frac{\alpha}{2}}\), called the critical values , are marked on the bottom panel as the \(z\) values associated with the probability the analyst has set as the level of significance in the test, \(\alpha\). The probabilities in the tails of both panels are, therefore, the same.
Notice that for each \(\overline x\) there is an associated \(z_ {} \) or \(z_c\), called the observed or calculated \(z\), that comes from solving the equation above. This observed \(z\) is nothing more than the number of standard errors that the sample mean is from the hypothesized mean. If the sample mean falls "too many" standard errors from the hypothesized mean we conclude that the sample mean could not have come from the distribution with the hypothesized mean, given our pre-set required level of significance. It could have come from \(H_0\), but it is deemed just too unlikely. In Figure \(\PageIndex{1}\) both \(\overline x_1\) and \(\overline x_3\) are in the tails of the distribution. They are deemed "too far" from the hypothesized value of the mean given the chosen level of alpha. If in fact this sample mean it did come from \(H_0\), but from in the tail, we have made a Type I error: we have rejected a good null. Our only real comfort is that we know the probability of making such an error, \(\alpha\), and we can control the size of \(\alpha\).
Figure \(\PageIndex{2}\) shows the third possibility for the location of the sample mean, \(\overline x\). Here the sample mean is within the two critical values. That is, within the probability of 1 - \(\alpha\) and we cannot reject the null hypothesis.
This gives us the decision rule for testing a hypothesis for a two-tailed test:
| Decision rule: two-tailed test |
|---|
| If \(\left|\mathrm{z}_{obs}\right| \leq \mathrm{z}_{\frac{\alpha}{2}}\) : then do not reject \(H_0\) |
| If \(\left|\mathrm{z}_{obs}\right|>\mathrm{z}_{\frac{\alpha}{2}}\) : then reject \(H_0\) |
Table \(\PageIndex{1}\)
This rule will always be the same no matter what hypothesis we are testing or what formulas we are using to make the test. The only change will be to change the \(z_ {obs} \) to the appropriate symbol for the test statistic for the parameter being tested. Stating the decision rule another way: if the sample mean is unlikely to have come from the distribution with the hypothesized mean we must reject the null hypothesis. Here we define "unlikely" as having a probability less than alpha of occurring.
\(p\)-value Approach
An alternative decision rule can be developed by calculating the probability of observing the sample mean \(\overline x\) if we assume the null hypothesis is true . This probability would, in other words, equal the tail probability that the t-test statistic is equal or greater than our observed test statistic \(z_{obs}\) from sample data (under the assumption that \(H_0\) is true).
Here the notion of "likely" and "unlikely" is defined by the probability of drawing a sample with a mean \(\overline x\) from a population with the hypothesized mean that is either larger or smaller than that found in the sample data. Simply stated, the \(p\)-value approach compares the desired significance level, \(\alpha\), to the \(p\)-value which is the probability of drawing a sample mean further from the hypothesized value than the actual sample mean, assuming \(H_0\) is true. A large \(p\)-value calculated from the data indicates this is a likely outcome and that we should not reject the null hypothesis. The smaller the \(p\)-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. The relationship between the decision rule of comparing the observed (calculated) test statistic, \(z_{obs}\), and the critical value, \(z_\alpha\), and using the \(p\)-value can be seen in Figure \(\PageIndex{3}\).
The observed (calculated) value of the test statistic is \(z_c\) in this figure and is marked on the bottom graph of the standard normal distribution because it is a \(z\) value. In this case the calculated value is in the tail and thus we must reject the null hypothesis. The associated \(\overline x\) is just unusually large (and thus too unlikely) to believe that it came from the distribution with a mean of \(\mu_0\) with a significance level of \(\alpha\).
If we use the \(p\)-value decision rule we need one more step. We need to find in the standard normal table the probability associated with the observed test statistic, \(z_ {obs}\). We then compare that to the \(\alpha\) associated with our selected level of confidence. In Figure \(\PageIndex{3}\) we see that the \(p\)-value is less than \(\alpha\) and therefore we must reject the null. We know that the \(p\)-value is less than \(\alpha\) because the area under the \(p\)-value is smaller than \(\alpha/ 2\). When our sample size is small (having less than 100 observations), we can find an approximate \(p\)-value from the Student's \(t\) table. In the relevant \(df\) row based on our sample size, we first locate the test statistic value(s) closest to our observed \(t\)-statistic, \(t_{obs}\). We then find the one-tailed or two-tailed probabilities associated with those test statistic value(s) depending on the type of test we are conducting (see next section). In case of a two-tailed test, the one-tailed probabilities from the Student's \(t\) table need to be multiplied by two. The obtained \(p\)-value interval will give us the range that we will then compare to \(\alpha\) and make our decision.
It is important to note that two researchers drawing randomly from the same population may find two different \(p\)-values from their samples. This occurs because the \(p\)-value is calculated as the probability in the tail beyond the sample mean assuming that the null hypothesis is correct. Because the sample means will in all likelihood be different this will create two different \(p\)-values.
Here is a systematic way to make a decision of whether you reject or do not reject a null hypothesis if using the \(p\)-value and a preset or preconceived \(\alpha\) (the " significance level "). A preset \(\alpha\) is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. In any case, the value of \(\alpha\) is the decision of the analyst. When you make a decision to reject or not reject \(H_0\), do as follows:
- If \(p\)-value < \(\alpha\), reject \(H_0\). There is sufficient evidence to conclude that \(H_0\) is an incorrect belief and that the alternative hypothesis , \(H_a\), may be correct.
- If \(p\)-value \(\geq \alpha\), cannot reject \(H_0\). There is not sufficient evidence to conclude that the alternative hypothesis, \(H_a\), may be correct. In this case the status quo - the baseline set by the null hypothesis - stands.
- When you "cannot reject \(H_0\)", it does not mean that you should believe that \(H_0\) is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of \(H_0\); and we must remember that the null is only a baseline comparison point, set up for testing our alternative or research hypothesis, \(H_a\).
One and Two-tailed Tests
The discussion of Figure \(\PageIndex{1}\) - Figure \(\PageIndex{3}\) was based on the null and alternative hypothesis presented in Figure \(\PageIndex{1}\). This was called a two-tailed test because the alternative hypothesis allowed that the mean could have come from a population which was either larger or smaller than the hypothesized mean in the null hypothesis. This could be seen by the statement of the alternative hypothesis as \(\mu \neq 100\), in this example.
It may be that the analyst has an interest in whether the sample differs in a certain direction from (for example, is higher than) the hypothesized value. If this is the case, it becomes a one-tailed test and all of the alpha probability is placed in just one tail and not split into \(\alpha /2\) as in the above case of a two-tailed test. For example, a car manufacturer claims that their Model 17B provides gas mileage of greater than 25 miles per gallon. The null and alternative hypothesis would be:
- \(H_0: \mu \leq 25\)
- \(H_a: \mu > 25\)
The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent significance that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo.
Figure \(\PageIndex{4}\) shows the two possible cases and the form of the null and alternative hypothesis that give rise to them, where \(\mu_0\) is the hypothesized value of the population mean.
Effects of Sample Size on Test Statistic
In developing the confidence intervals for the mean from a sample, we found that most often we would not have the population standard deviation, \(\sigma\). If the sample size were less than 100, we could simply substitute the point estimate for \(\sigma\), the sample standard deviation, \(s\), and use the Student's \(t\)-distribution to correct for this lack of information.
When testing hypotheses we are faced with this same problem and the solution is exactly the same. Namely, if the population standard deviation is unknown, and the sample size is less than 100, substitute \(s\), the point estimate for the population standard deviation, \(\sigma\), in the formula for the test statistic and use the Student's \(t\) distribution. All the formulas and figures above are unchanged except for this substitution and changing the \(Z\) distribution to the Student's \(t\) distribution on the graph. Remember that the Student's \(t\) distribution can only be computed knowing the proper degrees of freedom for the problem. In this case, the degrees of freedom is computed as before with confidence intervals: \(df = (n-1)\). The observed \(t\)-value is compared to the \(t\)-value associated with the pre-set level of confidence required in the test, \(t_{\alpha, df}\) found in the Student's \(t\) tables. If we do not know \(\sigma\), but the sample size is 100 or more, we simply substitute \(s\) for \(\sigma\) and use the normal distribution.
Table \(\PageIndex{2}\) summarizes these rules.
| Sample size | Test statistic |
|---|---|
|
< 100
(\(\sigma\) unknown) |
\(t_{obs}=\frac{\overline{x}-\mu_{0}}{s / \sqrt{n}}\) |
|
< 100
(\(\sigma\) known) |
\(z_{obs}=\frac{\overline{x}-\mu_{0}}{\sigma / \sqrt{n}}\) |
|
> 100
(\(\sigma\) unknown) |
\(z_{obs}=\frac{\overline{x}-\mu_{0}}{s / \sqrt{n}}\) |
|
> 100
(\(\sigma\) known) |
\(z_{obs}=\frac{\overline{x}-\mu_{0}}{\sigma / \sqrt{n}}\) |
Table \(\PageIndex{2}\) Test Statistics for Test of Means, Varying Sample Size, Population Standard Deviation Known or Unknown
A Systematic Approach for Testing a Hypothesis
A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test.
- Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test?
-
Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for businesses are 90, 95, and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error.
Next, on the basis of the hypotheses and sample size, select the appropriate test statistic and find the relevant critical value: \(z_\alpha\), \(t_\alpha\), etc. Drawing the relevant probability distribution and marking the critical value is always big help. Be sure to match the graph with the hypothesis, especially if it is a one-tailed test.
- Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated.
-
Compare the calculated test statistic and the critical value. Marking these on the graph will give a good visual picture of the situation. There are now only two situations:
- The test statistic is in the tail: Reject the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data.
- The test statistic is not in the tail: Cannot reject the null, the sample data are compatible with the hypothesized population parameter.
- Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 5 % level of significance we must reject the null hypothesis that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”.
All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question.


31.4: The Total Scattering Intensity is Related to the Periodic Structure of the Electron Density in the Crystal
Systematic Absences
Diffraction data from a face-centered cubic crystal will be missing reflections (spots) from all hkl planes in which h + k , h + l , and k + l are odd numbers. The reason for these absent reflections is that the arrangement of the lattice points leads to total destructive interference of the diffracted X-rays. Although the lack of data may seem like a hindrance to determining a crystal structure, the systematic absence of reflections from specific families of planes actually reveals the presence of symmetry features in the crystal lattice that help identify the lattice type. Crystals with centering features (body-centered cels, face-centered cells, etc.), glide planes, and screw axes will produce diffraction patterns with systematic absences.
The Structure Factor
Systematic reflection absences, as well as variations in reflection intensity, are described mathematically by the structure factor, a mathematical function that incorporates all of the variables that affect how X-rays are scattered by the electrons of the atoms in the crystal lattice. (We do not concern ourselves with the effect of the nucleus on scattering because the mass of the nucleus is so large relative to the mass of an electron.)
The Atomic/Ionic Scattering Factor
The construction of a structure factor begins with the fundamental assumption that the total scattering intensity is related to the electron density of the atoms in the crystal. But the scattering also depends on the scattering angle and the wavelength of the radiation. To account for all of these parameters, an atomic scattering factor , \(f\), has been developed:
\[f = 4\pi \int_0^{\infty} \rho(r)\dfrac{\sin \, kr}{kr}r^2 \, dr \nonumber \]
In this equation, \(r\) is the radius of an idealized spherical atom, \(\rho(r)\) is the electron density of this idealized atom. Because the X-ray wavelength used in diffraction experiments is of similar size as the size of the atoms it scatters from, there will be interference among the waves scattered by the electrons on a single atom. The more electrons in an atom and the more diffuse the electron cloud around the atom, the greater the destructive interference among the scattered waves. To account for this intra-atomic interference, the \(\dfrac{\sin \, kr}{kr}\) term is included in the scattering factor. Here, \(k\) is equal to \(\dfrac{4\pi \sin \theta}{\lambda}\), where \(\theta\) is the scattering angle and \(\lambda\) is the wavelength of the X-ray. A plot of \(f\) versus \(\dfrac{\sin \theta}{\lambda}\) is shown in figure \(\PageIndex{1}\). Notice that at \(\theta\) = 0, \(f\) is equal to the number of electrons in the particle. Also notice that, although the isoelectronic ions Ca 2+ and Cl - have an identical \(f\) value at \(\theta\) = 0, their \(f\) values differ at larger angles because the ions have different electron densities. Finally, notice that H atoms do not scatter a great deal, thus making them difficult to "locate" among the scattering from all of the larger atoms.
Scattering by Neighboring Atoms and Neighboring Planes
Along with the scattering factor for individual atoms, we must also take into consideration the interferences caused by neighboring atoms in the lattice. As our example, we will analyze a crystal that contains atoms of two elements, here designated P and Q . We will assume that the respective atoms have scattering factors \(f\) P and \(f\) Q . We will also assume that the distance between successive P atoms is the same as the distance between successive Q atoms. This distance is \(\dfrac{a}{h}\), where \(a\) is the length of the unit cell along the a axis. Finally, we will assume that the distance between P and Q atoms is \(x\). In this arrangement, the P atoms lie in neighboring hkl planes, and the Q atoms lie in neighboring hkl planes that are interleaved between the hkl planes of the P atoms. Figure \(\PageIndex{2}\) shows the arrangement of the atoms lying along the a axis.
In this arrangement, the difference in path length of X-rays scattered by successive P atoms, \(\Delta\) PP , (and the difference in path length of X-rays scattered by successive Q atoms, (\(\Delta\) QQ ) is described by the van Laue equation
\(\Delta\) PP = \(\Delta\)QQ = \(\dfrac{a}{h}(cos\alpha - cos\alpha_0) = n\lambda\)
The difference in path length for X-rays diffracted by neighboring P and Q atoms is thus
\(\Delta\)PQ = \(x(\cos\alpha - \cos\alpha_0) = \dfrac{nh\lambda x}{a}\)
Because of this difference in path length, there is a resulting a phase difference between the scattered beams from neighboring P and Q atoms due to the difference in the time required to travel the different distances:
\(\phi = 2\pi \dfrac{\Delta_{PQ}}{\lambda} = 2\pi \dfrac{h\lambda x}{\lambda a} = \dfrac{2\pi h x}{a} \)
The light scattered from neighboring P and Q atoms will have an amplitude dependent on the scattering factors of each atom and the angular frequency of the X-ray radiation, \(\omega\)
\[A = f_P \, cos \omega t + f_Q \, cos(\omega t + \phi) \nonumber \]
Replacing the cosine functions with exponential functions will simplify the following mathematical calculation, thus
\[A = f_P \, e^{i\omega t} + f_Q \, e^{i(\omega t + \phi)} \label{1} \]
Because intensity is proportional to amplitude squared,
\( I \propto |A|^2 = (f_P \, e^{i\omega t} + f_Q \, e^{i(\omega t + \phi)})(f_P \, e^{-i\omega t} + f_Q \, e^{-i(\omega t + \phi)} \)
which can be simplified to
\(f_P^2 + f_Pf_Qe^{i\phi} + f_Pf_Qe^{-i\phi} + f_Q^2 \)
Switching back to cosine functions, we get
\(f_P^2 + f_Q^2 + 2 f_Pf_Qcos\phi \)
The \(f_P^2\) term represents the constructive interference of the x-rays scattered from the parallel planes through the P atoms. The \(f_Q^2\) term represents the constructive interference of the x-rays scattered from the parallel planes through the Q atoms. The \(2 f_Pf_Qcos\phi \) term represents the interference from the scattering from the two sets of interleaved P and Q planes. From this equation, we can see that the intensity of the scattered light is independent of the angular frequency of the X-ray radiation because there is no term involving \(\omega\). Thus, in equation \(\ref{1}\) the frequency-dependent \(e^{i\omega t}\) must have a value of 1, and the equation can be rewritten and redefined as the structure factor along the a axis of the crystal, \(F(h)\)
\[F(h) = f_P + f_Q \, e^{i\phi)} = f_P + f_Q \, e^{\dfrac{2\pi h x}{a}} \label{2} \]
For a unit cell with sides of length \(a, \, b,\) and \(c\) that contains atoms of type \(j\) positioned at points \(x_j, y_j \) and \(z_j\), equation \(\ref{2}\) becomes
\[F(hkl) = \sum_j f_{j} e^{2\pi (hx_{j}/a + ky_{j}/b + lz_{j}/c)} \label{3} \]
Here \(f_j\) is the scattering factor for type \(j\) atoms, and hkl are the Miller indicies of the diffracting planes. By convention, \(x_j, y_j, \) and \(z_j\) are expressed in units of \(a, b,\) and \(c\), so that \(x_j^{'} =\dfrac{x_j}{a}\), \(y_j^{'} =\dfrac{y_j}{b}\), and \(z_j^{'} =\dfrac{z_j}{c}\), to give
\[F(hkl) = \sum_j f_j e^{2\pi (hx_{j}^{'} + ky_{j}^{'} + lz_{j}^{'})} \label{4} \]
Combining the Influences
\(F(hkl)\) is the structure factor of a crystal, and the intensity of reflections from this crystal are proportional to \(|F(hkl)|^2\). The strcuture factor can thus be used to explain the systematic absences described in the introduction to this section. Whenever \(F(hkl)\) = 0 for a set of Miller planes, the reflection will be absent.
Derive an expression for the structure factor of a face-centered cubic unit cell of identical atoms. Determine which hkl planes will not produce reflections.
Solution
The coordinates of the lattice points in a face-centered cubic unit cell are (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,1/2,1/2), (1/2,1/2,0), (1/2,0,1/2), (1,1/2,1/2), (1/2,1,1/2), (1/2,1/2,1). The length of the sides of the cell is \(a\). Each corner lattice point is shared by 8 cells, so only 1/8 of each of these atoms is considered to be contributing to the scattering. Each face lattice point is shared by 2 cells, so only 1/2 of each of these atoms is considered to be contributing to the scattering. We will use equation \(\ref{4}\) to solve the problem, remembering that \(a = b = c\) for a cubic cell:
\[F(hkl) = \dfrac{1}{8} f(e^{2\pi i(0+0+0)} + e^{2\pi i(h+0+0)} + e^{2\pi i(0+k+0)} + e^{2\pi i(0+0+l)} + e^{2\pi i(h+0+l)} + e^{2\pi i(h+k+0)} + e^{2\pi i(h+k+l)} + e^{2\pi i(0+k+l)}) + \dfrac{1}{2}f(e^{2\pi i(0+k/2+l/2)} + e^{2\pi i(h/2+k/2+0)}+ e^{2\pi i(h/2+0+l/2)} + e^{2\pi i(h+k/2+l/2)} + e^{2\pi i(h/2+k+l/2)} + e^{2\pi i(h/2+k/2+l)}) \label{5} \]
Because \(e^{2\pi i} \) = 1 and \(e^{\pi i}\) = -1, equation \(\ref{5}\) becomes
\[F(hkl) = \dfrac{1}{8}f(1^0 + 1^h + 1^k + 1^l + 1^{h+l} + 1^{h+k} + 1^{h+k+l} + 1^{k+l}) + \dfrac{1}{2}f((-1)^{k+l} + (-1)^{h+k} + (-1)^{h+l} + (-1)^{2h+k+l} + (-1)^{h+2k+l} + (-1)^{h+k+2l}) \nonumber \]
Because \(1^n = 1\) for all \(n\),
\[F(hkl) = \dfrac{1}{8}f(8) + \dfrac{1}{2}f((-1)^{k+l} + (-1)^{h+k} + (-1)^{h+l} + (-1)^{2h+k+l} + (-1)^{h+2k+l} + (-1)^{h+k+2l}) \nonumber \]
Upon testing a few hkl planes, you will see that if \(h,k,l\) are a mixture of odd and even numbers, then \(F(hkl) = 0\), but if \(h,k,l\) are all odd or all even, the value of \(F(hkl) = 4f\). Three examples:
All odd \(F(111) = f (1 + \dfrac{1}{2}((-1)^2 + (-1)^2 + (-1)^2 + (-1)^4 + (-1)^4 + (-1)^4 = f(1 + 3) = 4f \)
All even \(F(222) = f (1 + \dfrac{1}{2}((-1)^4 + (-1)^4 + (-1)^4 + (-1)^8 + (-1)^8 + (-1)^8 = f(1 + 3) = 4f \)
Mix \(F(110) = f (1 + \dfrac{1}{2}((-1)^1 + (-1)^2 + (-1)^1 + (-1)^3 + (-1)^3 + (-1)^2 = f(1 + \dfrac{1}{2}((-1) + 1 + (-1) + (-1) + (-1) + 1)) = 1+ (-1) = 0 \)
For crystals comprised of atoms of more than one element, there are often reflections that are weaker than other reflections. These lower intensities are sometimes created when the scattering factors of different elements are not identical, leading to incomplete destructive interference, a smaller \(F(hkl)\), and thus a reflection of lower intensity. The lower intensity reflections are also a direct result of the structure factor. For example, for an ionically-bonded compound with the formula MX, which crystallizes in a face-centered cubic lattice, the structure factor can be shown to be
\[f(hkl) = f_+[1 + (-1)^{h+k} + (-1)^{h+l} + (-1)^{k+l}] + f_-[(-1)^{h+k+l} + (-1)^{h} + (-1)^{k} + (-1)^{l}] \nonumber \]
Inspection of this equation reveals that \(F(hkl) = 4(f_+ + f_-)\) if \(h, k, l\) are all even, and \(F(hkl) = 4(f_+ - f_-)\)if \(h, k, l\) are all odd. Thus, reflections from the all-even \(hkl\) planes are more intense than reflections from the all-odd \(hkl\) planes.
See also
The structure factor. P. Coppens. International Tables for Crystallography (2006). Vol. B, ch. 1.2, pp. 10-24


5.1: Gas Pressure
- Define the property of pressure
- Define and convert among the units of pressure measurements
- Describe the operation of common tools for measuring gas pressure
- Calculate pressure from manometer data
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure \(\PageIndex{1}\)). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.
Pressure is defined as the force exerted on a given area:
\[P=\dfrac{F}{A} \label{9.2.1}\]
Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area.
Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure \(\PageIndex{2}\).—the elephant or the figure skater?
A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in 2 ), so the pressure exerted by each foot is about 14 lb/in 2 :
\[\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2}\]
The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in 2 , so the pressure exerted by each blade is about 30 lb/in 2 :
\[\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3}\]
Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:
\[\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4}\]
The SI unit of pressure is the pascal (Pa) , with 1 Pa = 1 N/m 2 , where N is the newton, a unit of force defined as 1 kg m/s 2 . One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch— pounds per square inch (psi) —for example, in car tires. Pressure can also be measured using the unit atmosphere (atm) , which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table \(\PageIndex{1}\) provides some information on these and a few other common units for pressure measurements
| Unit Name and Abbreviation | Definition or Relation to Other Unit | Comment |
|---|---|---|
| pascal (Pa) | 1 Pa = 1 N/m 2 | recommended IUPAC unit |
| kilopascal (kPa) | 1 kPa = 1000 Pa | |
| pounds per square inch (psi) | air pressure at sea level is ~14.7 psi | |
| atmosphere (atm) | 1 atm = 101,325 Pa | air pressure at sea level is ~1 atm |
| bar (bar, or b) | 1 bar = 100,000 Pa (exactly) | commonly used in meteorology |
| millibar (mbar, or mb) | 1000 mbar = 1 bar | |
| inches of mercury (in. Hg) | 1 in. Hg = 3386 Pa | used by aviation industry, also some weather reports |
| torr | \(\mathrm{1\: torr=\dfrac{1}{760}\:atm}\) | named after Evangelista Torricelli, inventor of the barometer |
| millimeters of mercury (mm Hg) | 1 mm Hg ~1 torr |
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
- torr
- atm
- kPa
- mbar
Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1.
- \(\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}\)
- \(\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}\)
- \(\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}\)
- \(\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}\)
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure \(\PageIndex{5}\)). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure— the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure— the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).
Meteorology, Climatology, and Atmospheric Science
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure \(\PageIndex{5}\)) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.
In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.
The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure \(\PageIndex{7}\): the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.
Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.
Contributors and Attributions
Summary
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
Key Equations
- \(P=\dfrac{F}{A}\)
- p = hρg
Glossary
- atmosphere (atm)
- unit of pressure; 1 atm = 101,325 Pa
- bar
- (bar or b) unit of pressure; 1 bar = 100,000 Pa
- barometer
- device used to measure atmospheric pressure
- hydrostatic pressure
- pressure exerted by a fluid due to gravity
- manometer
- device used to measure the pressure of a gas trapped in a container
- pascal (Pa)
- SI unit of pressure; 1 Pa = 1 N/m 2
- pounds per square inch (psi)
- unit of pressure common in the US
- pressure
- force exerted per unit area
- torr
- unit of pressure; \(\mathrm{1\: torr=\dfrac{1}{760}\,atm}\)
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


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30.9: Collaborative Group Activities
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If one of the rocks from Mars examined by a future mission to the red planet does turn out to have unambiguous signs of ancient life that formed on Mars, what does your group think would be the implications of such a discovery for science and for our view of life elsewhere? Would such a discovery have any long-term effects on your own thinking?
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Suppose we receive a message from an intelligent civilization around another star. What does your group think the implications of this discovery would be? How would your own thinking or personal philosophy be affected by such a discovery?
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A radio message has been received from a civilization around a star 40 light-years away, which contains (in pictures) quite a bit of information about the beings that sent the message.  How would you advise the president? Does your group agree on your answer or do you also have a minority view to present?
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If there is no evidence that UFOs are extraterrestrial visitors, why does your group think that television shows, newspapers, and movies spend so much time and effort publicizing the point of view that UFOs are craft from other worlds? Make a list of reasons. Who stands to gain by exaggerating stories of unknown lights in the sky or simply fabricating stories that alien visitors are already here?
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Does your group think scientists should simply ignore all the media publicity about UFOs or should they try to respond? If so, how should they respond? Does everyone in the group agree?
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Suppose your group is the team planning to select the most important sights and sounds of Earth to record and put on board the next interstellar spacecraft. What pictures (or videos) and sounds would you include to represent our planet to another civilization?
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Let’s suppose Earth civilization has decided to broadcast a message announcing our existence to other possible civilizations among the stars. Your group is part of a large task force of scientists, communications specialists, and people from the humanities charged with deciding the form and content of our message. What would you recommend? Make a list of ideas.
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Think of examples of contact with aliens you have seen in movies and on TV. Discuss with your group how realistic these have been, given what you have learned in this class. Was the contact in person (through traveling) or using messages? Why do you think Hollywood does so many shows and films that are not based on our scientific understanding of the universe?
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Go through the Drake equation with your group and decide on values for each factor in the estimate. (If you disagree on what a factor should be within the group, you can have a “minority report.”) Based on the factors, how many intelligent, communicating civilizations do you estimate to be thriving in our Galaxy right now?


2: Motivation for Your Journey
Career exploration and decision-making is built upon the foundation of personal self-knowledge. If you do not know yourself, how can you possibly know what you want to do for a career? In this part of the course, you will be taking several career assessments to understand yourself better and make informed choices when researching your career options. These assessments will help identify your personality preferences and workplace strengths and then direct you toward professions that best complement your unique personality. You also will identify your values, underlying needs and motivations then make connections between your internal self-knowledge and your future career vision (Martinez & Shaker, 2021).
Thumbnail: Career goal word graphic. (Pixabay License via Pixabay )


17.2B: Vitamins and Amino Acids
- Describe how microorganisms and plants can synthesize many uncommon amino acids and vitamins
Amino Acids
Amino acids are biologically important organic compounds made from amine (-NH2) and carboxylic acid (-COOH) functional groups, along with a side-chain specific to each amino acid. The key elements of an amino acid are carbon, hydrogen, oxygen, and nitrogen. About 500 amino acids are known which can be classified in many ways.
Microorganisms and plants can synthesize many uncommon amino acids. For example, some microbes make 2-aminoisobutyric acid and lanthionine, which is a sulfide-bridged derivative of alanine. Both of these amino acids are found in peptidic lantibiotics such as alamethicin. While in plants, 1-aminocyclopropane-1-carboxylic acid is a small disubstituted cyclic amino acid that is a key intermediate in the production of the plant hormone ethylene.
Vitamins
Vitamin K is a group of structurally similar, fat-soluble vitamins that are needed for the posttranslational modification of certain proteins required for blood coagulation and in metabolic pathways in bone and other tissue. They are 2-methyl-1,4-naphthoquinone (3-) derivatives. This group of vitamins includes two natural vitamers: vitamin K1 and vitamin K2.
Vitamin K1, also known as phylloquinone or phytomenadione (also called phytonadione), is synthesized by plants, and is found in highest amounts in green leafy vegetables because it is directly involved in photosynthesis.
Vitamin K2 has several subtypes, one of which is involved in bone metabolism. Vitamin K2 homologs (menaquinones) are characterized by the number of isoprenoid residues in their side chain. Menaquinones are abbreviated MK-n, where n represents the number of isoprenoid side chain residues. For example, menaquinone-4 (abbreviated MK-4), has four isoprene residues in its side chain. Bacteria in the colon (large intestine) can produce a range of vitamin K2 forms, and can also convert K1 into K2 (MK-7 homolog). No known toxicity exists for vitamins K1 or K2.
Three synthetic types of vitamin K are known: vitamins K3, K4, and K5. Although the natural K1 and K2 forms are nontoxic, the synthetic form K3 (menadione) has shown toxicity.
Key Points
- Amino acids are organic compounds made from amine (-NH2) and carboxylic acid (-COOH) functional groups, and a specific side-chain.
- Vitamin K is a group of structurally similar, fat-soluble vitamins that are needed for the posttranslational modification of certain proteins.
- Bacteria in the colon (large intestine) can produce a range of vitamin K2 forms, and can also convert vitamin K1 into vitamin K2.
Key Terms
- amino acid : Any organic compound containing both an amino and a carboxylic acid functional group.
- vitamin : Any of a specific group of organic compounds essential in small quantities for healthy human growth, metabolism, development, and body function; found in minute amounts in plant and animal foods or sometimes produced synthetically; deficiencies of specific vitamins produce specific disorders.
- synthesize : To combine two or more things to produce a new, more complex product.
- dichotomous key : a tool for identification of plants and animals, written as a sequence of paired questions, the choice of which determines the next pair of questions until a name or identification is reached


4.4: Google Docs- Enhanced Formatting Features
By the end of this section, you will be able to:
- Insert and format page numbers
- Insert and modify a header/footer
- Insert and format a list
The market trends report has been composed thus far in Microsoft Word. However, it could have just as easily been composed in Google Docs . Docs has many of the same features that we see in Word. Some people prefer working in Docs as it can be more user-friendly than Word, particularly its features involving collaborating with others.
Many of Docs’s advanced formatting features are similar to those in Word. Just like in Word, you can insert and format page numbers, headers, footers, and multilevel lists. Some customizable features that are present in Word may not be present in Docs, but some users might prefer fewer choices, as it facilitates a more user-friendly program. Here, we will revisit many of the tools we covered earlier and apply them to the market trends report in Docs. We will start with an earlier draft of the report where we began the chapter (Figure 4.2).
Mac computers come with Safari as the default internet browser. While Docs works just fine in Safari, it functions better in Google Chrome. Chrome is also a Google product and, as such, comes with integrations that work seamlessly with Docs, such as browser extensions and notifications, as well as supports Chrome-only features, like voice typing.
Page Numbers
The tool for placing page numbers into a Google document has fewer customizable formatting options than the tool in Word. There are only two page number styles: top right or bottom right. These are both available in the Insert menu. Let’s start with the market trends draft and add page numbers (Figure 4.65). If you select More options, a few more customization options appear. You can choose to have your page numbers start counting from the second page of your document, which could be useful if your first page is a cover page or a table of contents. You can also start your page numbering at a certain number (i.e., other than the number 1), as shown in Figure 4.66. The page number is always placed in the header or footer.
One convenient feature of Docs is its ability to easily add a page count to your page number. First, you insert your page number, then manually type the word “of” after it (Figure 4.67). Then, go to the Insert menu, click Page numbers , and select Page count (Figure 4.68). This will add a field that gives the total number of pages in the document. Now, your page count should appear as “# of #” (e.g., “1 of 5”) (Figure 4.69).
Headers and Footers
You can add headers and footers to your document in Docs. The Headers & footers command is also located in the Insert menu. Once you have added your header or footer, you can choose to further configure it by choosing Options, the blue command that is located on the header or footer itself. This Options button opens to a drop-down menu with a few choices. If you select Header format or Footer format, you will see a dialog box with the option of applying certain header/footer settings to a section of the entire document, as seen in Figure 4.70. You can also control the height of each header/footer in inches, as well as choose to not have the header or footer appear on the first page. This latter choice is useful in documents for which you have a cover page. Simply tick the box Different first page. When you are finished configuring the headers/footers, select Apply.
You can edit and format the text contained in the header/footer as you would text in any other part of the document: by manually changing the font type; applying bold, italic, or underlined format; adding an image; or choosing a different alignment.
You can also add a horizontal line that visually separates your header or footer from the rest of the text. This can add a neat and professional look to your document. Just put the cursor where you want the line, go to the Insert menu, and choose Horizontal line , as shown in Figure 4.71.
Once you have made all your adjustments in the header/footer, just place your cursor anywhere in the body of the document to exit from the header/footer editing mode, or, alternatively, press the Esc (escape) key on the keyboard.
Remember that using the View menu, you can change the preview of the header/footer. Make sure that you are viewing the document with Print layout checked so that you can view your headers and footers. If you choose to work without the Print layout option checked, you won’t see the headers/footers.
Lists
The chapter Creating and Working in Documents briefly discussed how to insert a bulleted or numbered list in Docs. In this section, we will use the document outline we used earlier in this chapter to create a multilevel list in Docs.
To access numbered, lettered, or bulleted lists, go to the Format menu and choose the Bullets & numbering option. This will open a drop-down menu, where you can see your choices for customization. You can also access lists (numbered and bulleted) from the action bar (Figure 4.73).
Because adding lists to your document is generally done for organization and visual purposes, you want to make sure they are easy to read and showcase the most important information. For all list types, you can increase or decrease the space between the lines. This can help with readability. To do this, select the whole list and go to the Line and Paragraph spacing command and choose a wider or narrower space, as shown in Figure 4.74. You can also change the color of the text, which could be useful for helping your list stand out from the rest of your document. Select the text you want to format and use the action bar command for Text color.
Numbered and Lettered Lists
To create a multilevel list , choose the style of list you want, then use the Tab key to indent your lines. We want to create an outline of the headings and subheadings for our marketing trends report. This will give collaborators an idea of the format for the report, as well as allow us to use the outline to allocate sections to certain people or departments to fill in the content. For example, Figure 4.74 shows a straightforward Numbered list from the Insert menu, but Figure 4.75 shows one with multiple levels. You can create these list levels by using the Tab key, which will increase the indent on the line, creating a sublevel. If you want a further sublevel, press the Tab key twice, making it a subcategory of the category. You can also do this by using the Increase indent command in the action bar.
Bulleted Lists
Unlike numbered and lettered lists, bulleted lists do not have an obvious sequence. Docs offers different types of bullets to indicate the multilevel layers, which you can customize (Figure 4.76).
The process for creating a multilevel bulleted list is the same as for numbered and lettered lists: To create another sublevel below your current level, press Enter to go to the next level, then press Tab or use the Increase indent command to add it. In Figure 4.77, you can see the result of the new multilevel list using bullets instead of numbers. The best practice is to use bulleted lists only if they are no longer than half a page; after that, a numbered list is more effective. You can change the bullet type by clicking in the bulleted list, going to the Format menu, then choosing List options from the Bullets & numbering tool (Figure 4.78).
Checklists
A checklist is a useful type of list option in Docs that has several everyday applications. You may use them to make a printed to-do list, such as a list of tasks or errands to carry with you when you are away from a computer. Or you may want to add a checklist to a guidelines document for other people to be able to print out and reference, or use digitally. For example, it could also be useful for the multiple rounds of editing that will be needed for the market trends report. As the document progresses through the various departments for editing and revision, each department could tick a box when their round is complete, indicating that the document has been approved by the various departments.
A checklist is inserted the same way as any other type of list, except that it is technically listed as a subtype of a bulleted list in Docs. That is, you follow the same process as adding a bulleted list, but then just make sure to choose the option that shows the checklist. As you can see in Figure 4.79, the to-do checklist is complete.


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arterial blood sampling
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blood is obtained via puncture into an artery
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bronchoscopy collection
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invasive procedure where a thin, flexible tube with a camera is inserted into the airways, and sputum samples are collected from specific areas within the respiratory system
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butterfly needle
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winged infusion set with a vacutainer attached
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capillary blood collection
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obtaining blood from capillaries
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catheterized urine sample
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a catheter is inserted into the bladder to obtain a urine sample directly from the bladder
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clean-catch urine sample
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patient is instructed to start urinating into the toilet, stop briefly, and then continue urinating into the collection container; most common method of urine collection
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complete urinalysis
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examines the physical, chemical, and microscopic composition of a sample of urine
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dipstick
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special test strip containing chemically infused pads
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fecal occult blood card
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card containing a testing area that changes color when it comes into contact with blood
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fecal occult blood test (FOBT)
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test used to detect small amounts of blood in the stool
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feces (also, stool or bowel movement)
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solid waste product that is expelled from the body through the rectum and anus during the process of defecation
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first morning void (FMV) sample
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collecting the first urine voided in the morning
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hematochezia
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bloody stool
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hematuria
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blood in the urine
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induced sputum collection
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a hypertonic saline or other aerosolized agent is used to induce sputum production
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lancet
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device used to puncture the skin to collect a capillary blood sample
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melena
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dark, tarry stools, often associated with upper gastrointestinal bleeding
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multiple-draw needle
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needle that connects to a separate vacutainer
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nasal swab
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a swab is used to collect a sample from the inside of the nasal passage
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nasopharyngeal swab
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a swab is used to collect a sample from the nasopharynx, which is the upper part of the throat that connects to the back of the nasal cavity
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nasotracheal suction collection
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a nasotracheal suction catheter is used to collect sputum directly from the lower respiratory tract
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occult blood
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blood that is hidden from plain sight
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oropharyngeal swab
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a swab is used to collect a sample from the oropharynx, which is the part of the throat at the back of the mouth
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pediatric collection device
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a urine collection bag for infants and young children
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plasma
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the dominant component of blood that contains water, proteins, electrolytes, lipids, and glucose
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random stool sample
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a single stool sample collected at any time without specific timing or preparation
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random urine sample
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a urine sample collected at any time without specific timing or preparation
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reference range
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includes typical values for a specific laboratory test
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spontaneous sputum collection
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patient coughs up sputum into a sterile container
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sputum
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mucus and other secretions that are coughed up from the mouth
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sputum trap
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a specialized collection device that attaches to the patient’s breathing equipment (such as ventilators) to collect sputum samples
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steatorrhea
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fat in the stool
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stool culture
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used to identify and isolate specific bacteria or pathogens causing gastrointestinal infections
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stool fat test
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used to measure the amount of fat in the stool
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stool for
Clostridium difficile
(
C. diff
) toxin
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detects the presence of
C. difficile
toxins in the stool
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stool for ova and parasites (O&P)
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testing used to identify the presence of parasites or their eggs in the stool
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stool specimen (also, fecal specimen or fecal sample)
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a sample of feces (solid waste) collected from a person for laboratory analysis
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suprapubic aspiration
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invasive method involving using a needle and syringe to aspirate urine directly from the bladder through the abdominal wall
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therapeutic range
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includes target values for certain tests when managing specific conditions
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three-day stool collection
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stool samples are collected over a period of three consecutive days
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timed urine collection
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collect urine at specific times throughout the day
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tourniquet
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tightly tied band applied around a limb to temporarily constrict or restrict blood flow
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turbidity
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cloudiness or haziness
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twenty-four-hour urine collection
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collection of all urine produced over a twenty-four-hour period
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urinary casts
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small particles shaped like a tube that may contain different types of cells or substances
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vacutainer adapter
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a plastic device that holds the vacutainer tube in place and connects to the needle for blood collection
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vacutainer needle
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a needle with a built-in vacutainer adapter
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vacutainer tube
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a sterile, color-coded tube used to collect and transport a blood sample
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venipuncture
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the process of introducing a needle into a patient’s vein to collect a blood sample or insert an IV catheter


4.3E: Exercises
Practice Makes Perfect
Direct Translation Applications
In the following exercises, translate to a system of equations and solve.
1. The sum of two number is 15. One number is 3 less than the other. Find the numbers.
2. The sum of two number is 30. One number is 4 less than the other. Find the numbers.
- Answer
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13 and 17
3. The sum of two number is −16. One number is 20 less than the other. Find the numbers.
4. The sum of two number is \(−26\). One number is 12 less than the other. Find the numbers.
- Answer
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\(−7\) and \(−19\)
5. The sum of two numbers is 65. Their difference is 25. Find the numbers.
6. The sum of two numbers is 37. Their difference is 9. Find the numbers.
- Answer
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\(14\) and \(23\)
7. The sum of two numbers is \(−27\). Their difference is \(−59\). Find the numbers.
8. The sum of two numbers is \(−45\). Their difference is \(−89\). Find the numbers.
- Answer
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\(22\) and \(−67\)
9. Maxim has been offered positions by two car companies. The first company pays a salary of $10,000 plus a commission of $1000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?
10. Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?
- Answer
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Eighty cable packages would need to be sold to make the total pay the same.
11. Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal?
12. Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?
- Answer
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Mitchell would need to sell 120 stoves for the companies to be equal.
13. Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?
14. June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
- Answer
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8 and 40 gallons
15. Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?
16. Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?
- Answer
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1000 calories playing basketball and 400 calories canoeing
17. Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.
18. Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?
- Answer
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Oranges cost $2 per pound and bananas cost $1 per pound
19. Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?
20. Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?
- Answer
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Package of paper $4, stapler $7
21. The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?
22. The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?
- Answer
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Hot dog 150 calories, cup of cottage cheese 220 calories
23. Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?
24. Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?
- Answer
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Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade.
Solve Geometry Applications
In the following exercises, translate to a system of equations and solve.
25. The difference of two complementary angles is 55 degrees. Find the measures of the angles.
26. The difference of two complementary angles is 17 degrees. Find the measures of the angles.
- Answer
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\(53.5\) degrees and \(36.5\) degree.
27. Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.
28. Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.
- Answer
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16 degrees and 74 degrees
29. The difference of two supplementary angles is 8 degrees. Find the measures of the angles.
30. The difference of two supplementary angles is 88 degrees. Find the measures of the angles.
- Answer
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134 degrees and 46 degrees
31. Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.
32. Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.
- Answer
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37 degrees and 143 degrees
33. The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.
34. The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.
- Answer
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\(16°\) and \(74°\)
35. The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.
36. The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.
- Answer
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\(45°\) and \(45°\)
37. Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.
38. Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.
- Answer
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Width is 41 feet and length is 118 feet.
39. A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.
40. The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.
- Answer
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Width is 10 feet and length is 40 feet.
Solve Uniform Motion Applications
In the following exercises, translate to a system of equations and solve.
41. Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?
42. College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?
- Answer
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11 hours
43. At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?
44. Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?
- Answer
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\(1.5\) hour
45. The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.
46. A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.
- Answer
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Boat rate is 16 mph and current rate is 4 mph.
47. A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.54.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.)
48. A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.
- Answer
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Boat rate is 18 mph and current rate is 2 mph.
49. A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
50. A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
- Answer
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Jet rate is 265 mph and wind speed is 22 mph.
51. A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
52. A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
- Answer
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Jet rate is 415 mph and wind speed is 25 mph.
Writing Exercises
53. Write an application problem similar to Example . Then translate to a system of equations and solve it.\
54. Write a uniform motion problem similar to Example that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.
- Answer
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Answers will vary.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?


8.15: Works Consulted and Further Reading
Berkey, Jonathan P., The Formation of Islam: Religion and Society in the Near East 600 – 800 . Cambridge: Cambridge University Press, 2003.
Donner, Fred, The Early Islamic Conquests. Princeton: Princeton University Press, 1981.
Esposito, John L., Islam: The Straight Path. Oxford: Oxford University Press, 1998.
Halm, Heinz, The Empire of the Mahdi: The rise of the Fatimids. Leiden: E.J. Brill, 1997.
Lassner, Jacob, The Shaping of Abbasid Rule. Princeton: Princeton University Press, 1980.
Kennedy, Hugh, The Prophet and the Age of the Caliphates: The Islamic Near East from the Sixth to the Eleventh Century. London: Longman, 2004.
Lewis, Bernard, The Arabs in History. Oxford: Oxford University Press, 1993.
Momen, Moojan, An Introduction to Shi‘I Islam. New Haven: Yale University Press, 1985.
Powell, James M., ed. Muslims Under Latin Rule, 1100 – 1300. Princeton: Princeton University Press, 1990.
Watt, W. Mongomery, Muhammad: Prophet and Statesman . Oxford: Oxford University Press, 1974.


8.4: Adding Tables
Tables are adding by via the menu "Elements" in toolbar and selecting Table or directly via the editor toolbar (see below).
You can choose the number of rows and columns you need (can be changed later). This gives you an empty table structure that you can start to fill out. If you want to delete a row, select an entry in the row you want to delete and click on the x that appears to the left. If you want to add a row above or below, you can do that by clicking the > that appears above or below the x respectively.
If you are copying a table from a website you will likely be able to copy its structure via a simple copy-paste. You can also add tables directly by copy-pasting a section of a spreadsheet although this varies with the spreadsheet application you are using and the complexity of the table. The same is true of tables in word processing applications
Right click on the table and select table properties. Here you can choose whether the first row and/or column is headings and add a caption to the table. Making the first row/column headings is useful for screen readers and strongly recommended if possible. Allow table to be responsive, this ensures proper viewing on all types of viewing devices.
It is a good idea to select the "make table responsive" section (see Figure \(\PageIndex{2}\)) of any new table so that it will be easily viewable on mobile devices and computers.


3.5.3: Modern Primates
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Anthropology is the study of humans and the human condition. Humans are primates. We belong to the taxonomic order Primates (pronounced pry may tees ). This order encompasses humans as well as what we call non-human primates. Non-human primates are our closest biological and evolutionary relatives. So, we study them in anthropology in order to learn more about ourselves.
Figure \(\PageIndex{1}\) : Golden Lion Tamarin. Woodland Park Zoo, Seattle, WA 9/30/2011
General Mammalian Characteristics
The earliest evidence of mammals is from the Mesozoic era, however, there is limited fossil evidence and the fossils that have been found are mouse-like forms with quadrupedal locomotion. Primates evolved from an ancestral mammal during the Cenozoic era and share many characteristics with other mammals. The Cenozoic is the era of the adaptive radiation of mammals with thirty different mammalian orders evolving.
Some of the general characteristics of mammals includes:
- mammary glands: females produce milk to feed young during their immediate post-natal growth period
- hair: mammals have hair (sometimes called fur) that covers all or parts of their body
- jaw: the lower jaw is a single bone
- ear: the middle ear contains three bones: stapes (stirrup), incus (anvil), and malleus (hammer)
- heart: four-chambered heart
- aortic artery: main artery leaving the heart curves to the left to form the aortic arch
- diaphragm: mammals have one
- endothermy: mammals regulate their body temperature to maintain homeostais (a constant body temperature)
- diphyodonty: teeth are replaced only once during lifetime
General primate characteristics
While primates share traits with other mammals, such as mammary glands and endothermy, there are a number of derived traits that all primates share. Because humans share an evolutionary history with non-human primates (we share an ancestral primate), we share certain biological and behavioral traits with them.
Body
Primates have a flexible and generalized limb structure that's able to move readily in many directions. Compare the flexibility of the spider monkeys on the left with the horse on the right.
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Figure \(\PageIndex{2}\) : Spider monkeys |
Figure \(\PageIndex{3}\) : Horse |
Hands
Primates have prehensile hands (and most of them have prehensile feet also). This means that they have the ability to grasp and manipulate objects. Primates have 5 digits ( pentadactyly ) on their hands and feet. Note that a few primates, like spider monkeys, have what are called vestigial thumbs. This means that they have either a very small or non-existent external thumb (but in that case, they will still have a small internal thumb bone).
Figure \(\PageIndex{4}\) : Primate hands and feet
Primates have flattened nails instead of claws. Primates also have very sensitive tactile pads on their digits with fingerprints ( dermatoglyphs ). Some monkeys, like woolly monkeys, have similar pads with fingerprints on the ends of their prehensile tails. Therefore, it functions just like another hand. Check out the similarities and differences of the various primate hands in Figure \(\PageIndex{4}\).
Teeth
Primates have heterodonty , meaning that they have different teeth that perform different tasks when processing food via biting or chewing. With this, they have an ability to have a more generalized diet (compared to a specialist diet), meaning that they have more dietary flexibility. Why would this be a good thing to have?
Compare the teeth of the cayman (a relative of crocodiles and alligators) with the human teeth on the right. While the cayman's teeth do vary in size, they don't vary in structure. However, the human has incisors, canines, premolars, and molars -- all of which perform different food processing tasks.
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Figure \(\PageIndex{5}\) : Cayman |
Figure \(\PageIndex{6}\) : Human jawbone |
Senses
Compared to most other mammals, primates have an increased reliance on vision and a decreased reliance on their senses of smell and hearing. Associated with this are their smaller, flattened noses, loss of whiskers, and relatively small, hairless ears. Also associated with this are their forward facing eyes with accompanying binocular or stereoscopic vision . This type of vision means that both eyes have nearly the same field of vision with a lot of overlap between them. It provides wonderful depth perception (but a loss of peripheral vision).
Note first how the eyes of the monkey on the left are more front-facing than the eyes of the cow on the right. Also note how flat the monkey's nose is and how small its ears are, when compared to those of the cow.
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Figure \(\PageIndex{7}\) : Monkey |
Figure \(\PageIndex{8}\) : Cow |
Brain
In relation to other mammals, primates have a more expanded and elaborate brain. Compare the complexity of the human brain on the left to the cat brain on the right.
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Figure \(\PageIndex{9}\) : Human brain |
Figure \(\PageIndex{10}\) : Cat brain |
Life history
Life history refers to the pattern that an organism takes from conception to death. When compared to other mammals, primates have:
- longer gestation (pregnancy) periods
- reduced number of offspring (usually one, but some species commonly have twins)
- delayed maturation, with a long infancy and juvenile learning period
- extension of the entire lifespan
Figure \(\PageIndex{11}\) : Chimpanzee mother and baby
Behavior
The behavioral traits we share with other primates include:
- a greater dependence on flexible, learned behavior
- a tendency to live in social groups
Figure \(\PageIndex{12}\): Baboons crossing the road
References
- Jurmain R, Kilgore L, Trevathan W. Essentials of physical anthropology, 4th edition. Belmont (CA): Wadsworth, Cengage Learning; 2013. 437 p.
- Larsen CS. Our origins: discovering physical anthropology, 3rd edition. New York (NY): W.W Norton & Company, Inc.; 2014. 478 p.
- Thies ML. Mammalian characteristics [Internet] [cited 2015 Aug 2]. Available from: http://www.shsu.edu/~bio_mlt/mammals.html


17.2: How Pathogens Cause Disease
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- Ying Liu
- City College of San Francisco
Learning Objectives
- Summarize Koch’s postulates and molecular Koch’s postulates, respectively, and explain their significance and limitations
- Explain the concept of pathogenicity (virulence) in terms of infectious and lethal dose
- Distinguish between primary and opportunistic pathogens and identify specific examples of each
For most infectious diseases, the ability to accurately identify the causative pathogen is a critical step in finding or prescribing effective treatments. Today’s physicians, patients, and researchers owe a sizable debt to the physician Robert Koch (1843–1910), who devised a systematic approach for confirming causative relationships between diseases and specific pathogens.
Koch’s Postulates
In 1884, Koch published four postulates that summarized his method for determining whether a particular microorganism was the cause of a particular disease. Each of Koch’s postulates represents a criterion that must be met before a disease can be positively linked with a pathogen. In order to determine whether the criteria are met, tests are performed on laboratory animals and cultures from healthy and diseased animals are compared (Figure \(\PageIndex{1}\)).
Koch’s Postulates
- The suspected pathogen must be found in every case of disease and not be found in healthy individuals.
- The suspected pathogen can be isolated and grown in pure culture.
- A healthy test subject infected with the suspected pathogen must develop the same signs and symptoms of disease as seen in postulate
- The pathogen must be re-isolated from the new host and must be identical to the pathogen from postulate 2.
In many ways, Koch’s postulates are still central to our current understanding of the causes of disease. However, advances in microbiology have revealed some important limitations in Koch’s criteria. Koch made several assumptions that we now know are untrue in many cases. The first relates to postulate 1, which assumes that pathogens are only found in diseased, not healthy, individuals. This is not true for many pathogens. For example, H. pylori , described earlier in this chapter as a pathogen causing chronic gastritis, is also part of the normal microbiota of the stomach in many healthy humans who never develop gastritis. It is estimated that upwards of 50% of the human population acquires H. pylori early in life, with most maintaining it as part of the normal microbiota for the rest of their life without ever developing disease.
Koch’s second faulty assumption was that all healthy test subjects are equally susceptible to disease. We now know that individuals are not equally susceptible to disease. Individuals are unique in terms of their microbiota and the state of their immune system at any given time. The makeup of the resident microbiota can influence an individual’s susceptibility to an infection. Members of the normal microbiota play an important role in immunity by inhibiting the growth of transient pathogens. In some cases, the microbiota may prevent a pathogen from establishing an infection; in others, it may not prevent an infection altogether but may influence the severity or type of signs and symptoms. As a result, two individuals with the same disease may not always present with the same signs and symptoms. In addition, some individuals have stronger immune systems than others. Individuals with immune systems weakened by age or an unrelated illness are much more susceptible to certain infections than individuals with strong immune systems.
Koch also assumed that all pathogens are microorganisms that can be grown in pure culture (postulate 2) and that animals could serve as reliable models for human disease. However, we now know that not all pathogens can be grown in pure culture, and many human diseases cannot be reliably replicated in animal hosts. Viruses and certain bacteria, including Rickettsia and Chlamydia , are obligate intracellular pathogens that can grow only when inside a host cell. If a microbe cannot be cultured, a researcher cannot move past postulate 2. Likewise, without a suitable nonhuman host, a researcher cannot evaluate postulate 2 without deliberately infecting humans, which presents obvious ethical concerns. AIDS is an example of such a disease because the human immunodeficiency virus (HIV) only causes disease in humans.
Molecular Koch’s Postulates
In 1988, Stanley Falkow (1934–) proposed a revised form of Koch’s postulates known as molecular Koch’s postulates. These are listed in the left column of Table \(\PageIndex{1}\). The premise for molecular Koch’s postulates is not in the ability to isolate a particular pathogen but rather to identify a gene that may cause the organism to be pathogenic.
Falkow’s modifications to Koch’s original postulates explain not only infections caused by intracellular pathogens but also the existence of pathogenic strains of organisms that are usually nonpathogenic. For example, the predominant form of the bacterium Escherichia coli is a member of the normal microbiota of the human intestine and is generally considered harmless. However, there are pathogenic strains of E. coli such as enterotoxigenic E. coli (ETEC) and enterohemorrhagic E. coli (O157:H7) (EHEC). We now know ETEC and EHEC exist because of the acquisition of new genes by the once-harmless E. coli , which, in the form of these pathogenic strains, is now capable of producing toxins and causing illness. The pathogenic forms resulted from minor genetic changes. The right-side column of Table \(\PageIndex{1}\) illustrates how molecular Koch’s postulates can be applied to identify EHEC as a pathogenic bacterium.
| Molecular Koch’s Postulates | Application to EHEC |
|---|---|
| (1) The phenotype (sign or symptom of disease) should be associated only with pathogenic strains of a species. | EHEC causes intestinal inflammation and diarrhea, whereas nonpathogenic strains of E. coli do not. |
| (2) Inactivation of the suspected gene(s) associated with pathogenicity should result in a measurable loss of pathogenicity. | One of the genes in EHEC encodes for Shiga toxin, a bacterial toxin (poison) that inhibits protein synthesis. Inactivating this gene reduces the bacteria’s ability to cause disease. |
| (3) Reversion of the inactive gene should restore the disease phenotype. | By adding the gene that encodes the toxin back into the genome (e.g., with a phage or plasmid), EHEC’s ability to cause disease is restored. |
As with Koch’s original postulates, the molecular Koch’s postulates have limitations. For example, genetic manipulation of some pathogens is not possible using current methods of molecular genetics. In a similar vein, some diseases do not have suitable animal models, which limits the utility of both the original and molecular postulates.
Key Concepts and Summary
- Koch’s postulates are used to determine whether a particular microorganism is a pathogen. Molecular Koch’s postulates are used to determine what genes contribute to a pathogen’s ability to cause disease.
Footnotes
- 1 Food and Drug Administration. “Bad Bug Book, Foodborne Pathogenic Microorganisms and Natural Toxins.” 2nd ed. Silver Spring, MD: US Food and Drug Administration; 2012.
- 2 M. Otto. “ Staphylococcus epidermidis --The ‘Accidental’ Pathogen.” Nature Reviews Microbiology 7 no. 8 (2009):555–567.
- 3 The O in TORCH stands for “other.”
- 4 D. Davies. “Understanding Biofilm Resistance to Antibacterial Agents.” Nature Reviews Drug Discovery 2 (2003):114–122.


10.5: Cancer and the Cell Cycle
- Describe how cancer is caused by uncontrolled cell growth
- Understand how proto-oncogenes are normal cell genes that, when mutated, become oncogenes
- Describe how tumor suppressors function
- Explain how mutant tumor suppressors cause cancer
Cancer comprises many different diseases caused by a common mechanism: uncontrolled cell growth. Despite the redundancy and overlapping levels of cell cycle control, errors do occur. One of the critical processes monitored by the cell cycle checkpoint surveillance mechanism is the proper replication of DNA during the S phase. Even when all of the cell cycle controls are fully functional, a small percentage of replication errors (mutations) will be passed on to the daughter cells. If changes to the DNA nucleotide sequence occur within a coding portion of a gene and are not corrected, a gene mutation results. All cancers start when a gene mutation gives rise to a faulty protein that plays a key role in cell reproduction. The change in the cell that results from the malformed protein may be minor: perhaps a slight delay in the binding of Cdk to cyclin or an Rb protein that detaches from its target DNA while still phosphorylated. Even minor mistakes, however, may allow subsequent mistakes to occur more readily. Over and over, small uncorrected errors are passed from the parent cell to the daughter cells and amplified as each generation produces more non-functional proteins from uncorrected DNA damage. Eventually, the pace of the cell cycle speeds up as the effectiveness of the control and repair mechanisms decreases. Uncontrolled growth of the mutated cells outpaces the growth of normal cells in the area, and a tumor (“-oma”) can result.
Proto-oncogenes
The genes that code for the positive cell cycle regulators are called proto-oncogenes . Proto-oncogenes are normal genes that, when mutated in certain ways, become oncogenes , genes that cause a cell to become cancerous. Consider what might happen to the cell cycle in a cell with a recently acquired oncogene. In most instances, the alteration of the DNA sequence will result in a less functional (or non-functional) protein. The result is detrimental to the cell and will likely prevent the cell from completing the cell cycle; however, the organism is not harmed because the mutation will not be carried forward. If a cell cannot reproduce, the mutation is not propagated and the damage is minimal. Occasionally, however, a gene mutation causes a change that increases the activity of a positive regulator. For example, a mutation that allows Cdk to be activated without being partnered with cyclin could push the cell cycle past a checkpoint before all of the required conditions are met. If the resulting daughter cells are too damaged to undergo further cell divisions, the mutation would not be propagated and no harm would come to the organism. However, if the atypical daughter cells are able to undergo further cell divisions, subsequent generations of cells will probably accumulate even more mutations, some possibly in additional genes that regulate the cell cycle.
The Cdk gene in the above example is only one of many genes that are considered proto-oncogenes. In addition to the cell cycle regulatory proteins, any protein that influences the cycle can be altered in such a way as to override cell cycle checkpoints. An oncogene is any gene that, when altered, leads to an increase in the rate of cell cycle progression.
Tumor Suppressor Genes
Like proto-oncogenes, many of the negative cell cycle regulatory proteins were discovered in cells that had become cancerous. Tumor suppressor genes are segments of DNA that code for negative regulator proteins, the type of regulators that, when activated, can prevent the cell from undergoing uncontrolled division. The collective function of the best-understood tumor suppressor gene proteins, Rb, p53, and p21, is to put up a roadblock to cell cycle progression until certain events are completed. A cell that carries a mutated form of a negative regulator might not be able to halt the cell cycle if there is a problem. Tumor suppressors are similar to brakes in a vehicle: Malfunctioning brakes can contribute to a car crash.
Mutated p53 genes have been identified in more than one-half of all human tumor cells. This discovery is not surprising in light of the multiple roles that the p53 protein plays at the G 1 checkpoint. A cell with a faulty p53 may fail to detect errors present in the genomic DNA (Figure \(\PageIndex{1}\)). Even if a partially functional p53 does identify the mutations, it may no longer be able to signal the necessary DNA repair enzymes. Either way, damaged DNA will remain uncorrected. At this point, a functional p53 will deem the cell unsalvageable and trigger programmed cell death (apoptosis). The damaged version of p53 found in cancer cells, however, cannot trigger apoptosis.
The loss of p53 function has other repercussions for the cell cycle. Mutated p53 might lose its ability to trigger p21 production. Without adequate levels of p21, there is no effective block on Cdk activation. Essentially, without a fully functional p53, the G 1 checkpoint is severely compromised and the cell proceeds directly from G 1 to S regardless of internal and external conditions. At the completion of this shortened cell cycle, two daughter cells are produced that have inherited the mutated p53 gene. Given the non-optimal conditions under which the parent cell reproduced, it is likely that the daughter cells will have acquired other mutations in addition to the faulty tumor suppressor gene. Cells such as these daughter cells quickly accumulate both oncogenes and non-functional tumor suppressor genes. Again, the result is tumor growth.
Summary
Cancer is the result of unchecked cell division caused by a breakdown of the mechanisms that regulate the cell cycle. The loss of control begins with a change in the DNA sequence of a gene that codes for one of the regulatory molecules. Faulty instructions lead to a protein that does not function as it should. Any disruption of the monitoring system can allow other mistakes to be passed on to the daughter cells. Each successive cell division will give rise to daughter cells with even more accumulated damage. Eventually, all checkpoints become nonfunctional, and rapidly reproducing cells crowd out normal cells, resulting in a tumor or leukemia (blood cancer).
Review Questions
___________ are changes to the order of nucleotides in a segment of DNA that codes for a protein.
- Proto-oncogenes
- Tumor suppressor genes
- Gene mutations
- Negative regulators
- Answer
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Answer: C
A gene that codes for a positive cell cycle regulator is called a(n) _____.
- kinase inhibitor.
- tumor suppressor gene.
- proto-oncogene.
- oncogene.
- Answer
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Answer: C
A mutated gene that codes for an altered version of Cdk that is active in the absence of cyclin is a(n) _____.
- kinase inhibitor.
- tumor suppressor gene.
- proto-oncogene.
- oncogene.
- Answer
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Answer: D
Which molecule is a Cdk inhibitor that is controlled by p53?
- cyclin
- anti-kinase
- Rb
- p21
- Answer
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Answer: D
Human papillomavirus can cause cervical cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity?
- E6 activates p53
- E6 inactivates p53
- E6 mutates p53
- E6 binding marks p53 for degradation
- Answer
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Answer: D
Free Response
Outline the steps that lead to a cell becoming cancerous.
- Answer
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Answer: If one of the genes that produces regulator proteins becomes mutated, it produces a malformed, possibly non-functional, cell cycle regulator, increasing the chance that more mutations will be left unrepaired in the cell. Each subsequent generation of cells sustains more damage. The cell cycle can speed up as a result of the loss of functional checkpoint proteins. The cells can lose the ability to self-destruct and eventually become “immortalized.”
Explain the difference between a proto-oncogene and a tumor suppressor gene.
- Answer
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Answer: A proto-oncogene is a segment of DNA that codes for one of the positive cell cycle regulators. If that gene becomes mutated so that it produces a hyperactivated protein product, it is considered an oncogene. A tumor suppressor gene is a segment of DNA that codes for one of the negative cell cycle regulators. If that gene becomes mutated so that the protein product becomes less active, the cell cycle will run unchecked. A single oncogene can initiate abnormal cell divisions; however, tumor suppressors lose their effectiveness only when both copies of the gene are damaged.
List the regulatory mechanisms that might be lost in a cell producing faulty p53.
- Answer
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Answer: Regulatory mechanisms that might be lost include monitoring of the quality of the genomic DNA, recruiting of repair enzymes, and the triggering of apoptosis.
p53 can trigger apoptosis if certain cell cycle events fail. How does this regulatory outcome benefit a multicellular organism?
- Answer
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Answer: If a cell has damaged DNA, the likelihood of producing faulty proteins is higher. The daughter cells of such a damaged parent cell would also produce faulty proteins that might eventually become cancerous. If p53 recognizes this damage and triggers the cell to self-destruct, the damaged DNA is degraded and recycled. No further harm comes to the organism. Another healthy cell is triggered to divide instead.
Glossary
- oncogene
- mutated version of a normal gene involved in the positive regulation of the cell cycle
- proto-oncogene
- normal gene that when mutated becomes an oncogene
- tumor suppressor gene
- segment of DNA that codes for regulator proteins that prevent the cell from undergoing uncontrolled division
Contributors and Attributions
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Connie Rye (East Mississippi Community College), Robert Wise (University of Wisconsin, Oshkosh), Vladimir Jurukovski (Suffolk County Community College), Jean DeSaix (University of North Carolina at Chapel Hill), Jung Choi (Georgia Institute of Technology), Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax (CC BY 4.0; Download for free at<EMAIL_ADDRESS>).