BIO 315: Invertebrate Zoology



Integrating Concepts in Biology IIAnswer Key for Sample Exam covering Chs 22-24 on populations/ecological systems of CellsInstructions for Instructors: This sample exam KEY covering Chapters 22-24 of ICB can be used as a guide to grading. You might be looking for something more specific and because some of the questions are open-ended, other answers not provided here might also be correct. See the ICB approach to assessment guide in the Instructor Resource area for more information (CH22) What differences do you observe in the fingerprint tracings of normal and sickle-cell hemoglobin? Use data from Data Gallery #1 to support your answer.The two hemoglobin fingerprints can be almost completely superimposed on each other (FIGURE 22.3). One peptide (#4) is in different positions in normal and sickle-cell hemoglobin. It appears along the set of peptides that are slightly positively charged. The altered peptide has a stronger positive charge in sickle-cell hemoglobin relative to normal hemoglobin, which caused it to migrate toward the negatively charged electrode during electrophoresis. (CH22) Show how scientists reconstructed the peptide sequences for normal and sickle-cell hemoglobin. Once you reconstruct the sequences, describe the difference between the two peptide chains in terms of the positions and amino acid substitutions that were observed. Use data from Data Gallery #1 to support your answer.Isolate peptide #4 for the two hemoglobins, break it down into smaller chains and perform the two dimensional fingerprinting technique (FIGURE 22.4). In determining the amino acid chain for this peptide, use fact that trypsin cuts proteins wherever it finds a lysine or an arginine; thus lysine must be at the end of each chain. Use empirical data and knowledge of the structures and charges of each amino acid to deduce the peptide structure of each spot on the fingerprints. Taking into account the overlap in sequences in Figure 22.4, you can line up the sequences as in TABLE 22.2. Conclude that the seventh position was altered, such that in sickle-cell hemoglobin a valine was substituted for a glutamic acid. (CH22) What are the conditions under which hemoglobin precipitates out of solution, and how are normal and sickle-cell hemoglobin different in that regard? What do you predict will happen to red blood cells if hemoglobin precipitates out of solution and crystallizes? Use data from Data Gallery #1 to support your answer.TABLE 22.3 shows that is much lower for both normal and sickle-cell hemoglobin when oxygen was lacking, but the sickle-cell hemoglobin was even less soluble under that condition than normal hemoglobin. Sickle-cell hemoglobin crystallizes in the absence of oxygen, and this can be very damaging to red blood cells and capillaries.(CH22) Sickle cell disease was the first disease for which anyone had found the molecular cause. Answer the following questions using data from Data Gallery #1 to support your answers.What causes the red blood cells to change their shape? Limit your answer to a maximum of two sentences.TABLE 22.3 shows that is much lower for both normal and sickle-cell hemoglobin when oxygen was lacking, but the sickle-cell hemoglobin was even less soluble under that condition than normal hemoglobin. Sickle-cell hemoglobin crystallizes in the absence of oxygen, and the precipitation causes the concentration of dissolved solutes to decrease inside the cell, which causes water to diffuse out of the cell, collapsing the cell and causing anemia.What is the specific difference between wild-type and the sickle cell protein? Limit your answer to a maximum of two sentences.Ingram found the seventh position to be altered (FIGURE 22.4 and TABLE 22.2), such that in sickle-cell hemoglobin a valine was substituted for a glutamic acid. Ingram correctly deduced that the sickle-cell peptide had one less carboxylic acid group than normal hemoglobin. While scientists later determined that Ingram misinterpreted the fingerprint—the H-V-L-L sequences turned out to be V-H-L, the data in TABLE 22.2 indicate a point mutation in the DNA sequence of individuals with sickle-cell anemia leads to the altered hemoglobin. (CH22) If homozygotes do not live long enough to reproduce, how is it that the mutated hemoglobin gene persists in a population? Why would evolution not lead to elimination of the mutation? Use data from Data Gallery #1 to support your answer.The mild sickling form is known to be heterozygous and individuals that are heterozygous at the hemoglobin gene are partially protected from malaria (TABLE 22.4). (CH22) Answer the following questions regarding Lyme disease using data from Data Gallery #1 to support your answers. Limit your answers to a maximum of two sentences for each part.What cellular function does the pathogen disrupt to cause the terrible consequences? The OSP-C protein is required to cause infection (TABLE 22.5) and mount an immune response from the host. However, the mechanism of infection results in lower attraction of neutrophils to the site of infection (FIGURE 22.7). Perhaps B. burgdorferi takes advantage of this mechanism that ticks use to evade the immune system. There is a normal inflammatory response that causes neutrophils to be recruited early on in the infection, but neutrophils disappeared quickly in mice infected with normal B. burgdorferi, allowing the bacteria to establish within the host.Why does Lyme disease cause so many different symptoms given your answer to part A above? Once the immune system is evaded through lower attraction of neutrophils to the site of infection (FIGURE 22.7), B. burgdorferi can travel throughout the body and then be found in different tissues (TABLE 22.6) suggesting that a variety of symptoms may result depending on where the pathogen ends up.(CH22) Describe briefly what the three treatments in the table below represent, and what conclusion you come to regarding the data obtained by this experiment. Write up to 5 sentences and incorporate at least one of the themes of the Cell Big Idea, that structure relates to function and cells communicate with other cells, into your answer.The wild-type is unmalipulated B. burgdorferi. OSP-C negative is a genetically engineered strain that had the OSP-C gene removed from its genome. It thus lacked the ability to produce the OSP-C protein. OSP-C- reinserted was developed from the former OSP-C negative strain where the OSP-C gene was reinserted into another part of the genome. This served as a comparison for the wild-type to ensure that OSP-C was causing the observed effects. The data in the table indicate that OSP-C is necessary for infection, because OSP-C negative strains were unable to infect mice or cause an immune response in mice, whereas high proportions of wild-type and OSP-C reinserted strains could do both. Infection route may also be important; in both wild-type and OSP-C reinserted B. burgdorferi more ticks were re-infected from the infected mice when the strain was delivered through a tick bite than injection. This may indicate a role of tick saliva or the importance of B. burgdorferi living in ticks prior to infecting a host.Table 22.5(CH22) Examine the cells in Figure 22.6. Using the descriptions of color coding in the text and figure legend to locate cell nuclei, describe the distribution of FUS/TLS in the cells for the familial ALS patients and for the control individuals. Does FUS/TLS distribute similarly?Figure 22.5 Figure 22.6 FUS/TLS is found in the nuclei of normal and ALS individuals, but the mutated protein in the ALS individual also occurred in the cytoplasm (two left panels). In the figure, bright white coloration denotes areas that stained for the marker protein, the nucleus and FUS/TLS. Green areas stained only for FUS/TLS, red areas stained only for the marker, and blue areas stained only the nucleus. The large yellow area in the top left panel is an area that stained for both the marker and FUS/TLS, outside of the nucleus. Several cells in both left panels had neither FUS/TLS nor the marker in their nuclei. Ubiquitin localized in ALS patients’ neuronal nuclei but not in controls (two right panels). In these images, ubiquitin is green, nuclei are blue, nuclei with FUS/TLS are pink, nuclei with ubiquitin and FUS/TLS are whitish-pink. Ubiquitin is a protein that covalently binds to other proteins, which it does to “tag” the proteins so that degradation enzymes can find them. Proteins that don’t function properly are often tagged for elimination by ubiquitin. Antibodies to ubiquitin are used to identify abnormal accumulations of protein inside cells, and you can see that ubiquitin is everywhere in the normal cell, but much more likely to be found in nuclei with FUS/TLS in the ALS patient. (CH22) Answer the following questions about Giardia in one (1) sentence each. Be sure to reference one piece of supporting evidence for each question from Data Gallery #1.What is at least one non-human animal that Giardia can infect, and what is the advantage of having multiple hosts? Rats (FIGURE 22.14), gerbils (FIGURE 22.15), beavers, and nutria (FIGURE 22.17) can all be infected by Giardia. The advantage may be that the pathogen does not have to wait for one specific host species to come along and pick it up, especially given its mode of dispersal in desiccation-resistant cysts.Which life stage might be involved in transmission from host to host? The cyst is likely involved in transmission because it is more prevalent in the large intestines of gerbils, suggesting it is expelled in feces as the resistant, transmitted dispersal stage (FIGURES 22.13 and 22.15).What might be the function of the other life stage?FIGURE 22.15 shows trophozoites are found primarily in small intestine (duodenum, jejunum, ileum), suggesting they are the infectious stage, infecting the small intestine and causing disruption of digestion and assimilation, leading to symptoms of giardiasis. DATA GALLERY #1Figure 22.2Figure 22.3Figure 22.4Table 22.2Table 22.1Table 22.4Table 22.3Table 22.7Table 22.5Figure 22.7Figure 22.7Figure 22.8Figure 22.11Figure 22.14Figure 22.19Figure 22.20Figure 22.13Figure 22.12Figure 22.23Figure 22.22Figure 22.15Figure 22.18Table 22.8Table 22.11Table 22.10Table 22.9(CH22) Answer the following questions regarding infection of rice plants by a fungal pathogen. Use data from Data Gallery #1 to support your answers. Limit your answers to each part to no more than 2 sentences.Describe the effect of incubation time and external pressure on cell collapse. What is the relevance of this to fungal infection?Internal pressure increased with longer incubation times (FIGURE 22.8), such that a higher external osmotic pressure was needed to cause 100% collapse of the cells. The external pressure needed to be higher than the internal pressure for collapse to occur, and that pressure was lower for shorter incubation times, leading to the conclusion that pressure builds in the infection cells prior to penetration. High pressure in infection cells allowed penetration of even the hardest substrate, but the harder the substrate, the more incubation time was required (FIGURE 22.9A). Further, when the external osmotic pressure increased, penetration was inhibited severely even for softer surfaces (FIGURE 22.9B).Describe the effect of increasing external pressure on infection cells’ ability to penetrate substrates of various hardness (1 to 6). What is the relevance of this to fungal infection?When the external osmotic pressure increased, penetration was inhibited severely even for softer surfaces (FIGURE 22.9B). For hard surfaces, a high percentage of penetration can occur, but only at low external osmotic pressures. Fungal infection can then occur more easily in watery environments where the osmotic pressure outside the cell is low.(CH22) Answer the following questions regarding Giardia. Use data from Data Gallery #1 to support your answers. Answer each question in no more than 2 sentences.What is the significance of Giardia having two life stages? The significance is that each stage may specialize, one to infect and the other to disperse to new hosts. Depending upon how the pathogen has evolved to disperse, this may be required, and it seems to be so for Giardia (FIGURE 22.15). Which life stage might be involved in transmission? The cyst is likely involved in transmission because it is more prevalent in the large intestines of gerbils, suggesting it is expelled in feces as the resistant, transmitted dispersal stage (FIGURES 22.13 and 22.15).What might be the function of the other life stage? FIGURE 22.15 shows trophozoites are found primarily in small intestine (duodenum, jejunum, ileum), suggesting they are the infectious stage, infecting the small intestine and causing disruption of digestion and assimilation, leading to symptoms of giardiasis. How do these life stages relate to the fact that this pathogen can infect multiple host species? Because of the way the pathogen disperses as a desiccation-resistant cyst into a body of water, there is not direct transmission from one host to another. Being able to infect multiple host species is advantageous for that type of dispersal/transmission. Rats (FIGURE 22.14), gerbils (FIGURE 22.15), beavers, and nutria (FIGURE 22.17) can all be infected by Giardia. (CH22) Below are data from a 2007 cholera outbreak in one city in southern Sudan. Cholera is an infectious disease caused by a bacterium that causes severe watery diarrhea, which can lead to dehydration and even death if untreated. Based on what you know about determining disease transmission and your ability to interpret new data, answer the following questions. Given the rapid spike in the number of cases, hypothesize as to the route of transmission. Answer in one (1) sentence.Given the rapid spike along with the fact that it produces watery diarrhea, it is likely to be passed through close contact or fecal to oral transmission. Unhygienic water sources could likely lead to rapid spread.What additional data are required in order to test your hypothesis? Provide a list (do NOT write sentences).We would need to know something about the water sources of the city, whether there were water treatment facilities, what kind of treatment/health care was given/available to individuals, and relationships among individuals who became ill, among other things.Figure not in text:(CH22) Reported Lyme disease cases in the US have more than tripled since the Centers for Disease Control and Prevention (CDC) began recording cases in 1991. One conclusion we could make about these data is that the number of cases is actually increasing. Another is that doctors are just getting better at detecting and diagnosing Lyme disease. Can you think of a way to distinguish between the two alternatives? Answer in no more than three (3) sentences.Figure not in text:One of the conclusions is that Lyme disease incidence is actually increasing. The other is that it is not increasing, but we are better at detecting it now than when it first emerged. Medical records could be searched for individuals with similar symptoms that were misdiagnosed to determine if there was a bias towards undercounting cases. Often, medical records can reveal much or there may even be tissue samples or patients with chronic diseases that can be tested for Lyme, but when they were first afflicted were diagnosed with something like MS (which may actually happen).(CH22) Is the coronavirus in animals the same coronavirus in humans that causes SARS? What evidence supports that conclusion? Use data from Data Gallery #1 to support your answer. Answer in no more than three sentences.You cannot conclude that coronavirus in animals is the same or that either species in which it is found is the reservoir for the SARS coronavirus (FIGURE 22.22 shows they are related, but not necessarily the same). However, its presence in market animals suggests a connection between animals and humans (FIGURE 22.20). The positive animals could have been infected from another untested source in the market, so the true reservoir cannot be unequivocally identified. The virus could have passed from the humans to the animals, although this is less likely given that a higher percentage of market workers who deal with animals possessed SARS coronavirus antibody compared to those working exclusively with vegetables (TABLE 22.10).(CH22) What do the data below tell you about how SARS is transmitted to uninfected people? How does the transmission mechanism relate to the global distribution of a disease? Answer in no more than two (2) sentences.Figure 22.23The data suggest that SARS spreads rapidly, as indicated by the spikes in the frequency distribution (# of cases increases rapidly). This suggests a transmission mechanism such as close contact, coughing or sneezing. This relates to the global spread of this epidemic because traveling health workers and other travelers can bring it to a new country and rapidly infect many people.(CH23) Describe an experiment that helped elucidate the function of parietal cells? How does this experiment help you understand how parietal cells function? Use data from Data Gallery #2 to support your answer. There are many options here. Generally, look to the set of experiments done in TABLE 23.2, and whether the experiment is described accurately by the student.(CH23) Answer the following questions regarding glucose transport in the small intestines. Use data from Data Gallery #2 to support your answer. What is the effect of the sodium-dependent glucose co-transporter (SGLT1) on glucose accumulation in epithelial cells? In the blood?By knocking out the sglt1 gene in mice and then comparing responses to wild-type mice, scientists can determine the effect of the SGLT1 transmembrane protein. Mutant sglt1?/? mice take up very little glucose into epithelial cells and show very little glucose in the blood relative to wild-type mice (FIGURE 23.7A and B). What is the effect of glucose transporter 2 (GLUT2) on glucose accumulation in epithelial cells? In the blood?Glucose accumulation in epithelial cells is higher in mutant glut2?/? mice than wild-type mice. But glucose concentrations are lower in the blood of those mutant mice than in the wild-type mice (FIGURE 23.7C and D). This suggests, as you should have concluded, that glucose can enter the epithelial cell but cannot get out. So it enters but doesn’t leave the epithelial cells in mutant glut2?/? mice. Where do the sodium-dependent glucose co-transporter (SGLT1) and glucose transporter 2 (GLUT2) localize on the epithelial cell membranes relative to the lumen of the small intestine? The localization of SGLT1 is near the lumen side of the epithelial cell membranes (FIGURE 23.8A). GLUT2 localizes in membranes on the sides and away from the lumen, appearing to surround the nucleus of epithelial cells (FIGURE 23.8C). (CH23) Is there a period of time after blood feeding in R. prolixus where the head is necessary for molting? Is there a period of time after blood feeding where the head can be removed with no effect on molting? What do you conclude from this? Use data from Data Gallery #2 to support your answer.Yes, there is what Wigglesworth called a critical period—a period of time during which the head is necessary for molting (TABLE 23.3). After that critical period is over, the head is no longer necessary for molting. Because the length of time of each stage varies, the critical period also varies. Furthermore, the epidermis begins to change shortly after blood feeding. Epidermal changes are necessary for molting; the epidermis secretes the new exoskeleton that replaces the old one. Mitosis in the epidermis begins several days after blood feeding, and other epidermal changes begin to occur. Insects decapitated 1 day after blood feeding did not exhibit those changes. Epidermal changes and molting occurred in fifth stage individuals that were decapitated after 7 days. Conclusion: the head is necessary for a growth or molting hormone and that this hormone increases in concentration over time. After it reaches a critical concentration that initiates mitosis in the epidermis, the head is no longer necessary for molting to occur.(CH23) which is the key experiment in the table below demonstrating the source of molting hormone in insects? What do the data suggest about the source? Answer in two (2) sentences. Table 23.6The key experiment is really the last one, although reference to other experiments is required to make the argument. The last row demonstrates that the small dorsal fragment of the protocerebrum is important to molting (3rd experiment), and that is a subsection of the dorsal half of the brain (2nd experiment). That narrows it down to a small part of the brain. However, in the last experiment, it’s demonstrated that if the first thoracic segment of the transplantee is missing, no molting occurs, suggesting that a signal comes from the protocerebrum and is received by a gland in the first thoracic segment. Stating that the 4th experiment eliminating the corpus allatum as the source of the molting hormone, while important to Wigglesworth’s initial hypothesis, does not answer the question. It only eliminates one gland as the source – it does not demonstrate the actual source. (CH23) If a mutation were to lead to a deficiency in the gene that produces both p16 and p19 proteins, what would you predict would occur to stem cell self-renewal based on the schematic on the left? Which set of data best addresses this deficiency and how does your prediction relate to the data? Answer in no more than three (3) sentences.Walk through at least one example of what happens, specifically, when p16 or p19 is mutated/missing. For instance, in the absence of p16 there will be less/no inhibition of the cyclin D/CDK4/6 complex, and then allow that complex to phosphorylate Rb. That would then increase movement into the cell cycle. If p19 is eliminated, p53 may still be active, but perhaps not as much as if p19 is present. Thus, less Cip/Kip activiation is expected, so less inhibition or suppression of the cyclin E/CDK2 complex would be expected. Less suppression of that complex means there would be more second phosphorylation of Rb. In all, loss of p16 and p19 should lead to more cell proliferation. The graph on the right (FIGURE 23.16) best addresses this, and supports the predictions from the schematic on the left. Figure 23.18 Components of cell cycle regulation in tissue-specific stem cells. Figure 23.15 Effects of loss of Bmi-1 on stem cells.Figure 23.16 Effects of loss of p16Ink4a on stem cell(CH23) If a mutation were to lead to a deficiency in the Bmi-1 protein, what would you predict would occur to stem cell self-renewal based on the schematic on the left? How does your answer relate to the data on the right, and which panel(s) in that figure is (are) important to drawing that conclusion? Answer in no more than three (3) sentences.Walk through at least one example of what happens, specifically, when Bmi-1 is mutated/missing. For instance, in the absence of Bmi-1, suppression or inhibition of p16 and p19 would be lost or decreased. Loss of p16 inhibition would then allow p16 to increase and more strongly inhibit or suppress the cyclin D/CDK4/6 complex, which would reduce or eliminate phosphorylation of Rb. That would help reduce the movement of the cell into the cell cycle, as unphosphorylated Rb keeps the cell in the gap phase. Then link that to the figure. This, and other predictions of what would happen, are mirrored in the figures. Bmi-1 deficient mice do indeed have fewer cells than normal mice, when stem cells are taken from these mice and cultured. BrdU incorporation is also lower in the Bmi-1 deficient mice, indicating that fewer stem cells are synthesizing DNA than those from normal mice. Figure 23.18 Components of cell cycle regulation in tissue-specific stem cells. Figure 23.15 Effects of loss of Bmi-1 on stem cells.(CH23) Which of the following is (are) the key experiment(s) in the table below demonstrating that there is a juvenile hormone produced in 1st through 4th stage juvenile insects (rather than a metamorphosis hormone produced by 5th stage juvenile insects)? Answer in two (2) sentences. The experiments described in TABLE 23.5 suggest that the head, specifically the corpus allatum, is the source of the hormone that regulates metamorphosis and that the hormone is different from molting hormone. If just the tip of the head is cut off in a post-critical period individual, juvenile hormone is still produced, as indicated by the effects on individuals decapitated prior to reaching the critical period (row #2). It appears that the head is still necessary after the molting critical period in order to prevent metamorphosis, and this is further supported by decapitation experiments on fourth stages at compared to beyond the critical period. Row 4 shows that fourth stage individuals decapitated right at the molting critical period leads to molting, but also causes fourth stage individuals to metamorphose into an abnormal adult stage, suggesting JH and molting hormone are not the same. And then in rows 5 and 6, presence of a JH in fourth stage post critical period individuals is demonstrated when pre-critical period fifth stage individuals molt but fail to metamorphose. Finally, if there is no brain or corpus allatum in fifth stage individuals, even if they’re post critical period, no metamorphosis occurs when connected to a post critical period fourth stage individual with a corpus allatum. While this is a long answer, the detail is provided for the instructor. The student will have to choose one or more of these to succinctly discuss.(CH23) What effect does loss of Bmi-1 have on expression of p16Ink4a and p19Arf? What would you predict these changes in expression would have on self-renewal, based on the scheme in Figure 23.14? Use data from Data Gallery #2 to support your answer. Answer in no more than two (2) sentences.In the absence of Bmi-1, suppression or inhibition of p16 and p19 would be lost or decreased. Loss of p16 inhibition would then allow p16 to increase and more strongly inhibit or suppress the cyclin D/CDK4/6 complex, which would reduce or eliminate phosphorylation of Rb. That would help reduce the movement of the cell into the cell cycle, as unphosphorylated Rb keeps the cell in the gap phase. Bmi-1 deficient mice do indeed have fewer cells than normal mice, when stem cells are taken from these mice and cultured (FIGURES 23.15 and 23.17). BrdU incorporation is also lower in the Bmi-1 deficient mice, indicating that fewer stem cells are synthesizing DNA than those from normal mice.(CH23) What effects did the loss of p16Ink4a have on adult mouse neural stem cells? Did you predict this based on the scheme in Figure 23.14? Use data from Data Gallery #2 to support your answer. Answer in no more than two (2) sentences.Loss of p16 should lead to more cell proliferation. FIGURE 23.16 best addresses this, and supports the predictions from FIGURE 23.14. In the absence of p16 there will be less/no inhibition of the cyclin D/CDK4/6 complex, and then allow that complex to phosphorylate Rb. That would then increase movement into the cell cycle. If p19 is eliminated, p53 may still be active, but perhaps not as much as if p19 is present. Thus, less Cip/Kip activiation is expected, so less inhibition or suppression of the cyclin E/CDK2 complex would be expected. Less suppression of that complex means there would be more second phosphorylation of Rb. (CH23) Which treatments in Figure 23.17 (in Data Gallery #2) are significantly different from the control, the normal mice? Does the loss of p16Ink4a restore self-renewal in mice that have also lost Bmi-1? Answer in no more than three sentences.For CNS stem cells, all the Bmi-1 deficient cells show significantly lower self-renewal than the control. The loss of p16 partially restores self-renewal in mice that are also Bmi-1 deficient. This experiment has the potential to directly determine whether Bmi-1 promotes neural stem cell self-renewal by suppressing expression of p16Ink4a, as is shown in FIGURE 23.18. Higher self-renewal in mice that lost both genes, as compared to mice that lost only Bmi-1, is consistent with the result that p16Ink4a expression increases in Bmi-1 deficient mice, which leads to strong suppression of Rb phosphorylation. Then in mice that lost both Bmi-1 and p16Ink4a, suppression of Rb phosphorylation is lessened, allowing partial recovery of the cell cycle and self-renewal. The recovery is partial because there are other pathways involved in regulating stem cell self-renewal, but this example points out the complex controls on stem cell self-renewal.(CH24) Explain how you can use data regarding dividing cells to estimate the growth rate and the doubling time of the population of cells. Use data from Data Gallery #2 to support your answer. Answer in no more than two (2) sentences.Use weighted averages to calculate the mean. Either state that or describe the process and then rank them in B.a., B.c., and S.e. (FIGURE 24.2).(CH24) What differences do you note between the distributions of dividing times of the species shown in Figure 24.2B? Answer in no more than two (2) concise sentences.Figure 24.2There are differences in the range of dividing times, with some species having much narrower ranges than others. Some species have multiple peaks in the frequency distributions and there are certainly differences in the average or median dividing times, suggesting different intrinsic growth rate differences.(CH24) In the following formula, where t is in hours, explain how to determine both doubling time and growth rate: x = 8 x 22t.For a population that starts with x0 cells and divides every T minutes, the number of cells after t hours is modeled by x = x0 × 2(60/T )t. For example, when a population starts with 8 cells and divides every 20 minutes, x = 8 × 2(60/20) t = 8 x 23t, and when it divides every 45 minutes, x = 8 x 2(60/45) t ~ 8 × 21.33 t. In each case, the multiple of t in the exponent is the number of generations per hour, or growth rate. The doubling time is the number of hours per generation, or T/60, which is the reciprocal of the number multiplying t in the exponent. For example, if x = 8 × 25t, then the doubling time is 1/5 of an hour. If x = 8 × 20.2t, then the doubling time is 1/0.2 = 5 hours.(CH24) Do blooms of algae lead to negative effects on other creatures? Use data from Data Gallery #2 to support your answer. Answer in no more than two (2) sentences.Yes, you can cite the manipulated lab experiment (FIGURE 24.21), which reveals that toxic effects of two dinoflagellate species on various marine species, some of which are affected more adversely than others.(CH24) Why might there be a lag phase in cultures of single-celled organisms that are grown from individuals transferred from cultures already in the exponential phase? After a period of growth, why does growth rate of populations decline from the maximum exponential growth rate? Use data from Data Gallery #2 to support your answer. Answer in no more than three sentences.The lag observed in FIGURE 24.3 was an artifact from placing a small population in a large flask so that there was a lag before absorbance was affected. In all likelihood, the bacteria in conditions with high levels of resources and low population densities were all dividing in a manner similar to that observed by Kelly and Rahn. After a period of growth, growth rate declines because of limited resources in the cultures.Data Gallery #2Figure 23.3Figure 23.5Figure 23.4Table 23.1Table 23.2Figure 23.6Figure 23.7Figure 23.8Figure 23.10Table 23.4Table 23.3Table 23.5Table 23.6Figure 23.17Table 23.7Figure 23.15Figure 23.16Table 23.8Table 23.9Table 23.10Table 23.11Figure 24.1Figure 24.3Figure 24.4 Figure 24.2Figure 24.6Figure 24.8Figure 24.10Table 24.1Table 24.2Figure 24.11Figure 24.13Figure 24.14Figure 24.15 Figure 24.18Figure 24.17Figure 24.20Figure 24.19Figure 24.21Figure 24.22(CH24) Explain how environmental conditions can support algal growth and ultimately facilitate algal blooms. Use data from Data Gallery #2 to support your answer. Answer in no more than three (3) sentences.FIGURE 24.15: For the most part, higher deposition of nutrients in summer months leads to a higher percentage of annual algal blooms occurring in those months. Twenty percent of all toxic bloom events occur in July, which is the month with the highest percentage of annual deposition of all nutrients tested. FIGURE 24.17 shows relationships between certain conditions, temperature, salinity, ammonium, and maximum density during algal blooms.(CH24) Are toxic algal blooms related to deposition of any nutrients? If so, which ones? Use data from Data Gallery #2 to support your answer. Answer in no more than three (3) sentences.FIGURE 24.15: For the most part, higher deposition of nutrients in summer months leads to a higher percentage of annual algal blooms occurring in those months. Twenty percent of all toxic bloom events occur in July, which is the month with the highest percentage of annual deposition of all nutrients tested (nitrogen, phosphate, and silicon dioxide). (CH24) Use data from Data Gallery #2 to illustrate how one connection in the nitrogen cycle, depicted below in Figure 24.12, was determined. Answer in no more than two (2) sentences.Several connections can be shown. We know from the data examined that rhizobia fix nitrogen from the atmosphere (FIGURE 10). We also examined data that plants can take up nitrogen directly from the soil, both leguminous and non-leguminous (FIGURE 24.8, TABLES 24.1 and 24.2). Finally, we examined accumulation of nitrite by soil bacteria when provided with ammonium sulfate, showing that these bacteria can convert ammonium to nitrite (FIGURE 24.13).Figure 24.12(CH24) Use data from Data Gallery #1 to illustrate how different conditions affect the growth of populations of both prokaryotes and eukaryotes. Answer in no more than two (2) sentences. You can use any data that show how population growth is affected by different conditions. For instance, Figure 24.4 shows that Lactobacillus plantarum, a prokaryote, populations grow at different rates depending upon the temperature. You must include both a prokaryote and eukaryote example. The dinoflagellates in 24.6 can be used for your eukaryote example. Other possibilities exist, but the data must indicate how population growth is affected by the different conditions. (CH24) Explain the data that show legumes grown in the same amount of nitrogen fertilizer as non-legumes. What conclusions do you come to regarding their growth? Also explain the data that show that legumes can grow in the absence of nitrogen. Use data from Data Gallery #2 to support your answer. Answer in no more than three (3) sentences.When legumes are grown in the same amount of fertilizer as non-legumes, the legumes end up with more nitrogen (FIGURE 24.8) and yet use less of the soil nitrogen (TABLE 24.1). This means that nitrogen is fixed from another source, the atmosphere. And it is shown in FIGURE 24.10 that rhizobia are fixing the nitrogen, as plants grown with no nitrogen and rhizobia end up with almost as much nitrogen as plants in the high N control with no rhizobia.(CH24) What are the effects of rhizobia in comparison to fertilizer on growth, nitrogen content, and nitrogen fixation of chickpeas or other legumes? Use data from Data Gallery #2 to support your answer. Answer in no more than three (3) sentences.FIGURE 24.10 shows that rhizobia are fixing nitrogen, as plants grown with no nitrogen and rhizobia end up with almost as much nitrogen as plants in the high N control with no rhizobia. Rhizobia added to soil leads to plants with higher mass of root nodules than plants grown with ambient rhizobia and added nitrogen (FIGURE 24.11). This figure also shows mass of the plants is relatively equal.(CH24) Both of the figures below graph the same model [Pt+1 = Pt + r Pt (1-Pt/K)]. Explain which term in the formula, when different, leads to the differences observed in the two figures. If these two graphs represent the growth of two species, or of one species under two different sets of environmental conditions, what is the biological interpretation of the comparison between the two species? Answer in no more than three (3) sentences.This is the logistic model (BME 24.2) and it is r that is different in the two graphs. r represents the growth rate and the larger it is, the faster the population grows towards K, the carrying capacity. In some cases, when r is very large K may be exceeded, leading to rapid population decline, as shown in the graph on the right. The biological interpretation is that if these are two different species they have different intrinsic growth rates, or if they represent one species, the environment differs, for instance the amount of food or nutrients available may be different in the two situations. ................
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