Solution Manual for Microbiology with Diseases by Body System, 4th Edition

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207Answers to End-of-ChapterQuestions for ReviewNote that the End-of-Chapter Concept Maps are available online in the Study Area ofMastering Microbiology (www.masteringmicrobiology.com) so that students can completethem interactively. Concept mapping answers are also found in the Answers section in theback of the textbook. Instructors can access completed concept maps and teaching tips in theResource section of Mastering Microbiology.CHAPTER 1A Brief History of MicrobiologyMultiple Choice1.a3.d5.c7.a9.d2.c4.a6.d8.b10.dFill in the Blanks1.Martinus Beijerinck and Sergei Winogradsky2.Louis Pasteur and Eduard Buchner3.Paul Ehrlich4.Edward Jenner5.John Snow6.Robert Koch7.John Snow8.Louis Pasteur9.Louis PasteurVisualize It!1.1. cilium; 2. flagellum; 3. pseudopod; 4. nucleus2.Microbes were only in the dust in the lowest portion of the swan neck tube.Matching1.J4.C, H, K7.C10.D2.H5.B8.E11.I3.C6.A9.D12.L

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208MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4eShort Answer1.The theory of spontaneous generation hindered the development of the field of microbi-ology because the theoryseemedto explain observed phenomena such as food spoilage,so further investigations were not made. Leeuwenhoek’s microscopic observationsrevealed previously unseen microorganisms and advanced interest in microbiology.2.Spallanzani’s flasks contained sterilized nutrients but were sealed from air, whereasPasteur’s flasks had sterilized nutrients in flasks with curves (“swan-necked” flasks),which allowed air to enter but trapped microorganisms that fell in the tube. AlthoughPasteur’s controlled investigation provided all the environmental and nutritional require-ments for life to arise spontaneously, the flasks remained sterile. This settled the dispute.3.Six types of microorganisms: archaea, bacteria, protozoa, fungi, algae, and worms.4.Leeuwenhoek’s investigations changed the world by revealing the presence of microor-ganisms. Subsequent investigations were directed at microorganisms, their taxonomy,their effect on disease, disease control, and so on; and the result not only is of academicinterest but also affects the quality and duration of human life.5.Macroscopictapeworms are studied inmicrobiologybecause their eggs and infectivestages are microscopic and because of historical tradition.6.“The Golden Age of Microbiology” was a 50-year period during which scientistssearched for the answers to four questions: (1) Is spontaneous generation of microbial lifepossible? (2) What causes fermentation? (3) What causes disease? (4) How can weprevent infections and disease? Applying the scientific method to their investigations,biologists pioneered discoveries in disciplines such as industrial microbiology, genetics,environmental microbiology, antiseptic medical technology, serology, immunology, andchemotherapy. The season of discovery was “golden” to the field of microbiology.7.Today, microbiological investigations are propelled by four major questions: (1) Whatare the basic chemical reactions of life? (2) How do genes work? (3) What roles do mi-croorganisms play in the environment? (4) How do we defend against disease?8.Pasteur’s fermentation experiments followed the scientific method. (1) Pasteur observedNeedham’s investigations and questioned the source of microorganisms. (2) He hypothe-sized that dust in the air contained microbes that would reproduce in nutrient broth. (3)His experimental design (with controls) included heated infusion in “swan-neckedflasks,” which remained free of dust and therefore microbes unless the flasks were tiltedto allow the infusion to touch the dust in the curve of the flask. Those flasks that weretilted showed microbial life. (4) Based on the results, Pasteur concluded that microbes inthe infusion were not spontaneous but rather were descendants of microbes in the air oron dust particles.9.Koch’s postulates are: (1) The suspected causative agent must be found in every case ofthe disease and be absent from healthy hosts. (2) The agent must be isolated and grownoutside the host. (3) When the agent is introduced to a healthy, susceptible host, the hostmust get the disease. (4) The same agent must be re-isolated from the diseased experi-mental host. These postulates are significant because when they are satisfied, the cause ofan infectious disease is proven.10.HAI is an acronym for healthcare-associated infections (formerly called nosocomial in-fections). These terms refer to infections acquired in a healthcare setting. Their frequencyemphasizes the need for good hygiene in healthcare facilities.

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Answers to End-of-Chapter Questions for Review209Critical Thinking1.Koch would have been extremely frustrated by attempts to identify the agents of viraldiseases. Methods to detect viruses did not exist, and the means to grow viruses in the la-boratory had not been developed. His initial success with anthrax provided the basis forthe development of his postulates, for the bacteriologic methodology his laboratorydevised, and for his renown. Had Koch pursued viral diseases, his lifetime accomplish-ments would have been greatly reduced: fewer etiologic agents would have been identi-fied, and doubt would have been cast on the power of his postulates.2.Under the influence of Pasteur’s and Koch’s work, the Germ Theory of Disease hadbecome widely accepted in the early 1900s, and the prevailing belief in the scientificcommunity was that microorganisms caused all diseases. In this atmosphere, explana-tions for disease that did not involve microbial infection were ridiculed. Biochemistrywas a very young science, and metabolism was just beginning to be investigated.Although Funk’s interpretation of his observations was correct, he could not provide amechanism for how polished rice might promote disease.3.Checking for the presence ofH. influenzaein a large number of diagnosed cases of fluwould likely have cast doubt on the conclusion thatH. influenzaewas the cause of flu. Insome cases,H. influenzaeis not present, and thus Koch’s first postulate would not be sat-isfied. Further, inoculation ofH. influenzaeinto an experimental organism would nothave produced flu, violating the third postulate.4.The relatively low temperature used in pasteurization is not high enough, nor is heatinglong enough in duration to destroy bacterial endospores or fungal spores. Any suchcontaminants would have a chance to germinate and ferment the milk over winter break.5.Pour the milk into sterile containers through a filter that will trap bacteria, endospores,and spores, and then incubate in a sterile chamber or in a chamber with filtered vents thatallow air exchange but no microbial entry. If the sterile milk remains sterile, then sponta-neous generation did not occur.6.Identifying cholera cases in persons who visited the Broad Street area only during theday would demonstrate that nocturnal exposure to “fermented vapors” was not the sourceof cholera.7.Redi showed that flies laid eggs that hatched onto larvae on meat by covering jars ofmeat with gauze upon which the events took place. Needham’s experiments with boiledinfusions and gravy in sealed containers suggested that microbes arise spontaneously, butSpallanzani repeated Needham’s experiments more rigorously and concluded that spon-taneous generation of microorganisms does not occur. The absence of air exchange inSpallanzani’s experiments caused some doubt about his conclusions. Pasteur demon-strated the growth of microorganisms in supposedly sterile infusions in swan-neckedflasks, which remained sterile unless contaminated with nonsterile dust.8.There are a number of possible answers, including infection control (hygiene and thecontrol of disease in health care settings); bioremediation (the use of microbes to removepollutants); pharmaceuticals (design and development of medications for infectious dis-ease) and recombinant DNA technology (the alteration of microbial genes to synthesizeuseful products).9.Algae have simple nutritional requirements (carbon dioxide, water, light, a few salts) thatare easily obtained from the environment; therefore, they do not need to obtain nutrientsfrom other living things.

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210MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4e10.Buchner’s experiments demonstrated that intact living cells were not required for fermen-tation of sugars. Had the experiments been done in the 1850s, prior to Pasteur’s experi-ments on spontaneous generation, one argument for spontaneous generation (theappearance of yeast cells in fermenting sugar solutions) would have been weakened.Spallanzani’s results might have been more widely accepted as demonstrating that spon-taneous generation does not occur. When scientists accepted that animals do not arisefrom spontaneous generation, they might have been less likely to exempt microorgan-isms.11.Many bacteria are very small and may be easily overlooked during examination with alight microscope. In contrast, Koch’s method allows a single microbe to reproduce into amuch more visible population of cells—a colony—which is less likely to be overlooked.In addition, many microorganisms, especially pathogens, have stringent nutritional re-quirements; dilution in standard growth medium might be harmful to the microbes andtherefore prevent their detection.12.Koch’s postulates are not useful in determining the cause of diseases that are not infec-tions (e.g., Down syndrome, lung cancer). Some diseases (e.g., cholera) occur only underspecific conditions (high numbers, specific genetic component), so the microbes may bepresent in asymptomatic persons, and Koch’s first postulate is not met. Disease caused bymicroorganisms that are infectious only in humans is difficult to prove by Koch’s postu-lates because ethical considerations prohibit deliberate exposure of humans to potentialharm (e.g., HIV/AIDS), preventing the third postulate from being applied. Some syn-dromes are common to a variety of microbes (e.g., pneumonia), so identifying a singlecausative agent is not possible (first postulate again). Alzheimer disease may be a physi-ologic/genetic disorder and not suitable for the application of Koch’s postulates, or itmay be a syndrome caused by a nonliving agent (prion), which also precludes applicationof the postulates.13.Kluyver was referring to the metabolism/biochemistry common to all cellular life on theplanet, regardless of the number of cells per organism or size. The statement can beextended to shared cellular structures as well.14.Nitrogen compounds (NO3, NO2, NH3, etc.) are often a limiting resource for growingplants, but nitrogen gas (N2) is abundant in the air. Beijerinck’s discovery that some bac-teria can convert nitrogen gas to nitrogen compounds is important to the cultivation ofthese grains. It may be possible to (1) introduce into the soil bacteria capable of convert-ing N2to organic forms of nitrogen, (2) create soil conditions that promote the growth ofsuch bacteria, and/or (3) genetically modify crop plants with bacterial genes to convertnitrogen gas for themselves.CHAPTER 2The Chemistry of MicrobiologyMultiple Choice1.c3.c5.b7.a9.a2.d4.c6.c8.a10.c

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Answers to End-of-Chapter Questions for Review211Fill in the Blanks1.valence6.hydrolysis2.nonpolar covalent7.exothermic3.ATP8.products4.fats, waxes, amylose, glycogen9.pH5.functional groups10.riboseVisualize It!1.Primary structure: light blue strands. Secondary structure: greenα-helices, blueβ-pleatedsheets (flat ribbons). The entire molecule represents tertiary structure.2.See Figure 2.21. A carbon atom is at every angle formed by lines without a lettered des-ignation. The amino group (NH2) is in the purple box, the carboxyl group is in the brownarea, and the side group is the large group in the blue box.Short Answer1.Three chemical bonds are polar covalent bonds (e.g., water), ionic bonds (e.g., salt), andhydrogen bonds (e.g., in DNA). Non-polar covalent bonds (lipids) are a fourth type.2.Five properties of water that are vital to life are that: it is cohesive, it is an excellentsolvent, it absorbs heat energy to moderate temperatures, it is liquid over a broad rangeof temperature, and it participates in many chemical reactions in cells.3.In saturated fatty acids, all the carbon bonds are linked by single bonds and are cova-lently linked to H except on the ends. In unsaturated fatty acids, there is at least onecarbon-to-carbon double bond. In polyunsaturated fatty acids, there are several carbon-to-carbon double bonds.4.Atomic oxygen contains six electrons in its valence shell. Molecular oxygen forms whentwo atoms of oxygen bond covalently. Each oxygen atom shares two electrons tocomplete its valence shell complement of eight electrons.5.Water is an excellent solvent because its partial charges (O–, H+) interact with electricallycharged molecules and dissolve them (they go into solution).Critical Thinking1.Eukaryotes do not metabolizeD-stereoisomers of amino acids and therefore have noenzymes to digest glycoproteins containingD-glutamic acid.2.Dehydrogenation:de= minus, or removal;hydrogen= H; –ation= process. Therefore,dehydrogenationmeans “the process of removing hydrogen.” The formation of doublebonds in a fatty acid requires the removal of hydrogens.3.Both students are correct. H+is the symbol for hydrogen that lacks an electron (an ion)found in aqueous conditions. Hydrogen has only one proton and one electron and noneutrons, so loss of an electron leaves a proton.4.Gelatin can be changed back to a liquid because the gelatin solid is formed by relativelyweak hydrogen bonds, which can be broken by an increase in energy (heating). Heatingegg white, in contrast, involves changing covalent bonds and denaturing (unfolding)proteins. When the egg white cools, covalent bonds reform randomly, forming

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212MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4einterconnections between proteins that would otherwise not share covalent bonds. Break-ing covalent bonds requires much more energy than is required to break hydrogen bonds,but no amount of heat will restore order to the chaos of hard-boiled egg white proteins.5.Earth-like extraterrestrial life forms would exhibit a preference for one stereoisomer overthe other in the same manner as earthly life forms, and the sample from space would con-tain predominantly one stereoisomeric form of amino acids. An equal, or nearly equal,mix of stereoisomers would be consistent with nonmetabolic (nonliving) processes pro-ducing the amino acids.6.Baking soda (sodium bicarbonate, NaHCO3) will neutralize this poison.7.Neon has two electrons in the inner shell and eight electrons in the outer cloud/shell. Ar-gon has two in the innermost cloud/shell, eight electrons in the middle cloud, and eightelectrons in the outer cloud/shell. An eight-electron cloud is a highly stable (low-energy)state and has no more room to add electrons. Losing an electron greatly decreases thestability, so not sharing electrons with other atoms is a more stable (lower-energy) statethan “giving away” or “accepting” electrons. Because electron sharing (or loss) is the ba-sis for chemical interactions, these elements do not interact with other atoms.8.Chemists use the termorganicto mean compounds that contain carbon and hydrogen,which includes all biological molecules. Nonscientists use the termorganicto mean“from nature,” which is perceived as good or better than synthetic. There is thereforesome overlap: both groups use the termorganicto mean molecules of biological origin,but the latter position overlooks the fact that many natural processes produce toxins (e.g.,lecithinase, botulism toxin, rattlesnake venom).9.Hydrogen shares its single electron with carbon. The carbon shares three electrons withnitrogen.10.Both atoms in a triple bond have a more stable state when the valence shell is “full” witheight electrons. Removing three electrons simultaneously (breaking the triple bond) re-quires much more energy than removing a single electron.11.The polar nature of the bonds between H and O and the nonlinear structure of the watermolecule mean that each H2O molecule is usually participating in three or four hydrogenbond pairs. Energy is required to break these bonds, and three to four bonds need to bebroken for each water molecule to escape the liquid and evaporate.12.Magnesium hydroxide dissociates to release two OH–, whereas each molecule of HCldissociates to release only one H+.13.Linoleic acid is the least linear of the fatty acids in Table 2.4 and will best interfere withtight packing of the fatty acids in bacterial membranes that would otherwise occur at coldtemperatures.14.Glycine’s R group is hydrogen. Glycine therefore has two hydrogens attached to the cen-tral carbon, and its “mirror images” are identical, that is, there are no stereoisomers.15.Living organisms use five nucleotide bases (adenine, cytosine, guanine, thymine, anduracil). Cells combine A, C, or T nucleotide bases with either of two different sugars (de-oxyribose and ribose) to form nucleotides (for DNA and RNA, respectively). Thymine iscombined only with deoxyribose, and uracil is combined only with ribose; therefore,there are eight nucleotides [(3 × 2) + 1 + 1 = 8].

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Answers to End-of-Chapter Questions for Review213CHAPTER 3Cell Structure and FunctionMultiple Choice1.b4.c7.a10.b13.d2.b5.c8.a11.c14.b3.c6.d9.a12.d15.cMatching1.D, GlycocalyxB, H, I FlagellaF, Axial filamentsH, CiliaA, FimbriaeC, G PiliE, Hami2.A, RibosomeD, CytoskeletonF, CentrioleE, NucleusI, MitochondrionG, ChloroplastC, Endoplasmic reticulumH, Golgi bodyB, PeroxisomeVisualize It!1.a. cytoplasm—contains metabolic chemicals; b. nucleoid—site of DNA (genes);c. glycocalyx—adhesion; d. cell wall—protects against osmotic forces; e. inclusions—stored chemicals; f. flagellum—motility; g. cytoplasmic membrane—controls import andexport; h. nucleolus—site of RNA synthesis; i. cilium—motility; j. 80S ribosome—makeproteins; k. nuclear membrane—contains DNA (genes); l. mitochondrion—makes ATP(energy source); m. centriole—plays a role in cell division; n. Golgi body—packages se-cretions; o. rough endoplasmic reticulum (RER)—transports proteins; p. SER—lipid syn-thesis; q. cytoskeleton—helps maintain cell shape2.axial (endoflagella), peritrichous, polar (tuft), polar (single)3.a. Chemical A enters the cell via facilitated diffusion through a protein channel, which islikely specific for chemical A. Chemical B does not have such a route. b. At some con-centration of chemical A, all the protein channels are filled, such that the rate of diffusioncannot increase. c. The cell could increase the rate of diffusion of chemical A by insert-ing more of the channel protein into the membrane. d. The rate of diffusion of chemicalB could be increased by removing chemical B from the cytoplasm at an increased rate.This would have the effect of increasing its concentration gradient.

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214MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4eShort Answer1.Diffusion is the movement (spreading) of molecules from an area of high concentrationto an area of low concentration. An example is the spreading of perfume molecules inthe air.2.Three flagellar arrangements in prokaryotes are:3.Cytosol is the liquid portion of cytoplasm made of water with dissolved and suspendedsubstances. In the cytosol are inclusions and organelles.4.The cytoplasmic membrane is described as a fluid mosaic. It is a mosaic because itsproteins are arranged like tiles (“membrane rafts”), and it is fluid because proteins andlipids are free to flow laterally.5.Differences and similarities between living and nonliving things are examined in thequestion of whether viruses are alive or not. Biologists agree that all living things mustexhibit growth (increase in size), reproduction (increase in number), responsiveness totheir environment, and metabolism (controlled chemical reactions producing energy andstructure).6.Growth is an increase in size of an organism, which involves increase in cell mass,whereas reproduction is an increase in population by sexual means or asexual means(binary fission, budding).7.Four similarities shared by bacterial cells and algal cells are glycocalyx, flagella, cellwall, and cytosol. Four differences include nucleus and chloroplasts in algae, size andcomponents of ribosomes, and linear versus circular DNA, cell wall composition; singleversus multiple chromosomes. See also Table 3.6.SketchNameDescriptionSee Figure 3.7Polar, singleSingle flagellumPolar, tuftSeveral flagella at one endPeritrichousFlagella cover the surface of the cell8.Streptococcus pyogenesEntamoeba histolyticasmaller in sizelarger in sizeno nucleusmembranous nucleusmembranous organelles absentmembranous organellescell wall of peptidoglycanno cell wallsingle chromosomemultiple chromosomessmaller 70S ribosomes in cytoplasmlarger 80S ribosomes in cytoplasmcircular chromosomelinear chromosomeshistones absentHistones present9.StructureSketchDescriptionPiliFigure 3.11Long hollow tubules of pilin, used for conjugation,longer than fimbriaeFimbriaeFigure 3.10Bristlelike, sticky, proteinaceous, hundreds per cellCiliaFigure 3.31Figure 3.32Short hairlike structures composed of tubulin proteinin a 9 + 2 arrangement. There are usually 100–1000

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Answers to End-of-Chapter Questions for Review21510.The ability to metabolize is unique to living things. However, if a virus is considerednonliving, one could argue that it is involved in metabolism by taking over its host andcausing the host to make more viruses.11.Archaeal flagella are thinner, are solid, and grow basally, whereas bacterial flagella arethicker, are hollow, and grow at the tip. The proteins making up the flagellar subunitsdiffer between the two domains but are similar between species in the domains. Archaealflagella may have sugar molecules attached and are powered by ATP, in contrast to bac-terial flagella, which do not have sugars attached and are powered by the flow of hydro-gen or sodium ions across the cytoplasmic membrane at the basal body. Finally, archaealflagella always rotate together as a bundle. Eukaryotic flagella are hollow tubes com-posed of tubulin in a 9 + 2 arrangement which have no hook, and do not rotate but have awhip-like motion. Eukaryotic flagella are powered by ATP hydrolysis.12.See also Table 3.6 on p. 87.13.Masses of bacteria adhere to a substrate by means of sticky fimbriae and glycocalyces.14.A molecule has difficulty moving across a cell membrane if it is too large, electricallycharged, or hydrophilic.15.In the three types of passive transport (diffusion, facilitated diffusion, and osmosis), theelectrochemical gradient provides the source of energy rather than the cell having to ex-pend ATP. Diffusion is the movement of small or lipid-soluble molecules down the con-centration gradient through the spaces in the cell membrane. Facilitated diffusion occurswhen permeases provide a pathway for diffusion of large or electrically charged mole-cules. Osmosis is specifically the diffusion of water through the membrane.16.Active transport uses proteinaceous ports to move substances against their concentrationgradient using energy. In group translocation, the substance is chemically changed duringtransport across prokaryotic membranes. Endocytosis occurs when eukaryotic cells ex-tend the cell membrane around substances to bring them into the cell. Exocytosis occurswhen vesicles containing substances are fused with the cytoplasmic membrane, dumpingtheir contents outside the cell.per cell, and they provide motility by a coordinatedwavelike motion.CharacteristicBacteriaEukaryotesSize0.2–2.0μm10–100μmPresence of nucleusAbsentPresentPresence of membrane-bound organellesPresent in fewPresentStructure of flagellaSome have flagella composedof flagellin, with basal bodyand hook. RotateSome have flagellacomposed of tubulin, nohook, whip-like motionChemicals in cell wallsPeptidoglycanPresent in some,polysaccharideTypes of ribosomesSmall 70SLarge 80S (small 70S inmitochondria andchloroplast)Structure ofchromosomesUsually single circularMultiple linear

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216MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4e17.Symports simultaneously transport two substances in the same direction through the cellmembrane, whereas antiports simultaneously transport two chemicals in opposite direc-tions through the cell membrane.18.The endosymbiotic theory proposes an explanation for why mitochondria and chloro-plasts have 70S ribosomes, circular DNA, and two membranes. The theory states thatlarge prokaryotes internalized small aerobic prokaryotes that then lost the ability to existoutside their host, forming the early eukaryotes. The theory does not explain the twomembranes of the nuclear envelope nor why only a few polypeptides of mitochondriaand chloroplasts are made in the organelles while the bulk of their proteins come fromnuclear DNA and cytoplasmic ribosomes.Critical Thinking1.A chemical that prevents Golgi function would cause a variety of problems in the humanbody, all related to secretion. The respiratory tract would dry out, making breathing andgas exchange more difficult. Digestive system problems would include loss of lubricatingmucus as well as maladsorption due to lack of digestive enzyme secretion. These wouldbe among the most immediate problems. Problems with hormone secretion would soondevelop. Bacteria would be unaffected because they have no Golgi bodies.2.The nucleus and mitochondria contain DNA, so methylene blue would stain the nucleusand mitochondria.3.Bacterial ribosomes differ from eukaryotic ribosomes in structure and some aspects offunction. Drugs targeting those biochemical differences may be toxic to bacteria butharmless to humans. Bacteria possess metabolic pathways not present in humans(e.g., folic acid synthesis); drugs may inhibit such processes in bacteria without harm tohumans. Bacteria have cell walls critical to their survival whereas human cells have nocomparable structures, so drugs may target bacterial cell wall structure or synthesis with-out affecting human cellular function or structures.4.When the contents of the digestive system are hypertonic as a result of salt secretion,water leaves the digestive tract cells and enters the intestinal chamber. Water leaves thebloodstream to replace the water lost from the digestive tract cells. Continued loss of wa-ter into the intestines “pulls” more water from the blood, which then depletes the waterstored in the tissues, possibly leading to potentially dangerous dehydration.5.The organelle the electrode penetrated was likely the central vacuole, a prominent featureof many plant cells.6.As a eukaryote,Ostreococcusmust have a nucleus. Since it is photosynthetic, it musthave a chloroplast.7.Long projections from basal bodies are flagella, so this cell is probably motile. Becausethe basal body is in the wall rather than within the cytoplasm, the cell is a prokaryote.8.Positive phototaxis suggests these organisms need light, most likely for photosynthesis.They need to be incubated in the presence of light for at least part of the day. Dark incu-bators would not be suitable.9.A strain ofNeisseriafound in a biofilm may be pathogenic. Pathogens tend to have strin-gent environmental requirements, and the interior conditions of a biofilm may meet thoseneeds while other members of the biofilm community may provide nutritional needs.10.SER (smooth endoplasmic reticulum) functions in intracellular transport and in lipidsynthesis. If it does not function, then the affected cell will be unable to make

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Answers to End-of-Chapter Questions for Review217cytoplasmic membranes or organelle membranes and will be hindered in transportingsubstances and growth, since cell growth requires new cytoplasmic membrane.11.The antibiotic killed the bacteria, but as the dead bacteria were degraded, the bacterialcomponents, including lipopolysaccharides (LPS) of the outer membrane, were released.The lipid A component of LPS caused the body’s defense responses (fever and inflam-mation) to intensify.12.The hypertonic solutions have higher concentrations of solutes than the solute concentra-tions in the bacterial and fungal cells. Because the cell membranes largely prevent the so-lutes from entering the cells and achieving equilibrium, water tends to flow out of thecells. The resulting dehydration brings metabolic activity to a halt, both because metabo-lites would crystallize out of solution and because water is critical to most metabolic re-actions.13.Steaming mail for 30 seconds is not adequate treatment to protect people from anthrax.Anthrax is contracted by contact with endospores, and bacterial endospores are not de-stroyed by the relatively gentle heating of a 30-second steaming.14.Exchange rates across the surface of a cell and surface-to-volume ratios are consideredmajor factors that limit cell size. Eukaryotic cells have extensive internal membranes,particularly the endoplasmic reticulum, that function to vastly increase the functional sur-face area of the cell. Many of the critical energy production functions (ATP synthesis,photosynthesis) take place in the bacterial cell membrane, whereas in eukaryotes the mi-tochondria and chloroplasts are sites of these critical functions; for this reason, surface-to-volume ratios are not limiting for these functions in eukaryotes.Concept Mapping1.Gram-positive cell wall.6.Glycan chains2.Teichoic acids7.N-acetylglucosamine3.Gram-negative cell wall8.Lipopolysaccharide (LPS)4.Periplasm9.Porin5.Peptidoglycan10.Lipid ACHAPTER 4Microscopy, Staining, and ClassificationMultiple Choice1.c3.c5.d7.b9.a2.d4.d6.d8.a10.dFill in the Blanks1.600×4.Contrast2.heat fixation5.negatively3.increases, increases, moreVisualize It!1.a.scanning electron,b.bright-field light,c.phase-contrast light,d.fluorescent light,e.transmission electron,f.differential interference contrast (Nomarski)

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218MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4e2.See Figure 4.4.a.ocular;b.body;c.arm;d.objectives;e.stage;f.condenser;g.diaphragm;h.lightsource;i.course focus;j.fine focus;k.base.Short Answer1.Because electrons travel in smaller wavelengths than the wavelengths of visible light,electron microscopes can magnify with greater resolution.2.Magnification does notmakethings “bigger,” but it does cause the image toappearlarger.3.Because electron microscopes require a vacuum, only dead organisms are examined.4.The order of operations in a Gram stain is: primary stain, mordant, decolorizing agent,counterstain.5.Latin is used in taxonomic nomenclature because it was the scientific language of thetime of Linnaeus (the “father” of taxonomy) and because using Latin ensures that nocountry has priority in the language of taxonomy.6.A “specific epithet” is the second word in an organism’s scientific name. It is usually anadjective, and it is written in lowercase letters. It follows the genus name, and both areeither printed in italics or underlined.7.In the study of the taxonomic relationships among prokaryotic cells, the RNA of theirribosomes distinguishes relationships, as in the classification of the Eukaryotes, Bacteria,and Archaea into separate domains.8.Electron and scanning tunneling microscopes require vacuums and use electron beams.Organisms cannot live through the specimen preparation and procedures involved inelectron microscopy.Critical Thinking1.Miki was using malachite green dye, which is usually used to stain endospores.2.The term “successfully interbreeding organisms” refers to organisms that undergo sexualreproduction, in which haploid gametes are produced and exchanged. Bacteria and mostother microorganisms do not reproduce sexually, so the definition does not apply.3.No, it is not logical to assume that the current taxonomy will stay the same. Our under-standing of the interrelationships among organisms based on genetic sequences isundergoing rapid development and is still in the early stages. Also, only a small percent-age of microorganisms have been characterized and sequenced: as more organisms areidentified and characterized, our understanding and therefore our taxonomy are likely tocontinue to develop and change.4.Immersion oil has optical properties similar to those of the glass in slides and microscopelenses. Light passing from the slide to the oil and into the lens is not refracted (bent), somost of the light passing through the specimen continues into the lens, increasing theresolution.5.Such a microscope at 1000× magnification (10× times 100×) will be able to distinguishbetween objects that are 300 nm (0.3 μm) or farther apart. Therefore, the first two objectswill be resolved, but the objects 40 nm apart will appear as a single object.6.Both the Gram and the acid-fast staining procedures differentiate cells on the basis of dif-ferent cell wall structures (thick versus thin peptidoglycan in Gram stains, waxy versusnonwaxy in the acid-fast stain), and both procedures use a decolorizing step to remove

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Answers to End-of-Chapter Questions for Review219the initial dye from cells in which it has not been trapped, followed by a second dye tomake the decolorized cell visible. Heating followed by cooling in the acid-fast proceduretakes the place of the Gram’s iodine mordant: Heating allows the dye to penetrate the cellwall; cooling traps the dye inside the cell wall.7.Chondrus crispus:a curly alga with the consistency of cartilage (chondrus)Diploccocus pneumoniae:pair of coccus, and causes pneumoniaStreptococcus pneumoniae:spherical or berry-shaped bacterium (coccus) that grows inchains (strepto-) and causes pneumonia.Enterococcus faecalis:spherical bacterium (coccus) from the digestive system (enteric)found in feces.Enterococcus faecium:spherical bacterium (coccus) from the digestive system (enteric)found in feces; related to but not the same asE. faecalis.Escherichia coli:bacterium named to honor Escherich and that lives in the colon.Haemophilus influenzae:blood-loving (i.e., grows on blood agar) bacterium thatcauses flu.Homo sapiens:human species that fancies itself wise.Isabella abbottae:not a descriptive name; indicates that Isabella Abbott is worthy of thehonor of having an organism named after her.Pasteurella haemolytica:bacterium named to honor Pasteur and that destroys redblood cells.8.The bacteria were discovered by directly detecting the genetic material of the bacteria.Comparison of the new genetic sequences with existing genetic information indicatedthat the bacteria were new species. Ribosomal RNAs of prokaryotes and eukaryotes aresignificantly different, so the new organisms could be identified as bacteria on the basisof their ribosomal RNAs.Concept Mapping1.Iodine4.Primary stain7.Purple2.Ethanol and acetone5.Decolorizer8.Gram-negative bacteria3.Safranin6.Gram-positive bacteria9.Thin peptidoglycan layerCHAPTER 5Microbial MetabolismMultiple Choice1.c5.a9.a13.a17.a2.a6.c10.c14.a18.c3.c7.b11.d15.a19.a4.a8.d12.d16.c20.cMatching1.C3.E2.B4.A

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Solution Manual for Microbiology with Diseases by Body System, 4th Edition - Page 15 preview image

220MANUAL/TEST BANK FORMICROBIOLOGY WITH DISEASES BY BODY SYSTEM,4eFill in the Blanks1.the original reaction center chlorophyll2.23.pentose phosphate, Entner-Doudoroff4.The Krebs cycle5.Oxygen, ½ O26.NO3–, SO4–2, CO3–27.inorganic9.chemiosmosis10.NAD+, FADVisualize It!1.See Figure 5.12.2.a.glycolysis (cytosol),b.electron transport (cristae),c.Kreb’s cycle (matrix)3.Erythrose-4-phosphate, Pentose phosphate cycle; PEP, glycolysisShort Answer1.Amination reactions are synthetic reactions in which an amine from ammonia is added toa metabolite. Transaminations are exchange reactions in which amine groups are movedby transferases from one amino acid to another.2.For anabolic reactions to occur, reactants must collide with sufficient energy for bonds toform between them. Increasing the concentration of reactants or ambient temperatureswill cause more chemical reactions to occur; however, in living organisms neither reac-tant concentration nor temperature is usually high enough to ensure that bonds will form.Therefore, chemical reactions of life depend on catalysts. Organic catalysts are called en-zymes.3.One way in which organisms regulate metabolism is by controlling the quantity andtiming of enzyme synthesis. Also, some eukaryotic cells control some enzymatic activi-ties by compartmentalizing enzymes inside membranes so that certain metabolic reac-tions proceed physically separated from the rest of the cell. Enzymatic inhibitors alsomay block an enzyme’s active site, preventing substrate binding. Noncompetitiveinhibitors prevent enzymatic activity by binding to an allosteric site. Cells also controlenzymatic action through feedback inhibition.4.Binding at an allosteric site alters the shape of the active site, thereby altering the sub-strate binding. It affects the whole pathway of enzymatic reactions in either of two ways.In allosteric inhibition, binding of substrate is prevented, which halts any further activity.8.Category of EnzymeDescriptionHydrolaseCatabolizes substrate by adding waterIsomeraseRearranges atomsLigase/polymeraseJoins 2 molecules togetherTransferaseMoves functional groupsOxidoreductaseAdds or removes electronsLyaseSplits large molecules

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Solution Manual for Microbiology with Diseases by Body System, 4th Edition - Page 16 preview image

Answers to End-of-Chapter Questions for Review221Allosteric activation, which causes the shape of the active site to change, permits sub-strate binding, thereby activating an otherwise inactive enzyme. In the former case, thesubstrate for the next step is not produced, so the pathway is shut down. In the latter case,substrate molecules are supplied, so the metabolic pathway can proceed.5.In negative feedback, or feedback inhibition, the end product of a series of reactions is anallosteric inhibitor of an enzyme in an earlier part of the pathway. Because the product ofeach reaction in the pathway is the substrate for the next reaction, inhibition of the firstenzyme in the series inhibits the entire pathway.6.Microorganisms with fermentation pathways, in which organic molecules are the finalelectron acceptors instead of oxygen, can colonize anaerobic environments, which areunavailable to strictly aerobic organisms.7.A molecule may be oxidized in one of two ways in the absence of oxygen: by losing asingle electron or by losing a hydrogen atom. A molecule other than oxygen is the elec-tron acceptor.8.Cyanobacteria, purple sulfur bacteria, green sulfur bacteria, green nonsulfur bacteria,purple nonsulfur bacteria, a few protozoa, and algae are photosynthetic.9.We breathe oxygen to be used as the final electron acceptor from the electron transportchain, and we give off carbon dioxide from the decarboxylations of the Krebs cycle.10.Cyanobacteria and algae capture light energy from the sun and use it to drive thesynthesis of carbohydrates from CO2and H2O through photosynthesis. In the dissociationof water during photosynthesis, the molecular oxygen is a waste product.11.E. coliuses glycolysis to split a 6-carbon glucose molecule into two 3-carbon sugar mol-ecules, which are ultimately transformed into two molecules of pyruvic acid. The carbonatoms in pyruvic acid may be further metabolized to carbon dioxide, or used for the syn-thesis of other carbon compounds.12.Yeast produce alcohol and make bread rise through the process of fermentation, theend products of which are alcohol (ethanol) and CO2.13.The most significant production of ATP in both prokaryotic and eukaryotic cells occursin the electron transport chain.14.Vitamins are essential (“vital”) cofactors for enzymes and carrier molecules. For exam-ple, flavoproteins contain flavin, a coenzyme derived from riboflavin. One form of flavinis flavin mononucleotide (FMN), which is an initial carrier molecule of the electrontransport chains of the mitochondria. Some microbes can produce some cofactors andcarriers, but anything an organism cannot produce it must obtain as nutrients.15.Glucose may act as an allosteric inhibitor of one of the enzymes that metabolize lactose.Critical Thinking1.The scientists could introduceThermusand oxygen into the water so that arsenite isoxidized to arsenate.2.DAHAP is a substrate for three different synthetases, each of which is part of a differentmetabolic pathway. As long as one pathway is active, DAHAP is required and will beproduced. Only when all three pathways are shut down by feedback inhibition willDAHAP no longer be required and its synthesis shut down.3.The pentose phosphate pathway is an important source of metabolic intermediatesrequired for amino acid and nucleotide synthesis.

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