Microbiology Bacteria Growing Bacteria Microorganisms.
Stainings
Acetone/Alcohol
Crystal Violet
Iodine
Methlyene Blue
Dipping and Staining place a leaf or fruit piece into Methlyene Blue then wash in cold water and dry.
Growing Bacillus Anthracis Anthrax and other bacillus rods, spirals on blood agar or agar. Gelatine Unflavored, Sugar and mix into water.
cheese, some uncooked raw meat may contain bacillus anthracis food poisoning.

Bacteria Growing Experiments in Petri Plates Introduction Bacteria are microorganisms that grow everywhere. We can collect and grow them in specially prepared petri dishes. Blood agar or tryptic soy agar with 5% sheep's blood is an excellent medium for supplying bacteria with nutrients and an environment in which we can see them grow. Sterile powdered agar with nutrients can be mixed with water, heated and then poured into empty petri plates or ready-to-use dishes can be purchased. The undigestible agar is a gelatin-like substance with a semi solid surface on which the bacteria can grow while they consume the added nutrients (like sheep's blood). In fact, this is why gelatin itself does not make a good growing medium. Some bacteria can digest gelatin, which is a protein derived from animal tissue. This destroys the growing surface in the petri plate making it unsuitable as a bacteria growth medium. Caution Most bacteria collected in the environment will not be harmful. However, once they multiply into millions of colonies in a petri dish they become more of a hazard. Be sure to protect open cuts with rubber gloves and never ingest or breathe in growing bacteria. Keep growing petri dishes taped closed until your experiment is done. Then you should safely destroy the fuzzy bacteria colonies using bleach. Below are general outlines of three types of experiments involving bacteria growth. They are offered to assist in designing your own experiment or project. Let us know what you tested and how your experiment turned out. We'd be delighted to hear from you! Click here and use our Contact Us form. Experiment 1 Direct Contact Discussion. In this type of experiment, bacteria is transferred directly to the prepared petri plate via direct contact. You can test the effectiveness of different soaps by treating different petri dishes with "dirty" hands before washing and "clean" hands after washing. Or, you can press a variety of common objects like coins, combs, etc. on different plates and compare the bacteria growth that results. What you need. Prepared petri plates containing agar medium and nutrients. The Science Company has these available at this link. Bacteria on hands, paws, etc. Wax pencil for labeling dishes. Masking tape. Bleach. What to do. Prepared petri dishes should be refrigerated until used and always stored upside down (i.e media in upper dish, cover on bottom). This keeps condensation which forms in the lid from dropping onto and disrupting the bacteria growing surface. When ready to use, let dishes come to room temperature before taking samples (about one hour). Inoculate the dish by gently pressing fingers, finger nails, lips, coin, etc onto agar surface. Avoid tearing agar surface. Replace cover on dish, tape closed, and label each dish so you know the source of the bacteria. Store upside down. Let grow in undisturbed warm location, ideally in an environment around 100° F (37° C) - not in sunlight or on a heating register. You should see growth within a couple of days. The dishes will start to smell which means the bacteria are growing. Make observations and keep records of what you see growing in each dish. Can you make any conclusions about what objects had the most bacteria? Before disposing of dishes in the trash the bacteria should be destroyed. Pour a small amount of household bleach over the colonies while holding dish over sink. Caution - do not allow bleach to touch your skin, eyes or clothes. It will burn! Experiment 2 Collected bacteria samples Discussion. Use a sterilized inoculating loop or sterile swabs to collect bacteria from different locations and then streak each petri dish with your sample. This involves a bit more technique than Experiment 1 but offers a wider choice of bacteria sampling locations. Swabs can be run over doorknobs, bathroom fixtures, animal mouths, etc. What you need. Prepared petri dishes containing agar medium and nutrients. The Science Company has these available at this link. Bacteria collected from doorknobs, bathroom fixtures, etc. Wax pencil for labeling dishes. Masking tape. Sterile swabs or inoculating loop. Alcohol burner (source of flame to sterilize inoculating loop). Bleach. What to do. Prepared petri dishes should be refrigerated until used and always stored upside down (i.e media in upper dish, cover on bottom). This keeps condensation which forms in the lid from dropping onto and disrupting the bacteria growing surface. When ready to use, let dishes come to room temperature before taking samples (about one hour). Collect bacteria from each location using one swab or resterilized innoculating loop) for each new spot. Inoculate each dish by streaking a pattern gently across the entire agar surface without tearing into it. Another common technique is to divide each plate into four quadrants by marking the lid with a cross. Streak your sample in straight lines starting in quadrant 1. Generally, after a few days, quadrant one will show the most growth. Depending on bacteria abundance on the swab, quadrant 4 may show no grow or only a few colonies. It is sometimes easier to distinguish different bacteria types in this low growth, less cluttered area. Replace cover on dish, tape closed, and label each dish so you know the source of the bacteria. Store upside down. Let grow in undisturbed warm location, ideally in an environment around 100° F (37° C) - not in sunlight or on a heating register. You should see growth within a couple of days. The dishes will start to smell which means the bacteria are growing. Make observations and keep records of what you see growing in each dish. Can you make any conclusions about what locations had the most bacteria? Before disposing of dishes in the trash the bacteria should be destroyed. Pour a small amount of household bleach over the colonies while holding dish over sink. Caution - do not allow bleach to touch your skin, eyes or clothes. It will burn! Experiment 3 Testing the effectiveness of bacteria killing agents Discussion. In this type of project, a petri dish is inoculated with bacteria then a paper disk treated with an antiseptic agent is placed in the dish. After several days, a halo develops around the paper disk indicating a zone of no growth. Comparisons can be made between different antibacterial agents. What you need. Prepared petri dishes containing agar medium and nutrients. The Science Company has these available at this link. Bacteria collected from doorknobs, bathroom fixtures, etc. Wax pencil for labeling dishes. Masking tape. Sterile swabs or inoculating loop. Alcohol burner (source of flame to sterilize inoculating loop). Antibacterial agent (soaps, disinfectants, etc.). Sterile water. Test tubes, 12 x 75mm. Filter paper or paper towel. Small containers in which to soak paper disks. Hole punch. Tweezers. Ruler. Bleach. What to do. Prepared petri dishes should be refrigerated until used and always stored upside down (i.e media in upper dish, cover on bottom). This keeps condensation which forms in the lid from dropping onto and disrupting the bacteria growing surface. Prepare sterilized water by boiling water and letting cool to room temperature. When ready to use, let petri dishes come to room temperature before taking samples (about one hour). Prepare antiseptic disks by using a hole punch to create paper disks out of a piece of filter paper or paper towel. Soak one disk in each antibacterial agent to be tested. Set aside until step 6. Collect bacteria from each location using one swab for each new spot. Fill a small test tube partly full of sterilized water. Dip bacteria laden swab into water. This will transfer some of the bacteria you collected into the water. Now, inoculate a petri dish by pouring the water into the dish so the entire surface is covered. Pour out excess water. Repeat for each bacteria sample using fresh water and clean test tube each time. Place a pretreated antiseptic disk in each inoculated petri dish. Replace cover on dish, tape closed, store upside down. Be sure to label each petri dish with a name or number. Let grow in undisturbed warm location, ideally in an environment around 100° F (37° C) - not in sunlight or on a heating register. You should see growth within a couple of days. You should also see a "halo" around each disk indicating a no growth zone. Measure and compare the size of the kill zone to determine effectiveness of each antibacterial agent. Before disposing of dishes in the trash the bacteria should be destroyed. Pour a small amount of household bleach over the colonies while holding dish over sink. Caution - do not allow bleach to touch your skin, eyes or clothes.

Lab Supplies
microscopes
microscope slides
petri dishes
test tubes

Bacillus Anthracis (Anthrax) is a form of gram-positive rod shaped bacillus that occurs in chains and grows anerobically. These bacteria can be found in soil, water, and air in vegetation. Bacillus Anthracis (Anthrax) is the causative organism of anthrax in man and animals. Anthrax is primarily an infectious disease of domestic animals whereby causing various forms of septicemia. Man contracts the disease sporadically by coming into contact with infected animals or contaminated animal products. Bacillus Anthracis (Anthrax) is a non-motile, straight sporing bacillus, rectangular in shape and measuring 4 – 8 microns by 1 – 1.5 microns. The bacilli are arranged in chains, end to end (streptobacilli), but may also occur singly and in pairs.

The organisms can be stained strongly with grams stain and methylene – blue for the recognition of Bacillus Anthracis (Anthrax) in blood films. The bacilli grow will in the environment of aerobe and facultative anaebe; within the temperature range of 12 – 45 degrees Celsius. The organisms grow at an optimum temperature of 35 degrees Celsius on all ordinary media.

Hi I like to order the following and I have enclosed $ Quantity Description 1 Petri Seal Tape 408-303 White $28.00 1 High Five Latex E-Grip 410205 M E-Grip, Powder Free $69.00 You can mail to Robert Bio World Test Tube Support #221-610 $112.45 25mm white test tube rack Gram Stain Crystal Violet Iodine Safranin-O

Microbiology Chapter 1 Bacteria seen through a microscope fungi, yeasts, molds, protozoa microscopic algae. Viruses life and non life uncomfortable infections, diseases spoiled food. Microorganisms most are pathogenic disease producing. Bacteria are enclosed in cell walls composed of carbohydrate and protein complex called peptidoglycan. Bacteria shapes bacillus rod like coccus or spiral are formed in Chains, clusters, pairs.

Chapter 2 Chemical elements Hydrogen, Oxygen, Sodium, Iodine.

1. Atoms with the same atomic number and chemical behavior are classified as chemical elements. 2. Refer to Figure 2.1. 3. 14C and 12C are isotopes of carbon. 12C has 6 neutrons in its nucleus and 14C has 8 neutrons. 4. Hydrogen bonds. 5. a. Ionic b. Single covalent bond c. Double covalent bonds d. Peptide bond e. Hydrogen bond 6. 104 or 10,000 times. 7. Element Atomic Weight x Number of Atoms = Total Weight of That Element C 12 x 6 = 72 H 1 x 12 = 12 O 16 x 6 = 96 The molecular weight of C6H12O6 is 180 grams. 8.a. Synthesis reaction, condensation, or dehydration b. Decomposition reaction, digestion, or hydrolysis c. Exchange reaction d. Reversible reaction 9. The enzyme lowers the activation energy required for the reaction, and therefore speeds up this decomposition reaction. 10.a. Lipid b. Protein c. Carbohydrate d. Nucleic acid 11. > 12.a. Left to right b. Right to left 13. Breaking of bonds between phosphorus and oxygen. These are covalent bonds. 14. The entire protein shows tertiary structure. No quaternary structure. 15. Multiple-Choice Answers are in boldface. Radioisotopes are frequently used to label molecules in a cell. The fate of atoms and molecules in a cell can then be followed. This process is the basis for questions 1-3. 1. Assume E. coli are grown in a nutrient medium containing the radioisotope 16N. After a 48-hour incubation period, the 16N would most likely be found in the E. coli's a. carbohydrates. b. lipids. c. proteins. d. water. e. None of the above. 2. If Pseudomonas bacteria are supplied with radioactively-labeled cytosine, after a 24-hour incubation period, this cytosine would most likely be found in the cells' a. carbohydrates. b. DNA. c. lipids. d. water. e. protein. 3. If E. coli were grown in a medium containing the radio-active isotope 32P, the 32P would be found in all of the following molecules of the cell except a. ATP. b. carbohydrates. c. DNA. d. plasma membrane. e. None of the above. 4. A carbonated drink, pH3, is _____ times more acid than distilled water. a. 4 b. 10 c. 100 d. 1,000 e. 10,000 5. The best definition of ATP is: a. A molecule stored for food use. b. A molecule that supplies energy to do work. c. A molecule stored for an energy reserve. d. A molecule used as a source of phosphate. 6. Which of the following is an organic molecule? a. H2O (water) b. O2 (oxygen) c. C18H29SO3 (Styrofoam) d. FeO (iron oxide) e. F2C­CF2 (Teflon) Classify the molecules shown in questions 7-10 using the following choices. The dissociation products of the molecules are shown to help you. 7. HNO3 => H1 1 NO23 a. Acid b. Base c. Salt 8. H2SO4 => 2H1 1 SO2-4 a. Acid b. Base c. Salt 9. NaOH => Na1 1 OH2 a. Acid b. Base c. Salt 10. MgSO4 => Mg21 1 SO2-4 a. Acid b. Base c. Salt Critical Thinking 1.a. Synthesis reaction. b. H2CO3 is an acid. ATP and DNA have 5-carbon sugars. ATP has ribose, and DNA has deoxyribose; ATP and DNA contain the purine, adenine. 3. In order to maintain the proper fluidity, the percentage of unsaturated lipids decreases at the higher temperature. 4. These animals have cellulose-degrading bacteria in specialized structures in their digestive tracts. Clinical Applications 1. The enzymes will degrade organic molecules on the clothing. Any stains caused by these molecules will be removed when the molecules are digested. 2. T. ferrooxidans can oxidize sulfur ("thio") as well as iron ("ferro"). The oxidation of sulfide in pyrite produces sulfuric acid, which dissolves the limestone. Gypsum forms in a subsequent exchange reaction. 2S2- + 3O2 + 2H2O ----> 2SO42- + 4H+ 2CaCO3 + 4H+ + 2SO42- ----> 2CaSO4 + 2H+ + 2HCO3- 3. Since L-isomers are more common in nature, most cells, such as phagocytes, will be able to degrade the L-isomers. D-isomers will be resistant to metabolism by most cells. 4.Amphotericin B would not work against most bacteria because they lack sterols. Fungi have sterols and are generally susceptible to amphotericin B. Human cells have sterols.

Chapter 3 Microscopes Light microscopy Compound light microscope has lenses, Visible light examine very small specimens. Illuminator, condenser, objective lenses, ocular lens, resolution. Size shape, arrangement, of bacterial cells. 3 shapes, spherical, coccus, rod shaped bacillus, spiral.

Penicillin refers to a group of over 50 chemically related antibiotics. Containing a B-lactam ring called the nucleus. Penicillin molecules are differentiated by the chemical side chains attached to their nuclei. Penicillins can be produced either naturally or semisynthetically. Natural Penicillins Penicillin extracted from cultures of the mold Penicillum exists in several closely related forms. Natural penicillins.

Microbiology Chapter 3 Microbiology Chapter 3 Answers Review 1. 1 mm = 10-6m 1 nm = 10-9 m 1 mm = 103nm 2. a. Ocular lens b. Objective lens c. Diaphragm d. Condensor e. Illuminator 3. Ocular lens magnification ¥ oil immersion lens magnification total magnification of specimen. 10¥ = ¥100¥ = 1000¥ 4. a. Compound light microscope b. Darkfield microscope c. Phase-contrast microscope d. Fluorescence microscope e. Electron microscope f. Differential interference contrast microscope 5. …that a beam of electrons focused by magnets…on a television-like screen or photographic plate. 6. Type of Microscope Maximum Magnification Resolution Compound light 2,000¥ 0.2 mm Electron 100,000¥ 0.0025 mm 7. Bacterial cells have a slightly negative charge, and the colored positive ion of a basic dye is attracted to the negative charge of the cell. Acid dyes do not stain bacterial cells because the negatively charged colored ion is repelled by the like charge of the cell. 8. a. A simple stain is used to determine cell shape and arrangement. b. A differential stain is used to distinguish kinds of bacteria based on their reaction to the differential stain. c. A negative stain does not distort the cell and is used to determine cell shape, size, and the presence of a capsule. d. A flagella stain is used to determine the number and arrangement of flagella. 9. In a Gram stain, the mordant combines with the basic dye to form a complex that will not wash out of gram-positive cells. In a flagella stain, the mordant accumulates on the flagella so that they can be seen with a light microscope. 10. A counterstain stains the colorless non-acid-fast cells so that they are easily seen through a microscope. 11. In the Gram stain, the decolorizer removes the color from gram-negative cells. In the acid-fast stain, the decolorizer removes the color from non-acid-fast cells. 12. Endospore: safranin is the counterstain. Gram: safranin is the counterstain. 13. Appearance after this step of Steps Gram-positive cells Gram-negative cells Crystal violet Purple Purple Iodine Purple Purple Alcohol-acetone Purple Colorless Safranin Purple Red Multiple-Choice Answers are in boldface. 1. You are trying to identify an unknown gram-negative bacterium. Which of the following stains is not necessary? a. negative stain b. acid-fast stain c. flagella stain d. endospore stain e. Both b and d 2. Assume you stain Bacillus by applying malachite green with heat and then counterstain with safranin. Through the microscope, the green structures are a. cell walls. b. capsules. c. endospores. d. flagella. e. impossible to identify. 3. Carbolfuchsin can be used as a simple stain and a negative stain. As a simple stain, the pH is a. 2. b. higher than the negative stain. c. lower than the negative stain. d. the same as the negative stain. 4. Looking at the cell of a photosynthetic microorganism, you observe that the chloroplasts are green in brightfield microscopy and red in fluorescence microscopy. You conclude that a. chlorophyll is fluorescent. b. the magnification has distorted the image. c. you're not looking at the same structure in both microscopes. d. the stain masked the green color. e. None of the above. 5. Which of the following is not a functionally analogous pair of stains? a. nigrosin and malachite green b. crystal violet and carbolfuchsin c. safranin and methylene blue d. ethanol-acetone and acid-alcohol e. None of the above 6. Which of the following pairs is mismatched? a. capsule-negative stain b. cell arrangement-simple stain c. cell size-negative stain d. gram stain-bacterial identification e. None of the above 7. Assume you stain Clostridium by applying a basic stain, carbolfuchsin, with heat, decolorizing with acid-alcohol, and counterstaining with an acid stain, nigrosin. Through the microscope, the endospores are ____1____ and the cells are stained ____2____. a. 1-red- 2-black b. 1-black- 2-colorless c. 1-colorless- 2-black d. 1-red- 2-colorless e. 1-black- 2-red 8. Assume you are viewing a Gram-stained field of red cocci and blue baccili through the microscope. You can safely conclude that you have a. made a mistake in staining. b. two different species. c. old bacterial cells. d. young bacterial cells. e. None of the above. 9. In 1996, scientists described a new tapeworm parasite that has killed at least one person. The initial examination of the patient's abdominal mass was most likely made using a. brightfield microscopy. b. darkfield microscopy. c. electron microscopy. d. phase-contrast microscopy. e. fluorescence microscopy. Critical Thinking 1. The counterstain safranin can be omitted. Gram-positive bacteria will appear purple, and gram-negative bacteria will be colorless. 2. You would be able to discern two objects separated by the four distances given because each is equal to or greater than the resolving power of the microscope. 3. The high lipid content of acid-fast cell walls makes them impermeable to most stains. If the primary stain penetrates, the Gram stain decolorizer will not decolorize the cell. Therefore, acid-fast bacteria would be gram-positive if they could be Gram stained. 4. Inclusions as well as endospores may not stain in a Gram stain. The endospore stain will identify the unstained structure as an endospore. Clinical Applications 1. Ehrlich observed that mycobacteria could not be decolorized with acid-alcohol, so he reasoned that an acidic disinfectant would not be able to penetrate the cell wall. 2. N. gonorrhoeae bacteria are gram-negative (red) diplococci, often found in the large human cells (phagocytes). 3. These are called clue cells. The large red cells are human mucosal cells; gram-positive bacteria on the surface of the human cells. 4. The reporter was unfamiliar with the magnification of EMs and thought the magnified rods were worms. http://occawlonline.pearsoned.com/bookbind/pubbooks/tfc/chapter3/deluxe.html

Review Chapter 4 1. Metabolism is the sum of all chemical reactions that occur within a living organism. 2. Catabolic reactions break down organic compounds and release energy, while anabolic reactions use the products of catabolism and energy to build cell material. 3. 4. a. When the enzyme and substrate combine, the substrate molecule will be transformed. b. When the competitive inhibitor binds to the enzyme, the enzyme will not be able to bind with the substrate. c. When the noncompetitive inhibitor binds to the enzyme, the active site of the enzyme will be changed so the enzyme cannot bind with the substrate. d. The noncompetitive inhibitor. 5. The optimum temperature for an enzyme is one that favors movement of molecules so the enzyme can "find" its substrate. Lower temperatures will decrease the rate of collisions and the rate of reactions. Increased temperatures will denature the enzyme. 6. Ethyl alcohol, lactic acid, butyl alcohol, acetone, and glycerol are some of the possible products. Refer to Table 5.4 and Figure 5.18b. 7. Organism Carbon Source Energy Source Photoautotroph CO2 Light Photoheterotroph Organic molecules Light Chemoautotroph CO2 Inorganic molecules Chemoheterotroph Organic molecules Organic molecules 8. ATP generated by Reaction Photophosphorylation An electron, liberated from chlorophyll by light, is passed down an electron transport chain. Oxidative phosphorylation Cytochrome c passes two electrons to cytochrome a. Substrate-level phosphorylation Phosphoenolpyruvic acid Pyruvic acid 9. Oxidation-reduction: A coupled reaction in which one substance is oxidized and one is reduced. a. The final electron acceptor in aerobic respiration is molecular oxygen; in anaerobic respiration, it is another inorganic molecule. b. The final electron acceptor in respiration is usually an inorganic molecule; in fermentation, it is an organic molecule. c. In cyclic photophosphorylation, electrons are returned to chlorophyll. In noncyclic photophosphorylation, chlorophyll receives electrons from hydrogen atoms. 10. The pentose phosphate pathway produces pentoses for the synthesis of nucleic acids, precursors for the synthesis of glucose by photosynthesizing organisms, precursors in the synthesis of certain amino acids, and NADPH. 11. a. is the Calvin-Benson cycle, b. is glycolysis, and c. is the Krebs cycle. 12. Glycerol is catabolized by pathway b. as dihydroxyacetone phosphate. Fatty acids by pathway c. as acetyl groups. 13. In pathway (c) at a-ketoglutaric acid. 14. Glyceraldehyde-3-phosphate from the Calvin-Benson cycle enters glycolysis. Pyruvic acid from glycolysis is decarboxylated to produce acetyl for the Krebs cycle. 15. In (a), between glucose and glyceraldehyde-3-phosphate. 16. The conversion of pyruvic acid to acetyl, isocitric acid to a-ketoglutaric acid, and a-ketoglutaric acid to succinyl~CoA. 17. By pathway (c) as acetyl groups. 18. Uses Produces Calvin-Benson cycle 6 NADPH Glycolysis 2 NADH Pyruvic acid ----> acetyl 1 NADH Isocitric acid ----> a-ketoglutaric acid 1 NADH a-ketoglutaric acid ----> Succinyl~CoA 1 NADH Succinic acid ----> Fumaric acid 1 FADH2 Malic acid ----> Oxaloacetic acid 1 NADH 19. Dihydroxyacetone phosphate; acetyl; oxaloacetic acid; a-ketoglutaric acid. 20. Reactions requiring ATP are coupled with reactions that produce ATP. 21. The reaction rate will increase until the enzymes are saturated. Multiple Choice Answers are in boldface. 1. Which substance in the following reaction is being reduced? a. Acetaldehyde b. NADH c. Ethanol d. NAD1 2. Which of these reactions produces the most molecules of ATP during aerobic metabolism? a. glucose => glucose-6-P b. phosphoenolpyruvic acid => pyruvic acid c. glucose => pyruvic acid d. acetyl => CO2 1 H2O e. succinic acid => fumaric acid 3. Which of the following processes does not generate ATP? a. photophosphorylation b. the Calvin-Benson cycle c. oxidative phosphorylation d. substrate-level phosphorylation e. none of the above 4. Which of the following compounds has the greatest amount of energy for a cell? a. CO2 b. ATP c. glucose d. O2 e. lactic acid 5. Which of the following is the best definition of the Krebs cycle? a. The oxidation of pyruvic acid. b. The way cells produce CO2 c. A series of chemical reactions in which NADH is produced from the oxidation of pyruvic acid. d. A method of producing ATP by phosphorylating ADP. e. A series of chemical reactions in which ATP is produced from the oxidation of pyruvic acid. 6. Which of the following is the best definition of respiration? a. A sequence of carrier molecules with O2 as the final electron acceptor. b. A sequence of carrier molecules with an inorganic molecule as the final electron acceptor. c. A method of generating ATP. d. The complete oxidation of glucose to CO2 and H2O. e. A series of reactions in which pyruvic acid is oxidized to CO2 and H2O. 7. Which culture produces the most lactic acid? a. E. coli growing in glucose broth at 35°C with O2 for 5 days. b. E. coli growing in glucose broth at 35°C without O2 for 5 days. c. Both a and b. d. Neither a nor b. 8. Which culture produces the most ATP? a. E. coli growing in glucose broth at 35°C with O2 for 5 days. b. E. coli growing in glucose broth at 35°C without O2 for 5 days. c. Both a and b. d. Neither a nor b. 9. Which culture uses NAD1? a. E. coli growing in glucose broth at 35°C with O2 for 5 days. b. E. coli growing in glucose broth at 35°C without O2 for 5 days. c. Both a and b. d. Neither a nor b. 10. Which culture uses the most glucose? a. E. coli growing in glucose broth at 35°C with O2 for 5 days. b. E. coli growing in glucose broth at 35°C without O2 for 5 days. c. Both a and b. d. Neither a nor b. Critical Thinking 1. Answers should include the following: Protons are pumped from one side of the membrane to the other; transfer of protons back across the membrane generates ATP. a. Outer portion is acidic and b. Has a positive electric charge. c. Energy-conserving sites are the three loci where H+ molecules are pumped out. d. Kinetic energy is realized at ATP synthetase when protons cross the membrane. 2. Streptococcus is only capable of fermentation, which yields two molecules of ATP for each molecule of glucose consumed. Most of the energy that cells obtain from catabolism is from respiration. 3. NADH must be reoxidized so it can pick up hydrogen atoms again. NADH is usually reoxidized in respiration when electrons are transferred to the electron transport chain carrier molecules. In fermentation, NADH is reoxidized when electrons are transferred to pyruvic acid (reducing pyruvic acid to lactic acid, for example). 4. Once the enzyme combines with substrate, its affinity for substrate increases. The reaction proceeds at a linear rate (black). When the enzyme becomes saturated with competitive inhibitor, the reaction will completely stop (red). 5. Carbohydrate catabolism: a. Oxidation of glucose (glycolysis, Krebs, aerobic ETC). b. CO2 fixation and oxidation of glucose. c. CO2 fixation and oxidation of glucose (anaerobic ETC). Energy production. All three use chemiosmotic mechanisms. In addition to oxidative phosphorylation, Spirulina and Ectothiorhodospira use photophosphorylation. 6. Glucose = 38 ATP. Butterfat = 217 ATP. Glycerol goes into glycolysis to produce 1 ATP. Six acetyl groups are produced from each of the 12-carbon chains by beta-oxidation. Each acetyl can be used to produce 12 ATPs in the Krebs cycle. 7. Two electrons removed from As3+ are picked up by NAD+ for use in the electron transport chain. Thiobacillus could be used to remove arsenic from industrial waste water and groundwater. Clinical Applications 1. X factor is necessary to synthesize cytochromes. V factor is used as an electron acceptor in oxidation reactions. 2. The drug ddC is missing an O atom on C3 so it cannot be joined to the phosphate group of another nucleotide. 3. Live bacteria or whole bacterial cells are not being injected. Enzymes are specific for their substrates, so streptokinase will react only with its substrate, fibrin.

Microbiology Chapter 6 1. Binary fusion is the normal reproductive method of bacteria is binary fusion in which a single cell divides into two identical cells. 2. Bacteria growth curve is a graph indicating the growth of a bacterial population over time. The bacterial growth curve: Lag phase Exponential phase Stationary phase Death phase. Shown below is a graph. 3. CHONPS letters indicate Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, Sulfur. Living organisms composed of CHONPS. Salt water environment is inside and/or outside. 4. Most bacteria grow best at pH 6.5 and 7.5. 5. Why can high concentrations of salt or sugar be used to preserve food? Sweeteners are sugars, or sugar substitutes, used in food preparation to give foods a sweet taste and to enhance the flavor of other ingredients. 6. Catalase is a enzyme that catalyzes the breakdown of hydrogen peroxide to water and oxygen. Hydrogen peroxide can be used to destroy some bacteria. Peroxidase is a enzyme that breaksdown hydrogen peroxide. Superoxide free radical is a toxic form of oxygen O2 formed during aerobic respiration. Superoxide dismutase is a enzyme that destroys superoxide free radicals. 7. Anaerobic incubator Culture Techniques Incubator for growth of some microorganisms. 8. Clostridium 8 Methods of bacteria counting A standard plate count reflects the number of viable microbes. A plate count maybe done by either the pour plate method or the spread plate method. The aerobic plate count is intended to indicate the level of microorganism in product. The conventional plate count method for examining frozen, chilled, precooked, or prepared foods.

Microbiology Answers Chapter 7 Review 1. a. Lysis. b. Altered permeability and leakage of cell contents. c. Destruction of enzymes and structural proteins such as those in the plasma membrane. d. Interference with protein synthesis and cell division. 2. Autoclave. Due to the high specific heat of water, moist heat is readily transferred to cells. 3. Most organisms that cause disease or rapid spoilage of food are destroyed by pasteurization. 4. Variables that affect determination of the thermal death point are a. The innate heat resistance of the strain of bacteria. b. The past history of the culture, whether it was freeze-dried, wetted, etc. c. The clumping of the cells during the test. d. The amount of water present. e. The organic matter present. f. Media and incubation temperature used to determine viability of the culture after heating. 5. a. Ionizing radiation can break DNA directly. However, due to the high water content of cells, the formation of free radicals (H· and OH·) that break DNA strands is likely to occur. b. Ultraviolet radiation damages DNA by the formation of thymine dimers. 6. Microorganisms tend to die at a constant rate over a period of time. The constant rate is indicated by the straight line after exposure to the bactericidal compound. 7. Sterilization Method Temp. Time Type Preferred Use Mechanism of Action Autoclaving 121°C 15 min Moist Media, equipment Protein denaturation Hot air 170°C 2 hr Dry Glassware Oxidation Pasteurization 72°C 15 sec Moist Milk, alcoholic drinks Protein denaturation 8. All three processes kill microorganisms; however, as moisture and/or temperatures are increased, less time is required to achieve the same result. 9. Salts and sugars create a hypertonic environment. Salts and sugars (as preservatives) do not directly affect cell structures or metabolism; instead, they alter the osmotic pressure. Jams and jellies are preserved with sugar; meats are usually preserved with salt. Molds are more capable of growth in high osmotic pressure than bacteria. 10. 1. Acts rapidly. 2. Attacks all, or a wide range of, microbes. 3. Is able to penetrate. 4. Readily mixes with water. 5. Is not hampered by organic matter. 6. Stable. 7. Does not stain or corrode. 8. Nontoxic. 9. Pleasant odor. 10. Economical. 11. Safe to transport. 11. Method of Action Standard Use a. Disrupts plasma membrane Skin surfaces b. Inhibits protein function Antiseptic c. Oxidation Disinfect water d. Denatures proteins, destroys lipids Skin surfaces e. Oligodynamic AgNO3 to prevent gonococcal eye infections f. Inactivation of proteins Chemical sterilizer g. Denatures proteins Chemical sterilizer h. Oxidation Antiseptic 12. Disinfectant B is preferable because it can be diluted more and still be effective. 13. Quaternary ammonium compounds are most effective against gram-positive bacteria. Gram-negative bacteria that were stuck in cracks or around the drain of the tub would not have been washed away when the tub was cleaned. These gram-negative bacteria could survive the washing procedure. Some pseudomonads can grow on quats that have accumulated. Multiple-Choice Answers are in boldface. 1. Which of the following does not kill endospores? a. autoclaving b. incineration c. hot-air sterilization d. pasteurization e. none of the above 2. Which of the following is most effective for sterilizing mattresses and plastic Petri dishes? a. chlorine b. ethylene oxide c. glutaraldehyde d. autoclaving e. nonionizing radiation 3. Which of these disinfectants does not act by disrupting the plasma membrane? a. phenolics b. phenol c. quaternary ammonium compounds d. halogens e. biguanides 4. Which of the following cannot be used to sterilize a heat labile solution stored in a plastic container? a. gamma radiation b. ethylene oxide c. nonionizing radiation d. autoclaving e. short-wavelength radiation 5. Which of the following is not true about quaternary ammonium compounds? a. bactericidal against gram-positive bacteria b. sporicidal c. amoebicidal d. fungicidal e. kills enveloped viruses 6. A classmate is trying to determine how a disinfectant might kill cells. You observed that when he spilled the disinfectant in your reduced litmus milk, the litmus turned blue again. You suggest to your classmate that a. the disinfectant might inhibit cell wall synthesis. b. the disinfectant might oxidize molecules. c. the disinfectant might inhibit protein synthesis. d. the disinfectant might denature proteins. e. he take his work away from yours. 7. Which of the following is most likely to be bactericidal? a. membrane filtration b. ionizing radiation c. freeze-drying d. deep-freezing e. all of the above 8. Which of the following is used to control microbial growth in foods? a. organic acids b. alcohols c. aldehydes d. heavy metals e. all of the above 9. Which disinfectant is the most effective? a. A b. B c. C d. D 10. Which disinfectant(s) is (are) bactericidal? a. A, B, C, and D b. A, C, and D c. A and B d. C and D e. B only Critical Thinking 1. a. Z. b. No. A culture medium would have to be inoculated from the zone of inhibition to determine the presence of viable bacteria. 2. a. Acid-resistant cell wall. b. Metabolizes many organic molecules. c. Endospores. 3. a. Disinfectant B diluted with distilled water. b. Can't tell; test was done on Salmonella. 4. Bactericidal effects of microwave radiation are due to heat. Clinical Applications 1. a. Hot water does not achieve sterilization, and b. There were no check valves to prevent backflow. Also, adapter and other invasive items should be disposable whenever possible. 2. Iodophors are approved for antiseptic uses and not for disinfection. 3. Serratia from the environment entered the jar from the air, wash water, or hands. The effectiveness of the quat was reduced by the cotton, and quat-resistant Serratia were able to survive. The bacteria were introduced into the methylprednisolone by swabbing the top of the vial. The disinfectant needs to be changed. Microbiology Answers Chapter 8 Review 1. DNA consists of a strand of alternating sugars (deoxyribose) and phosphate groups with a nitrogenous base attached to each sugar. The bases are adenine, thymine, cytosine, and guanine. DNA exists in a cell as two strands twisted together to form a double helix. The two strands are held together by hydrogen bonds between their nitrogenous bases. The bases are paired in a specific, complementary way: A-T and C-G. The information held in the sequence of nucleotides in DNA is the basis for synthesis of RNA and proteins in a cell. 2. a. b. DNA polymerases synthesize a complementary strand of DNA from a DNA template. RNA polymerase starts each fragment of the lagging strand with an RNA primer. c. Each new double-stranded DNA molecule contains one original strand and one new strand. 3. a. ATATTACTTTGCATGGACT. b. met-lys-arg-thr (end). c. TATAATGAAACGTACCTGA. d. No change. e. Cysteine substituted for arginine. f. Proline substituted for threonine (missense mutation). g. Frameshift mutation. h. Adjacent thymines might polymerize. i. ACT. 4. One end of the mRNA molecule becomes associated with a ribosome. Ribosomes are composed of rRNA and protein. The anticodon of a tRNA with its activated amino acid pairs with the mRNA codon at the ribosome. 5. A mutant is isolated by direct selection because it grows on a particular medium. The colonies on an antibiotic-containing medium can be identified as resistant to that antibiotic. A mutant is isolated by indirect selection because it does not grow on a particular medium. Replica plating could be employed to inoculate an antibiotic-containing medium. Colonies that did not grow on this medium can be isolated from the original plate and are antibiotic sensitive. 6. Matching b A mutagen that is incorporated into DNA in place of a normal base. d A mutagen that causes the formation of highly reactive ions. c A mutagen that alters adenine so that it base-pairs with cytosine. a A mutagen that causes insertions. e A mutagen that causes the formation of pyrimidine dimers. 7. The basis for the Ames test is that a mutated cell can revert to a cell that resembles the original, nonmutant cell by undergoing another mutation. The reversion rate of histidine auxotrophs of Salmonella in the presence of a mutagen will be higher than the spontaneous rate (in the absence of a mutagen). 8. Plasmids are small, self-replicating circles of DNA that are not associated with the chromosome. The F plasmid can be integrated into the chromosome. The F plasmid can be transferred from a donor to a recipient cell in conjugation. When the F plasmid becomes integrated into the chromosome, the cell is called an Hfr. During conjugation between an Hfr and an F cell, the chromosome of the Hfr cell, with its integrated F factor, replicates, and the new copy of the chromosome is transferred to the recipient cell. 9. a. …a repressor protein must be bound tightly to the operator site…it will bind to the repressor so that transcription can occur. b. …called a corepressor, causes the repressor to bind to the operator. Derepression is by removal of the corepressor, C in this case, when the corepressor is needed in the cell. c. None; constitutive enzymes are produced at certain necessary levels regardless of the amount of substrate or end-product. 10. Light repair; dark repair; proofreading by DNA polymerase. 11. a. The genetic makeup of an organism. b. The external manifestations of the genotype. c. Rearrangement of genes to form new combinations; in nature, this usually occurs between members of the same species; in vitro, recombinant DNA is made from genes of different species. 12. CTTTGA. Endospores and pigments offer protection against UV radiation. Additionally, repair mechanisms can remove and replace thymine polymers. 13. a. Culture 1 will remain the same. Culture 2 will be converted to F+ but will have its original genotype. b. The donor and recipient cells' DNA can recombine to form combinations of A+B+C+ and A-B-C-. If the F plasmid also is transferred, the recipient cell may become F+. 14. Semiconservative replication ensures that the offspring cell will have one correct strand of DNA. Any mutations that may have occurred during DNA replication have a greater chance of being correctly repaired. 15. Mutation and recombination provide genetic diversity. Environmental factors select for the survival of organisms through natural selection. Genetic diversity is necessary for the survival of some organisms through the processes of natural selection. Organisms that survive may undergo further genetic change, resulting in the evolution of the species. Multiple-Choice Answers are in boldface. 1. The transfer of DNA from a donor to a recipient cell by a bacteriophage. a. conjugation b. transcription c. transduction d. transformation e. translation 2. The transfer of DNA from a donor to a recipient as naked DNA in solution. a. conjugation b. transcription c. transduction d. transformation e. translation 3. Feedback inhibition differs from repression because feedback inhibition a. is less precise. b. is slower acting. c. stops the action of preexisting enzymes. d. stops the synthesis of new enzymes e. All of the above. 4. Bacteria can acquire antibiotic resistance by a. mutation. b. insertion of transposons. c. acquiring plasmids. d. All of the above. e. None of the above. 5. Suppose you inoculate three flasks of minimal salts broth with E. coli. Flask A contains glucose. Flask B contains glucose and lactose. Flask C contains lactose. After a few hours of incubation, you test the flasks for the presence of b-galactosidase. Which flask(s) do you predict will have this enzyme? a. A b. B c. C d. A and B e. B and C 6. Plasmids differ from transposons because plasmids a. become inserted into chromosomes. b. are self-replicated outside the chromosome. c. move from chromosome to chromosome. d. carry genes for antibiotic resistance. e. None of the above. Critical Thinking 1. Cancerous cells are growing faster than normal cells. Mutations have a greater effect when a cell is growing because it is synthesizing DNA and enzymes. The probability of a lethal mutation also is increased in rapidly growing cells. 2. The cell does not regulate the rate at which DNA is synthesized, but it regulates the rate at which replication forks on the chromosomes are initiated. The cell initiates multiple forks so that a daughter cell will inherit a complete chromosome plus additional portions from multiple replication forks. Chromosome replication begins during or immediately after division. 3. a. Mercuric ion. b. To detoxify it. c. Detoxifying mercuric ion will allow the cell to live in areas where other organisms may not be able to. Clinical Applications 1. a. Chloroquine interferes with transcription; erythromycin interferes with translation; acyclovir interferes with DNA replication. b. Erythromycin is specific for bacterial ribosomes. c. Acyclovir is effective against viruses because a viral enzyme tries to use it as a DNA nucleotide. d. Chloroquine and acyclovir will have the most effects on the host because they affect eucaryotic DNA. The effects of erythromycin on mitochondrial ribosomes are small for short-term use. e. Chloroquine is used against malaria. Malaria is caused by a eucaryote (protozoan). f. Acyclovir is used against Herpesvirus infections. Erythromycin affects bacterial (70S) ribosomes, not viruses or eucaryotes. 2. The amino acid sequence reflects the RNA (genome) of the virus. Sequence B is the most dissimilar and, therefore, probably not from the dentist. Microbiology Answers Chapter 9 Review 1. Recombinant DNA (rDNA) is DNA that is combined from different sources. In nature, rDNA results from conjugation, transduction, and tranformation. Genetic engineering is the artifical making of rDNA. 2. a. Both are DNA. cDNA is a segment of DNA made by RNA-dependent DNA polymerase. It is not necessarily a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. b. Both are DNA. A restriction fragment is a segment of DNA produced when a restriction endonuclease hydrolyzes DNA. It is not usually a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. c. Both are DNA. A DNA probe is a short, single-stranded piece of DNA. It is not a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. d. Both are enzymes. DNA polymerase synthesizes DNA one nucleotide at a time using a DNA template; DNA ligase joins pieces (strands of nucleotides) together. e. Both are DNA. Recombinant DNA results from joining DNA from two different sources; cDNA results from copying a strand of RNA. 3. a. A desired gene can be spliced into a plasmid and inserted into a cell by transformation. b. A desired gene can be spliced into a viral genome and inserted into a cell by transduction. c. Antibiotic resistance genes are used as markers or labels on plasmids so that the cell containing the plasmid can be found by direct selection on an antibiotic-containing medium. d. A genetically engineered bacterium should be producing a new protein product. Radioactively labeled antibodies against a specific protein can be used to locate the bacterial colony producing the protein. 4. Restriction fragments from one source can be cloned in microbial cells to make a gene library. Synthetic DNA is made in a lab. 5. In protoplast fusion, two wall-less cells fuse together to combine their DNA. A variety of genotypes can result from this process. In b, c, and d, specific genes are inserted directly into the cell. 6. BamH I, EcoR I, and Hind III make sticky ends. Fragments of DNA produced with the same restriction enzyme will spontaneously anneal to each other at their sticky ends. 7. The gene can be spliced into a plasmid and inserted into a bacterial cell. As the cell grows, the number of plasmids will increase. The polymerase chain reaction can make copies of a gene using DNA polymerase in vitro. 8. In a eucaryotic cell, RNA polymerase copies DNA and removes the introns, leaving the exons in the mRNA. cDNA can be made from the mRNA by reverse transcriptase. 9. Large colonies are ampicillin-resistant because they are growing. Smaller ampicillin-sensitive colonies may appear later after the antibiotic has been degraded by the ampicillin-resistant bacteria. White colonies have the new gene. 10. See Tables 9.1 and 9.2. 11. You probably used a few plant cells in a Petri plate for your experiment. How will you select the plant cells that actually have the new Ti plasmid? You can grow these cells on plant-cell culture media with tetracycline. Only the cells with the new plasmid will grow. Multiple-Choice Answers are in boldface. 1. Restriction enzymes were first discovered with the observation that a. DNA is restricted to the nucleus. b. phage DNA is destroyed in a host cell. c. foreign DNA is kept out of a cell. d. foreign DNA is restricted to the cytoplasm. e. All of the above. 2. The DNA probe, GGCTTA, will hybridize with DNA containing a. CCGUUA. b. CCGAAT. c. GGCTTA. d. GGCUUA e. GGCAAU. 3. The basic steps of genetic engineering are listed below. What is the fourth step? a. transformation b. ligation c. plasmid cleavage d. restriction-enzyme digestion of gene e. isolation of gene 4. The following steps are used to make cDNA. What is the second step? a. reverse transcription b. RNA processing to remove introns c. transcription d. translation 5. If you put a gene in a virus, the next step in genetic engineering would be a. insertion of a plasmid. b. transformation. c. transduction. d. PCR. e. Southern blotting. Critical Thinking 1. EcoR I: 2 fragments Hind III: 2 fragments Both enzymes: 4 fragments The smallest fragment containing the tet gene is cut at one end by Hind III and at the other end by EcoR I. 2. Isolate cDNA (or synthesize DNA) for the desired gene from HIV. Insert the HIV gene into vaccinia virus's DNA. Infect the host cells with vaccinia virus. As the virus reproduces, it will cause production of the HIV protein. 3. "Normal" DNA polymerase is denatured by the heating step, so a technician would have to add DNA polymerase to the reaction vessel every 2 minutes. DNA polymerase from Thermus is not denatured by 90°C, so fresh DNA polymerase does not have to be added every 2 minutes. Clinical Applications 1. Sample 2 was positive for V. cholerae because the probe paired with DNA in lane 2. V. cholerae ingested by oysters is a source of disease for humans. The PCR doesn't require isolation and additional incubation for test results. 2. The vector and new gene fragments appear in the fifth lane; therefore, transformation did occur. Microbiology Chapter 10 Answers Review 1. Taxonomy is the science of classifying organisms to establish the relatedness between groups of organisms. 2. Living organisms cannot be grouped into two groups. For example, plant and animal is not acceptable because if fungi are grouped with plants, the definition of plants can't include cellulose and photosynthesis. If fungi are grouped with animals, the definition of animals can't include no cell wall and ingestive. The goal is to look for a "natural" scheme; that is, what criteria can be used to characterize all organisms. 3. Monera or Procaryotae: All procaryotic organisms. Protista: Unicellular (or multicellular without tissue organization) eucaryotic organisms; usually flagellated at some time. Fungi: Unicellular or multicellular organisms that absorb organic nutrients; noncellulose cell walls; lack flagella. Plantae: Multicellular eucaryotes with tissue formation; cellulose cell walls; generally photosynthetic. Animalia: Multicellular eucaryotes with tissue formation; develop from an embryo (gastrula); lacking cell walls; ingest organic nutrients through a mouth of some kind. 4. The three distinct chemical types of cells (see Table 10.2). In a five-kingdom system, four of the kingdoms have eucaryotic cells and the fifth is based on procaryotic structure alone. 5. a. Both are procaryotic. They differ in composition of their cell walls, plasma membranes, and rRNAs. b. Both are bound by ester-linked plasma membranes. Eucarya have membrane-bound organelles. c. Both use methionine as the start signal. Eucarya have membrane-bound organelles and ester-linked membranes. 6. Binomial nomenclature is the system of assigning a genus and specific epithet to each organism. 7. Common names are not specific and can be misleading. According to the rules of scientific nomenclature, each organism has only one binomial. 8. The genus name must be written out so the reader knows what organism is being discussed, since the abbreviation for both of these species is E. coli. 9. Kingdom, division (phylum), class, order, family, genus, species. 10. A species is a group of closely related organisms having limited geographical distribution that interbreeds but does not breed with other species. Species can be distinguished morphologically. Because of the distinct differences between eucaryotic organisms and bacteria, a bacterial species is defined as a population of cells with similar characteristics. A viral species is a population of viruses with similar characteristics that occupies a particular ecological niche. 11. (See Table 10.4) Used primarily for identification: morphological characteristics differential staining biochemical tests serology phage typing Used primarily for taxonomic classification: amino acid sequencing fatty acid profiles flow cytometry DNA base composition DNA fingerprinting rRNA sequencing PCR nucleic acid hybridization Data obtained from laboratory tests employing any (or all) of these thirteen techniques can be assimilated using numerical taxonomy to provide information on classification. 12. Most microorganisms do not contain structures that are readily fossilized, making it difficult to obtain information on the evolution of microorganisms. Recent developments in molecular biology have provided techniques for determining evolutionary relationships amongst bacteria. 13. A and D appear to be most closely related because they have similar G-C moles %. No two are the same species. Multiple Choice Answers are in boldface. 1. Bergey's Manual of Systematic Bacteriology differs from Bergey's Manual of Determinative Bacteriology in that the former a. groups bacteria into 19 species. b. groups bacteria according to shared properties. c. groups bacteria according to pathogenic properties. d. is based on phylogenetic relationships between bacteria. e. All of the above. 2. Bacillus and Lactobacillus are not in the same family. This indicates that which one of the following is not sufficient to assign an organism to a taxon? a. biochemical characteristics b. amino acid sequencing c. phage typing d. serology e. morphological characteristics 3. Which of the following is used to classify organisms into the Kingdom Fungi? a. ability to photosynthesize; possess a cell wall b. unicellular; possess cell wall; procaryotic c. unicellular; lacking cell wall; eucaryotic d. absorptive; possess cell wall; eucaryotic e. ingestive; lacking cell wall; multicellular; procaryotic 4. Which of the following is not true about scientific nomenclature? a. Each name is specific. b. Names vary with geographical location. c. The names are standardized. d. Each name consists of a genus and specific epithet. e. It was first designed by Linnaeus. 5. You could identify an unknown bacterium by all of the following except a. hybridizing a DNA probe from a known bacterium with the unknown's DNA. b. making a fatty acid profile of the unknown. c. specific antiserum agglutinating the unknown. d. ribosomal RNAsequencing. e. percentage of guanine 1cytosine. 6. The wall-less mycoplasmas are considered to be related to gram-positive bacteria. Which of the following would provide the most compelling evidence for this? a. They share common rRNA sequences. b. Some gram-positive bacteria and some mycoplasmas produce catalase. c. Both groups are procaryotic. d. Some gram-positive bacteria and some mycoplasmas have coccus-shaped cells. e. Both groups contain human pathogens. Critical Thinking 1. A and D are most closely related. 2. Based on the nucleic acid composition, Micrococcus and Staphylococcus probably are not related. The current classification scheme uses morphological and biochemical similarities instead of phylogenetic relatedness to assign bacteria to taxa. 3. DNA probe. Labeled DNA will hybridize with homologous DNA indicating (a) identity if the probe is known DNA and the homologous DNA is from an unknown bacterium, or (b) relatedness when the two organisms are known. PCR. The primer used in PCR will hybridize with homologous DNA so unrelated DNA will not be copied. Or, after making copies by PCR, a DNA probe can be used to locate specific DNA. 4. SF = Streptococcus faecalis. SF broth is used to culture Enterococcus faecalis. Clinical Applications 1. The patient had plague. The Yersinia (gram-negative rod) was missed in the first Gram stain. After culturing gram-negative rods, biochemical tests were not conclusive because Yersinia is biochemically inactive. Plague can be transmitted by the respiratory route so the patient's contacts were given prophylactic antibiotic treatment. 2. Incorrect identification could be due to mutation to sucrose+, misreading of the indicator in the sucrose fermentation test, or contamination by a sucrose+ organism. 3. One possible key is shown below. Alternate keys could be made starting with morphology or glucose fermentation. 4.

Chapter 16 Nonspecific Defenses of the Host body's defenses
Nonspecific resistance refers to defenses against any pathogen regardless of species and specific resistance refers to defenses against a specific pathogen.

First line of defense
Intact skin Normal microbiota

Second line of defense
Phagocytic white blood cells
Inflammation
Fever
Antimicrobial substances
Third line of defense
Specialized lymphocytes B cells and T cells
Antibodies

Growing Bacteria in prepared media petri plates 1. Petri dishes should be refrigerated until they will be used and should always be upside down. Keeps condensation from dropping from lid onto the agar. 2. When ready to use let dishes come to room temperature before taking samples about 1 hour. 3. If growing bacteria from hands press firmly against the agar and replace dish in the lid upside down. Let bacteria grow in undisturbed warm location. 4. Collect bacteria with toothpicks or inoculating loop. 5. Bacteria grows in 2 to 3 days.

Microbiology Chapter 21 Answers Review 1. Bacteria usually enter through inapparent openings in the skin. Fungal pathogens (except subcutaneous) often grow on the skin itself. Viral infections of the skin (except warts and herpes simplex) most often gain access to the body through the respiratory tract. 2. Staphylococcus aureus; Streptococcus pyogenes. 3. Disease Etiology Symptoms Treatment Notes Impetigo Staphylococcus aureus Vesicles that rupture and crust over Hexachlorophene May be epidemic Erysipelas Streptococcus pyogenes Thickened red patches, swollen at margins Penicillin May be endogenous 4. Disease Etiological Agent Clinical Symptoms Method of Transmission Acne P. acnes Infected oil glands Direct contact Pimples S. aureus Infected hair follicles Direct contact Warts Papovavirus Benign tumor Direct contact Chickenpox Herpesvirus Vesicular rash Respiratory route Fever blisters Herpesvirus Recurrent "blisters" Direct contact Measles Paramyxovirus Papular rash, Koplik spots Respiratory route Rubella Togavirus Macular rash Respiratory route 5. Both are fungal infections. Sporotrichosis is a subcutaneous mycosis; athlete's foot is a cutaneous mycosis. 6. a. Conjunctivitis is an inflammation of the conjunctiva, and keratitis is an inflammation of the cornea. b. See pp. 574-575. 7. Candidiasis is caused by Candida albicans. The yeast is able to grow when the normal microbiota are suppressed or when the immune system is suppressed. The yeast can be transferred from another person or be transient microflora. White patches in the mouth or bright red areas of the skin and mucous membranes are signs of infection. Antifungal agents such as miconazole are used to treat candidiasis. Systemic infections are treated with oral ketoconazole. 8. The test determines the woman's susceptibility to rubella. If the test is negative, she is susceptible to the disease. If she acquires the disease during pregnancy the fetus could become infected. A susceptible woman should be vaccinated. 9. Symptoms Disease Koplik spots Measles Macular rash Measles Vesicular rash Chickenpox Small, spotted rash German measles "Blisters" Cold sore Corneal ulcer Keratoconjunctivitis 10. The central nervous system can be invaded following keratoconjunctivitis; this results in encephalitis. 11. Attenuated measles, mumps, and rubella viruses. 12. Varicella-zoster virus appears to remain latent in nerve cells following recovery from a childhood infection of chickenpox. Later, the virus may be activated and cause a vesicular rash (shingles) in the area of the nerve. 13. To prevent neonatal gonorrheal ophthalmia. This is caused by N. gonorrhoeae contracted by the newborn during passage through the birth canal. 14. Trachoma. 15. Scabies is an infestation of mites in the skin. It is treated with permethrin insecticide or gamma benzene hexachloride. Multiple Choice Answers are in boldface. Use the following information to answer questions 1 and 2. A 6-year- old girl was taken to the physician for evaluation of a slowly growing bump on the back of her head. The bump was a raised, scaling lesion 4 cm in diameter. Wood's lamp examination of the scalp showed fluorescence. A fungal culture of material from the lesion was positive for a fungus with numerous microconidia. 1. The girl's disease was a. rubella b. candidiasis c. dermatomycosis d. a cold sore e. none of the above. 2. Besides the scalp, this disease can occur on all of the following except a. feet. b. nails. c. the groin. d. subcutaneous tissue. e. none of the above. Use the following information to answer questions 3 and 4. A 12-year-old boy had a fever, rash, headaches, sore throat, and cough. He also had a macular rash on his trunk, face, and arms. A throat culture was negative for Streptococcus pyogenes. 3. The boy most likely had a. streptococcal sore throat. b. measles. c. rubella. d. smallpox. e. none of the above. 4. All of the following are complications of this disease except a. middle ear infections. b. pneumonia. c. birth defects. d. encephalitis. e. none of the above. 5. A patient has conjunctivitis. If you isolated Pseudomonas from the patient's mascara, you would most likely conclude all of the following except that a. the mascara was the source of the infection. b. Pseudomonas is causing the infection. c. Pseudomonas has been growing in the mascara. d. the mascara was contaminated by the manufacturer. e. none of the above. Critical Thinking 1. S. aureus is adapted for surviving on the human skin, which has a high concentration of NaCl. Microorganisms that are not adapted to this hypertonic environment will not be able to tolerate the 7.5% NaCl in mannitol salt agar. 2. Most warts regress spontaneously. Removal of warts is usually for cosmetic reasons. Occasionally warts are painful when they are located where pressure is placed on them (e.g., plantar warts on the sole of the foot). 3. The infections were transmitted by the contact lenses or cosmetics. Cosmetics are inoculated with microbes each time they are used. Some of the microbes grow, resulting in large inoculations of the eyes. Contact lenses can be improperly cleaned (i.e., not using an antiseptic) or contaminated by fingers. 4. The virus had one host-humans. It was not found in soil, water, or nonhuman organisms. Polio and measles meet this criterion. Clinical Applications 1. Pseudomonas aeruginosa. This bacterium is common in soil and is resistant to many antibiotics. 2. Toxic shock syndrome due to growth of Staphylococcus at the injection site. 3. The symptoms of toxic shock syndrome were caused by toxins produced from the secondary infection (S. aureus). Microbiology Chapter 22 Answers Review 1. Meningitis is an infection of the meninges; encephalitis is an infection of the brain itself. 2. Causative Agent Susceptible Population Mode of Transmission Treatment N. meningitidis Children; military recruits Respiratory Penicillin H. influenzae Children Respiratory Rifampin S. pneumoniae Children; elderly Respiratory Penicillin L. monocytogenes Anyone Foodborne Penicillin C. neoformans Immunosuppressed individuals Respiratory Amphotericin B 3. "Hemophilus" refers to the requirement of this genus for growth factors found in blood (X and V factors). "Influenzae" because it was thought to be the causative agent of influenza. 4. The symptoms of tetanus are not due to bacterial growth (infection and inflammation) but to neurotoxin. 5. Salk Sabin Composition Formalin-inactivated viruses Live, attenuated viruses Advantages No reversion to virulence Oral administration Disadvantages Booster dose needed; injected Reversion to virulence 6. a. Vaccination with tetanus toxoid. b. Immunization with antitetanus toxin antibodies. 7. "Cleaned" because C. tetani is found in soil that might contaminate a wound. "Deep puncture" because it is likely to be anaerobic. "No bleeding" because a flow of blood ensures an aerobic environment and some cleansing. 8. Clostridium botulinum. Canned foods. Paralysis. Supportive respiratory care; antitoxin. Anaerobic, non-acidic environment. Diagnosis is made by detecting toxin in foods or patient by inoculating mice with suspect samples. Prevention: use of adequate heat in canning; boiling food before consumption to inactivate toxin. 9. Etiology-Mycobacterium leprae. Transmission-Direct contact. Symptoms-Nodules on the skin; loss of sensation. Treatment-Dapsone and rifampin. Prevention-BCG vaccine. Susceptible-People living in the tropics; genetic predisposition. 10. Etiology-Picornavirus (poliovirus). Transmission-Ingestion of contaminated water. Symptoms-Headache, sore throat, fever, nausea; rarely paralysis. Prevention-Sewage treatment. These vaccinations provide artificially acquired active immunity because they cause the production of antibodies, but they do not prevent or reverse damage to nerves. 11. Etiology-Rhabdovirus. Transmission-Bite of infected animal; inhalation. Reservoirs-Skunks, bats, foxes, raccoons. Symptoms-Muscle spasms, hydrophobia, CNS damage. 12. Postexposure treatment-Passive immunization with antibodies followed by active immunization with HDCV. Preexposure treatment-Active immunization with HDCV. Following exposure to rabies, antibodies are needed immediately to inactivate the virus. Passive immunization provides these antibodies. Active immunization will provide antibodies over a longer period of time, but they are not formed immediately. 13. Disease Etiology Vector Symptoms Treatment Arboviral encephalitis Togaviruses, Arboviruses Mosquitoes (Culex) Headache, fever, coma Immune serum African trypanosomiasis Trypanosoma brucei gambiense,T. b. rhodesiense Tsetse fly Decreased physical activity and mental acuity Suramin; melarsoprol 14. Most antibiotics cannot cross the blood-brain barrier. 15. The causative agent of Creutzfeldt-Jakob disease (CJD) is transmissible. Although there is some evidence for an inherited form of the disease, it has been transmitted by transplants. Similarities with viruses are (1) the prion cannot be cultured by conventional bacteriological techniques and (2) the prion is not readily seen in patients with CJD. Multiple Choice Answers are in boldface. 1. Which of the following is not true? a. Only puncture wounds by rusty nails result in tetanus. b. Rabies is seldom found in rodents (e. g., rats, mice). c. Polio is transmitted by the fecal-oral route. d. Arboviral encephalitis is rather common in the United States. e. None of the above. 2. Which of the following does not have an animal reservoir or vector? a. listeriosis b. cryptococcosis c. Naegleria meningoencephalitis d. rabies e. African trypanosomiasis 3. A 12-year-old girl hospitalized for Guillain-Barré syndrome had a 4-day history of headache, dizziness, fever, sore throat, and weakness of legs. Seizures began 2 weeks later. Bacterial cultures were negative. She died 3 weeks after hospitalization. An autopsy revealed inclusions in brain cells that tested positive in an immunofluoresence test. She probably had a. rabies. b. Creutzfeldt-Jakob disease. c. botulism. d. tetanus. e. leprosy. 4. After receiving a corneal transplant, a woman developed dementia and loss of motor function; she then became comatose and died. Cultures were negative. Serological tests were negative. Autopsy revealed spongiform degeneration of her brain. She most likely had a. rabies. b. Creutzfeldt-Jakob disease. c. botulism. d. tetanus. e. leprosy. 5. The endotoxin is responsible for symptoms caused by which of the following organisms? a. N. meningitidis b. S. pyogenes c. L. monocytogenes d. C. tetani e. C. botulinum 6. The increased incidence of encephalitis in the summer months is due to a. maturation of the viruses. b. increased temperature. c. the presence of adult mosquitoes. d. an increased population of birds. e. an increased population of horses. 7. Produces the highest antibody titer. a. antirabies antibodies b. HDVC 8. Used for passive immunization. a. antirabies antibodies b. HDVC 9. Microscopic examination of cerebrospinal fluid reveals gram-positive rods. a. Cryptococcus b. Hemophilus c. Listeria d. Naegleria e. Neisseria 10. Microscopic examination of cerebrospinal fluid from a person who washes windows reveals ovoid cells. a. Cryptococcus b. Hemophilus c. Listeria d. Naegleria e. Neisseria Critical Thinking 1. The term rusty nail implies that the sharp object has been contaminated with soil and possibly C. tetani. C. tetani can grow in deep puncture wounds, and a nail is capable of producing such a wound. 2. Both diseases are caused by species of Mycobacterium. 3. The only cases of polio in the United States during the last few years have been caused by OPV. Clinical Applications 1. Hemophilus influenzae meningitis, treated with rifampin. 2. Cryptococcus neoformans. Need microscopic observation of the fungus from cerebrospinal fluid or culture. 3. Naegleria fowleri meningitis; treated with amphotericin B, miconazole, and rifampin.

Chapter 5 Fermentations Starting material Ethanol use for wine, grape juice. Micro organism Saccharomyces cerevisioevar ellipsoideus Lactic acid cheese, yogurt, milk. Lacto bacillus Streptococcus bacteria Aerobic respiration Aerobic Anaerobic. Growth media blood agar for growing bacteria. Chapter 6 Microbial Growth number of cells Microbes are growing, increasing in numbers. Grow into colonies, groups of cells large enough to be seen without using a Microscope. Growth temperature. 80 degrees 100 degrees. Chapter 13 Viruses Viroids Prions. Measles Virus, Viral infection fatal. 1. produced by fungi conidiospore or conidium a unicellular or multicellular spore is not enclosed in a sac. Conidia are produced in a chain at the end of a condiophore. A ascospore results from the fusion of the nuclei of two cells that can be either morphologically similar or dissimilar. These spores are produced in a sac like structure called a ascus. 5. When algal partner is cultured separately in vitro about 1% of the carbohydrates produced during photosynthesis are released into the culture medium. 6. Lichens Algae

Chapter 16 Microbiology Answers Review 1. a. The ability of the human body to ward off diseases. b. The lack of resistance to an infectious disease. c. Host defenses that tend to protect the body from any kind of pathogen. 2. Mechanical Chemical Skin Dry, packed cells Sebum Eyes Tears Lysozyme Digestive tract Movement out HCl Respiratory tract Ciliary escalator Urinary tract Movement out Genital tract Movement out Acidic in female 3. See Table 16.1. 4. Phagocytosis is the ingestion of a microorganism or any foreign particulate matter by a cell. 5. Granulocytes have granules in the cytoplasm. Among the granulocytes, neutrophils have the most prominent phagocytic activity. Monocytes are agranulocytes (without granules) that develop into macrophages. When an infection occurs, granulocytes migrate to the infected area. Monocytes follow the granulocytes to the infected tissue. During migration, monocytes enlarge and develop into actively phagocytic cells called macrophages. Macrophages phagocytize dead or dying bacteria. 6. Phagocytic cells that migrate to the infected area are called wandering macrophages. Fixed macrophages remain in certain tissues and organs. 7. Refer to Figures 16.7 and 16.8. 8. Inflammation is the body's response to tissue damage. The characteristic symptoms of inflammation are redness, pain, heat, and swelling. 9. The functions of inflammation are: (1) To destroy the injurious agent, if possible, and to remove it and its by-products from the body; (2) If destruction is not possible, to confine or wall off the injurious agent and its by-products by forming an abscess; (3) To repair or replace tissues damaged by the injurious agent or its by-products. 10. Leukocytic pyrogen, released from phagocytic granulocytes, has the ability to raise body temperature. The higher temperature is believed to inhibit the growth of some microorganisms. The higher temperature speeds up body reactions and may help body tissues to repair themselves more quickly. 11. The chill is an indication that body temperature is rising. Shivering and cold skin are mechanisms for increasing internal temperature. Crisis indicates body temperature is falling. The skin becomes warm as circulation is returned to it when the body attempts to dissipate extra heat. 12. Complement is a group of proteins found in normal blood serum. See Figures 16.10 and 16.11. 13. Activation of complement can result in immune adherence and phagocytosis, local inflammation, and cell lysis. 14. Endotoxin binds C3b, which activates C5-C9 to cause cell lysis. This can result in free cell wall fragments, which bind more C3b, resulting in C5-C9 damage to host cell membranes. 15. Interferons are antiviral proteins produced by infected cells in response to viral infections. Alpha-IFN and b-IFN induce uninfected cells to produce antiviral proteins. Gamma-IFN is produced by lymphocytes and activates neutrophils to kill bacteria. Chapter Sixteen Answers are in boldface. 1. Legionella uses C3b receptors to enter monocytes. This a. prevents phagocytosis. b. degrades complement. c. inactivates complement. d. prevents inflammation. e. prevents cytolysis. 2. Chlamydia can prevent the formation of phagolysosomes, and therefore Chlamydia can a. avoid being phagocytized. b. avoid destruction by complement. c. prevent adherence. d. avoid being digested. e. None of the above. 3. If the following are placed in the order of occurrence, which would be the third step? a. emigration b. digestion c. formation of a phagosome d. formation of a phagolysosome e. margination 4. If the following are placed in the order of occurrence, which would be the third step? a. activation of C5-C9 b. cell lysis c. antigen-antibody reaction d. activation of C3 e. activation of C2-C4 5. A human host can prevent a pathogen from getting enough iron by a. reducing dietary intake of iron. b. binding iron with transferrin. c. binding iron with hemoglobin. d. excreting excess iron. e. binding iron with siderophores. 6. A decrease in the production of C3 would result in a. increased susceptibility to infection. b. increased numbers of white blood cells. c. increased phagocytosis. d. activation of C5-C9. e. None of the above. 7. In 1884, Elie Metchnikoff observed blood cells collected around a splinter inserted in a sea star embryo. This was the discovery of a. blood cells. b. sea stars. c. phagocytosis. d. immunity. e. None of the above. 8. Helicobacter pylori uses the enzyme urease to counteract a chemical defense in the human organ in which it lives. This chemical defense is a. lysozyme. b. hydrochloric acid. c. superoxide radicals. d. sebum. e. complement. 9. Which of the following statements about a-IFN is not true? a. It interferes with viral replication. b. It is host specific. c. It is released by fibroflasts. d. It is virus-specific. e. It is released by lymphocytes. 10. Which of the following does not stimulate phagocytes? a. cytokines b. g-IFN c. C3b d. lipid A e. histamine Critical Thinking 1. Transferrin binds available iron so bacteria can't have it to grow. A bacterium might respond with increased siderophores to take up iron. 2. The inflammatory response is usually a beneficial response. Exceptions to this are hypersensitivities and autoimmune diseases, which are discussed in Chapter 18. Each of the drugs has side effects; while reducing inflammation, another undesirable condition might result. 3. Organism How does this strategy avoid destruction by the complement? Disease Group A streptococci No C5-C9 Streptococcal sore throat Hemophilus influenzae type b Hides LPS, which can activate C Meningitis Pseudomonas aeruginosa Binds C in solution instead of on cell surface Septicemia; pyelonephritis Trypanosoma cruzi C5-C9 doesn't get activated Chagas' disease 4. Microorganisms Effect Disease Influenzavirus Kills host cell Influenza M. tuberculosis Prevents digestion in phagocytes Tuberculosis T. gondii Prevents digestion in phagocytes Toxoplasmosis Trichophyton Digests keratin Athlete's foot T. cruzi Prevents digestion in phagocytes Chagas' disease Clinical Applications 1. Kinins cause vasodilation and increased permeability of blood vessels. Symptoms should include increased secretions from the nose and eyes. Rhinoviruses cause the common cold. 2. The proportions of white blood cells may change during diseases. The results of a differential count can be used to diagnose diseases. A patient with mononucleosis will have an increased number of monocytes. Neutropenia: decreased neutrophils. Eosinophilia: increased eosinophils. 3. Phagocytosis is inhibited. 4. Neutrophils will not phagocytize and they will die prematurely. Chapter 17 Microbiology Answers Review 1. The ability to produce antibodies against microorganisms and their toxins provides a type of resistance called immunity. 2. a. Nonspecific defenses are designed to protect you against any kind of microorganism. Immunity or specific resistance involves the production of antibodies. Antibodies are directed against specific microorganisms. b. Humoral immunity is due to antibodies (and B cells). Cell-mediated immunity is due to T cells. c. Active immunity refers to antibodies produced by the individual who carries them. Passive immunity refers to antibodies produced by another source and then transferred to the individual who needs the antibodies. d. Acquired immunity is the resistance to infection obtained during the life of the individual. Acquired immunity results from the production of antibodies. Innate resistance refers to the resistance of species or individuals to certain diseases that is not dependent on antigen-specific immunity such as antibodies. e. Natural immunity is acquired naturally, i.e., from mother to newborn, or following an infection. Artificial immunity is acquired from medical treatment, i.e., by injection of antibodies or by vaccination. f. T-dependent antigens: Certain antigens must combine with self-antigens to be recognized by TH cells and then by B cells. T-independent antigens can elicit an antibody response without T cells. g. T cells can be classified by their surface antigens: TH cells possess the CD4 antigen; TC and TS cells have the CD8 antigen. 3. a. Artificially acquired active immunity. b. Naturally acquired active immunity. c. Naturally acquired passive immunity. d. Artificially acquired passive immunity. 4. An antigen is a chemical substance that causes the body to produce specific antibodies and can combine with these antibodies. A hapten is a low-molecular-weight substance that is not antigenic unless it is attached to a carrier molecule. Once an antibody has been formed against the hapten, the hapten alone will react with the antibodies independently of its carrier. 5. An antibody is a protein produced by the body in response to the presence of an antigen; it is capable of combining specifically with that antigen. Antibodies are proteins and usually consist of four polypeptide chains. Two of the chains are identical and are called heavy (H) chains. The other two chains are identical to each other but are of lower molecular weight and are called light (L) chains. The variable portions of the H and L chains are where antigen binding occurs. The variable portion is different for each kind of antibody. The remaining constant portions of each chain are identical for all of the antibodies in one class of antibody. Refer to Figure 17.5 for the structure of IgG antibodies. 6. Each person has a population of B cells with receptors for different antigens. When the appropriate antigen contacts the antigen receptor on a B cell, the cell proliferates to produce a clone of cells. Plasma cells in this clone produce antibodies specific to the antigen that caused their formation. 7. See Figures 17.7, 17.12, 17.13, 17.15, 17.17, and 17.18. 8. Cytotoxic T cells (TC) destroy target cells upon contact. Delayed hypersensitivity T cells (TD)produce lymphokines. Helper T cells (TH) interact with an antigen to "present" it to a B cell for antibody formation. Suppressor T cells (TS) inhibit the conversion of B cells into plasma cells. Lymphokines cause an inflammatory response. An example of a cytokine is macrophage chemotactic factor, which attracts macrophages to the infection site. See Table 17.2 for functions of other cytokines. 9. a. Area a shows the primary response to the antigen. Area b shows the anamnestic response, in which the antibody titer is greater and remains high longer than in the primary response. The booster dose stimulated the memory cells to respond to the antigen. 10. Neutralize toxins, inactivate viruses, fix complement to initiate cytolysis. 11. Surface recognition sites for antigen peptides and MHC proteins. 12. NK cells lyse target cells (usually tumor cells and virus-infected cells) on contact. 13. Both would prevent attachment of the pathogen; (a) interferes with the attachment site on the pathogen and (b) interferes with the pathogen's receptor site. 14. See Figure 17.10. 15. The person recovered because s/he produced antibodies against the pathogen. The memory response will continue to protect the person against that pathogen. 16. Human gamma globulin is the fraction of human serum in which antibodies are found. If antibodies against hepatitis are in the gamma globulin, this would be artificially acquired passive immunity. Multiple Choice Answers are in boldface. 1. The type of protection provided by the injection of a toxoid. a. innate resistance b. naturally acquired active immunity c. naturally acquired passive immunity d. artificially acquired active immunity e. artificially acquired passive immunity 2. The type of protection provided by the injection of an antitoxin. a. innate resistance b. naturally acquired active immunity c. naturally acquired passive immunity d. artificially acquired active immunity e. artificially acquired passive immunity 3. The type of protection resulting from recovery from an infection. a. innate resistance b. naturally acquired active immunity c. naturally acquired passive immunity d. artificially acquired active immunity e. artificially acquired passive immunity 4. Antibodies that protect the fetus and newborn. a. IgA b. IdD c. IgE d. IgG e. IgM 5. The first antibodies synthesized; especially effective against microorganisms. a. IgA b. IdD c. IgE d. IgG e. IgM 6. Antibodies that are bound to mast cells and involved in allergic reactions. a. IgA b. IdD c. IgE d. IgG e. IgM 7. Put the following in the correct sequence to elicit an antibody response: 1. TH cell recognizes B cell; 2. APC phagocytizes antigen; 3. antigen digest goes to surface of APC; 4. TH recognizes antigen digest and MHC; 5. B cell recognizes antigen. a. 1, 2, 3, 4, 5 b. 5, 4, 3, 2, 1 c. 3, 4, 5, 1, 2 d. 2, 3, 4, 1, 5 e. 4, 5, 3, 1, 2 8. Responsible for the differentiation of T and NK cells. a. antigen b. hapten c. IL-1 d. IL-2 e. perforin 9. Activates a T cell to bind IL-2. a. antigen b. hapten c. IL-1 d. IL-2 e. perforin 10. Patients with Chediak-Higashi syndrome suffer from various types of cancer. These patients are most likely lacking which of the following: a. TD cells b. TH cells c. B cells d. NK cells e. TS cells Critical Thinking 1. TC cells secrete TNFg-IFN, which diffuse through liver cells and stimulate these cells to produce antiviral proteins. 2. a. IL-2 stimulates proliferation and differentiation of T cells and natural killer (NK) cells. NK cells and some T cells are effective against cancer cells. b. IL-2 stimulates the immune response, which increases the unwanted immune response as well. 3. Having had an M. tuberculosis infection and recovered (naturally acquired active immunity); vaccination with BCG (artificially acquired passive immunity). Clinical Applications 1. Antibiotics and immunity can cause gram-negative cells to lyse, releasing cell wall fragments. This exposes the body to more endotoxin. The woman's life-threatening condition was due to endotoxin shock. Monoclonal antibodies removed the cell walls. 2. Increased susceptibility to infection due to decreased antibody formation. 3. He could not produce the secretory component of IgA. 4. The mechanism is called antibody enhancement. Immune complexes of antibodies and viruses attach to cells, facilitating viral penetration. Chapter 18 Microbiology Answers Review 1. a. Whole-agent. Live, avirulent virus that can cause the disease if it mutates back to its virulent state. b. Whole-agent; (heat-) killed bacteria. c. Subunit; (heat- or formalin-) inactivated toxin. d. Subunit e. Subunit f. Conjugated g. Nucleic acid 2. If excess antibody is present, an antigen will combine with several antibody molecules. Excess antigen will result in an antibody combining with several antigens. Refer to Figure 18.2. 3. Particulate antigens react in agglutination reactions. The antigens can be cells or soluble antigens bound to synthetic particles. Soluble antigens take part in precipitation reactions. 4. a. Some viruses are able to agglutinate red blood cells. This is used to detect the presence of large numbers of virions capable of causing hemagglutination (e.g., influenza virus). b. Antibodies produced against viruses that are capable of agglutinating red blood cells will inhibit the agglutination. Hemagglutination inhibition can be used to detect the presence of antibodies against these viruses. c. This is a procedure to detect antibodies that react with soluble antigens by first attaching the antigens to insoluble latex spheres. This procedure may be used to detect the presence of antibodies that develop during certain mycotic or helminthic infections. 5. See Figure 18.10. 6. a. Direct test (see Figure 18.10a) b. Indirect test (see Figure 18.10b) 7. An indirect ELISA test is used to detect the presence of antibodies. A known antigen is fixed to a small well, and the patient's serum is added. Patient's antibodies will react with the antigen in the well. Antihuman immunoglobulins bound to an enzyme are added to the well. The antihuman immunoglobulins will bind to the patient's antibodies. Substrate for the enzyme is then added and a positive reaction indicating presence of the antibody in the patient's serum is shown by the enzyme-substrate reaction. A direct ELISA test is used to detect the presence of an antigen. Antibodies are fixed to a small well, and the unknown antigen is added. If the antigen reacts with the antibodies, the antigen will be bound to the well. Antibodies specific for the antigen are then added to the well. This second layer of antibodies is bound to an enzyme. Substrate for the enzyme is then added, and a positive reaction indicating the identity of the antigen is shown by the enzyme-substrate reaction (see Figure 18.12). 8. a. Direct test b. Indirect test The direct test provides definitive proof. 9. Matching e Precipitation d, f Immunoelectrophoresis a Agglutination g Radioimmunoassay c Complement fixation f Neutralization b, d ELISA 10. Matching e Agglutination c Complement fixation a ELISA f FA b Neutralization d Precipitation Multiple Choice Answers are in boldface. 1. Patient's serum, influenza virus, sheep red blood cells, and antisheep red blood cells are mixed in a tube. What happens if the patient has antibodies against influenza? a. hemolysis b. hemagglutination c. hemagglutination-inhibition d. no hemolysis e. precipitin ring forms 2. Patient's serum, Chlamydia, guinea pig complement, sheep red blood cells, and antisheep red blood cells are mixed in a tube. What happens if the patient has antibodies against Chlamydia? a. hemolysis b. hemagglutination c. hemagglutination-inhibition d. no hemolysis e. precipitin ring forms 3. The examples in questions 1 and 2 are a. direct tests. b. indirect tests. 4. Which is the third step in a direct ELISA test? a. anti-Brucella b. Brucella c. substrate for the enzyme 5. Which item is from the patient in an indirect ELISA test? a. anti-Brucella b. Brucella c. substrate for the enzyme 6. In an immunodiffusion test, a strip of filter paper containing diphtheria antitoxin is placed on a solid culture medium. Then bacteria are streaked perpendicular to the filter paper. If the bacteria are toxigenic, a. the filter paper will turn red. b. a line of antigen-antibody precipitate will form. c. the cells will lyse. d. the cells will fluoresce. e. None of the above. Critical Thinking 1. The live vaccine may revert to a more virulent form; exogenous protein contaminants in viral vaccines; the inherent instability of certain live viral preparations. 2. Viruses multiply only in cells of a particular species. Protection of that species through immunization could prevent further growth of the virus. Many pathogenic bacteria are capable of growth in different species or in nonliving reservoirs. 3. Traditionally, by vaccinating a large animal such as a horse or a goat, and purifying the antibodies from its blood. Now these antibodies can be obtained in vitro by monoclonal antibody techniques. 4. The antibodies (called reagin) are not specific. The disease is syphilis. Clinical Applications 1. (a) is proof of a disease state. (b) could indicate disease, prior disease and recovery, or vaccination. The disease is tuberculosis. 2. No reaction; the antibodies will neutralize the toxin. This is a neutralization reaction. The disease is scarlet fever. 3. Patient A probably has the disease. Patient B does not have and never had the disease. Patient C recovered from the disease. Patient D acquired the disease between days 7 and 14; an example of seroconversion. The disease is legionellosis. 4. A high level of radioactive Ag in Ag-Ab complexes means the Ab combined with the known estrone because there was no estrone in the soil; the horses are not pregnant. A low level of radioactive Ag in Ag-Ab complexes means the Ab combined with estrone in the soil; the horses are pregnant. Begin Microbiology Chapters

Student Resources Chapter 1 Answers to End of Chapter Study Questions. Review 1. The observations of flies coming out of manure and maggots coming out of dead animals, and the appearance of microorganisms in liquids after a day or two, led people to believe that living organisms arose from nonliving matter. 2. Pasteur's S-neck flasks allowed air to get into the beef broth, but the curves of the S trapped bacteria before they could enter the broth. 3.a. Certain microorganisms cause diseases in insects. Microorganisms that kill insects can be effective biological control agents because they are specific for the pest and do not persist in the environment. b. Carbon, oxygen, nitrogen, sulfur, and phosphorus are required for all living organisms. Microorganisms convert these elements into forms that are useful for other organisms. Many bacteria decompose material and release carbon dioxide into the atmosphere for plants to use. Some bacteria can take nitrogen from the atmosphere and convert it into a form that can be used by plants and other microorganisms. c. Normal microbiota are microorganisms that are found in and on the human body. They do not usually cause disease, and can be beneficial. d. Organic matter in sewage is decomposed by bacteria into carbon dioxide, nitrates, phosphates, sulfate, and other inorganic compounds in a wastewater treatment plant. e. Recombinant DNA techniques have resulted in insertion of the gene for insulin production into bacteria. These bacteria can produce human insulin inexpensively. f. Microorganisms can be used as vaccines. Some microbes can be genetically engineered to produce components of vaccines. 4. Matching b Studies biodegradation of toxic wastes. g Studies the causative agent of Hantavirus pulmonary syndrome. c, e Studies the production of human proteins by bacteria. a Studies the symptoms of AIDS. d Studies the production of toxin by E. coli. b Studies the life cycle of Cryptosporidium. c Develops gene therapy for a disease. f Studies the fungus Candida albicans. 5. Matching l Avery, MacLeod and McCarty o Beadle and Tatum p Berg b Dubos r Ehrlich d Fleming j Hooke k Iwanowski c Jacob and Monod a Jenner m Koch s Lancefield e Lederberg and Tatum h Lister f Pasteur g Stanley i van Leeuwenhoek n Virchow q Weizmann 6. Erwinia carotovora is the correct way to write this scientific name. Scientific names can be derived from the names of scientists. In this case, Erwinia is derived from Erwin F. Smith, an American plant pathologist. Scientific names also can describe the organism, its habitat, or its niche. E. carotovora is a pathogen of carrots (vora = "eat"). 7. Matching d Algae c Bacteria b Fungi f Helminths e Protozoa a Viruses 8.a. B. thuringiensis is sold as a biological insecticide. b. Saccharomyces is the yeast sold for making bread, wine, and beer. Multiple Choice Answers are in boldface. 1. Which of the following is a scientific name? a) Mycobacterium tuberculosis b) Tubercle bacillus 2. Which of the following is not a characteristic of the eubacteria? a) Procaryotic b) Have peptidoglycan cell walls c) Have the same shape d) Grow by binary fission e) Have the ability to move 3. Which of the following is the most important element of Koch's germ theory of disease? The animal shows disease symptoms when a) the animal has been in contact with a sick animal. b) the animal has a lowered resistance. c) a microorganism is observed in the animal. d) a microorganism is inoculated into the animal. e) microorganisms can be cultured from the animal. 4. Recombinant DNA is a) DNA in bacteria. b) the study of how genes work. c) the DNA resulting when genes of two different organisms are mixed. d) the use of bacteria in the production of foods. e) the production of proteins by genes. 5. Which of the following statements is the best definition of biogenesis? a) Nonliving matter gives rise to living organisms. b) Living cells can only arise from preexisting cells. c) A vital force is necessary for life. d) Air is necessary for living organisms. e) Microorganisms can be generated from nonliving matter. 6. Which of the following is a beneficial activity of microorganisms? a) Some microorganisms are used as food for humans. b) Some microorganisms produce oxygen. c) Some microorganisms provide nitrogen for plant growth. d) Some microorganisms are used in sewage treatment processes. e) All of the above. 7. It has been said that bacteria are essential for the existence of life on Earth. Which of the following would be the essential function performed by bacteria? a) Control of insects. b) Directly provide food for humans. c) Decompose organic material and recycle elements. d) Cause disease. e) Produce human growth hormones such as insulin. 8. Which of the following is an example of bioremediation? a) Application of oil-degrading bacteria to an oil spill. b) Application of bacteria to a crop to prevent frost damage. c) Fixation of gaseous nitrogen into usable nitrogen. d) Bacteria producing a human protein such as interferon. e) All of the above. 9. Spallanzani's conclusion about spontaneous generation was challenged because Lavoisier had just shown that oxygen was the vital component of air. Which of the following statements is true? a) All life requires air. b) Only disease-causing organisms required air. c) Some microbes do not require air. d) Pasteur kept air out of his biogenesis experiments. e) Lavoisier was mistaken. 10. Which of the following statements about E. coli is not true? a) E. coli was the first disease-causing bacterium identified by Koch. b) E. coli is part of the normal microbiota of humans. c) E. coli is beneficial in human intestines. d) A disease-causing strain of E. coli causes bloody diarrhea. e) None of the above. Critical Thinking 1. Pasteur showed that life comes from preexisting life. The microorganisms that produced chemical and physical changes in beef broth and wine came from a few cells that entered the liquids from dust, containers, or the air. After showing that microorganisms could both grow on and change organic matter, Pasteur and others began to suspect that diseases were the result of microorganisms growing on living organic matter. 2. Semmelweis had observed an increased incidence of fever when medical students worked in obstetrics, as compared to the incidence during the students' summer break. The medical students were carrying bacteria from the autopsy room. Lister observed that compound bone fractures could result in death while recovery from simple fractures was without incident. 3. There are many! Check the dairy section for fermented products such as sour cream, yogurt, and cheese. Protein supplements often are yeasts. Bread, wine, and beer are products of yeasts and some bacteria. Sauerkraut is cabbage that has been fermented by lactobacilli. Vinegar is produced by bacterial growth on ethyl alcohol (wine). Xanthan, a thickener in many foods, is made by Xanthomonas bacteria. 4. Factors contributing to infectious disease include: mutations in existing organisms, spread of diseases to new areas, ecological disturbances such as deforestation, lack of immunization, pesticide resistance, and antibiotic resistance. Clinical Applications 1.a. Treatment with penicillin suggests a bacterial cause because only bacterial diseases are treatable with this drug. The summer onset also suggested an infectious disease perhaps related to an outdoor activity such as swimming or contact with mosquitoes or ticks. b. Lyme disease. c. The tick vector is more active during these months. Additionally, people spend more time outdoors and potentially in contact with ticks during these months. 2. Pasteur showed that microbes were omnipresent and were responsible for "diseases" (i.e., spoilage) of food; Lister reasoned that these microbes might be responsible for diseases of people. Neither Lister nor Pasteur proved that microbes caused diseases. Koch provided a repeatable proof to demonstrate that a microbe causes a disease.

Chapter 2 outline The science of the interaction between atoms and molecules is called chemistry. Chemical reactions is a change of energy occurs durng chemical reactions. Water Water is the most abundant substance in cells. Because water is a polar molecule it is a solvent. Water is a reactant in many of the decomposition reactions of digestion. Water is a temperature buffer. Acids Bases and Salts. A acid dissociates into H+ and anions. A base dissociates into OH- and cations. A salt dissociates into negative and positive ions neither of which is H+ or OH-. The structure of ATP is high energy phosphate bonds are indicated by wavy lines. When ATP breaks down to ADP and inorganic phosphate a large amount of chemical energy is released for use in other chemical reactions. Student Resources, Chapter 2 Answers to End of Chapter Study Questions Review 1. Atoms with the same atomic number and chemical behavior are classified as chemical elements. 2. Refer to Figure 2.1. 3. 14C and 12C are isotopes of carbon. 12C has 6 neutrons in its nucleus and 14C has 8 neutrons. 4. Hydrogen bonds. 5. a. Ionic b. Single covalent bond c. Double covalent bonds d. Peptide bond e. Hydrogen bond 6. 104 or 10,000 times. 7. Element Atomic Weight x Number of Atoms = Total Weight of That Element C 12 x 6 = 72 H 1 x 12 = 12 O 16 x 6 = 96 The molecular weight of C6H12O6 is 180 grams. 8.a. Synthesis reaction, condensation, or dehydration b. Decomposition reaction, digestion, or hydrolysis c. Exchange reaction d. Reversible reaction 9. The enzyme lowers the activation energy required for the reaction, and therefore speeds up this decomposition reaction. 10.a. Lipid b. Protein c. Carbohydrate d. Nucleic acid 11. > 12.a. Left to right b. Right to left 13. Breaking of bonds between phosphorus and oxygen. These are covalent bonds. 14. The entire protein shows tertiary structure. No quaternary structure. 15. Multiple-Choice Answers are in boldface. Radioisotopes are frequently used to label molecules in a cell. The fate of atoms and molecules in a cell can then be followed. This process is the basis for questions 1-3. 1. Assume E. coli are grown in a nutrient medium containing the radioisotope 16N. After a 48-hour incubation period, the 16N would most likely be found in the E. coli's a. carbohydrates. b. lipids. c. proteins. d. water. e. None of the above. 2. If Pseudomonas bacteria are supplied with radioactively-labeled cytosine, after a 24-hour incubation period, this cytosine would most likely be found in the cells' a. carbohydrates. b. DNA. c. lipids. d. water. e. protein. 3. If E. coli were grown in a medium containing the radio-active isotope 32P, the 32P would be found in all of the following molecules of the cell except a. ATP. b. carbohydrates. c. DNA. d. plasma membrane. e. None of the above. 4. A carbonated drink, pH3, is _____ times more acid than distilled water. a. 4 b. 10 c. 100 d. 1,000 e. 10,000 5. The best definition of ATP is: a. A molecule stored for food use. b. A molecule that supplies energy to do work. c. A molecule stored for an energy reserve. d. A molecule used as a source of phosphate. 6. Which of the following is an organic molecule? a. H2O (water) b. O2 (oxygen) c. C18H29SO3 (Styrofoam) d. FeO (iron oxide) e. F2C­CF2 (Teflon) Classify the molecules shown in questions 7-10 using the following choices. The dissociation products of the molecules are shown to help you. 7. HNO3 => H1 1 NO23 a. Acid b. Base c. Salt 8. H2SO4 => 2H1 1 SO2-4 a. Acid b. Base c. Salt 9. NaOH => Na1 1 OH2 a. Acid b. Base c. Salt 10. MgSO4 => Mg21 1 SO2-4 a. Acid b. Base c. Salt Critical Thinking 1.a. Synthesis reaction. b. H2CO3 is an acid. ATP and DNA have 5-carbon sugars. ATP has ribose, and DNA has deoxyribose; ATP and DNA contain the purine, adenine. 3. In order to maintain the proper fluidity, the percentage of unsaturated lipids decreases at the higher temperature. 4. These animals have cellulose-degrading bacteria in specialized structures in their digestive tracts. Clinical Applications 1. The enzymes will degrade organic molecules on the clothing. Any stains caused by these molecules will be removed when the molecules are digested. 2. T. ferrooxidans can oxidize sulfur ("thio") as well as iron ("ferro"). The oxidation of sulfide in pyrite produces sulfuric acid, which dissolves the limestone. Gypsum forms in a subsequent exchange reaction. 2S2- + 3O2 + 2H2O ----> 2SO42- + 4H+ 2CaCO3 + 4H+ + 2SO42- ----> 2CaSO4 + 2H+ + 2HCO3- 3. Since L-isomers are more common in nature, most cells, such as phagocytes, will be able to degrade the L-isomers. D-isomers will be resistant to metabolism by most cells. 4.Amphotericin B would not work against most bacteria because they lack sterols. Fungi have sterols and are generally susceptible to amphotericin B. Human cells have sterols.

Student Resources, Chapter 3 microscope and stains Answers to End-of-Chapter Study Questions Review 1. 1 mm = 10-6m 1 nm = 10-9 m 1 mm = 103nm 2. a. Ocular lens b. Objective lens c. Diaphragm d. Condensor e. Illuminator 3. Ocular lens magnification ¥ oil immersion lens magnification = total magnification of specimen. 10¥ ¥ 100¥ = 1000¥ 4. a. Compound light microscope b. Darkfield microscope c. Phase-contrast microscope d. Fluorescence microscope e. Electron microscope f. Differential interference contrast microscope 5. …that a beam of electrons focused by magnets…on a television-like screen or photographic plate. 6. Type of Microscope Maximum Magnification Resolution Compound light 2,000¥ 0.2 mm Electron 100,000¥ 0.0025 mm 7. Bacterial cells have a slightly negative charge, and the colored positive ion of a basic dye is attracted to the negative charge of the cell. Acid dyes do not stain bacterial cells because the negatively charged colored ion is repelled by the like charge of the cell. 8. a. A simple stain is used to determine cell shape and arrangement. b. A differential stain is used to distinguish kinds of bacteria based on their reaction to the differential stain. c. A negative stain does not distort the cell and is used to determine cell shape, size, and the presence of a capsule. d. A flagella stain is used to determine the number and arrangement of flagella. 9. In a Gram stain, the mordant combines with the basic dye to form a complex that will not wash out of gram-positive cells. In a flagella stain, the mordant accumulates on the flagella so that they can be seen with a light microscope. 10. A counterstain stains the colorless non-acid-fast cells so that they are easily seen through a microscope. 11. In the Gram stain, the decolorizer removes the color from gram-negative cells. In the acid-fast stain, the decolorizer removes the color from non-acid-fast cells. 12. Endospore: safranin is the counterstain. Gram: safranin is the counterstain. 13. Appearance after this step of Steps Gram-positive cells Gram-negative cells Crystal violet Purple Purple Iodine Purple Purple Alcohol-acetone Purple Colorless Safranin Purple Red Multiple-Choice Answers are in boldface. 1. You are trying to identify an unknown gram-negative bacterium. Which of the following stains is not necessary? a. negative stain b. acid-fast stain c. flagella stain d. endospore stain e. Both b and d 2. Assume you stain Bacillus by applying malachite green with heat and then counterstain with safranin. Through the microscope, the green structures are a. cell walls. b. capsules. c. endospores. d. flagella. e. impossible to identify. 3. Carbolfuchsin can be used as a simple stain and a negative stain. As a simple stain, the pH is a. 2. b. higher than the negative stain. c. lower than the negative stain. d. the same as the negative stain. 4. Looking at the cell of a photosynthetic microorganism, you observe that the chloroplasts are green in brightfield microscopy and red in fluorescence microscopy. You conclude that a. chlorophyll is fluorescent. b. the magnification has distorted the image. c. you're not looking at the same structure in both microscopes. d. the stain masked the green color. e. None of the above. 5. Which of the following is not a functionally analogous pair of stains? a. nigrosin and malachite green b. crystal violet and carbolfuchsin c. safranin and methylene blue d. ethanol-acetone and acid-alcohol e. None of the above 6. Which of the following pairs is mismatched? a. capsule-negative stain b. cell arrangement-simple stain c. cell size-negative stain d. gram stain-bacterial identification e. None of the above 7. Assume you stain Clostridium by applying a basic stain, carbolfuchsin, with heat, decolorizing with acid-alcohol, and counterstaining with an acid stain, nigrosin. Through the microscope, the endospores are ____1____ and the cells are stained ____2____. a. 1-red- 2-black b. 1-black- 2-colorless c. 1-colorless- 2-black d. 1-red- 2-colorless e. 1-black- 2-red 8. Assume you are viewing a Gram-stained field of red cocci and blue baccili through the microscope. You can safely conclude that you have a. made a mistake in staining. b. two different species. c. old bacterial cells. d. young bacterial cells. e. None of the above. 9. In 1996, scientists described a new tapeworm parasite that has killed at least one person. The initial examination of the patient's abdominal mass was most likely made using a. brightfield microscopy. b. darkfield microscopy. c. electron microscopy. d. phase-contrast microscopy. e. fluorescence microscopy. Critical Thinking 1. The counterstain safranin can be omitted. Gram-positive bacteria will appear purple, and gram-negative bacteria will be colorless. 2. You would be able to discern two objects separated by the four distances given because each is equal to or greater than the resolving power of the microscope. 3. The high lipid content of acid-fast cell walls makes them impermeable to most stains. If the primary stain penetrates, the Gram stain decolorizer will not decolorize the cell. Therefore, acid-fast bacteria would be gram-positive if they could be Gram stained. 4. Inclusions as well as endospores may not stain in a Gram stain. The endospore stain will identify the unstained structure as an endospore. Clinical Applications 1. Ehrlich observed that mycobacteria could not be decolorized with acid-alcohol, so he reasoned that an acidic disinfectant would not be able to penetrate the cell wall. 2. N. gonorrhoeae bacteria are gram-negative (red) diplococci, often found in the large human cells (phagocytes). 3. These are called clue cells. The large red cells are human mucosal cells; gram-positive bacteria on the surface of the human cells. 4. The reporter was unfamiliar with the magnification of EMs and thought the magnified rods were worms.

Student Resources, Chapter 4 Answers to End of Chapter Study Questions Review 1. 2. Endospore formation is called sporogenesis. It is initiated by certain adverse environmental conditions. Formation of a new cell from an endospore is called germination. This process is triggered by favorable growth conditions. 3. 4. Matching d Cell wall f Endospore a Fimbriae c Flagella a, e Glycocalyx i Pili b, h Plasma membrane g Ribosomes 5. An endospore is called a resting structure because it is a method of one cell "resting," or surviving, as opposed to growing and reproducing. The protective endospore wall allows a bacterium to withstand adverse conditions in the environment. 6. a. Both allow materials to cross the plasma membrane from a high concentration to a low concentration without expending energy. Facilitated diffusion requires carrier proteins. b. Both require enzymes to move materials across the plasma membrane. In active transport, energy is expended. c. Both move materials across the plasma membrane with an expenditure of energy. In group translocation, the substrate is changed after it crosses the membrane. 7. Mycoplasmas do not have cell walls. 8. a. Diagram (a) refers to a gram-positive bacterium because the lipopolysaccharide-phospholipid- lipoprotein layer is absent. b. The gram-negative bacterium initially retains the violet stain, but it is released when the outer membrane is dissolved by the decolorizing agent. After the dye-iodine complex enters, it becomes trapped by the peptidoglycan of gram-positive cells. c. The outer layer of the gram-negative cells prevents penicillin from entering the cells. d. Essential molecules diffuse through the gram-positive wall. Porins and specific channel proteins in the gram-negative outer membrane allow passage of small water-soluble molecules. e. Gram-negative. 9. An extracellular enzyme (amylase) hydrolyzes starch into disaccharides (maltose) and monosaccharides (glucose). A carrier enzyme (maltase) hydrolyzes maltose and moves one glucose into the cell. Glucose can be transported by group translocation as glucose-6-phosphate. 10. Matching b Chloroplasts a Endoplasmic reticulum e Golgi complex d Lysosomes c Mitochondria 11. A mitochondrion is an example of an organelle that resembles a procaryotic cell. The inner membrane of a mitochondrion is arranged in folds similar to mesosomes. ATP is generated on this membrane just as it is in procaryotic plasma membranes. Mitochondria can reproduce by binary fission, and they contain circular DNA and 70S ribosomes. 12. Phagocytosis. Pinocytosis. 13. Erythromycin inhibits protein synthesis in a procaryotic cell; it will inhibit protein synthesis in mitochondria and chloroplasts. Multiple-Choice Answers are in boldface. 1. Which of the following is not a distinguishing characteristic of procaryotic cells? a. They have a single, circular chromosome. b. They lack membrane-enclosed organelles. c. They have cell walls containing peptidoglycan. d. Their DNA is not associated with histones. e. They lack a plasma membrane. 2. Which statement best describes what happens when a gram-positive bacterium is placed in distilled water and penicillin? a. No change will result; the solution is hypotonic. b. Water will move into the cell. c. Water will move out of the cell. d. The cell will undergo osmotic lysis. e. Sucrose will move into the cell from an area of higher to lower concentration. 3. Which statement best describes what happens when a gram-negative bacterium is placed in distilled water and penicillin? a. No change will result; the solution is hypotonic. b. Water will move into the cell. c. Water will move out of the cell. d. The cell will undergo osmotic lysis. e. Sucrose will move into the cell from an area of higher to lower concentration. 4. Which statement best describes what happens when a gram-positive bacterium is placed in an aqueous solution of lysozyme and 10% sucrose? a. No change will result; the solution is isotonic. b. Water will move into the cell. c. Water will move out of the cell. d. The cell will undergo osmotic lysis. e. Sucrose will move into the cell from an area of higher to lower concentration. 5. Which of the following statements best describes what happens to a cell exposed to polymyxins that destroy phospholipids? a. In an isotonic solution, nothing will happen. b. In a hypotonic solution, the cell will lyse. c. Water will move into the cell. d. Intracellular contents will leak from the cell. e. Any of the above might happen. 6. Which of the following is not true about fimbriae? a. They are composed of protein. b. They may be used for attachment. c. They are found on gram-negative cells. d. They are composed of pilin. e. They may be used for motility. 7. Which of the following pairs is mismatched? a. glycocalyx-adherence b. pili-reproduction c. membrane-DNA synthesis d. cell wall-protection e. plasma membrane-transport 8. Which of the following pairs is mismatched? a. metachromatic granules-stored phosphates b. polysaccharide granules-stored starch c. lipid inclusions-poly-B-hydroxybutyric acid d. sulfur granules-energy reserve e. ribosomes-protein storage 9. You have isolated a motile, gram-positive cell with no visible nucleus. You can assume this cell has a. ribosomes. b. mitochondria. c. an endoplasmic reticulum. d. a Golgi complex. e. all of the above. 10. The antibiotic amphothericin B disrupts plasma mem-branes by combining with sterols; it will affect all of the following cells except a. animal cells. b. eubacterial cells. c. fungal cells. d. Mycoplasma cells e. plant cells. Critical Thinking 1. Eucaryotic cells must be large enough to hold a nucleus and a mitochondrion (the minimum number of organelles). Procaryotic cells contain molecules needed to carry on metabolic activities, but do not contain membrane-enclosed organelles, which require extra space. 2. Micromonas has a nucleus, one mitochondrion, one chloroplast, one Golgi complex, and one flagellum. 3. Like eubacteria, archaea lack organelles. However, archaea also lack peptidoglycan cell walls. A more complete list of differences is in Table 10.2. 4. The large size of the organism caused the misidentification. Electron microscopy would reveal that this is a procaryotic cell; chemical analysis of the cell wall would reveal peptidoglycan. 5. Water would passively leave the cell in a hypertonic environment. If a cell pumps K+ in, water will follow, thus preventing plasmolysis. Clinical Applications 1. Cell death released cell wall fragments. The gram-negative cell wall is responsible for the symptoms of septic shock. 2. The endospores allow survival in the presence of oxygen and during heating. 3. Enterobacter, Pseudomonas, and Klebsiella are gram-negative. Their cell walls contain lipid A endotoxin. 4. The bacteria were adhering to the inside of the pipes as a biofilm. Fimbriae and the glycocalyx allow the bacteria to adhere. 5. Bacterial endospores allow these bacteria to survive in products on store shelves. B. thuringiensis is sold as an insecticide, and B. subtilis as a fungicide.

Student Resources, Chapter 5 Answers to End-of-Chapter Study Questions Review 1. Metabolism is the sum of all chemical reactions that occur within a living organism. 2. Catabolic reactions break down organic compounds and release energy, while anabolic reactions use the products of catabolism and energy to build cell material. 3. 4. a. When the enzyme and substrate combine, the substrate molecule will be transformed. b. When the competitive inhibitor binds to the enzyme, the enzyme will not be able to bind with the substrate. c. When the noncompetitive inhibitor binds to the enzyme, the active site of the enzyme will be changed so the enzyme cannot bind with the substrate. d. The noncompetitive inhibitor. 5. The optimum temperature for an enzyme is one that favors movement of molecules so the enzyme can "find" its substrate. Lower temperatures will decrease the rate of collisions and the rate of reactions. Increased temperatures will denature the enzyme. 6. Ethyl alcohol, lactic acid, butyl alcohol, acetone, and glycerol are some of the possible products. Refer to Table 5.4 and Figure 5.18b. 7. Organism Carbon Source Energy Source Photoautotroph CO2 Light Photoheterotroph Organic molecules Light Chemoautotroph CO2 Inorganic molecules Chemoheterotroph Organic molecules Organic molecules 8. ATP generated by Reaction Photophosphorylation An electron, liberated from chlorophyll by light, is passed down an electron transport chain. Oxidative phosphorylation Cytochrome c passes two electrons to cytochrome a. Substrate-level phosphorylation Phosphoenolpyruvic acid Pyruvic acid 9. Oxidation-reduction: A coupled reaction in which one substance is oxidized and one is reduced. a. The final electron acceptor in aerobic respiration is molecular oxygen; in anaerobic respiration, it is another inorganic molecule. b. The final electron acceptor in respiration is usually an inorganic molecule; in fermentation, it is an organic molecule. c. In cyclic photophosphorylation, electrons are returned to chlorophyll. In noncyclic photophosphorylation, chlorophyll receives electrons from hydrogen atoms. 10. The pentose phosphate pathway produces pentoses for the synthesis of nucleic acids, precursors for the synthesis of glucose by photosynthesizing organisms, precursors in the synthesis of certain amino acids, and NADPH. 11. a. is the Calvin-Benson cycle, b. is glycolysis, and c. is the Krebs cycle. 12. Glycerol is catabolized by pathway b. as dihydroxyacetone phosphate. Fatty acids by pathway c. as acetyl groups. 13. In pathway (c) at a-ketoglutaric acid. 14. Glyceraldehyde-3-phosphate from the Calvin-Benson cycle enters glycolysis. Pyruvic acid from glycolysis is decarboxylated to produce acetyl for the Krebs cycle. 15. In (a), between glucose and glyceraldehyde-3-phosphate. 16. The conversion of pyruvic acid to acetyl, isocitric acid to a-ketoglutaric acid, and a-ketoglutaric acid to succinyl~CoA. 17. By pathway (c) as acetyl groups.18. Uses Produces Calvin-Benson cycle 6 NADPH Glycolysis 2 NADH Pyruvic acid ----> acetyl 1 NADH Isocitric acid ----> a-ketoglutaric acid 1 NADH a-ketoglutaric acid ----> Succinyl~CoA 1 NADH Succinic acid ----> Fumaric acid 1 FADH2 Malic acid ----> Oxaloacetic acid 1 NADH 19. Dihydroxyacetone phosphate; acetyl; oxaloacetic acid; a-ketoglutaric acid. 20. Reactions requiring ATP are coupled with reactions that produce ATP. 21. The reaction rate will increase until the enzymes are saturated. Multiple ChoiceAnswers are in boldface.1. Which substance in the following reaction is being reduced?a. Acetaldehydeb. NADHc. Ethanold. NAD12. Which of these reactions produces the most molecules of ATP during aerobic metabolism?a. glucose => glucose-6-Pb. phosphoenolpyruvic acid => pyruvic acidc. glucose => pyruvic acidd. acetyl => CO2 1 H2Oe. succinic acid => fumaric acid3. Which of the following processes does not generate ATP?a. photophosphorylationb. the Calvin-Benson cyclec. oxidative phosphorylationd. substrate-level phosphorylatione. none of the above4. Which of the following compounds has the greatest amount of energy for a cell?a. CO2b. ATP c. glucosed. O2e. lactic acid 5. Which of the following is the best definition of the Krebs cycle?a. The oxidation of pyruvic acid.b. The way cells produce CO2c. A series of chemical reactions in which NADH is produced from the oxidation of pyruvic acid.d. A method of producing ATP by phosphorylating ADP.e. A series of chemical reactions in which ATP is produced from the oxidation of pyruvic acid. 6. Which of the following is the best definition of respiration?a. A sequence of carrier molecules with O2 as the final electron acceptor.b. A sequence of carrier molecules with an inorganic molecule as the final electron acceptor.c. A method of generating ATP.d. The complete oxidation of glucose to CO2 and H2O.e. A series of reactions in which pyruvic acid is oxidized to CO2 and H2O.7. Which culture produces the most lactic acid?a. E. coli growing in glucose broth at 35°C with O2 for 5 days.b. E. coli growing in glucose broth at 35°C without O2 for 5 days.c. Both a and b.d. Neither a nor b.8. Which culture produces the most ATP?a. E. coli growing in glucose broth at 35°C with O2 for 5 days.b. E. coli growing in glucose broth at 35°C without O2 for 5 days.c. Both a and b.d. Neither a nor b. 9. Which culture uses NAD1?a. E. coli growing in glucose broth at 35°C with O2 for 5 days.b. E. coli growing in glucose broth at 35°C without O2 for 5 days.c. Both a and b.d. Neither a nor b. 10. Which culture uses the most glucose?a. E. coli growing in glucose broth at 35°C with O2 for 5 days.b. E. coli growing in glucose broth at 35°C without O2 for 5 days.c. Both a and b.d. Neither a nor b. Critical Thinking 1. Answers should include the following:Protons are pumped from one side of the membrane to the other; transfer of protons back across the membrane generates ATP. a. Outer portion is acidic and b. Has a positive electric charge. c. Energy-conserving sites are the three loci where H+ molecules are pumped out. d. Kinetic energy is realized at ATP synthetase when protons cross the membrane. 2. Streptococcus is only capable of fermentation, which yields two molecules of ATP for each molecule of glucose consumed. Most of the energy that cells obtain from catabolism is from respiration. 3. NADH must be reoxidized so it can pick up hydrogen atoms again. NADH is usually reoxidized in respiration when electrons are transferred to the electron transport chain carrier molecules. In fermentation, NADH is reoxidized when electrons are transferred to pyruvic acid (reducing pyruvic acid to lactic acid, for example). 4. Once the enzyme combines with substrate, its affinity for substrate increases. The reaction proceeds at a linear rate (black). When the enzyme becomes saturated with competitive inhibitor, the reaction will completely stop (red). 5. Carbohydrate catabolism: a. Oxidation of glucose (glycolysis, Krebs, aerobic ETC). b. CO2 fixation and oxidation of glucose. c. CO2 fixation and oxidation of glucose (anaerobic ETC). Energy production. All three use chemiosmotic mechanisms. In addition to oxidative phosphorylation, Spirulina and Ectothiorhodospira use photophosphorylation. 6. Glucose = 38 ATP.Butterfat = 217 ATP. Glycerol goes into glycolysis to produce 1 ATP. Six acetyl groups are produced from each of the 12-carbon chains by beta-oxidation. Each acetyl can be used to produce 12 ATPs in the Krebs cycle. 7. Two electrons removed from As3+ are picked up by NAD+ for use in the electron transport chain. Thiobacillus could be used to remove arsenic from industrial waste water and groundwater.Clinical Applications 1. X factor is necessary to synthesize cytochromes. V factor is used as an electron acceptor in oxidation reactions. 2. The drug ddC is missing an O atom on C3 so it cannot be joined to the phosphate group of another nucleotide. 3. Live bacteria or whole bacterial cells are not being injected. Enzymes are specific for their substrates, so streptokinase will react only with its substrate, fibrin.

Student Resources, Chapter 6 Answers to End-of-Chapter Study Questions Review1. In binary fission, the cell elongates and the chromosome replicates. Next, the nuclear material is evenly divided. The plasma membrane invaginates toward the center of the cell. The cell wall thickens and grows inward between the membrane invaginations; two new cells result. 2. Refer to Figure 6.14. The period of no cell division is called lag phase. During lag phase, the bacteria are synthesizing enzymes that are necessary for growth. In log phase, the cells are dividing at the maximum rate under the conditions provided. The number of cells dividing equals the number of cells dying in stationary phase. When the number of deaths exceeds the number of divisions, death phase is observed. 3. Carbon (C) is required for synthesis of molecules that make up a living cell. Carbon-containing compounds also are required as an energy source for heterotrophs. 4. Most bacteria grow best between pH 6.5 and 7.5. 5. The addition of salt or sugar to foods increases the osmotic pressure for microorganisms on the food. The resulting hypertonic environment causes plasmolysis of the microbial cells. 6. a. Catalyzes the breakdown of H2O2 to O2 and H2O. b. H2O2; peroxide ion is O22-. c. Catalyzes the breakdown of H2O2; NADH + H+ + H2O2 NAD+ + 2H2O d. O-; this diatom has one unpaired electron. e. Converts superoxide to O2 and H2O2; 2O2- + 2H+ O2 + H2O2 The enzymes are important in protecting the cell from the strong oxidizing agents, peroxide and superoxide, that form during respiration. 7. Both environments prevent molecular oxygen from reaching the bacterial cells. In reducing media, thioglycolate combines with dissolved oxygen, thereby removing it from the medium. In an anaerobic incubator, air is replaced with an atmosphere of CO2 (and N2). Clostridium is an obligate anaerobe that lacks superoxide dismutase and catalase. Consequently, the accumulation of superoxides and peroxides will kill the cell in an aerobic environment. 8. Direct methods are those in which the microorganisms are seen and counted. Direct methods are direct count, standard plate count, filtration, and most probable number. 9. The growth rate of bacteria slows down with decreasing temperatures. Mesophilic bacteria will grow slowly at refrigeration temperatures and will remain dormant in a freezer. Bacteria will not spoil food quickly in a refrigerator. 10. Number of cells x 2n generations = Total number of cells 6 x 27 = 768 11. Petroleum can meet the carbon and energy requirements for an oil-degrading bacterium; however, nitrogen and phosphate are usually not available in large quantities. Nitrogen and phosphate are essential for making proteins, phospholipids, nucleic acids, and ATP. 12. A chemically defined medium is one in which the exact chemical composition is known. A complex medium is one in which the exact chemical composition is not known. 13. Multiple ChoiceAnswers are in boldface.1. Medium 1 isa. selective.b. differential.c. both selective and differential. 2. Medium 2 isa. selective.b. differential.c. both selective and differential. 3. Which of the lines best depicts the log phase of a thermophile incubated at room temperature?a. 1b. 2c. 3 4. Which of the lines best depicts the log phase of Listeria monocytogenes growing in a human?a. 1b. 2c. 3 5. Assume you inoculated 100 facultatively anaerobic cells onto nutrient agar and incubated the plant aerobically. You then inoculated 100 cells of the same species onto nutrient agar and incubated the second plant anaerobically. After incubation for 24 hours, you should havea. more colonies on the aerobic plate.b. more colonies on the anaerobic plate.c. the same number of colonies on both plates.6. The term trace elements refers toa. the elements CHONPS.b. vitamins.c. nitrogen, phosphorus, and sulfur.d. small mineral requirements.e. toxic substances.7. Which one of the following temperatures would most likely kill a mesophile?a. 250°C b. 0°C c. 9°Cd. 37°C e. 60°C 8. All of the following are true about agar except:a. It is a source of nutrients in culture media.b. It is a polysaccharide.c. It liquefies at 100°C.d. It solidifies at approximately 40°C.e. It is metabolized by few bacteria. 9. Which of the following types of media would not be used to culture aerobes?a. selective mediab. reducing mediac. enrichment mediad. differential mediae. complex media 10. An organism that has peroxidase and superoxide dismutase but lacks catalase is most likely ana. aerobe.b. aerotolerant anaerobe.c. obligate anaerobe. Critical Thinking 1. a. At x, the bacteria began a second lag phase during which they synthesized enzymes required to use the second carbon source. b. The first substrate provided the better growth conditions. The slope of the line is steeper, indicating that the bacteria grew faster. 2. In the presence of oxygen, H2O2 forms in Clostridium. The H2O2 accumulates in these catalase-negative cells and kills them. H2O2 does not form in Streptococcus. 3. Glucose provides a fermentable carbohydrate for chemoheterotrophs. Glucose is the carbon and energy source. Other macronutrients including nitrogen are provided in inorganic compounds in the "minimal salts." 4. a. A b. B c. A d. A e. A Clinical Applications 1. 1.68 x 108 They are the progeny of the original 10. 2. At least 53°C; 60°C was recommended after this study. The bacteria get in the food during preparation and those buried inside do not get hot enough to be killed. 3. Product 2 decreased bacterial numbers by 89% compared to a 17% decrease for both Products 1 and 2. All the bacteria probably did not grow. Only those that could grow aerobically on nutrient agar were counted in the experiment.

Student Resources, Chapter 7 Microbiology Answers to End-of-Chapter Study Questions Review 1. a. Lysis. b. Altered permeability and leakage of cell contents. c. Destruction of enzymes and structural proteins such as those in the plasma membrane. d. Interference with protein synthesis and cell division. 2. Autoclave. Due to the high specific heat of water, moist heat is readily transferred to cells. 3. Most organisms that cause disease or rapid spoilage of food are destroyed by pasteurization. 4. Variables that affect determination of the thermal death point are a. The innate heat resistance of the strain of bacteria. b. The past history of the culture, whether it was freeze-dried, wetted, etc. c. The clumping of the cells during the test. d. The amount of water present. e. The organic matter present. f. Media and incubation temperature used to determine viability of the culture after heating. 5. a. Ionizing radiation can break DNA directly. However, due to the high water content of cells, the formation of free radicals (H· and OH·) that break DNA strands is likely to occur. b. Ultraviolet radiation damages DNA by the formation of thymine dimers. 6. Microorganisms tend to die at a constant rate over a period of time. The constant rate is indicated by the straight line after exposure to the bactericidal compound. 7. Sterilization Method Temp. Time Type Preferred Use Mechanism of Action Autoclaving 121°C 15 min Moist Media, equipment Protein denaturation Hot air 170°C 2 hr Dry Glassware Oxidation Pasteurization 72°C 15 sec Moist Milk, alcoholic drinks Protein denaturation 8. All three processes kill microorganisms; however, as moisture and/or temperatures are increased, less time is required to achieve the same result. 9. Salts and sugars create a hypertonic environment. Salts and sugars (as preservatives) do not directly affect cell structures or metabolism; instead, they alter the osmotic pressure. Jams and jellies are preserved with sugar; meats are usually preserved with salt. Molds are more capable of growth in high osmotic pressure than bacteria. 10. 1. Acts rapidly. 2. Attacks all, or a wide range of, microbes. 3. Is able to penetrate. 4. Readily mixes with water. 5. Is not hampered by organic matter. 6. Stable. 7. Does not stain or corrode. 8. Nontoxic. 9. Pleasant odor. 10. Economical. 11. Safe to transport. 11. Method of Action Standard Use a. Disrupts plasma membrane Skin surfaces b. Inhibits protein function Antiseptic c. Oxidation Disinfect water d. Denatures proteins, destroys lipids Skin surfaces e. Oligodynamic AgNO3 to prevent gonococcal eye infections f. Inactivation of proteins Chemical sterilizer g. Denatures proteins Chemical sterilizer h. Oxidation Antiseptic 12. Disinfectant B is preferable because it can be diluted more and still be effective. 13. Quaternary ammonium compounds are most effective against gram-positive bacteria. Gram-negative bacteria that were stuck in cracks or around the drain of the tub would not have been washed away when the tub was cleaned. These gram-negative bacteria could survive the washing procedure. Some pseudomonads can grow on quats that have accumulated. Multiple-Choice Answers are in boldface. 1. Which of the following does not kill endospores? a. autoclaving b. incineration c. hot-air sterilization d. pasteurization e. none of the above 2. Which of the following is most effective for sterilizing mattresses and plastic Petri dishes? a. chlorine b. ethylene oxide c. glutaraldehyde d. autoclaving e. nonionizing radiation 3. Which of these disinfectants does not act by disrupting the plasma membrane? a. phenolics b. phenol c. quaternary ammonium compounds d. halogens e. biguanides 4. Which of the following cannot be used to sterilize a heat labile solution stored in a plastic container? a. gamma radiation b. ethylene oxide c. nonionizing radiation d. autoclaving e. short-wavelength radiation 5. Which of the following is not true about quaternary ammonium compounds? a. bactericidal against gram-positive bacteria b. sporicidal c. amoebicidal d. fungicidal e. kills enveloped viruses 6. A classmate is trying to determine how a disinfectant might kill cells. You observed that when he spilled the disinfectant in your reduced litmus milk, the litmus turned blue again. You suggest to your classmate that a. the disinfectant might inhibit cell wall synthesis. b. the disinfectant might oxidize molecules. c. the disinfectant might inhibit protein synthesis. d. the disinfectant might denature proteins. e. he take his work away from yours. 7. Which of the following is most likely to be bactericidal? a. membrane filtration b. ionizing radiation c. freeze-drying d. deep-freezing e. all of the above 8. Which of the following is used to control microbial growth in foods? a. organic acids b. alcohols c. aldehydes d. heavy metals e. all of the above 9. Which disinfectant is the most effective? a. A b. B c. C d. D 10. Which disinfectant(s) is (are) bactericidal? a. A, B, C, and D b. A, C, and D c. A and B d. C and D e. B only Critical Thinking 1. a. Z. b. No. A culture medium would have to be inoculated from the zone of inhibition to determine the presence of viable bacteria. 2. a. Acid-resistant cell wall. b. Metabolizes many organic molecules. c. Endospores. 3. a. Disinfectant B diluted with distilled water. b. Can't tell; test was done on Salmonella. 4. Bactericidal effects of microwave radiation are due to heat. Clinical Applications 1. a. Hot water does not achieve sterilization, and b. There were no check valves to prevent backflow. Also, adapter and other invasive items should be disposable whenever possible. 2. Iodophors are approved for antiseptic uses and not for disinfection. 3. Serratia from the environment entered the jar from the air, wash water, or hands. The effectiveness of the quat was reduced by the cotton, and quat-resistant Serratia were able to survive. The bacteria were introduced into the methylprednisolone by swabbing the top of the vial. The disinfectant needs to be changed.

Student Resources, Chapter 8 Answers to End-of-Chapter Study Questions Review 1. DNA consists of a strand of alternating sugars (deoxyribose) and phosphate groups with a nitrogenous base attached to each sugar. The bases are adenine, thymine, cytosine, and guanine. DNA exists in a cell as two strands twisted together to form a double helix. The two strands are held together by hydrogen bonds between their nitrogenous bases. The bases are paired in a specific, complementary way: A-T and C-G. The information held in the sequence of nucleotides in DNA is the basis for synthesis of RNA and proteins in a cell. 2. a. b. DNA polymerases synthesize a complementary strand of DNA from a DNA template. RNA polymerase starts each fragment of the lagging strand with an RNA primer. c. Each new double-stranded DNA molecule contains one original strand and one new strand. 3. a. ATATTACTTTGCATGGACT. b. met-lys-arg-thr (end). c. TATAATGAAACGTACCTGA. d. No change. e. Cysteine substituted for arginine. f. Proline substituted for threonine (missense mutation). g. Frameshift mutation. h. Adjacent thymines might polymerize. i. ACT. 4. One end of the mRNA molecule becomes associated with a ribosome. Ribosomes are composed of rRNA and protein. The anticodon of a tRNA with its activated amino acid pairs with the mRNA codon at the ribosome. 5. A mutant is isolated by direct selection because it grows on a particular medium. The colonies on an antibiotic-containing medium can be identified as resistant to that antibiotic. A mutant is isolated by indirect selection because it does not grow on a particular medium. Replica plating could be employed to inoculate an antibiotic-containing medium. Colonies that did not grow on this medium can be isolated from the original plate and are antibiotic sensitive. 6. Matching b A mutagen that is incorporated into DNA in place of a normal base. d A mutagen that causes the formation of highly reactive ions. c A mutagen that alters adenine so that it base-pairs with cytosine. a A mutagen that causes insertions. e A mutagen that causes the formation of pyrimidine dimers. 7. The basis for the Ames test is that a mutated cell can revert to a cell that resembles the original, nonmutant cell by undergoing another mutation. The reversion rate of histidine auxotrophs of Salmonella in the presence of a mutagen will be higher than the spontaneous rate (in the absence of a mutagen). 8. Plasmids are small, self-replicating circles of DNA that are not associated with the chromosome. The F plasmid can be integrated into the chromosome. The F plasmid can be transferred from a donor to a recipient cell in conjugation. When the F plasmid becomes integrated into the chromosome, the cell is called an Hfr. During conjugation between an Hfr and an F cell, the chromosome of the Hfr cell, with its integrated F factor, replicates, and the new copy of the chromosome is transferred to the recipient cell. 9. a. …a repressor protein must be bound tightly to the operator site…it will bind to the repressor so that transcription can occur. b. …called a corepressor, causes the repressor to bind to the operator. Derepression is by removal of the corepressor, C in this case, when the corepressor is needed in the cell. c. None; constitutive enzymes are produced at certain necessary levels regardless of the amount of substrate or end-product. 10. Light repair; dark repair; proofreading by DNA polymerase. 11. a. The genetic makeup of an organism. b. The external manifestations of the genotype. c. Rearrangement of genes to form new combinations; in nature, this usually occurs between members of the same species; in vitro, recombinant DNA is made from genes of different species. 12. CTTTGA. Endospores and pigments offer protection against UV radiation. Additionally, repair mechanisms can remove and replace thymine polymers. 13. a. Culture 1 will remain the same. Culture 2 will be converted to F+ but will have its original genotype. b. The donor and recipient cells' DNA can recombine to form combinations of A+B+C+ and A-B-C-. If the F plasmid also is transferred, the recipient cell may become F+. 14. Semiconservative replication ensures that the offspring cell will have one correct strand of DNA. Any mutations that may have occurred during DNA replication have a greater chance of being correctly repaired. 15. Mutation and recombination provide genetic diversity. Environmental factors select for the survival of organisms through natural selection. Genetic diversity is necessary for the survival of some organisms through the processes of natural selection. Organisms that survive may undergo further genetic change, resulting in the evolution of the species.Multiple-ChoiceAnswers are in boldface. 1. The transfer of DNA from a donor to a recipient cell by a bacteriophage.a. conjugationb. transcriptionc. transductiond. transformatione. translation 2. The transfer of DNA from a donor to a recipient as naked DNA in solution.a. conjugationb. transcriptionc. transductiond. transformatione. translation 3. Feedback inhibition differs from repression because feedback inhibitiona. is less precise. b. is slower acting. c. stops the action of preexisting enzymes.d. stops the synthesis of new enzymes e. All of the above. 4. Bacteria can acquire antibiotic resistance bya. mutation.b. insertion of transposons.c. acquiring plasmids.d. All of the above.e. None of the above. 5. Suppose you inoculate three flasks of minimal salts broth with E. coli. Flask A contains glucose. Flask B contains glucose and lactose. Flask C contains lactose. After a few hours of incubation, you test the flasks for the presence of b-galactosidase. Which flask(s) do you predict will have this enzyme?a. A b. Bc. C d. A and B e. B and C 6. Plasmids differ from transposons because plasmidsa. become inserted into chromosomes.b. are self-replicated outside the chromosome.c. move from chromosome to chromosome.d. carry genes for antibiotic resistance.e. None of the above. Critical Thinking 1. Cancerous cells are growing faster than normal cells. Mutations have a greater effect when a cell is growing because it is synthesizing DNA and enzymes. The probability of a lethal mutation also is increased in rapidly growing cells. 2. The cell does not regulate the rate at which DNA is synthesized, but it regulates the rate at which replication forks on the chromosomes are initiated. The cell initiates multiple forks so that a daughter cell will inherit a complete chromosome plus additional portions from multiple replication forks. Chromosome replication begins during or immediately after division. 3. a. Mercuric ion. b. To detoxify it. c. Detoxifying mercuric ion will allow the cell to live in areas where other organisms may not be able to.Clinical Applications 1. a. Chloroquine interferes with transcription; erythromycin interferes with translation; acyclovir interferes with DNA replication. b. Erythromycin is specific for bacterial ribosomes. c. Acyclovir is effective against viruses because a viral enzyme tries to use it as a DNA nucleotide. d. Chloroquine and acyclovir will have the most effects on the host because they affect eucaryotic DNA. The effects of erythromycin on mitochondrial ribosomes are small for short-term use. e. Chloroquine is used against malaria. Malaria is caused by a eucaryote (protozoan). f. Acyclovir is used against Herpesvirus infections. Erythromycin affects bacterial (70S) ribosomes, not viruses or eucaryotes. 2. The amino acid sequence reflects the RNA (genome) of the virus. Sequence B is the most dissimilar and, therefore, probably not from the dentist.

Student Resources, Chapter 9 Answers to End-of-Chapter Study Questions Review 1. Recombinant DNA (rDNA) is DNA that is combined from different sources. In nature, rDNA results from conjugation, transduction, and tranformation. Genetic engineering is the artifical making of rDNA. 2. a. Both are DNA. cDNA is a segment of DNA made by RNA-dependent DNA polymerase. It is not necessarily a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. b. Both are DNA. A restriction fragment is a segment of DNA produced when a restriction endonuclease hydrolyzes DNA. It is not usually a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. c. Both are DNA. A DNA probe is a short, single-stranded piece of DNA. It is not a gene; a gene is a transcribable unit of DNA that codes for protein or RNA. d. Both are enzymes. DNA polymerase synthesizes DNA one nucleotide at a time using a DNA template; DNA ligase joins pieces (strands of nucleotides) together. e. Both are DNA. Recombinant DNA results from joining DNA from two different sources; cDNA results from copying a strand of RNA. 3. a. A desired gene can be spliced into a plasmid and inserted into a cell by transformation. b. A desired gene can be spliced into a viral genome and inserted into a cell by transduction. c. Antibiotic resistance genes are used as markers or labels on plasmids so that the cell containing the plasmid can be found by direct selection on an antibiotic-containing medium. d. A genetically engineered bacterium should be producing a new protein product. Radioactively labeled antibodies against a specific protein can be used to locate the bacterial colony producing the protein. 4. Restriction fragments from one source can be cloned in microbial cells to make a gene library. Synthetic DNA is made in a lab. 5. In protoplast fusion, two wall-less cells fuse together to combine their DNA. A variety of genotypes can result from this process. In b, c, and d, specific genes are inserted directly into the cell. 6. BamH I, EcoR I, and Hind III make sticky ends. Fragments of DNA produced with the same restriction enzyme will spontaneously anneal to each other at their sticky ends. 7. The gene can be spliced into a plasmid and inserted into a bacterial cell. As the cell grows, the number of plasmids will increase. The polymerase chain reaction can make copies of a gene using DNA polymerase in vitro. 8. In a eucaryotic cell, RNA polymerase copies DNA and removes the introns, leaving the exons in the mRNA. cDNA can be made from the mRNA by reverse transcriptase. 9. Large colonies are ampicillin-resistant because they are growing. Smaller ampicillin-sensitive colonies may appear later after the antibiotic has been degraded by the ampicillin-resistant bacteria. White colonies have the new gene. 10. See Tables 9.1 and 9.2. 11. You probably used a few plant cells in a Petri plate for your experiment. How will you select the plant cells that actually have the new Ti plasmid? You can grow these cells on plant-cell culture media with tetracycline. Only the cells with the new plasmid will grow.Multiple-ChoiceAnswers are in boldface.1. Restriction enzymes were first discovered with the observation thata. DNA is restricted to the nucleus.b. phage DNA is destroyed in a host cell.c. foreign DNA is kept out of a cell.d. foreign DNA is restricted to the cytoplasm.e. All of the above. 2. The DNA probe, GGCTTA, will hybridize with DNA containinga. CCGUUA.b. CCGAAT.c. GGCTTA.d. GGCUUAe. GGCAAU.3. The basic steps of genetic engineering are listed below. What is the fourth step?a. transformation b. ligationc. plasmid cleavage d. restriction-enzyme digestion of genee. isolation of gene 4. The following steps are used to make cDNA. What is the second step?a. reverse transcription b. RNA processingto remove intronsc. transcriptiond. translation 5. If you put a gene in a virus, the next step in genetic engineering would bea. insertion of a plasmid.b. transformation.c. transduction.d. PCR.e. Southern blotting. Critical Thinking 1. EcoR I: 2 fragmentsHind III: 2 fragmentsBoth enzymes: 4 fragmentsThe smallest fragment containing the tet gene is cut at one end by Hind III and at the other end by EcoR I. 2. Isolate cDNA (or synthesize DNA) for the desired gene from HIV. Insert the HIV gene into vaccinia virus's DNA. Infect the host cells with vaccinia virus. As the virus reproduces, it will cause production of the HIV protein. 3. "Normal" DNA polymerase is denatured by the heating step, so a technician would have to add DNA polymerase to the reaction vessel every 2 minutes. DNA polymerase from Thermus is not denatured by 90°C, so fresh DNA polymerase does not have to be added every 2 minutes.Clinical Applications 1. Sample 2 was positive for V. cholerae because the probe paired with DNA in lane 2. V. cholerae ingested by oysters is a source of disease for humans. The PCR doesn't require isolation and additional incubation for test results. 2. The vector and new gene fragments appear in the fifth lane; therefore, transformation did occur.

Student Resources, Chapter 10 Answers to End-of-Chapter Study Questions Review 1. Taxonomy is the science of classifying organisms to establish the relatedness between groups of organisms. 2. Living organisms cannot be grouped into two groups. For example, plant and animal is not acceptable because if fungi are grouped with plants, the definition of plants can't include cellulose and photosynthesis. If fungi are grouped with animals, the definition of animals can't include no cell wall and ingestive. The goal is to look for a "natural" scheme; that is, what criteria can be used to characterize all organisms. 3. Monera or Procaryotae: All procaryotic organisms. Protista: Unicellular (or multicellular without tissue organization) eucaryotic organisms; usually flagellated at some time. Fungi: Unicellular or multicellular organisms that absorb organic nutrients; noncellulose cell walls; lack flagella. Plantae: Multicellular eucaryotes with tissue formation; cellulose cell walls; generally photosynthetic. Animalia: Multicellular eucaryotes with tissue formation; develop from an embryo (gastrula); lacking cell walls; ingest organic nutrients through a mouth of some kind. 4. The three distinct chemical types of cells (see Table 10.2). In a five-kingdom system, four of the kingdoms have eucaryotic cells and the fifth is based on procaryotic structure alone. 5. a. Both are procaryotic. They differ in composition of their cell walls, plasma membranes, and rRNAs. b. Both are bound by ester-linked plasma membranes. Eucarya have membrane-bound organelles. c. Both use methionine as the start signal. Eucarya have membrane-bound organelles and ester-linked membranes. 6. Binomial nomenclature is the system of assigning a genus and specific epithet to each organism. 7. Common names are not specific and can be misleading. According to the rules of scientific nomenclature, each organism has only one binomial. 8. The genus name must be written out so the reader knows what organism is being discussed, since the abbreviation for both of these species is E. coli. 9. Kingdom, division (phylum), class, order, family, genus, species. 10. A species is a group of closely related organisms having limited geographical distribution that interbreeds but does not breed with other species. Species can be distinguished morphologically. Because of the distinct differences between eucaryotic organisms and bacteria, a bacterial species is defined as a population of cells with similar characteristics. A viral species is a population of viruses with similar characteristics that occupies a particular ecological niche. 11. (See Table 10.4) Used primarily for identification:morphological characteristicsdifferential stainingbiochemical testsserologyphage typing Used primarily for taxonomic classification:amino acid sequencingfatty acid profilesflow cytometryDNA base compositionDNA fingerprintingrRNA sequencingPCRnucleic acid hybridization Data obtained from laboratory tests employing any (or all) of these thirteen techniques can be assimilated using numerical taxonomy to provide information on classification. 12. Most microorganisms do not contain structures that are readily fossilized, making it difficult to obtain information on the evolution of microorganisms. Recent developments in molecular biology have provided techniques for determining evolutionary relationships amongst bacteria. 13. A and D appear to be most closely related because they have similar G-C moles %. No two are the same species.Multiple ChoiceAnswers are in boldface.1. Bergey's Manual of Systematic Bacteriology differs from Bergey's Manual of Determinative Bacteriology in that the formera. groups bacteria into 19 species.b. groups bacteria according to shared properties.c. groups bacteria according to pathogenic properties.d. is based on phylogenetic relationships between bacteria.e. All of the above. 2. Bacillus and Lactobacillus are not in the same family. This indicates that which one of the following is not sufficient to assign an organism to a taxon?a. biochemical characteristicsb. amino acid sequencingc. phage typingd. serologye. morphological characteristics 3. Which of the following is used to classify organisms into the Kingdom Fungi?a. ability to photosynthesize; possess a cell wallb. unicellular; possess cell wall; procaryoticc. unicellular; lacking cell wall; eucaryoticd. absorptive; possess cell wall; eucaryotice. ingestive; lacking cell wall; multicellular; procaryotic 4. Which of the following is not true about scientific nomenclature?a. Each name is specific.b. Names vary with geographical location.c. The names are standardized.d. Each name consists of a genus and specific epithet.e. It was first designed by Linnaeus. 5. You could identify an unknown bacterium by all of the following excepta. hybridizing a DNA probe from a known bacterium with the unknown's DNA.b. making a fatty acid profile of the unknown.c. specific antiserum agglutinating the unknown.d. ribosomal RNAsequencing.e. percentage of guanine 1cytosine. 6. The wall-less mycoplasmas are considered to be related to gram-positive bacteria. Which of the following would provide the most compelling evidence for this?a. They share common rRNA sequences.b. Some gram-positive bacteria and some mycoplasmas produce catalase.c. Both groups are procaryotic.d. Some gram-positive bacteria and some mycoplasmas have coccus-shaped cells.e. Both groups contain human pathogens. Critical Thinking 1. A and D are most closely related. 2. Based on the nucleic acid composition, Micrococcus and Staphylococcus probably are not related. The current classification scheme uses morphological and biochemical similarities instead of phylogenetic relatedness to assign bacteria to taxa. 3. DNA probe. Labeled DNA will hybridize with homologous DNA indicating (a) identity if the probe is known DNA and the homologous DNA is from an unknown bacterium, or (b) relatedness when the two organisms are known. PCR. The primer used in PCR will hybridize with homologous DNA so unrelated DNA will not be copied. Or, after making copies by PCR, a DNA probe can be used to locate specific DNA. 4. SF = Streptococcus faecalis. SF broth is used to culture Enterococcus faecalis.Clinical Applications 1. The patient had plague. The Yersinia (gram-negative rod) was missed in the first Gram stain. After culturing gram-negative rods, biochemical tests were not conclusive because Yersinia is biochemically inactive. Plague can be transmitted by the respiratory route so the patient's contacts were given prophylactic antibiotic treatment. 2. Incorrect identification could be due to mutation to sucrose+, misreading of the indicator in the sucrose fermentation test, or contamination by a sucrose+ organism. 3. One possible key is shown below. Alternate keys could be made starting with morphology or glucose fermentation. 4.

Student Resources, Chapter 11 Answers to End-of-Chapter Study Questions Review 1. Key I.A. Endospore-forming rods and cocciB.1. Gram-positive cocci2.a. Regular, non-sporing gram-positive rodsb. Irregular, non-sporing gram-positive rodsc. Mycobacteria and Nocardioforms C.1. Nocardioforms2. Actinomycetes II.A.1. Spirochetes2.a. Aerobic/microaerophilic, motile, helical/vibroid gram-negative bacteriab. Anaerobic, gram-negative straight, curved, or helical rods B.1. Gram-negative aerobic rods and cocci2. Facultatively anaerobic, gram-negative rods3. Anaerobic gram-negative cocciC. Rickettsias and chlamydias III. Mycoplasmas 2. a. Both are oxygenic photoautotrophs. Cyanobacteria are procaryotes; algae are eucaryotes. b. Both are chemoheterotrophs capable of forming mycelia; some form conidia. Actinomycetes are procaryotes; fungi are eucaryotes. c. Both are large rod-shaped bacteria. Bacillus forms endospores, Lactobacillus is a fermentative non-endospore-forming rod. d. Both are small rod-shaped bacteria. Pseudomonas has an oxidative metabolism; Escherichia is fermentative. Pseudomonas has polar flagella; Escherichia has peritrichous flagella. e. Both are helical bacteria. Leptospira (a spirochete) has an axial filament. Spirillum has flagella. f. Both are gram-negative, anaerobic bacteria. Veillonella are cocci, and Bacteroides are rods. g. Both are obligatory intracellular parasites. h. Both lack peptidoglycan cell walls. Ureaplasma are archaea; Mycoplasma are eubacteria (see Table 10.2). 3. MatchingNitrogen-fixing (e) FrankiaAnoxygenic (j) Purple bacteriaOxygenic (b) CyanobacteriaOxidize NO (i) NitrobacterReduce CO2 (g) Methanogenic bacteria*Slime layer (c) Cytophaga and (h) MyxobacteriaMyxocysts (h) MyxobacteriaAnaerobic (d) DesulfovibrioThermophilic (l) Sulfolobus*Filaments (k) SphaerotilusProjections (f) HyphomicrobiumUnusual (a) Archaea*These are genera within (a) Archaea Multiple ChoiceAnswers are in boldface.1. To which of the following groups does an acid-fast rod belong?a. regular nonsporing gram-positive rodsb. Mycobacteriac. gram-negative aerobic rods and coccid. anaerobic gram-negative rodse. irregular nonsporing gram-positive rods 2. Members of the endospore-forming group of bacteria can have which of the following oxygen requirements?a. aerobicb. obligately anaerobicc. facultatively anaerobicd. microaerophiliee. all of the above 3. Pathogenic bacteria can bea. motile.b. rods.c. cocci.d. anaerobic.e. All of the above. 4. Which of the following is an intracellular parasite?a. Rickettsiab. Mycobacteriumc. Bacillusd. Staphylococcuse. Streptococcus 5. Which of the following terms is the most specific?a. Bacillusb. Bacillusc. gram-positived. endospore-forming rods and coccie. anaerobic 6. Which of the following produce most of the antibiotics that are used to treat diseases?a. actinomycetes and related organismsb. endospore-forming rods and coccic. Streptococcusd. aerobic gram-negative bacteriae. fungi 7. Which of the following pairs is mismatched?a. Facultatively anaerobic gram-negative rods-Escherichiab. Facultatively anaerobic gram-negative rods-Shigellac. Endospore-forming gram-positive rods and cocci-Clostridiumd. Irregular nonsporing gram-positive rods-Corynebacteriume. Spiral and curved bacteria-Leptospira 8. Spirilla is not classified as a spirochete because spirochetesa. do not cause disease.b. possess axial filaments.c. possess flagella.d. are procaryotes.e. None of the above. 9. After Legionella was discovered, it was classified with the pseudomonads becausea. it is a pathogen.b. it is an aerobic gram-negative rod.c. it is difficult to culture.d. it is found in water.e. None of the above. 10. Cyanobacteria differ from purple and green phototrophic bacteria because cyanobacteriaa. produce oxygen during photosynthesis.b. do not require light.c. use H2S as an electron donor.d. have a membrane-enclosed nucleus.e. All of the above. Critical Thinking 1. See Appendix A. a. Gram-positive are listed in volumes 2 and 4. b. Gram-negative are listed in volume 1. c. Wall-less are Mycoplasma. d. Unusual walls are Archaeobacteria. 2. a. Gardnerella is included in the facultatively anaerobic gram-negative rods and the irregular, non- sporing, gram-positive rods. They have gram-positive cell walls but stain gram-negative. They are gram-positive bacteria but are listed under gram-negative bacteria to facilitate laboratory identification. b. Desulfotomaculum is classified based on its (1) metabolism, in the Dissimilatory sulfate reducers and (2) morphology, in the Endospore-formers. c. Gallionella is classified based on its (1) metabolism, in the Aerobic chemolithotrophic bacteria and (2) morphology, in the Budding and/or appendaged bacteria. 3. a. Pseudomonas b. Bacillus c. Thiobacillus d. Methane-producing archaeaClinical Applications 1. Neisseria (meningitidis) 2. Salmonella (hadar) 3. Listeria (monocytogenes)

Student Resources, Chapter 12 Answers to End-of-Chapter Study Questions Review 1. Conidiospores are asexual spores formed by the aerial mycelia of one organism. Ascospores are sexual spores resulting from the fusion of the nuclei of two opposite mating strains of the same species of fungus. 2.Spore Type(s) PhylumAsexual SexualZygomycotaZygospore SporangiosporeAscomycotaConidiospore, Arthrospore AscosporeBasidiomycotaConidiospore BasidiosporeDeuteromycotaConidiospore, Chlamydospore, Arthrospore None3. Sexual spores. 4.GenusMycosis BlastomycesSystemic SporothrixSubcutaneous MicrosporumCutaneous TrichosporonSuperficial AspergillusSystemic 5. a. E. coli b. P. chrysogenum 6. LichensThe alga produces carbohydrates for the lichen, and the fungus provides both the holdfast and protection from dessication. 7. As the first colonizers on newly exposed rock or soil, lichens are responsible for the chemical weathering of large inorganic particles and the consequent accumulation of soil. 8.PhylumCell wall composition Special featuresDinoflagellates*Cellulose and silica Some produce neurotoxinsEuglenoids*None Facultative heterotrophsDiatoms*Pectin and silica Produce much O2; fossilized inclusions form petroleum Red algaeCellulose Grow in deep water; source of agarBrown algaeCellulose and alginic acid Source of alginGreen algaeCellulose Plantlike*Unicellular The green algae (Chlorophyta) could be placed in the plant kingdom. They have chlorophyll b, as do land plants, and have colonial forms. In the most advanced colonial form (Volvox), groups of Chlamydomonas-like cells live together; some are specialized for reproductive functions, which suggests a possible evolutionary route for the formation of plant tissue. The other algae are most often classified as protists. 9. Cellular slime molds exist as individual amoeboid cells. Plasmodial or acellular slime molds are multinucleate masses of protoplasm. Both survive adverse environmental conditions by forming spores. 10. Ingestion. 11. Trichomonas cannot survive for long outside of a host because it does not form a protective cyst. Trichomonas must be transferred from host to host quickly. 12. Asexual reproduction occurs in the human host and sexual reproduction takes place in the mosquito. The definitive host and the vector are the mosquito. 13. This is a cestode. The encysted larva is called a cysticercus. Tapeworms are dorsoventrally flattened and have an incomplete digestive system. 14. The male reproductive organs are in one individual, and the female reproductive organs in another. Nematodes belong to the Phylum Aschelminthes. 15.Vector TypeExampleDiseaseMechanicalHousefly SalmonellosisSuitable for reproduction of parasite IxodesLyme disease As a hostAnopheles MalariaMultiple ChoiceAnswers are in boldface.1. How many kingdoms are represented in the following list of organisms: Echinococcus, Anopheles, Aspergillus, Taenia, Toxoplasma, Trichinella.a. 1b. 2c. 3d. 4e. 5 2. Which of the following pairs is most closely related?a. Entamoeba and Giardiab. Giardia and Balantidiumc. Plasmodium and Entamoebad. Toxoplasma and Trichomonase. None of the above Use the following choices to answer questions 3 and 4:(1) metacercaria(2) redia(3) adult(4) miracidium(5) cercaria 3. Put the above stages in order of development, beginning with the egg.a. 5, 4, 1, 2, 3b. 4, 2, 5, 1, 3c. 2, 5, 4, 3, 1d. 3, 4, 5, 1, 2e. 2, 4, 5, 1, 3 4. If a snail is the first intermediate host of a parasite with these stages, which stage would be found in the snail?a. 1b. 2c. 3d. 4e. 5 5. Which of the following statements about yeasts are true?(1) Yeasts are fungi.(2) Yeasts can form pseudohyphae.(3) Yeasts reproduce asexually by budding.(4) Yeasts are facultatively anaerobic.(5) All yeasts are pathogenic.(6) All yeasts are dimorphic.a. 1, 2, 3, 4b. 3, 4, 5, 6c. 2, 3, 4, 5d. 1, 3, 5, 6e. 2, 3, 4 6. Which of the following events follows cell fusion in an ascomycete?a. conidiophore formationb. conidiospore germinationc. ascus openingd. ascospore formatione. conidiospore release 7. The definitive host for Plasmodium vivax isa. human.b. Anopheles.c. a sporocyte.d. a gametocyte. 8. Fleas are the intermediate host for Dipylidium caninum tapeworm and dogs are the definitive host. Which stage of the parasite could be found in the flea?a. cysticerus larvab. proglottidsc. scolexd. adult 9. These are obligate intracellular parasites that lack mitochondria.a. Apicomplexab. Ciliophorac. Microsporad. Sarcomastigophora 10. These are nonmotile parasites with special organelles for penetrating host tissue.a. Apicomplexab. Ciliophorac. Microsporad. Sarcomastigophora Critical Thinking 1. Intermediate hosts: snail, fishDefinitive host: humanPhylum: PlatyhelminthesClass: Trematode 2. Plasmodial slime molds have an internal transport system, called protoplasmic streaming, to ensure circulation. 3. Mammals (e.g., bears) are a more likely part of the freshwater ecosystem, so parasites would evolve to use mammalian hosts. 4. Information on related organisms can provide models for the laboratory culture and clinical treatment of infections. Knowledge about relationships can provide insight regarding evolution. 5. Phylum: ProtozoaClass: MastigophoraHost: HumanVector: tsetse flyClinical Applications 1. Taenia solium; ingestion of tapeworm eggs excreted by a household member. Prevention: handwashing to break the fecal-oral cycle. 2. Coccidioides immitis; inhalation of arthrospores. Prevention: avoid working in contaminated soils. 3. Malaria; transmitted by bite of Anopheles mosquito. 4. a. From the Elastoplast bandage. b. Fungal spores can survive on the dry bandage. 5. Blastomyces dermatitidis, which may be treated with amphotericin B or an imidazole.

Student Resources, Chapter 13 Answers to End-of-Chapter Study Questions Review 1. The term filterable describes the property of passing through filters that retain bacteria. Viruses are too small to be seen with a light microscope, but their presence is known because material passed through a filter is still capable of causing a disease. 2. Viruses absolutely require living host cells to multiply. 3. A virus: (1) Contains DNA or RNA; (2) Has a protein coat surrounding the nucleic acid; (3) Multiplies inside a living cell using the synthetic machinery of the cell; and (4) Causes the synthesis of virions. A virion is a fully developed virus particle that transfers the viral nucleic acid to other cells and initiates multiplication. 4. The capsid of a helical virus is a hollow cylinder with a helical shape, which surrounds the nucleic acid (See Figure 13.3). An example of a helical virus is tobacco mosaic virus. Polyhedral viruses are many-sided (Figure 13.4). A polyhedral virus in the shape of an icosahedron is adenovirus. Polyhedral or helical viruses surrounded by an envelope are called enveloped viruses. An example of an enveloped helical virus is influenza (Figure 13.5), and herpes simplex is an enveloped polyhedral virus (Figure 13.6). 5. A sample of bacteriophage is mixed with host bacteria and melted nutrient agar. The mixture is then poured over a layer of nutrient agar in a Petri plate. Each phage infects a bacterium, multiplies, and releases new phages. These newly produced phages infect other bacteria, and more new viruses are produced. Following multiplication, the bacteria are destroyed. This produces a number of clearings or plaques in the layer of bacteria. The number of phages in the original sample can be estimated by counting the number of plaques. 6. Primary cell lines tend to die after a few generations. Continuous cell lines can be maintained through an indefinite number of generations. Continuous cell lines then allow long-term observations of viruses. Continuous cell lines are transformed cells. 7. A prophage gene codes for the cholera toxin. When phage DNA is incorporated into the cell's DNA, toxin can be produced. 8. Adsorption: The virus attaches to the cell membrane by means of spikes located on its envelope. Penetration: The virus gains entrance by pinocytosis, or its envelope may fuse with the plasma membrane of the host cell, allowing the virus to enter the cell. Uncoating: Uncoating refers to the separation of the capsid from the viral DNA. Biosynthesis: Viral DNA is released into the cell's nucleus, and transcription and translation from viral DNA occur. Viral DNA is synthesized. Maturation: Capsids form around strands of viral DNA. Release: The assembled capsid-containing nucleic acid pushes through the plasma membrane; a portion of the plasma membrane adheres to the capsid, thus forming the envelope. 9. a. Viruses cannot easily be observed in host tissues. Viruses cannot easily be cultured in order to be inoculated into a new host. Additionally, viruses are specific for their hosts and cells, making it difficult to substitute a laboratory animal for the third step of Koch's postulates. b. Some viruses can infect cells without inducing cancer. Cancer may not develop until long after infection. Cancers do not seem to be contagious. 10. Subacute sclerosing panencephalitis…common viruses…Students will have to suggest a mechanism to fill in the last blank; some examples are latent, in an abnormal tissue. 11. ProvirusTSTA appear on the host cell surface, or T antigens appear in the nucleus. Transformed cells do not exhibit contact inhibition. RNA-containing oncogenic viruses produce a double-stranded DNA molecule using reverse transcriptase. The DNA is integrated into the host cell's DNA as a provirus. The provirus may transform the host cell into a tumor cell. 12. Prions are infectious proteins that appear to lack any nucleic acid. Viroids are infectious RNAs that do not have a protein coat. A prion causes scrapie. A viroid causes potato spindle tuber viroid disease. 13. Of the rigid cell walls…vectors such as sap-sucking insects…plant protoplasts and insect cell cultures. Multiple ChoiceAnswers are in boldface.1. Place the following in the most likely order for biosynthesis of a bacteriophage: (1) phage lysozyme (2) mRNA(3) DNA (4) viral proteins(5) DNA polymerase.a. 5, 4, 3, 2, 1b. 1, 2, 3, 4, 5c. 5, 3, 4, 2, 1d. 3, 5, 2, 4, 1e. 2, 5, 3, 4, 1 2. The molecule serving as mRNA can be incorporated in the newly synthesized virus capsids of all of the following excepta. 1 strand RNA picornaviruses.b. 1 strand RNA togaviruses.c. 2 strand RNA rhabdoviruses.d. double-stranded RNA reoviruses.e. double-stranded DNA herpesviruses. 3. A virus with RNA-dependent RNA polymerasea. synthesizes DNA from an RNA template.b. synthesizes double-stranded RNA from an RNA template.c. synthesizes double-stranded RNA from a DNA template.d. transcribes mRNA from DNA.e. None of the above. 4. Which of the following would be the first step in the biosynthesis of a virus with reverse transcriptase?a. A complementary strand of RNA must be synthesized.b. Double-stranded RNA must be synthesized.c. A complementary strand of DNA must be synthesized from an RNA template.d. A complementary strand of DNA must be synthesized from a DNA template.e. None of the above. 5. An example of lysogeny in animals could bea. slow viral infections.b. latent viral infections.c. T-even bacteriophages.d. infections resulting in cell death.e. None of the above. 6. The ability of a virus to infect an organism is regulated bya. the host species.b. the type of cells.c. the availability of an attachment site.d. cell factors necessary for viral replication.e. All of the above. 7. Which of the following statements is not true?a. Viruses contain DNA or RNA.b. The nucleic acid of a virus is surrounded by a protein coat.c. Viruses multiply inside living cells using viral mRNA, tRNA, and ribosomes.d. Viruses cause the synthesis of specialized infectious elements.e. Viruses multiply inside living cells. 8. Place the following in the order in which they are found in a host cell: (1) capsid proteins(2) infective phage particles(3) phage nucleic acid.a. 1, 2, 3b. 3, 2, 1c. 2, 1, 3d. 3, 1, 2e. 1, 3, 2 9. Which of the following does not initiate DNA synthesis?a. a double-stranded DNA virus (Poxviridae)b. a DNA virus with reverse transcriptase (Hepadnaviridae)c. an RNA virus with reverse transcriptase (Retroviridae)d. a single)stranded RNA virus (Togaviridae)e. none of the above.10. A viral species is not defined on the basis of the disease symptoms it causes. The best example of this isa. polio.b. rabies.c. hepatitis.d. chickenpox and shingles.e. measles. Critical Thinking 1. Outside of living cells, viruses are inert. They cannot ingest and metabolize nutrients, and they cannot reproduce. These are descriptions one might use for chemicals, not living organisms. However, inside a living cell, viruses can multiply. Clinically, since they cause infection and disease, they might be considered alive. 2. A virus is small and cannot hold as much DNA as a cell. Genes that code for proteins that serve two functions conserve space on a viral nucleic acid. 3. These two diseases provide animal models for the study of acquired immunodeficiencies and treatments. Study of the viruses (SIV and FIV) can provide more information regarding the evolution of retroviruses. 4. A prophage, provirus, or plasmid begins as a strand of DNA outside the cell's chromosome that can be integrated into the chromosome. Like a plasmid, a prophage carries genes that can be used by the cell but are not essential. Prophages and proviruses are replicated with the cell's chromosome and remain in progeny cells. Prophage DNA will form a circle and replicate itself in the cell's cytoplasm. Unlike a plasmid, prophages and proviruses are not transferred in conjugation, and when they replicate themselves, viruses are produced that can destroy the host cell.Clinical Applications 1. Cytomegalovirus. The negative staining results indicated that the cause of her disease was not bacterial. The viral culture revealed the cause of her symptoms. 2. Herpes simplex virus. Presence of antibodies against this virus would confirm the etiology. 3. Hepatitis; these people acquired hepatitis A virus from contaminated ice-slushes. Picornaviridae Hepatitis A Virus Ingestion +RNA, ss Nonenveloped Hepadnaviridae Hepatitis B Virus Injection DNA, ds Enveloped Uses reverse transcriptase Flaviviridae Hepatitis C Virus Injection +RNA, ss Enveloped

Student Resources, Chapter 14 Answers to End-of-Chapter Study Questions Review 1. a. Etiology is the study of the cause of a disease, whereas pathogenesis is the manner in which the disease develops. b. Infection refers to the colonization of the body by a microorganism. Disease is any change from a state of health. A disease may, but does not always, result from infection. c. A communicable disease is a disease that is spread from one host to another, whereas a noncommunicable disease is not transmitted from one host to another. 2. Microorganisms that reside more or less permanently on the body are called normal microbiota. Microorganisms that are present for a few days or weeks are transient microbiota. 3. Symbiosis refers to different organisms living together. Commensalism is a symbiotic relationship in which one of the organisms is benefited and the other is unaffected. Corynebacteria living on the surface of the eye are commensals. Mutualism is a symbiosis in which both organisms are benefited. E. coli receives nutrients and a constant temperature in the large intestine and produces vitamin K and certain B vitamins that are useful for the human host. In parasitism, one organism benefits while the other is harmed. Salmonella typhi receives nutrients and warmth in the large intestine, and the human host experiences typhoid fever. 4. A reservoir of infection is a source of continual infection. Matching b Influenza c Rabies a Botulism 5. Koch's postulates establish the etiology of an infectious disease because the microorganism is removed from a sick organism, grown in a laboratory culture, and introduced into a healthy, susceptible organism. This shows that the microorganism caused the disease-not contact with a sick individual or environmental conditions. Some organisms are not easily seen in a host. Some microorganisms cannot be cultured on laboratory media. And some microorganisms are specific for one host. In a human, these pathogens will give rise to a group of signs and symptoms, but they will not cause the same disease in laboratory animals. 6. a. In transmission by direct contact, some kind of body contact between an infected individual and a susceptible host is required. b. Pathogens are transmitted from one host to another by fomites via indirect contact. c. Arthropod vectors can transmit pathogens mechanically where the pathogen is carried on external body parts. When an arthropod ingests a pathogen and the pathogen reproduces in the vector, it is called biological transmission. d. Droplet transmission also is a method of contact transmission. Pathogens are transmitted by droplets of saliva or mucus. e. Pathogens can be transmitted to a large number of individuals by food or water. This is called common-vehicle transmission. f. Airborne transmission refers to the spread of pathogens by droplet nuclei or dust. 7. Nutrition, fatigue, age, habits, lifestyle, occupation, preexisting illness, chemotherapy. 8. a. Acute b. Chronic c. Subacute 9. Hospital patients may be in a weakened condition and therefore predisposed to infection. Pathogenic microorganisms are generally transmitted to patients by contact and airborne transmission. The reservoirs of infection are the hospital staff, visitors, and other patients. 10. A disease constantly present in a population is an endemic disease. When many people acquire the disease in a relatively short time, it is an epidemic disease. 11. Epidemiology is the science dealing with when and where diseases occur and how they are transmitted in the human population. The Centers for Disease Control and Prevention (CDC) is a central source of epidemiological information. 12. Changes in body function felt by the patient are called symptoms. Symptoms such as weakness or pain are not measurable by a physician. Objective changes that the physician can observe and measure are called signs. 13. When microorganisms causing a local infection enter a blood or lymph vessel and are spread throughout the body, a systemic infection can result. 14. Mutualistic microorganisms are providing a chemical or environment that is essential for the host. In a commensal relationship, the microorganisms are obtaining nutrients from sloughed-off cells and secretions, which benefits the host by removing materials that might be invaded by pathogens. These organisms, however, are not essential; another microorganism might serve the function as well. 15. Incubation period, prodromal period, period of illness, period of decline (may be crisis), period of convalescence. Multiple ChoiceAnswers are in boldface.1. The emergence of new infectious diseases is probably due to all of the following except a. the need of bacteria to cause disease.b. the ability of humans to travel by air.c. changing environments (e.g., flood, drought, pollution).d. a pathogen crossing the species barrier.e. the increasing human population. 2. All members of a group of ornithologists studying barn owls in the wild have had salmonellosis (Salmonella gastroenteritis). One birder is experiencing her third infection. What is the most likely source of their infections?a. The ornithologists are eating the same food.b. They are contaminating their hands while handling the owls and nests.c. One of the workers is a Salmonella carrier.d. Their drinking water is contaminated. 3. Which of the following statements is not true?a. E. coli never causes disease.b. E. coli provides vitamin K for its host.c. E. coli often exists in a mutual relationship with humans.d. E. coli gets nutrients from intestinal contents. 4. Which of the following is not one of Koch's postulates?a. The same pathogen must be present in every case of the disease.b. The pathogen must be isolated and grown in pure culture from the diseased host.c. The pathogen from pure culture must cause the disease when inoculated into a healthy, susceptible laboratory animal.d. The disease must be transmitted from a diseased animal to a healthy, susceptible animal by some form of contact.e. The pathogen must be isolated in pure culture from an experimentally infected lab animal. Use the following information to answer questions 5-7. On September 6, a 6-year old boy experienced fever, chills, and vomiting. On September 7, he was hospitalized with diarrhea and swollen lymph nodes under both arms. On September 3, the boy had been scratched and bitten by a cat. The cat was found dead on September 5, and Yersinia pestis was isolated from the cat. Chloramphenicol was administered to the boy from September 7, when Y. pestis was isolated from him. On September 17, the boy's temperature returned to normal, and on September 22, he was released from the hospital. 5. Identify the incubation period for this case of bubonic plague.a. September 3-5.b. September 3-6.c. September 6-7.d. September 6-17. 6. Identify the prodromal period for this disease.a. September 3-5.b. September 3-6.c. September 6-7.d. September 6-17. 7. Identify the crisis during this disease.a. September 6.b. September 7.c. September 6-17.d. September 17. Use the following information to answer questions 8-10.A Maryland woman was hospitalized with dehydration; Vibrio cholerae and Plesiomonas shigelloides were isolated from the patient. She had neither traveled outside the United States nor eaten raw shellfish during the preceding month. She had attended a party 2 days before her hospitalization. Two other people at the party had acute diarrheal illness and elevated levels of serum antibodies against Vibrio. Everyone at the party ate crabs and rice pudding with coconut milk. Crabs left over from this party were served at a second party. One of the 20 people at the second party had onset of mild diarrhea; specimens from 14 of these people were negative for vibriocidal antibodies. 8. This is an example ofa. vehicle transmission.b. airborne transmission.c. transmission by fomites.d. direct contact transmission.e. nosocomial transmission. 9. The etiologic agent of the disease isa. Plesiomonas shigelloides.b. crabs.c. Vibrio cholerae.d. coconut milk.e. rice pudding. 10. The source of the disease wasa. Plesiomonas shigelloides.b. crabs.c. Vibrio cholerae.d. coconut milk.e. rice pudding. Critical Thinking 1. Koch provided reproducible steps using scientific methodology. De Bary was correct but did not provide proof or experimental procedures that would provide proof. Recall that Koch also developed culturing and staining procedures. 2. Nightingale collected data on infected persons (descriptive epidemiology) and learned that the place of most infections was in the combat area. She compared two groups of soldiers, those at home (control) and those in battle (experimental) (analytical epidemiology) to determine what factors contributed to disease. She then instituted sanitation measures in the Crimea, and the incidence of disease decreased (experimental epidemiology). Disease Transmission Prevention Cholera Water Proper disposal of sewage; disinfection of water. Typhus Body lice Washing; disinfection of clothes and bedding. Also see: I. B. Cohen, "Florence Nightingale." Scientific American 250:128-137, March 1984. 3. a. Malaria-Vector b. Tuberculosis-Airborne c. Nosocomial infections-Any method, although vectorborne is unlikely d. Salmonellosis-Common vehicle e. Streptococcal pharyngitis-Direct contact f. Mononucleosis-Droplet g. Measles-Direct contact; airborne h. Hepatitis A-Common vehicle; direct contact i. Tetanus-Indirect contact j. Hepatitis B-Indirect contact k. Chlamydial urethritis-Direct contact 4. Information for the world population would have to be provided to indicate whether a pandemic state exists. Typhoid fever is transmitted by the fecal-oral route, usually through water. 5. Cholera is transmitted by the fecal-oral route, usually through contaminated water. V. cholerae was brought in contaminated fish carried by tourists. People in the United States contracted cholera eating these fish. V. cholerae entered coastal waters when ships from South American waters released bilge water.Clinical Applications 1. Mistakes: Exposure to nasopharyngeal secretions and failure to get antibiotic therapy. Meningitis is transmitted by the respiratory route. 2. Probable source: contaminated cleaning water. Pseudomonads can tolerate a wide range of environmental conditions (e.g., low temperatures) and can grow on unusual carbon sources, including many detergents and disinfectants. 3. February 7 to March 9: incubation period.March 9: prodromal period.March 10 to March 17: period of illness.March 17: period of decline by crisis.Next 2 weeks: period of convalescence. Psittacosis is caused by Chlamydia psittaci. 4. The infection is probably transmitted from patient to patient on the unwashed hands of hospital staff. Hospital staff must wash their hands between patients, wear gloves when appropriate, and wash their hands after removing gloves. 5. Mycobacterium is usually transmitted by the respiratory route. The hospital source seems to be the water system. The boiler and pipes need to be cleaned and disinfected. Biofilms must be prevented from growing in the boiler and pipes.

Student Resources, Chapter 15 Answers to End-of-Chapter Study Questions Review 1. Mucous membranes: Microorganisms can adhere to and then penetrate mucous membranes. Skin: Microorganisms can penetrate unbroken skin through hair follicles and sweat ducts. Parenteral route: Pathogens can be introduced into tissues beneath the skin and mucous membranes by punctures, injections, bites, and cuts. 2. The ability of a microorganism to produce a disease is called pathogenicity. The degree of patho-genicity is virulence. 3. a. Would prevent adherence by making the mannose attachment site unavailable. b. Would prevent adherence of N. gonorrhoeae. c. S. pyogenes would not be able to attach to host cells and would be more susceptible to phagocytosis. 4. Cytopathic effects are observable changes produced in cells infected by viruses. Five examples are: a. Cessation of mitosis. b. Autolysis. c. The presence of inclusion bodies. d. Cell fusion producing multinucleated "giant" cells. e. Transformation. 5. Exotoxin Endotoxin Bacterial source Gram + Gram - Chemistry Proteins Lipid A Toxigenicity High Low Pharmacology Destroy certain cell parts or physiological functions Systemic, fever, weakness, aches, and shock Example Botulinum toxin Salmonellosis 6. Encapsulated bacteria can resist phagocytosis and continue growing. Streptococcus pneumoniae and Klebsiella pneumoniae produce capsules that are related to their virulence. M protein found in the cell walls of Streptococcus pyogenes and A protein in the cell walls of Staphylococcus aureus help these bacteria resist phagocytosis. 7. Hemolysins are enzymes that cause the lysis of red blood cells. Hemolysis might supply nutrients for bacterial growth. Leukocidins destroy neutrophils and macrophages that are active in phagocytosis. This decreases host resistance to infection. Coagulase is an enzyme that causes the fibrinogen in blood to clot. The clot may protect the bacterium from phagocytosis and other host defenses. Bacterial kinases break down fibrin. Kinases can destroy a clot that was made to isolate the bacteria, thus allowing the bacteria to spread. Hyaluronidase dissolves the hyaluronic acid that binds cells together. This could allow the bacteria to spread through tissues. 8. Pathogenic fungi do not have specific virulence factors; capsules, metabolic products, toxins, and allergic responses contribute to the virulence of pathogenic fungi. Some fungi produce toxins that, when ingested, produce disease. Protozoa and helminths elicit symptoms by destroying host tissues and producing toxic metabolic wastes. 9. Treponema. 10. Botulinum toxin is more potent than Salmonella toxin. A much smaller amount of botulinum toxin will kill 50 percent of the inoculated hosts. 11. Food infection refers to a disease that results from pathogens entering through the gastrointestinal route. The pathogens infect the gastrointestinal tract and produce endotoxins while they are growing. Food intoxication results from ingestion of a toxin formed in food. Pathogens grow in the food and excrete an exotoxin. The pathogens do not infect the host; symptoms are due to the toxin. 12. Viruses avoid the host's immune response by growing inside host cells; some can remain latent in a host cell for prolonged periods. Some protozoa avoid the immune response by mutations that change their antigens.Multiple ChoiceAnswers are in boldface.1. The removal of plasmids reduces virulence in which of the following organisms?a. Clostridium tetanib. Escherichia colic. Staphylococcus aureusd. Streptococcus mutanse. Clostridium botulinum 2. What is the LD50 for the bacterial toxin tested in the example below?Dilution, No. of Animals Died, No. of Animals Survived a. Dilution: 6 µg/kg, No of Animals Died: 0, Survived: 6b. Dilution: 12.5 µg/kg, No of Animals Died: 0, Survived: 6c. Dilution: 25 µg/kg, No of Animals Died: 3, Survived: 3d. Dilution: 50 µg/kg, No of Animals Died: 4, Survived: 2e. Dilution: 100 µg/kg, No of Animals Died: 6, Survived: 0 3. Which of the following is not a portal of entry for pathogens?a. mucous membranes of the respiratory tractb. mucous membranes of the gastrointestinal tractc. skind. bloode. parenteral route 4. All of the following can occur during bacterial infection. Which would prevent all of the others?a. vaccination against fimbriaeb. phagocytosisc. inhibition of phagocytic digestiond. destruction of adhesinse. alteration of cytoskeleton 5. The ID50 for bacterium A is 250 cells; the ID50 for bacterium B is 10 cells. Which of the following statements is not true?a. Both A and B are pathogens.b. Both bacteria produce infections in 50% of the inoculated hosts.c. B is more virulent than A.d. A and B are equally virulent; they cause infections in the same number of test animals.e. The severity of infections caused by A and B cannot be determined by the information provided. 6. An encapsulated bacterium can be virulent because the capsulea. resists phagocytosis.b. is an endotoxin.c. destroys host tissues.d. interferes with physiological processes.e. has no effect; since many pathogens do not have capsules, capsules do not contribute to virulence. 7. A drug that binds to mannose on human cells would preventa. the entrance of Vibrio enterotoxin.b. the attachment of pathogenic E. coli.c. the action of botulinum toxin.d. streptococcal pneumonia.e. the action of diphtheria toxin. 8. The earliest smallpox vaccines were infected tissue rubbed into the skin of a healthy person. The recipient of such a vaccine usually developed a mild case of smallpox, recovered, and was immune thereafter. The most likely reason this vaccine did not kill more people is:a. Skin is the wrong portal of entry for smallpox.b. The vaccine consisted of a mild form of the virus.c. Smallpox is normally transmitted by skin-to-skin contact.d. Smallpox is a virus.e. The virus mutated. 9. Which of the following does not represent the same mechanism for avoiding host defenses as the others?a. Rabies virus attaches to the receptor for the neurotransmitter acetylcholine.b. Salmonella attaches to the receptor for epidermal growth factor.c. EB virus binds to the host receptor for C3.d. Surface protein genes in Neisseria gonorrhoeae mutate frequently.e. None of the above. 10. Which of the following statements is true?a. The primary goal of a pathogen is to kill its host.b. Evolution selects for the most virulent pathogens.c. A successful pathogen doesnÕt kill its host before it is transmitted.d. A successful pathogen never kills its host. Critical Thinking 1. Toxins can be carried on plasmids. (Enterotoxins are carried on plasmids.) Bacteriophages could introduce chromosomally carried toxins via lysogeny. (Perhaps the Shigella-like toxins originated in this way.) Increased incidence in summer months suggests fecal-oral transmission. The disease is associated with the seasonal use of recreational waters and the lack of rainfall to provide clean water. 2. During the summer (highest light intensity), a smaller dose produces symptoms. 3. Yersinia Avoids destruction by complement Plague Helicobacter Neutralizes stomach acid Peptic Disease Syndrome Rhinovirus Gains access to host cells to avoid the immune reponse Common cold Salmonella Gains access to the host cells to avoid the immune response (mimics substrate for receptor on host cells) Gastroenteritis 4. When injected into rats, the ID50 for Salmonella typhimurium is 106 cells. If sulfonamides are injected with the salmonellae, the ID50 is 35 cells. Explain the change in ID50 value. Clinical Applications 1. Clostridium tetani growing at the site of the wound produced an exotoxin. Her pain and spasms were due to an infection. This disease, tetanus, cannot be transmitted to another person. 2. a. Infection. Vibrio parahaemolyticus. b. Intoxication. Ciguatera. 3. Pseudomonas is a gram-negative bacterium. Endotoxin from killed bacteria was present in the water. 4. Salmonella use the host's cytoskeleton to enter the cell. If the drug that inhibits cell division affects arrangement of the cytoskeleton, Salmonella will not be able to enter the cell.

Student Resources, Chapter 16 Answers to End-of-Chapter Study Questions Review 1. a. The ability of the human body to ward off diseases. b. The lack of resistance to an infectious disease. c. Host defenses that tend to protect the body from any kind of pathogen. 2. Mechanical Chemical Skin Dry, packed cells Sebum Eyes Tears Lysozyme Digestive tract Movement out HCl Respiratory tract Ciliary escalator Urinary tract Movement out Genital tract Movement out Acidic in female 3. See Table 16.1. 4. Phagocytosis is the ingestion of a microorganism or any foreign particulate matter by a cell. 5. Granulocytes have granules in the cytoplasm. Among the granulocytes, neutrophils have the most prominent phagocytic activity. Monocytes are agranulocytes (without granules) that develop into macrophages. When an infection occurs, granulocytes migrate to the infected area. Monocytes follow the granulocytes to the infected tissue. During migration, monocytes enlarge and develop into actively phagocytic cells called macrophages. Macrophages phagocytize dead or dying bacteria. 6. Phagocytic cells that migrate to the infected area are called wandering macrophages. Fixed macrophages remain in certain tissues and organs. 7. Refer to Figures 16.7 and 16.8. 8. Inflammation is the body's response to tissue damage. The characteristic symptoms of inflammation are redness, pain, heat, and swelling. 9. The functions of inflammation are: (1) To destroy the injurious agent, if possible, and to remove it and its by-products from the body; (2) If destruction is not possible, to confine or wall off the injurious agent and its by-products by forming an abscess; (3) To repair or replace tissues damaged by the injurious agent or its by-products. 10. Leukocytic pyrogen, released from phagocytic granulocytes, has the ability to raise body temperature. The higher temperature is believed to inhibit the growth of some microorganisms. The higher temperature speeds up body reactions and may help body tissues to repair themselves more quickly. 11. The chill is an indication that body temperature is rising. Shivering and cold skin are mechanisms for increasing internal temperature. Crisis indicates body temperature is falling. The skin becomes warm as circulation is returned to it when the body attempts to dissipate extra heat. 12. Complement is a group of proteins found in normal blood serum. See Figures 16.10 and 16.11. 13. Activation of complement can result in immune adherence and phagocytosis, local inflammation, and cell lysis. 14. Endotoxin binds C3b, which activates C5-C9 to cause cell lysis. This can result in free cell wall fragments, which bind more C3b, resulting in C5-C9 damage to host cell membranes. 15. Interferons are antiviral proteins produced by infected cells in response to viral infections. Alpha-IFN and b-IFN induce uninfected cells to produce antiviral proteins. Gamma-IFN is produced by lymphocytes and activates neutrophils to kill bacteria. Chapter SixteenAnswers are in boldface.1. Legionella uses C3b receptors to enter monocytes. Thisa. prevents phagocytosis.b. degrades complement.c. inactivates complement.d. prevents inflammation.e. prevents cytolysis. 2. Chlamydia can prevent the formation of phagolysosomes, and therefore Chlamydia cana. avoid being phagocytized.b. avoid destruction by complement.c. prevent adherence.d. avoid being digested.e. None of the above. 3. If the following are placed in the order of occurrence, which would be the third step?a. emigrationb. digestionc. formation of a phagosomed. formation of a phagolysosomee. margination 4. If the following are placed in the order of occurrence, which would be the third step?a. activation of C5-C9b. cell lysisc. antigen-antibody reactiond. activation of C3e. activation of C2-C4 5. A human host can prevent a pathogen from getting enough iron bya. reducing dietary intake of iron.b. binding iron with transferrin.c. binding iron with hemoglobin.d. excreting excess iron.e. binding iron with siderophores. 6. A decrease in the production of C3 would result ina. increased susceptibility to infection.b. increased numbers of white blood cells.c. increased phagocytosis.d. activation of C5-C9.e. None of the above. 7. In 1884, Elie Metchnikoff observed blood cells collected around a splinter inserted in a sea star embryo. This was the discovery ofa. blood cells.b. sea stars.c. phagocytosis.d. immunity.e. None of the above. 8. Helicobacter pylori uses the enzyme urease to counteract a chemical defense in the human organ in which it lives. This chemical defense isa. lysozyme.b. hydrochloric acid.c. superoxide radicals.d. sebum.e. complement. 9. Which of the following statements about a-IFN is not true?a. It interferes with viral replication.b. It is host specific.c. It is released by fibroflasts.d. It is virus-specific.e. It is released by lymphocytes. 10. Which of the following does not stimulate phagocytes?a. cytokinesb. g-IFNc. C3bd. lipid Ae. histamine Critical Thinking 1. Transferrin binds available iron so bacteria can't have it to grow. A bacterium might respond with increased siderophores to take up iron. 2. The inflammatory response is usually a beneficial response. Exceptions to this are hypersensitivities and autoimmune diseases, which are discussed in Chapter 18. Each of the drugs has side effects; while reducing inflammation, another undesirable condition might result. 3. Organism How does this strategy avoid destruction by the complement? Disease Group A streptococci No C5-C9 Streptococcal sore throat Hemophilus influenzae type b Hides LPS, which can activate C Meningitis Pseudomonas aeruginosa Binds C in solution instead of on cell surface Septicemia; pyelonephritis Trypanosoma cruzi C5-C9 doesn't get activated Chagas' disease 4. Microorganisms Effect Disease Influenzavirus Kills host cell Influenza M. tuberculosis Prevents digestion in phagocytes Tuberculosis T. gondii Prevents digestion in phagocytes Toxoplasmosis Trichophyton Digests keratin Athlete's foot T. cruzi Prevents digestion in phagocytes Chagas' disease Clinical Applications 1. Kinins cause vasodilation and increased permeability of blood vessels. Symptoms should include increased secretions from the nose and eyes. Rhinoviruses cause the common cold. 2. The proportions of white blood cells may change during diseases. The results of a differential count can be used to diagnose diseases. A patient with mononucleosis will have an increased number of monocytes. Neutropenia: decreased neutrophils. Eosinophilia: increased eosinophils. 3. Phagocytosis is inhibited. 4. Neutrophils will not phagocytize and they will die prematurely.

Student Resources, Chapter 17 Answers to End-of-Chapter Study Questions Review 1. The ability to produce antibodies against microorganisms and their toxins provides a type of resistance called immunity. 2. a. Nonspecific defenses are designed to protect you against any kind of microorganism. Immunity or specific resistance involves the production of antibodies. Antibodies are directed against specific microorganisms. b. Humoral immunity is due to antibodies (and B cells). Cell-mediated immunity is due to T cells. c. Active immunity refers to antibodies produced by the individual who carries them. Passive immunity refers to antibodies produced by another source and then transferred to the individual who needs the antibodies. d. Acquired immunity is the resistance to infection obtained during the life of the individual. Acquired immunity results from the production of antibodies. Innate resistance refers to the resistance of species or individuals to certain diseases that is not dependent on antigen-specific immunity such as antibodies. e. Natural immunity is acquired naturally, i.e., from mother to newborn, or following an infection. Artificial immunity is acquired from medical treatment, i.e., by injection of antibodies or by vaccination. f. T-dependent antigens: Certain antigens must combine with self-antigens to be recognized by TH cells and then by B cells. T-independent antigens can elicit an antibody response without T cells. g. T cells can be classified by their surface antigens: TH cells possess the CD4 antigen; TC and TS cells have the CD8 antigen. 3. a. Artificially acquired active immunity. b. Naturally acquired active immunity. c. Naturally acquired passive immunity. d. Artificially acquired passive immunity. 4. An antigen is a chemical substance that causes the body to produce specific antibodies and can combine with these antibodies. A hapten is a low-molecular-weight substance that is not antigenic unless it is attached to a carrier molecule. Once an antibody has been formed against the hapten, the hapten alone will react with the antibodies independently of its carrier. 5. An antibody is a protein produced by the body in response to the presence of an antigen; it is capable of combining specifically with that antigen. Antibodies are proteins and usually consist of four polypeptide chains. Two of the chains are identical and are called heavy (H) chains. The other two chains are identical to each other but are of lower molecular weight and are called light (L) chains. The variable portions of the H and L chains are where antigen binding occurs. The variable portion is different for each kind of antibody. The remaining constant portions of each chain are identical for all of the antibodies in one class of antibody. Refer to Figure 17.5 for the structure of IgG antibodies. 6. Each person has a population of B cells with receptors for different antigens. When the appropriate antigen contacts the antigen receptor on a B cell, the cell proliferates to produce a clone of cells. Plasma cells in this clone produce antibodies specific to the antigen that caused their formation. 7. See Figures 17.7, 17.12, 17.13, 17.15, 17.17, and 17.18. 8. Cytotoxic T cells (TC) destroy target cells upon contact. Delayed hypersensitivity T cells (TD)produce lymphokines. Helper T cells (TH) interact with an antigen to "present" it to a B cell for antibody formation. Suppressor T cells (TS) inhibit the conversion of B cells into plasma cells. Lymphokines cause an inflammatory response. An example of a cytokine is macrophage chemotactic factor, which attracts macrophages to the infection site. See Table 17.2 for functions of other cytokines. 9. a. Area a shows the primary response to the antigen. Area b shows the anamnestic response, in which the antibody titer is greater and remains high longer than in the primary response. The booster dose stimulated the memory cells to respond to the antigen. 10. Neutralize toxins, inactivate viruses, fix complement to initiate cytolysis. 11. Surface recognition sites for antigen peptides and MHC proteins. 12. NK cells lyse target cells (usually tumor cells and virus-infected cells) on contact. 13. Both would prevent attachment of the pathogen; (a) interferes with the attachment site on the pathogen and (b) interferes with the pathogen's receptor site. 14. See Figure 17.10. 15. The person recovered because s/he produced antibodies against the pathogen. The memory response will continue to protect the person against that pathogen. 16. Human gamma globulin is the fraction of human serum in which antibodies are found. If antibodies against hepatitis are in the gamma globulin, this would be artificially acquired passive immunity. Multiple ChoiceAnswers are in boldface.1. The type of protection provided by the injection of a toxoid.a. innate resistanceb. naturally acquired active immunityc. naturally acquired passive immunityd. artificially acquired active immunitye. artificially acquired passive immunity 2. The type of protection provided by the injection of an antitoxin.a. innate resistanceb. naturally acquired active immunityc. naturally acquired passive immunityd. artificially acquired active immunitye. artificially acquired passive immunity 3. The type of protection resulting from recovery from an infection.a. innate resistanceb. naturally acquired active immunityc. naturally acquired passive immunityd. artificially acquired active immunitye. artificially acquired passive immunity 4. Antibodies that protect the fetus and newborn. a. IgAb. IdDc. IgEd. IgGe. IgM 5. The first antibodies synthesized; especially effective against microorganisms.a. IgAb. IdDc. IgEd. IgGe. IgM 6. Antibodies that are bound to mast cells and involved in allergic reactions.a. IgAb. IdDc. IgEd. IgGe. IgM 7. Put the following in the correct sequence to elicit an antibody response: 1. TH cell recognizes B cell; 2. APC phagocytizes antigen; 3. antigen digest goes to surface of APC; 4. TH recognizes antigen digest and MHC; 5. B cell recognizes antigen.a. 1, 2, 3, 4, 5b. 5, 4, 3, 2, 1c. 3, 4, 5, 1, 2d. 2, 3, 4, 1, 5e. 4, 5, 3, 1, 2 8. Responsible for the differentiation of T and NK cells.a. antigenb. haptenc. IL-1d. IL-2e. perforin 9. Activates a T cell to bind IL-2.a. antigenb. haptenc. IL-1d. IL-2e. perforin 10. Patients with Chediak-Higashi syndrome suffer from various types of cancer. These patients are most likely lacking which of the following:a. TD cellsb. TH cellsc. B cellsd. NK cellse. TS cells Critical Thinking 1. TC cells secrete TNFg-IFN, which diffuse through liver cells and stimulate these cells to produce antiviral proteins. 2. a. IL-2 stimulates proliferation and differentiation of T cells and natural killer (NK) cells. NK cells and some T cells are effective against cancer cells. b. IL-2 stimulates the immune response, which increases the unwanted immune response as well. 3. Having had an M. tuberculosis infection and recovered (naturally acquired active immunity); vaccination with BCG (artificially acquired passive immunity). Clinical Applications 1. Antibiotics and immunity can cause gram-negative cells to lyse, releasing cell wall fragments. This exposes the body to more endotoxin. The woman's life-threatening condition was due to endotoxin shock. Monoclonal antibodies removed the cell walls. 2. Increased susceptibility to infection due to decreased antibody formation. 3. He could not produce the secretory component of IgA. 4. The mechanism is called antibody enhancement. Immune complexes of antibodies and viruses attach to cells, facilitating viral penetration.

Student Resources, Chapter 18 Answers to End-of-Chapter Study Questions Review 1. a. Whole-agent. Live, avirulent virus that can cause the disease if it mutates back to its virulent state. b. Whole-agent; (heat-) killed bacteria. c. Subunit; (heat- or formalin-) inactivated toxin. d. Subunit e. Subunit f. Conjugated g. Nucleic acid 2. If excess antibody is present, an antigen will combine with several antibody molecules. Excess antigen will result in an antibody combining with several antigens. Refer to Figure 18.2. 3. Particulate antigens react in agglutination reactions. The antigens can be cells or soluble antigens bound to synthetic particles. Soluble antigens take part in precipitation reactions. 4. a. Some viruses are able to agglutinate red blood cells. This is used to detect the presence of large numbers of virions capable of causing hemagglutination (e.g., influenza virus). b. Antibodies produced against viruses that are capable of agglutinating red blood cells will inhibit the agglutination. Hemagglutination inhibition can be used to detect the presence of antibodies against these viruses. c. This is a procedure to detect antibodies that react with soluble antigens by first attaching the antigens to insoluble latex spheres. This procedure may be used to detect the presence of antibodies that develop during certain mycotic or helminthic infections. 5. See Figure 18.10. 6. a. Direct test (see Figure 18.10a) b. Indirect test (see Figure 18.10b) 7. An indirect ELISA test is used to detect the presence of antibodies. A known antigen is fixed to a small well, and the patient's serum is added. Patient's antibodies will react with the antigen in the well. Antihuman immunoglobulins bound to an enzyme are added to the well. The antihuman immunoglobulins will bind to the patient's antibodies. Substrate for the enzyme is then added and a positive reaction indicating presence of the antibody in the patient's serum is shown by the enzyme-substrate reaction. A direct ELISA test is used to detect the presence of an antigen. Antibodies are fixed to a small well, and the unknown antigen is added. If the antigen reacts with the antibodies, the antigen will be bound to the well. Antibodies specific for the antigen are then added to the well. This second layer of antibodies is bound to an enzyme. Substrate for the enzyme is then added, and a positive reaction indicating the identity of the antigen is shown by the enzyme-substrate reaction (see Figure 18.12). 8. a. Direct test b. Indirect test The direct test provides definitive proof. 9. Matching e Precipitation d, f Immunoelectrophoresis a Agglutination g Radioimmunoassay c Complement fixation f Neutralization b, d ELISA 10. Matching e Agglutination c Complement fixation a ELISA f FA b Neutralization d Precipitation Multiple ChoiceAnswers are in boldface.1. Patient's serum, influenza virus, sheep red blood cells, and antisheep red blood cells are mixed in a tube. What happens if the patient has antibodies against influenza?a. hemolysisb. hemagglutinationc. hemagglutination-inhibitiond. no hemolysise. precipitin ring forms 2. Patient's serum, Chlamydia, guinea pig complement, sheep red blood cells, and antisheep red blood cells are mixed in a tube. What happens if the patient has antibodies against Chlamydia?a. hemolysisb. hemagglutinationc. hemagglutination-inhibitiond. no hemolysise. precipitin ring forms 3. The examples in questions 1 and 2 area. direct tests.b. indirect tests. 4. Which is the third step in a direct ELISA test?a. anti-Brucellab. Brucellac. substrate for the enzyme 5. Which item is from the patient in an indirect ELISA test?a. anti-Brucellab. Brucellac. substrate for the enzyme 6. In an immunodiffusion test, a strip of filter paper containing diphtheria antitoxin is placed on a solid culture medium. Then bacteria are streaked perpendicular to the filter paper. If the bacteria are toxigenic,a. the filter paper will turn red.b. a line of antigen-antibody precipitate will form.c. the cells will lyse.d. the cells will fluoresce.e. None of the above. Critical Thinking 1. The live vaccine may revert to a more virulent form; exogenous protein contaminants in viral vaccines; the inherent instability of certain live viral preparations. 2. Viruses multiply only in cells of a particular species. Protection of that species through immunization could prevent further growth of the virus. Many pathogenic bacteria are capable of growth in different species or in nonliving reservoirs. 3. Traditionally, by vaccinating a large animal such as a horse or a goat, and purifying the antibodies from its blood. Now these antibodies can be obtained in vitro by monoclonal antibody techniques. 4. The antibodies (called reagin) are not specific. The disease is syphilis. Clinical Applications 1. (a) is proof of a disease state. (b) could indicate disease, prior disease and recovery, or vaccination. The disease is tuberculosis. 2. No reaction; the antibodies will neutralize the toxin. This is a neutralization reaction. The disease is scarlet fever. 3. Patient A probably has the disease. Patient B does not have and never had the disease. Patient C recovered from the disease. Patient D acquired the disease between days 7 and 14; an example of seroconversion. The disease is legionellosis. 4. A high level of radioactive Ag in Ag-Ab complexes means the Ab combined with the known estrone because there was no estrone in the soil; the horses are not pregnant. A low level of radioactive Ag in Ag-Ab complexes means the Ab combined with estrone in the soil; the horses are pregnant.

Student Resources, Chapter 19 Answers to End-of-Chapter Study Questions Review 1. The immune state that results in altered immunologic reactions leading to pathogenic changes in tissue. 2. Mediator Function Histamine Increases blood capillary permeability, mucus secretion, and smooth muscle contraction. Leukotrienes Increase blood capillary permeability and smooth muscle contraction. Prostaglandins Increase smooth muscle contraction and mucus secretion. 3. Recipient's antibodies will react with donor's tissues. 4. The recipient will experience symptoms due to lysis of the donor RBCs. Hemolysis occurs because the antigen (donor RBCs)-antibody reaction fixes complement. 5. This condition develops when an Rh- mother becomes sensitized to the Rh+ antigens of her fetus. The mother's anti-Rh antibodies (IgG) can cross the placenta and react with fetal RBCs, causing their destruction. This condition can be prevented by passive immunization of the Rh- mother with anti-Rh antibodies shortly after birth. These anti-Rh antibodies combine with fetal Rh+ RBCs, which may have entered maternal circulation, and enhance their clearance, thereby reducing the sensitization of the mother's immune system to this antigen. 6. Refer to Figure 19.7. a. The observed symptoms are due to lymphokines. b. When a person contacts poison oak initially, the antigen (catechols on the leaves) binds to tissue cells, is phagocytized by macrophages, and is presented to receptors on the surface of T cells. Contact between the antigen and the appropriate T cell stimulates the T cell to proliferate and become sensitized. Subsequent exposure to the antigen results in sensitized T cells releasing lymphokines, and a delayed hypersensitivity occurs. c. Small repeated doses of the antigen are believed to cause the production of IgG (blocking) antibodies. 7. Autografts and isografts are the most compatible. Xenografts are the least compatible. 8. a. Compatible. There are no Rh antigens on the donor's RBCs. b. Incompatible. The recipient will produce anti-Rh antibodies. If the recipient receives Rh+ RBCs in a subsequent transfusion, a hemolytic reaction will develop. c. Incompatible. The recipient has anti-A antibodies that will result in lysis of the donor's RBCs. 9. Autoimmunity is a humoral (type II or type III) or cell-mediated (type IV) immune response against a person's own tissue antigen. During development, T cells that recognize self may not be eliminated. During adulthood, inactive T cells may become active or antibodies could cross-react with host cell antigens. New or altered antigens may be formed on the surface of host cells. These antigens may result from the use of certain drugs, or from infections by certain viruses. Antibodies to cell-membrane antigens of certain group A streptococci cross-react with human heart tissue. Severe, recurrent infections caused by b-hemolytic group A streptococci sometimes lead to the development of rheumatoid arthritis long after the streptococcal infection has subsided. 10. Type II Antibodies react with self. Type III Antibody-complement complexes deposit in tissues. Type IV T cells destroy self cells. See Table 19.3. 11. Natural Inherited Viral infections, most notably HIV Artificial Induced by immunosuppression drugs Result: Increased susceptibility to various infections depending on the type of immune deficiency. 12. Tumor cells have tumor-specific antigens such as TSTA and T antigen. Sensitized TC cells may react with tumor-specific antigens, initiating lysis of the tumor cells. 13. Some malignant cells can escape the immune system by antigen modulation or immunological enhancement. Immunotherapy might trigger immunological enhancement. The body's defense against cancer is cell-mediated and not humoral. Transfer of lymphocytes could cause graft-versus-host disease. 14. AIDS is the last stage of an HIV infection. HIV is transmitted by sexual contact, intravenous drug use, and across the placenta and mother's milk. HIV is prevented by using condoms for hetero- and homosexual intercourse and oral and anal copulation, and by not re-using needles.Multiple ChoiceAnswers are in boldface.1. Desensitization to prevent an allergic response can be accomplished by injecting small, repeated doses ofa. IgE antibodies.b. the antigen (allergen).c. histamine.d. IgG antibodies.e. antihistamine. 2. In vitro, recipient serum type B will agglutinate donor cells type A; while in vivo, recipient serum type B will lyse donor cells type A. The in vivo response is due toa. T cells.b. complement.c. antibodies.d. autoimmunity.e. None of the above. 3. Cytotoxic autoimmunity differs from immune complex autoimmunity in that cytotoxic reactionsa. involves antibodies.b. do not involve complement.c. are caused by T cells.d. do not involve IgE antibodies.e. None of the above. 4. Worldwide, the primary method of transmitting HIV isa. homosexual sex.b. heterosexual sex.c. injecting drug use.d. blood transfusions.e. kissing. 5. Which of the following is not the cause of a natural immune deficiency?a. a recessive gene resulting in no thymus glandb. a recessive gene resulting in few B cellsc. HIV infectiond. immunosuppressant drugs.e. None of the above Critical Thinking 1. During embryonic development, clones of lymphocytes called forbidden clones react with "self" antigens. These forbidden clones are suppressed by specific classes of T lymphocytes. Lymphocytes that can react with foreign ("nonself") antigens are left to function in the mature immune system. 2. When a human is passively immunized with horse serum, antibodies may be produced against the horse immunoglobulins (antibodies). Immune complexes form between horse immunoglobulins and the antibodies formed against them. Symptoms are due to complement fixation. 3. Yes, they make antibodies. They are more likely to have T-independent antibodies. Their anti-HIV antibodies are ineffective because the virus can remain in the host cell, can be transmitted by cell-to-cell fusion, and can undergo antigenic changes to its surface proteins. 4. Anti-AIDS drugs are nucleoside analogs or enzyme inhibitors. Clinical Applications 1. The infections are long-lasting, allowing sufficient time for sensitizing and shocking exposures to the fungal antigens. 2. a. An immediate hypersensitivity b. The mediators of anaphylaxis; refer to Review question 3 c. Skin tests d. Some workers will not produce IgE antibodies against the conidiospores. 3. Epinephrine is used to treat symptoms of type I hypersensitivity, systemic anaphylaxis. People with hypersensitivity to eggs may experience anaphylaxis from this vaccine. 4. The woman made IgG antibodies in response to the B antigen in the transfusion. IgG antibodies can cross the placenta. A normal type A+ person has anti-B antibodies of the IgM type that cannot cross the placenta.

Student Resources, Chapter 20 Answers to End-of-Chapter Study Questions Review 1. Antimicrobial Agents Synthetic or Antibiotic Method of Action Principal Use Isoniazid Synthetic Vitamin B6 analog Tuberculosis Sulfonamides Synthetic Inhibit folic acid synthesis Gram-negative bacteria Ethambutol Synthetic Competitive inhibitor Tuberculosis Trimethoprim Synthetic Inhibits folic acid synthesis Pneumocystis Fluoroquinolones Synthetic Inhibit DNA synthesis Urinary tract infections Penicillin, natural Antibiotic Inhibits cell wall synthesis Gram-positive bacteria Penicillin, semisynthetic Antibiotic Inhibits cell wall synthesis Broad spectrum; penicillin-resistant bacteria Cephalosporins Antibiotic Inhibit cell wall synthesis Penicillin-resistant bacteria Carbapenems Antibiotic Inhibit cell wall synthesis Broad spectrum Aminoglycosides Antibiotic Inhibit protein synthesis Gram-negative bacteria Tetracyclines Antibiotic Inhibit protein synthesis Broad spectrum Chloramphenicol Antibiotic Inhibits protein synthesis Salmonella Macrolides Antibiotic Inhibit protein synthesis Gram-negative bacteria Polypeptides Antibiotic Inhibit cell wall synthesis; injure plasma membrane Gram-positive bacteria; gram-negative bacteria Vancomycin Antibiotic Inhibits cell wall synthesis Penicillin-resistant Staphylococcus Rifamycins Antibiotic Inhibit mRNA synthesis Tuberculosis Polyenes Antibiotic Injure plasma membrane Fungicide Griseofulvin Antibiotic Inhibits mitosis Antifungal Amantadine Synthetic Blocks viral entry or uncoating Influenza A Zidovudine Synthetic Inhibits DNA synthesis AIDS Niclosamide Synthetic Inhibits oxidative phosphorylation Tapeworms 2. A chemotherapeutic agent is a substance taken into the body to combat disease. A synthetic chemotherapeutic agent is prepared in a laboratory, whereas antibiotics are produced naturally by bacteria and some fungi. 3. a. Ehrlich discovered the first chemotherapeutic agent (salvarsan, which was used to treat syphilis). b. Fleming discovered the antibiotic penicillin. 4. The drug should be toxic to the undesired microorganisms and not harmful to the host (selective toxicity).The drug should be active against many microorganisms (broad spectrum). The drug should not produce hypersensitivity in the host.The drug should not produce drug resistance in the host.The drug should not harm normal microbiota. 5. Because a virus uses the host cell's metabolic machinery, it is difficult to damage the virus without damaging the host. Fungi, protozoa, and helminths possess eucaryotic cells. Therefore, antiviral, antifungal, antiprotozoan, and antihelminthic drugs must also affect eucaryotic cells. 6. Pyrimidine (idoxuridine) and purine (acyclovir) analogs.Prevent release of nucleic acid from viruses into the host cell (amantadine).Inhibition of infection of cells (interferon). Enzyme inhibitors (indinavir). 7. In the broth dilution test, a series of cultures is prepared in a microtiter plate. To each well of liquid medium, the test organism and a different concentration of chemotherapeutic agent are added. The plate is incubated for 16-20 hours and observed for the presence of microbial growth. The minimal inhibitory concentration (MIC) is the lowest concentration of chemotherapeutic agent capable of preventing growth of the test organism. The lowest concentration of the agent that results in no growth in a subculture is the minimal bactericidal concentration (MBC). In the agar dilution method, bacterial colonies are replica-plated onto nutrient media plus varying concentrations of antimicrobial agents. The MIC is determined by measuring the colony growth. In the tube dilution test, both the MIC and the MBC can be determined. The agar dilution method has the advantage of ease of inoculation and media preparation. 8. In the disk-diffusion test, filter paper disks impregnated with chemotherapeutic agents are overlaid on an inoculated agar medium. During incubation, the agents diffuse from the disk and a zone of inhibition is observed in the area immediately around the disks. The zone of inhibition indicates susceptibility of the test organism to the agent tested. 9. Drug resistance is the lack of susceptibility of a microorganism to a chemotherapeutic agent. Drug resistance may develop when microorganisms are constantly exposed to an antimicrobial agent. The development of drug-resistant microorganisms can be minimized by judicious use of antimicrobial agents; following directions on the prescription; or by administering two or more drugs simultaneously. 10. a. Prevention of resistant strains of microorganisms; b. Take advantage of the synergistic effect; c. Provide therapy until a diagnosis is made; and d. Lessen the toxicity of individual drugs by reducing the dosage of each in combination. 11. a. Like polymyxin B, causes leaks in the plasma membrane. b. Interferes with translation. 12. a. Inhibits formation of peptide bond. b. Prevents translocation of ribosome along mRNA. c. Interferes with attachment of tRNA to mRNA-ribosome complex. d. Changes shape of 30S portion of ribosome, resulting in misreading mRNA. 13. DNA polymerase adds bases to the 3¢ -OH.Multiple ChoiceAnswers are in boldface.1. Which of the following pairs is mismatched?a. antihelminthic-inhibition of oxidative phosphorylationb. antihelminthic-inhibition of cell wall synthesisc. antifungal-injury to plasma membraned. antifungal-inhibition of mitosise. antiviral-inhibition of DNA synthesis 2. All of the following are modes of action of antiviral drugs excepta. inhibition of protein synthesis at 70S ribosomes.b. inhibition of DNA synthesis.c. inhibition of RNA synthesis.d. inhibition of uncoating.e. None of the above. 3. Which of the following modes of action would not be fungicidal?a. inhibition of peptidoglycan synthesisb. inhibition of mitosisc. injury to the plasma membraned. inhibition of nucleic acid synthesis.e. None of the above. 4. An antimicrobial agent should meet all of the following criteria excepta. selective toxicity.b. the production of hypersensitivities.c. a narrow spectrum of activity.d. no production of drug resistance.e. None of the above. 5. The most selective antimicrobial activity would be exhibited by a drug thata. inhibits cell wall synthesis.b. inhibits protein synthesis.c. injures the plasma membrane.d. inhibits nucleic acid synthesis.e. All of the above. 6. Antibiotics that inhibit translation have side effectsa. because all cells have proteins.b. only in the few cells that make proteins.c. because eucaryotic cells have 80S ribosomes.d. at the 70S ribosomes in eucaryotic cells.e. None of the above. 7. Which of the following will not affect eucaryotic cells?a. inhibition of the mitotic spindleb. binding with sterolsc. binding to 80S ribosomesd. binding to DNAe. none of the above 8. Cell membrane damage causes death becausea. the cell undergoes osmotic lysis.b. cell contents leak out.c. the cell plasmolyzes.d. the cell lacks a wall.e. None of the above. 9. A drug that intercalates into DNA has the following effects. Which one leads to the others?a. It disrupts transcription.b. It disrupts translation.c. It interferes with DNA replication.d. It causes mutations.e. It alters proteins. 10. Chloramphenicol binds to the 50S portion of a ribosome, which will interfere witha. transcription in procaryotic cells.b. transcription in eucaryotic cells.c. translation in procaryotic cells.d. translation in eucaryotic cells.e. DNA synthesis. Critical Thinking 1. a. No. Human cells lack cell walls. b. No. It inhibits a viral enzyme. c. Yes. It affects mitochondrial ribosomes. d. Probably not. Sterols protect human membranes. 2. Cells infected by viruses may be rapidly metabolizing (anabolizing) in order to synthesize viruses. These cells are more likely to incorporate base analogs than normal, uninfected cells. 3. The carbon source can't enter the cells. 4. S. griseus makes streptomycin during idiophase. The inactivating enzyme is necessary to protect the bacterium from the streptomycin. 5. a. A and D were equally effective. b. A or D would be recommended depending upon their respective side effects. c. You can't tell. A subculture from the zone of inhibition would have to be done to determine whether A was bactericidal or bacteriostatic. 6. MIC, 50 mg; MBC, 100 mg. Clinical Applications 1. Tetracycline induces lag phase, and penicillin requires log phase. In some infections, streptomycin enters more easily through a penicillin-damaged cell wall. 2. The bacteria were resistant to nalidixic acid but susceptible to sulfonamide. 3. Many bacteria were killed initially, so the patient started to feel better. Recall that bacteria die logarithmically, so it takes quite a while to kill an entire population. The concentration of penicillin dropped in his body when he stopped taking penicillin, so the surviving bacteria were able to grow.

Student Resources, Chapter 21 Answers to End-of-Chapter Study Questions Review 1. Bacteria usually enter through inapparent openings in the skin. Fungal pathogens (except subcutaneous) often grow on the skin itself. Viral infections of the skin (except warts and herpes simplex) most often gain access to the body through the respiratory tract. 2. Staphylococcus aureus; Streptococcus pyogenes. 3. Disease Etiology Symptoms Treatment Notes Impetigo Staphylococcus aureus Vesicles that rupture and crust over Hexachlorophene May be epidemic Erysipelas Streptococcus pyogenes Thickened red patches, swollen at margins Penicillin May be endogenous 4. Disease Etiological Agent Clinical Symptoms Method of Transmission Acne P. acnes Infected oil glands Direct contact Pimples S. aureus Infected hair follicles Direct contact Warts Papovavirus Benign tumor Direct contact Chickenpox Herpesvirus Vesicular rash Respiratory route Fever blisters Herpesvirus Recurrent "blisters" Direct contact Measles Paramyxovirus Papular rash, Koplik spots Respiratory route Rubella Togavirus Macular rash Respiratory route 5. Both are fungal infections. Sporotrichosis is a subcutaneous mycosis; athlete's foot is a cutaneous mycosis. 6. a. Conjunctivitis is an inflammation of the conjunctiva, and keratitis is an inflammation of the cornea. b. See pp. 574-575. 7. Candidiasis is caused by Candida albicans. The yeast is able to grow when the normal microbiota are suppressed or when the immune system is suppressed. The yeast can be transferred from another person or be transient microflora. White patches in the mouth or bright red areas of the skin and mucous membranes are signs of infection. Antifungal agents such as miconazole are used to treat candidiasis. Systemic infections are treated with oral ketoconazole. 8. The test determines the woman's susceptibility to rubella. If the test is negative, she is susceptible to the disease. If she acquires the disease during pregnancy the fetus could become infected. A susceptible woman should be vaccinated. 9. Symptoms Disease Koplik spots Measles Macular rash Measles Vesicular rash Chickenpox Small, spotted rash German measles "Blisters" Cold sore Corneal ulcer Keratoconjunctivitis 10. The central nervous system can be invaded following keratoconjunctivitis; this results in encephalitis. 11. Attenuated measles, mumps, and rubella viruses. 12. Varicella-zoster virus appears to remain latent in nerve cells following recovery from a childhood infection of chickenpox. Later, the virus may be activated and cause a vesicular rash (shingles) in the area of the nerve. 13. To prevent neonatal gonorrheal ophthalmia. This is caused by N. gonorrhoeae contracted by the newborn during passage through the birth canal. 14. Trachoma. 15. Scabies is an infestation of mites in the skin. It is treated with permethrin insecticide or gamma benzene hexachloride.Multiple ChoiceAnswers are in boldface.Use the following information to answer questions 1 and 2. A 6-year- old girl was taken to the physician for evaluation of a slowly growing bump on the back of her head. The bump was a raised, scaling lesion 4 cm in diameter. Wood's lamp examination of the scalp showed fluorescence. A fungal culture of material from the lesion was positive for a fungus with numerous microconidia.1. The girl's disease wasa. rubellab. candidiasisc. dermatomycosisd. a cold soree. none of the above. 2. Besides the scalp, this disease can occur on all of the following excepta. feet.b. nails.c. the groin.d. subcutaneous tissue.e. none of the above. Use the following information to answer questions 3 and 4. A 12-year-old boy had a fever, rash, headaches, sore throat, and cough. He also had a macular rash on his trunk, face, and arms. A throat culture was negative for Streptococcus pyogenes.3. The boy most likely hada. streptococcal sore throat.b. measles.c. rubella.d. smallpox.e. none of the above.4. All of the following are complications of this disease excepta. middle ear infections.b. pneumonia.c. birth defects.d. encephalitis.e. none of the above. 5. A patient has conjunctivitis. If you isolated Pseudomonas from the patient's mascara, you would most likely conclude all of the following except that a. the mascara was the source of the infection.b. Pseudomonas is causing the infection.c. Pseudomonas has been growing in the mascara.d. the mascara was contaminated by the manufacturer.e. none of the above. Critical Thinking 1. S. aureus is adapted for surviving on the human skin, which has a high concentration of NaCl. Microorganisms that are not adapted to this hypertonic environment will not be able to tolerate the 7.5% NaCl in mannitol salt agar. 2. Most warts regress spontaneously. Removal of warts is usually for cosmetic reasons. Occasionally warts are painful when they are located where pressure is placed on them (e.g., plantar warts on the sole of the foot). 3. The infections were transmitted by the contact lenses or cosmetics. Cosmetics are inoculated with microbes each time they are used. Some of the microbes grow, resulting in large inoculations of the eyes. Contact lenses can be improperly cleaned (i.e., not using an antiseptic) or contaminated by fingers. 4. The virus had one host-humans. It was not found in soil, water, or nonhuman organisms. Polio and measles meet this criterion. Clinical Applications 1. Pseudomonas aeruginosa. This bacterium is common in soil and is resistant to many antibiotics. 2. Toxic shock syndrome due to growth of Staphylococcus at the injection site. 3. The symptoms of toxic shock syndrome were caused by toxins produced from the secondary infection (S. aureus).

Student Resources, Chapter 22 Answers to End-of-Chapter Study Questions Review 1. Meningitis is an infection of the meninges; encephalitis is an infection of the brain itself. 2. Causative Agent Susceptible Population Mode of Transmission Treatment N. meningitidis Children; military recruits Respiratory Penicillin H. influenzae Children Respiratory Rifampin S. pneumoniae Children; elderly Respiratory Penicillin L. monocytogenes Anyone Foodborne Penicillin C. neoformans Immunosuppressed individuals Respiratory Amphotericin B 3. "Hemophilus" refers to the requirement of this genus for growth factors found in blood (X and V factors). "Influenzae" because it was thought to be the causative agent of influenza. 4. The symptoms of tetanus are not due to bacterial growth (infection and inflammation) but to neurotoxin. 5. Salk Sabin Composition Formalin-inactivated viruses Live, attenuated viruses Advantages No reversion to virulence Oral administration Disadvantages Booster dose needed; injected Reversion to virulence 6. a. Vaccination with tetanus toxoid. b. Immunization with antitetanus toxin antibodies. 7. "Cleaned" because C. tetani is found in soil that might contaminate a wound. "Deep puncture" because it is likely to be anaerobic. "No bleeding" because a flow of blood ensures an aerobic environment and some cleansing. 8. Clostridium botulinum. Canned foods. Paralysis. Supportive respiratory care; antitoxin. Anaerobic, non-acidic environment. Diagnosis is made by detecting toxin in foods or patient by inoculating mice with suspect samples. Prevention: use of adequate heat in canning; boiling food before consumption to inactivate toxin. 9. Etiology-Mycobacterium leprae.Transmission-Direct contact.Symptoms-Nodules on the skin; loss of sensation.Treatment-Dapsone and rifampin.Prevention-BCG vaccine.Susceptible-People living in the tropics; genetic predisposition. 10. Etiology-Picornavirus (poliovirus).Transmission-Ingestion of contaminated water.Symptoms-Headache, sore throat, fever, nausea; rarely paralysis. Prevention-Sewage treatment. These vaccinations provide artificially acquired active immunity because they cause the production of antibodies, but they do not prevent or reverse damage to nerves. 11. Etiology-Rhabdovirus.Transmission-Bite of infected animal; inhalation.Reservoirs-Skunks, bats, foxes, raccoons.Symptoms-Muscle spasms, hydrophobia, CNS damage. 12. Postexposure treatment-Passive immunization with antibodies followed by active immunization with HDCV. Preexposure treatment-Active immunization with HDCV. Following exposure to rabies, antibodies are needed immediately to inactivate the virus. Passive immunization provides these antibodies. Active immunization will provide antibodies over a longer period of time, but they are not formed immediately. 13. Disease Etiology Vector Symptoms Treatment Arboviral encephalitis Togaviruses, Arboviruses Mosquitoes (Culex) Headache, fever, coma Immune serum African trypanosomiasis Trypanosoma brucei gambiense,T. b. rhodesiense Tsetse fly Decreased physical activity and mental acuity Suramin; melarsoprol 14. Most antibiotics cannot cross the blood-brain barrier. 15. The causative agent of Creutzfeldt-Jakob disease (CJD) is transmissible. Although there is some evidence for an inherited form of the disease, it has been transmitted by transplants. Similarities with viruses are (1) the prion cannot be cultured by conventional bacteriological techniques and (2) the prion is not readily seen in patients with CJD.Multiple ChoiceAnswers are in boldface.1. Which of the following is not true?a. Only puncture wounds by rusty nails result in tetanus.b. Rabies is seldom found in rodents (e. g., rats, mice).c. Polio is transmitted by the fecal-oral route.d. Arboviral encephalitis is rather common in the United States.e. None of the above. 2. Which of the following does not have an animal reservoir or vector?a. listeriosisb. cryptococcosisc. Naegleria meningoencephalitisd. rabiese. African trypanosomiasis 3. A 12-year-old girl hospitalized for Guillain-Barré syndrome had a 4-day history of headache, dizziness, fever, sore throat, and weakness of legs. Seizures began 2 weeks later. Bacterial cultures were negative. She died 3 weeks after hospitalization. An autopsy revealed inclusions in brain cells that tested positive in an immunofluoresence test. She probably hada. rabies.b. Creutzfeldt-Jakob disease.c. botulism.d. tetanus.e. leprosy.4. After receiving a corneal transplant, a woman developed dementia and loss of motor function; she then became comatose and died. Cultures were negative. Serological tests were negative. Autopsy revealed spongiform degeneration of her brain. She most likely hada. rabies.b. Creutzfeldt-Jakob disease.c. botulism.d. tetanus.e. leprosy. 5. The endotoxin is responsible for symptoms caused by which of the following organisms?a. N. meningitidisb. S. pyogenesc. L. monocytogenesd. C. tetanie. C. botulinum 6. The increased incidence of encephalitis in the summer months is due toa. maturation of the viruses.b. increased temperature.c. the presence of adult mosquitoes.d. an increased population of birds.e. an increased population of horses. 7. Produces the highest antibody titer.a. antirabies antibodiesb. HDVC 8. Used for passive immunization.a. antirabies antibodiesb. HDVC 9. Microscopic examination of cerebrospinal fluid reveals gram-positive rods.a. Cryptococcusb. Hemophilusc. Listeriad. Naegleriae. Neisseria 10. Microscopic examination of cerebrospinal fluid from a person who washes windows reveals ovoid cells.a. Cryptococcusb. Hemophilusc. Listeriad. Naegleriae. Neisseria Critical Thinking 1. The term rusty nail implies that the sharp object has been contaminated with soil and possibly C. tetani. C. tetani can grow in deep puncture wounds, and a nail is capable of producing such a wound. 2. Both diseases are caused by species of Mycobacterium. 3. The only cases of polio in the United States during the last few years have been caused by OPV.Clinical Applications 1. Hemophilus influenzae meningitis, treated with rifampin. 2. Cryptococcus neoformans. Need microscopic observation of the fungus from cerebrospinal fluid or culture. 3. Naegleria fowleri meningitis; treated with amphotericin B, miconazole, and rifampin.

Student Resources, Chapter 23 Answers to End-of-Chapter Study Questions Review 1. Fever, decrease in blood pressure, and lymphangitis. 2. Bacteria can spread from an abscess with enzymes such as kinases and invade blood vessels. 3. Disease Causative Agent Predisposing Conditions p.s. S. pyogenes Abortion or childbirth s.b.e. a-hemolytic strep. Preexisting lesions a.b.e. S. aureus Abnormal heart valves 4. Rheumatic fever is an autoimmune disease that is precipitated by streptococcal sore throat. It is treated with anti-inflammatory drugs to relieve the symptoms. It is prevented by early diagnosis and treatment of streptococcal sore throat. 5. All are vectorborne rickettsial diseases. They differ from each other in (1) etiologic agent, (2) vector, (3) severity and mortality, and (4) incidence (e.g., epidemic, sporadic). 6. Causative Agent Vector Symptoms Treatment Plasmodium Anopheles Recurrent fever, chills Quinine derivative Arbovirus Aedes aegypti Fever, nausea, jaundice None Arbovirus Aedes aegypti Muscle and joint pain None Borrelia Soft ticks Recurrent fever Tetracycline 7. Francisella tularensis Animal reservoir, skin abrasions, ingestion, inhalation, bites Rabbits Small ulcer Careful handling of animals Brucella spp. Animal reservoir, ingestion of milk, direct contact Cattle Undulant fever Pasteurization of milk Bacillus anthracis Skin abrasions, inhalation, ingestion Soil, cattle Malignant pustule Surveillance and vaccination of cattle Borrelia burgdorferi Tick bites Deer, mice Rash, neurologic; arthritis Protection from ticks 8. PlagueCausative agent-Yersinia pestis.Vector-Rat flea.Reservoir-Rodents.Prognosis-Poor if untreated; good with antibiotic treatment.Treatment-Tetracycline, streptomycin.Control-Sanitation and ratproofing buildings. 9. Causative Agent Transmission Reservoir Endemic Area Schistosoma spp. Penetrate skin Aquatic snail Asia, South America Toxoplasma gondii Ingestion, inhalation Cats United States Trypanosoma cruzi "Kissing bug" Rodents Central America 10. Pasteurella multocida (pasteurellosis)Clostridium spp.Bacteroides spp.Fusobacterium spp.Streptobacillus (rat bite fever)Spirillum (rat bite fever)Rochalimaea (cat-scratch fever)Afipia (cat-scratch fever) 11. Gangrenous tissue is anaerobic and has suitable nutrients for C. perfringens. 12. Infectious mononucleosis is caused by EB virus and transmitted in oral secretions. 13. Bubonic plague-Proliferation of bacteria in lymph vessels and lymph nodes. Enlarged lymph nodes (buboes). Transmission-Flea bites.Pneumonic plague-Growth of bacteria in the lungs. Transmission-Respiratory route.Multiple ChoiceAnswers are in boldface.1. A patient presents with vomiting and diarrhea and a history of fever and headache. Bacterial cultures of blood, CSF, and stool are negative. What is your diagnosis?a. ehrlichiosisb. Lyme diseasec. septic shockd. toxoplasmosise. viral hemorrhagic fever 2. A patient was hospitalized because of continuing fever and progression of symptoms including headache, fatigue, and back pain. Tests for antibodies to Borrelia burgdorferi were negative. What is your diagnosis?a. ehrlichiosisb. Lyme diseasec. septic shockd. toxoplasmosise. viral hemorrhagic fever 3. A patient complained of headache. A CT (computed tomography) scan revealed cysts of varying size in her brain. What is your diagnosis?a. ehrlichiosisb. Lyme diseasec. septic shockd. toxoplasmosise. viral hemorrhagic fever 4. A patient presents with mental confusion, rapid breathing and heartbeat, and low blood pressure. What is your diagnosis?a. ehrlichiosisb. Lyme diseasec. septic shockd. toxoplasmosise. viral hemorrhagic fever 5. A patient has a red circular rash on his arm and fever, malaise, and arthralgia. The most appropriate treatment isa. penicillin.b. chloroquine.c. anti-inflammatory drugs.d. rifampin.e. no treatment. 6. Which of the following is not a tickborne disease?a. babesiosisb. ehrlichiosisc. Lyme diseased. relapsing fevere. tularemia 7. The patientÕs fever spikes each evening. Oxidase-positive, gram-negative cocci were isolated from a lesion on his arm. What is your diagnosis?a. brucellosisb. malariac. relapsing feverd. Rocky Mountain spotted fevere. Ebola hemorrhagic fever 8. The patient was hospitalized with fever and headache. Spirochetes were observed in her blood. What is your diagnosis?a. brucellosisb. malariac. relapsing feverd. Rocky Mountain spotted fevere. Ebola hemorrhagic fever 9. Which of the following diseases has the highest incidence in the United States?a. brucellosisb. Ebola hemorrhagic feverc. malariad. plaguee. Rocky Mountain spotted fever 10. Nineteen workers in a slaughterhouse developed fever and chills, with the fever spiking to 40 degrees celsius each evening. The most likely method of transmission of this disease isa. a vector.b. the respiratory route.c. a puncture wound.d. an animal bite.e. water. Critical Thinking 1. Patient B has a rising antibody titer, which indicates an active infection. A therapeutic abortion could be recommended to this woman. Patient C has no immunity to Toxoplasma and should be advised to avoid contact with reservoirs. Patient A has antibodies against Toxoplasma that should provide long-lasting immunity. 2. Mosquito eradication. 3. The presence of antibodies is causing the more severe disease; this is called antibody enhancement. Antibodies and live viruses form immune complexes that attach to macrophages and other cells, thus increasing the number of viruses that infect a cell.

Student Resources, Chapter 24 Answers to End-of-Chapter Study Questions Review 1. Droplet infection. Inhalation of cells and spores; ingestion of contaminated food. 2. Coarse hairs in the nose filter dust particles from inspired air. Mucus traps dust and microorganisms, and cilia move the trapped particles toward the throat for elimination. The ciliary escalator of the lower respiratory system moves particles toward the throat. Alveolar macrophages can phagocytize microorganisms that enter the lungs. IgA antibodies are found in mucus, saliva, and tears. 3. Beta-hemolytic streptococci inhibit growth of pneumococci; faster-growing organisms can compete with pathogens. 4. Mycoplasmal pneumonia is caused by Mycoplasma pneumoniae bacteria. Viral pneumonia can be caused by several different viruses. Mycoplasmal pneumonia can be treated with tetracyclines, whereas viral pneumonia cannot. 5. Bacteria infecting the nose and throat can move through the eustachian tube to the middle ear. Microorganisms can enter the ear directly via swimming pool water or injury to the eardrum or skull. The bacteria that most commonly cause otitis media are S. aureus, Streptococcus pneumoniae, b-hemolytic streptococci, and H. influenzae. The middle ear is connected to the nose and throat. 6. Upper Respiratory System Common cold Coronaviruses Sneezing, excessive nasal secretions, congestion Lower Respiratory System Viral pneumonia Several viruses Fever, shortness of breath, chest pains Influenza Influenzavirus Chills, fever, headache, muscular pains RSV Respiratory syncytial virus Coughing, wheezing Amantadine is used to treat influenza. Ribavirin may reduce RSV symptoms.7. Disease Symptoms Streptococcal pharyngitis Pharyngitis and tonsillitis Scarlet fever Rash and fever Diphtheria Membrane across throat Whooping cough Paroxysmal coughing Tuberculosis Tubercles, weight loss, and coughing Pneumococcal pneumonia Reddish lungs, fever H. influenzae pneumonia Similar to pneumococcal pneumonia Chamydial pneumonia Low fever, cough, and headache Legionellosis Fever and cough Psittacosis Fever and headache Q fever Chills and chest pain Epiglottitis Inflamed, abscessed epiglottis 8. Inhalation of large numbers of spores from Aspergillus or Rhizopus can cause infections in individuals with impaired immune systems, cancer, and diabetes. 9. No. Many different organisms (gram-positive bacteria, gram-negative bacteria, and viruses) can cause pneumonia. Each of these organisms is susceptible to different antimicrobial agents. 10. Disease Endemic Areas in the United States Histoplasmosis States adjoining the Mississippi and Ohio Rivers Coccidioidomycosis American Southwest Blastomycosis Mississippi Pneumocystis Ubiquitous Refer to Table 24.1. 11. In the tuberculin test, purified protein derivative (PPD) from M. tuberculosis is injected into the skin. Induration and reddening of the area around the injection site indicates an active infection or immunity to tuberculosis. 12. Hypothesis 1: Close indoor contact in winter promotes epidemic transmission.Hypothesis 2: The viruses grow best at slightly cooler temperatures. Hypothesis 3: A physiological change in humans during winter allows viral growth. 13. Gram-positive cocci Catalase-positive S. aureus b-hemolytic S. pyogenes a-hemolytic S. pneumoniae Gram-positive rods Not acid-fast C. diphtheriae Acid-fast M. tuberculosis Gram-negative cocci M. (B.) catarrhalis Aerobic gram-negative rods Coccobacilli B. pertussis Rods L. pneumophila Facultatively anaerobic gram-negative rodsCoccobacilli H. influenzae Rods K. pneumoniae IntracellularElementary bodies Chlamydia psittaci No elementary bodies Coxiella burnetii Wall-less M. pneumoniae Multiple ChoiceAnswers are in boldface. 1. A patient has fever, difficulty breathing, chest pains, fluid in the alveoli, and a positive tuberculin skin test. Gram-positive cocci are isolated from the sputum. The recommended treatment isa. penicillin.b. antitoxin.c. isoniazid.d. tetracyclines.e. none of the above. 2. No bacterial pathogen can be isolated from the sputum of a patient with pneumonia. Antibiotic therapy has not been successful. The next step should bea. culturing for Mycobacterium tuberculosis.b. culturing for Mycoplasma pneumoniae.c. culturing for fungi.d. a change in antibiotics.e. none; nothing more can be done. 3. Your culture from a pneumonia patient appears not to have grown. You do see colonies, however, when the plate is viewed at 1003.a. Chlamydiab. Coccidioidesc. Histoplasmad. Mycobacteriume. Mycoplasma 4. This pneumonia etiology requires cell culture.a. Chlamydiab. Coccidioidesc. Histoplasmad. Mycobacteriume. Mycoplasma 5. Microscopic examination of a lung biopsy shows ovoid cells in macrophages. You suspect these are the cause of the patient's symptoms, but your culture grows a filamentous organism.a. Chlamydiab. Coccidioidesc. Histoplasmad. Mycobacteriume. Mycoplasma 6. Microscopic examination of a lung biopsy shows spherules.a. Chlamydiab. Coccidioidesc. Histoplasmad. Mycobacteriume. Mycoplasma 7. In San Francisco, ten animal-health care technicians developed pneumonia 2 weeks after 130 goats were moved to the animal shelter where they worked. Which of the following is not true?a. Diagnosis is made by a blood agar culture of sputum.b. The cause is Coxiella burnetii.c. The cause is rickettsia.d. The disease was transmitted by aerosols.e. Diagnosis is made by complement-fixation tests for antibodies. 8. Which of the following leads to all the rest?a. catarrhal stageb. coughc. loss of ciliad. mucus accumulatione. trachaeal cytotoxin 9. Causes the formation of a membrane across the throat.a. Bordetella pertussisb. Corynebacterium diphtheriaec. Legionella pneumophilad. Mycobacterium tuberculosise. None of the above 10. Resistant to destruction by phagocytes.a. Bordetella pertussisb. Corynebacterium diphtheriaec. Legionella pneumophilad. Mycobacterium tuberculosise. None of the above Critical Thinking 1. a. When S. pyogenes causing streptococcal sore throat produces an erythrogenic toxin, the infection is called scarlet fever. b. Diphtheroids are nonpathogenic species of corynebacteria. C. diphtheriae, like S. pyogenes, produces an exotoxin when it is lysogenized by a phage. 2. There are many strains of influenza viruses because of antigenic drift and antigenic shift. 3. There are more than 200 viruses that cause colds. Influenzavirus changes its antigens every few months.Clinical Applications1. Coccidiodomycosis. Attempts to culture fungi after the initial diagnosis may have grown C. immitis. 2. AIDS patients were given pentamidine to prevent Pneumocystis pneumonia. Possible sources of infection: (1) the nurse diagnosed with tuberculosis,(2) aerosols created during pentamidine therapy, or (3) face-to-face exposure with TB patients. 3. The disease is psittacosis; transmitted by inhalation of particles from bird droppings. 4. Pneumococcal pneumonia. 5. Histoplasmosis; contracted by inhalation of conidia.

Student Resources, Chapter 25 Answers to End-of-Chapter Study Questions Review 1. Mouth-Streptococci, including S. mutans.Stomach and small intestine-None.Large intestine and rectum-Lactobacillus, Bacteroides and enterics. 2. S. mutans becomes established in the mouth when the teeth erupt from the gums. A sticky capsule enables the bacteria to adhere to teeth. The dextran capsule is produced when the bacteria grow on sucrose. These bacteria and others that become trapped in the dextran produce lactic acid, which erodes tooth enamel. 3. Disease Suspect Foods Symptoms Staph Not cooked prior to eating Vomiting and diarrhea Shigellosis Contaminated water Mucus and blood in stools Salmonellosis Poultry; contaminated water Fever, vomiting, and diarrhea Cholera Contaminated water Rice water stools Gastro. Contaminated water Vomiting and diarrhea Traveler's Contaminated water Vomiting and diarrhea Refer to Table 25.2. 4. Both are caused by Salmonella spp. However, typhoid fever is caused by a few strains of Salmonella that are invasive. The bacteria can cross the intestinal wall and can be disseminated throughout the body. Typhoid fever is characterized by fever and malaise without diarrhea. 5. Certain strains of E. coli may produce an enterotoxin or invade the epithelium of the large intestine. 6. At present there are no treatments for hepatitis. Exposed individuals can be given pooled immune globulin for hepatitis A or HBIG for passive immunity to hepatitis B. Vaccines can prevent hepatitis A and B. 7. Antibodies specific for HBSAg are used to screen blood for HBV. A viral protein is used to test blood for antibodies against HCV. 8. Toxins produced by fungi; see p. 677. 9. All four are caused by protozoa, although each etiologic agent is in a different phylum. The infections are acquired by ingesting protozoa in contaminated water. Giardiasis is a prolonged diarrhea. Amoebic dysentery is the most severe dysentery, with blood and mucus in the stools. Cryptosporidium and Cyclospora produce severe diseases in persons with immune deficiencies. 10. Disease Etiologic Agent Symptoms Amoebic Entamoeba histolytica Blood and mucus in stools, perforation of the intestinal wall, abscesses Bacillary Shigella spp. Leukocytes in feces in addition to the symptoms listed for amoebic dysentery 11. Fever, nausea, abdominal pain, cramps, diarrhea. The diagnosis is based on isolation of the etiologic agent from leftover food or the patient's stools. 12. Food intoxication: Microorganisms must be allowed to grow in food from the time of preparation to the time of ingestion. This usually occurs when foods are stored unrefrigerated or improperly canned. The etiologic agents (Staphylococcus aureus or Clostridium botulinum) must produce an exotoxin. Onset: 1 to 48 hours. Duration: A few days. Treatment: Antimicrobial agents are ineffective. The patient's symptoms may be treated. Food infection: Viable microorganisms must be ingested with food or water. The organisms could be present during preparation and survive cooking or be inoculated during later handling. The etiologic agents are usually gram-negative organisms (Salmonella, Shigella, Vibrio, and Escherichia) that produce endotoxins. Clostridium perfringens is a gram-positive bacterium that causes food infection. Onset: 12 hours to 2 weeks. Duration: Longer than intoxication because the microorganisms are growing in the patient. Treatment: Antimicrobial agents. 13. Disease Site Symptoms Mumps Parotid glands Inflammation of the parotid glands and fever CMV Liver, kidneys, other organs Liver and kidney malfunction Infectious hepatitis Liver Anorexia, fever, diarrhea Serum hepatitis Liver Anorexia, fever, joint pains, jaundice Viral gastroenteritis Lower GI tract Nausea, diarrhea, vomiting Refer to Table 25.2 to complete this question. 14. 15. Refer to Figure 25.24. 16. Adequate sewage treatment and sanitary living conditions. 17. Cook meat thoroughly. Eliminate the source of contamination to cattle and pigs.Multiple-ChoiceAnswers are in boldface.1. All of the following can be transmitted by recreational (i.e., swimming) water sources excepta. amoebic dysentery.b. cholera.c. giardiasis.d. hepatitis B.e. salmonellosis. 2. A patient with nausea, vomiting, and diarrhea within 5 hours after eating most likely hasa. shigellosis.b. cholera.c. E. coli gastroenteritis.d. salmonellosis.e. staphylococcal food poisoning. 3. Isolation of E. coli from a stool sample is diagnostic proof that the patient hasa. cholera.b. E. coli gastroenteritis.c. salmonellosis.d. typhoid fever.e. none of the above. 4. Gastric ulcers are caused bya. stomach acid.b. Helicobacter pylori.c. spicy food.d. acidic food.e. stress. 5. Microscopic examination of a patientÕs fecal culture shows comma-shaped bacteria. These bacteria require 2-4% NaCl to grow. The bacteria probably belong to the genusa. Campylobacter.b. Escherichia.c. Salmonella.d. Shigella.e. Vibrio. 6. A recent cholera epidemic in Peru had all of the following characteristics. Which one led to the others?a. eating raw fishb. sewage contamination of waterc. catching fish in contaminated waterd. Vibrio in fish intestinee. including fish intestines with edibles 7. Identification is based on the observation of oocysts in feces.a. Campylobacterb. Cryptosporidiumc. Escherichiad. Salmonellae. Trichinella 8. A characteristic disease symptom caused by this microorganism is swelling around the eyes. a. Campylobacterb. Cryptosporidiumc. Escherichiad. Salmonellae. Trichinella9. Microscopic observation of a stool sample reveals gram-negative helical cells.a. Campylobacterb. Cryptosporidiumc. Escherichiad. Salmonellae. Trichinella 10. This microbe is frequently transmitted to humans via raw eggs.a. Campylobacterb. Cryptosporidiumc. Escherichiad. Salmonellae. Trichinella Critical Thinking 1. Humans are not usually consumed by other animals. The larval stage of Trichinella is encysted in humans and must be ingested to mature in the intestines of a definitive host. 2. Disease Conditions Diagnosis Prevention Staph Lack of refrigeration Symptoms, presence of S. aureus in food Refrigeration Salmonellosis Fecal contamination, inadequate heating Isolation of Salmonella from stools Sanitation, thorough heating of foods 3. Beef c Deli d Chicken e Milk b Oysters a Pork f 4. The gram-negative bacterial infections, hepatitis A and E, viral gastroenteritis, and the protozoan diseases. The organisms are not likely to be salt-tolerant (except Vibrio), and an ocean swimmer swallows less water than a freshwater swimmer. Clinical Applications 1. Source of infection: Crabs.Bacteria: Vibrio cholerae.Prevention: Proper cooking temperature. 2. Source of infection: Chickens and improper cooking procedures. Bacteria: Salmonella heidelberg and S. stanley. Prevention: Refrigeration overnight; higher cooking temperature. 3. Fingersticks to draw blood samples. The disposable lancet was reused. The 2% and 0% did not receive fingersticks. 4. Bacteria: Shigella.Transmitted: Fecal-oral route.Source: Sewage-contaminated well water.

Student Resources, Chapter 26 Answers to End-of-Chapter Study Questions Review 1. Organs of the upper urinary tract are sterile. The resident microbiota of the urethra are Streptococcus, Bacteroides, Mycobacterium, Neisseria, and some enterobacteria. 2. Normal microbiota of the male genital system is the same as that of the urinary tract. During reproductive years, lactobacilli predominate in the vagina. 3. Urinary tract infections may be transmitted by improper personal hygiene and contamination during medical procedures. They are often caused by opportunistic pathogens. 4. The proximity of the anus to the urethra and the relatively short length of the urethra can allow contamination of the urinary bladder in females. Predisposing factors for cystitis in females are gastrointestinal infections and vaginal and urinary tract infections. 5. Escherichia coli causes about 75% of the cases. From lower urinary tract or systemic infections. 6. Disease Symptoms Diagnosis Gardnerella Fishy odor Odor, pH, clue cells Gonorrhea Painful urination Isolation of Neisseria Syphilis Chancre FTA-ABS PID Abdominal pain Culture of pathogen NGU Urethritis Absence of Neisseria LGV Lesion, lymph node enlargement Observation of Chlamydia in cells Chancroid Swollen ulcer Isolation of Hemophilus 7. Transmission-Water; enters via wounds.Activities-Water contact; contact with animals or rodent-infested places.Etiology-Leptospira interrogans. 8. Symptoms: Burning sensation, vesicles, painful urination.Etiology: Herpes simplex type 2 (sometimes type 1). When the lesions are not present, the virus is latent and noncommunicable. 9. Candida albicans: Severe itching; thick, yellow, cheesy discharge.Trichomonas vaginalis: Profuse yellow discharge with disagreeable odor. 10. Disease Prevention of Congenital Disease Gonorrhea Treatment of newborn's eyes Syphilis Prevention and treatment of mother's disease NGU Treatment of newborn's eyes Genital herpes Cesarean delivery during active infection Multiple-ChoiceAnswers are in boldface.1. Which of the following is usually transmitted by contaminated water?a. Chlamydiab. leptospirosisc. syphilisd. trichomoniasise. none of the above 2. Microscopic examination of vaginal smear shows flagellated eucaryotes.a. Candidab. Chlamydiac. Gardnerellad. Neisseriae. Trichomonas 3. Microscopic examination of vaginal smear shows ovoid eucaryotic cell.a. Candidab. Chlamydiac. Gardnerellad. Neisseriae. Trichomonas 4. Microscopic examination of vaginal smear shows epithelial cells covered with bacteria.a. Candidab. Chlamydiac. Gardnerellad. Neisseriae. Trichomonas 5. Microscopic examination of vaginal smear shows gram-negative cocci in phagocytes.a. Candidab. Chlamydiac. Gardnerellad. Neisseriae. Trichomonas 6. Difficult to treat with chemotherapya. candidiasisb. Gardnerella vaginosisc. genital herpesd. lymphogranuloma venereume. trichomoniasis 7. Fluid-filled vesiclesa. candidiasisb. Gardnerella vaginosisc. genital herpesd. lymphogranuloma venereume. trichomoniasis 8. Frothy, fishy dischargea. candidiasisb. Gardnerella vaginosisc. genital herpesd. lymphogranuloma venereume. trichomoniasis 9. The most common cause of cystitis.a. Chlamydia trachomatisb. Escherichia colic. Mycobacterium hominisd. Staphylococcus saprophyticus 10. In cases of NGU, diagnosis is made using PCR to detect microbial DNA.a. Chlamydia trachomatisb. Escherichia colic. Mycobacterium hominisd. Staphylococcus saprophyticus Critical Thinking 1. T. pallidum pertenue can survive on skin in the tropics because of the constant warm temperature and high humidity. In temperate regions, this type of infection (yaws) might not survive because of cooler, drier air. If T. pallidum pertenue successfully invaded the body, warmth and moisture are available. Sexual contact is a method of transmission that provides constant protection for the microorganisms. 2. Eliminates normal microbiota. Changes (increases) pH. 3. Nystatin will inhibit yeast without affecting bacteria. Chocolate agar and increased CO2 provide enriched conditions but are not selective. 4. Aerobic Leptospira Anaerobic Treponema Coccus Neisseria Requires X factor Hemophilus Gram-positive wall Gardnerella Parasite Chlamydia Urease-positive Ureaplasma Urease-negative Mycoplasma Fungus Candida Protozoan Trichomonas No organism Herpes simplex Clinical Applications 1. Her disease appears to be meningitis; however, the gram-negative cocci in the cervical culture indicate this is disseminated gonococcal infection. The N. gonorrhoeae was sexually transmitted. 2. Pathogen: Neisseria gonorrhoeaeTreatment: Cefoxitin The isolates all had the same antibiotic sensitivities. 3. The diagnosis of syphilis was made on November 8. Both the woman and the infant began syphilis therapy that day. Information on the baby's father was obtained in retrospect from a correctional facility. His condition was not diagnosed nor treated until after his wife was diagnosed and interviewed for contacts. Darkfield examination of the lesion and STS were not done on June 6 or July 1. Copies of the positive RPR results should have been forwarded to the local STD control office to ensure treatment and epidemiologic case management. The regular laboratory clerk was on leave at the time, and her replacement inadvertently forwarded all copies of the laboratory reports to the hospital. Thus, the STD Control Program was unaware that the serologic results were reactive. The hospital's Infection Control Nurse, who was responsible for reviewing STS results, also was away on leave; both patients' results were filed without being brought to the attention of their attending physicians.

Student Resources, Chapter 27 Answers to End-of-Chapter Study Questions Review 1. Extremophiles include thermophiles such as Thermus aquaticus, acidophiles such as Thiobacillus, halophiles such as Halobacterium, and endoliths. 2. The koala should have an organ housing a large population of cellulose-degrading microorganisms. 3. Penicillium might make penicillin to reduce competition from faster-growing bacteria. 4. 5. Amino acids; SO42-; plants and bacteria; H2S; carbohydrates; S0. 6. Phosphorus must be available for all organisms.7. Process Reactions Microorganisms Ammonification -NH2 ----> NH3 Proteolytic bacteria Nitrification NH3 ----> NO Nitrosomonas NO----> NO Nitrobacter Denitrification NO----> N2 Bacillus N fixation N2 ----> NH3 Rhizobium 8. Cyanobacteria: With fungi, cyanobacteria act as the photoautotrophic partner in a lichen; they may also fix nitrogen in the lichen. With Azolla, they fix nitrogen. Mycorrhizae: Fungi that grow in and on the roots of higher plants; increase absorption of nutrients. Rhizobium: In root nodules of legumes; fix nitrogen. Frankia: In root nodules of alders, roses, and other plants; fix nitrogen. 9. SettlingFlocculation treatmentSand filtration (or activated charcoal filtration)ChlorinationThe amount of treatment prior to chlorination depends on the amount of inorganic and organic matter in the water. 10. A coliform count is used to determine the bacteriologic quality of water; that is, the presence of human pathogens or evidence of fecal contamination. 11. b Leaching field a Removal of solids b Biological degradation b Activated sludge c Chemical precipitation of phosphorus b Trickling filter c Results in drinking water 12. Activated sludge is an aerobic process that can result in complete oxidation of organic matter. 13. Both require large areas of land and can result in the pollution of surface or groundwater if they are overloaded. 14. BOD Rate of Eutrophication Dissolved Oxygen Untreated 3+ 3+ + Primary 2+ 2+ 2+ Secondary + + 3+ Accumulation of BOD and loss of dissolved oxygen would be much less in a fast-moving river. Continual aeration caused by the river's movement would result in rapid oxidation of organic matter. 15. Biodegradation of sewage, herbicides, oil, or PCBs.Multiple-ChoiceAnswers are in boldface.1. Activated sludge system.a. the process takes place under aerobic conditions.b. the process takes place under anaerobic conditions.c. the amount of oxygen doesn't make any difference. 2. Denitrification.a. the process takes place under aerobic conditions.b. the process takes place under anaerobic conditions.c. the amount of oxygen doesn't make any difference. 3. Nitrogen fixation.a. the process takes place under aerobic conditions.b. the process takes place under anaerobic conditions.c. the amount of oxygen doesn't make any difference. 4. Methane production.a. the process takes place under aerobic conditions.b. the process takes place under anaerobic conditions.c. the amount of oxygen doesn't make any difference. 5. The water used to prepare intravenous solutions in a hospital contained endotoxins. Infection control personnel performed plate counts to find the source of the bacteria. All of the following conclusions about the bacteria can be drawn except which one?a. It was present as a biofilm in the pipes.b. It is gram-negative.c. It comes from fecal contamination.d. It comes from the city water supply.e. None of the above. 6. CO2 1 H2S Light C6H12O6 1 S0a. aerobic respirationb. anaerobic respirationc. anoxygenic photoauthotrophd. oxygenic photoautotroph 7. SO422 1 10 H1 1 10 e2 => H2S 1 4 H2Oa. aerobic respirationb. anaerobic respirationc. anoxygenic photoauthotrophd. oxygenic photoautotroph 8. CO2 1 8 H1 1 8 e2 => CH4 1 2 H2Oa. aerobic respirationb. anaerobic respirationc. anoxygenic photoauthotrophd. oxygenic photoautotroph Critical Thinking 1. The straight chain is readily degraded by beta oxidation (refer to "Lipid Catabolism," page 124). 2. Clinical Applications 1. The source of his recurrent infection was a biofilm on his pacemaker. The infection was cured with antibiotics because freed cells were sensitive to antibiotics; however, antibiotics couldn't penetrate the biofilm to eliminate the source. 2. Nitrates, phosphates, oxygen, or water may not be present in sufficient amounts for bacterial growth. The naturally occurring bacteria may be able to degrade the hydrocarbons. 3. Sewage infiltrated the municipal water supply via a pipe that was damaged by the flooding. Flow through the damaged pipe must be stopped; drinking water must be hyperchlorinated; infected individuals should be treated.

Tetanus [TET-nus] is a condition that affects the nervous system and causes painful, uncontrolled muscle spasms. People get tetanus when spores of the tetanus bacteria enter the body through an open wound and produce a powerful nerve poison. Tetanus spores are found throughout the environment, usually in soil, dust, and animal waste. Tetanus is preventable through immunization. Because of widespread use of tetanus vaccine, the condition is now rare. What is tetanus? Tetanus is a condition that affects the nervous system and causes painful, uncontrolled muscle spasms. Because of widespread immunization, tetanus is now rare. Another name for tetanus is lockjaw. What is the infectious agent that causes tetanus? Tetanus is caused by a toxin (poison) produced by spores of the bacterium Clostridium tetani. Spores are hardy forms of the bacteria that can survive in the environment in an inactive state for a long time. Where is tetanus found? Tetanus occurs worldwide. Tetanus spores are found throughout the environment, usually in soil, dust, and animal waste. How do people get tetanus? Tetanus spores can enter the body through a wound that is contaminated with soil, dust, or animal waste. Spores can get into the body through even a tiny pinprick or scratch, but they usually enter through deep puncture wounds or cuts, like those made by nails or knives. Tetanus spores can also get into the body when skin is damaged by burns or by injecting contaminated street drugs. Once the spores enter a wound, they produce a powerful nerve poison that spreads through the body and causes painful symptoms. What are the signs and symptoms of tetanus? The first signs of tetanus infection are usually a headache and spasms or cramping of the jaw muscles (lockjaw). As the poison spreads, it progressively attacks more groups of muscles, causing spasms in the neck, arms, legs, and stomach, and sometimes violent convulsions (seizures). How soon after exposure do symptoms appear? The time between the contamination of a wound and the first symptoms is usually less than 2 weeks but can range from 2 days to months. In general, the shorter the time between exposure and symptoms, the more severe the disease. How is tetanus diagnosed? Tetanus is diagnosed by its symptoms. Who is at risk for tetanus? In the United States, tetanus occurs mostly in newborns, children, young adults, and older adults who are either not immunized or not adequately immunized. Injecting drug users are also at risk. What complications can result from tetanus? In the United States, 3 of every 10 persons who get tetanus die from it. For those who survive, recovery can be long (1-2 months) and difficult. Muscle spasms usually decrease after about 2 weeks and disappear after another week or two, but the person may be weak and stiff for a long time. Other complications include breathing problems, bone fractures, high blood pressure, abnormal heartbeats, clotting in the blood vessels of the lung, pneumonia, and coma. What is the treatment for tetanus? Persons with tetanus usually must spend several weeks in the hospital in intensive care to manage the complications. How common is tetanus? Tetanus is a major problem in developing countries where immunization of children is not required or enforced. In the United States, most states require tetanus immunization for entry to school. Cases average between 50 and 100 per year, mostly in under-immunized older adults. Is tetanus an emerging infectious disease? No. Cases have been decreasing since the 1940s. However, getting children immunized and guaranteeing that they get the complete series of shots is still a challenge, especially among poor children and in areas of the country where under-immunization is a problem. How can tetanus be prevented? The most important way to prevent tetanus is through complete immunization and proper wound care. 1. Prevent tetanus through immunization -- An effective vaccine against tetanus has been available for many years. It is usually given to children combined with diphtheria and pertussis vaccines in a shot called DTP. A child needs five DTP shots, given at specified intervals, for complete protection. Tetanus booster shots are recommended every 10 years. 2. Prevent tetanus through proper care of wounds -- Cleaning all wounds, removing dead tissue, and using antibiotics for contaminated or infected wounds can reduce the likelihood of getting tetanus. Persons with wounds that are deep or dirty may need a tetanus booster shot if more than 5 years have passed since the last dose. An injection of tetanus immune globulin (TIG) given as soon as possible after a tetanus-prone injury can also help neutralize the poison that has not entered the nervous system.

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