micro test 3

14. A new medical research laboratory is being set up to handle various microorganisms. The lab will work with E. coli (nonpathogenic strains), HIV, Bacillus anthracis, and Ebola virus. How should the lab be organized in terms of biosafety levels, and which microbe belongs to each level? A. Biosafety Level 1 for E. coli, Level 2 for HIV, Level 3 for Bacillus anthracis, and Level 4 for Ebola virus. B. Biosafety Level 1 for Bacillus anthracis, Level 2 for E. coli, Level 3 for HIV, and Level 4 for Ebola virus. C. Biosafety Level 1 for HIV, Level 2 for E. coli, Level 3 for Ebola virus, and Level 4 for Bacillus anthracis. D. Biosafety Level 1 for E. coli, Level 2 for Bacillus anthracis, Level 3 for HIV, and Level 4 for Ebola virus.

-Answer: A. Biosafety Level 1 for E. coli, Level 2 for HIV, Level 3 for Bacillus anthracis, and Level 4 for Ebola virus. Rationale: E. coli (non-pathogenic strains): These are generally safe to handle and do not usually cause disease in healthy adults, so they are appropriate for Biosafety Level 1 (BSL-1), which is the basic level of protection suitable for agents not known to consistently cause disease in healthy adults. HIV: As a pathogen that can lead to a serious disease (AIDS) but requires specific conditions for transmission (like exchange of bodily fluids), HIV is typically handled at Biosafety Level 2 (BSL-2). BSL-2 is designed for work involving agents that pose moderate hazards to personnel and the environment. Bacillus anthracis: The causative agent of anthrax, Bacillus anthracis, can be lethal and is capable of aerosol transmission, necessitating a higher level of containment. It is handled at Biosafety Level 3 (BSL-3), which is for work with indigenous or exotic agents that may cause serious or potentially lethal disease through inhalation. Ebola virus: This is a highly infectious virus that causes severe disease and is often fatal. It requires the highest level of containment, Biosafety Level 4 (BSL-4), which is for work with dangerous and exotic agents that pose a high individual risk of aerosol-transmitted laboratory infections and life-threatening diseases.

7. In a hospital, two different protocols are used for cleaning: high level disinfectants for surgical instruments and low-level disinfectants for floors and surfaces. Given that high-level disinfectants eliminate all microorganisms except high levels of bacterial spores, while low-level disinfectants kill most bacteria, viruses, and fungi, what does this imply about the hospital's practices? A. Both protocols achieve sterilization but differ in application areas. B. The hospital uses sterilization for floors and surfaces, and disinfection for surgical instruments. C. High-level disinfectants are used for critical items requiring sterilization, while low-level disinfectants are sufficient for non-critical surfaces. D. The hospital doesn't employ antisepsis in either protocol.

-Answer: C. High-level disinfectants are used for critical items requiring sterilization, while low-level disinfectants are sufficient for non-critical surfaces. Rationale: High-Level Disinfectants: These are used for cleaning surgical instruments. High-level disinfectants are capable of eliminating all microorganisms, except for high levels of bacterial spores. While they do not achieve complete sterilization (which involves killing all forms of microbial life, including all spores), they provide a very high level of disinfection. This is critical for surgical instruments, which come into direct contact with sterile body sites or the vascular system and therefore require a higher standard of microbial control. Low-Level Disinfectants: These are used for floors and surfaces. Low-level disinfectants kill most bacteria, viruses, and fungi, but they may not be effective against more resistant organisms, such as bacterial spores. This level of disinfection is typically sufficient for general environmental surfaces in a hospital, which do not come into direct contact with sterile areas of the body.

16. Prion templating is a phenomenon in which a misfolded prion protein serves as a template to cause a normal, correctly folded version of the same protein to adopt an identical incorrect conformation. Prions are unique infectious agents that are composed solely of protein and do not contain any nucleic acid. They are thought to cause neurodegenerative diseases, such as Creutzfeldt-Jakob disease (CJD) and mad cow disease, in humans and animals, respectively. Match each step of the templating process to the correct label in the figure. (2 pts) 1. The cells make cellular prion protein (c-PrP), which is inserted into the cell's cytoplasmic membrane. 2. Erroneously folded p-PrP may infect the brain or may be produced by an altered c-PrP gene. 3. p-PrP (prion PrP protein) reacts with c-PrP (cellular PrP protein) on the cell surface. 4. p-PrP converts c-PrP to p-PrP. 5. p-PrP molecules continue to convert nor

1. A 2. B 3. C 4. D 5. E

9. Matching. Identify each of the events (1-7) that are occurring in the illustration of lytic versus lysogenic viral replication. (2 pts)

1. Phage attaches to cell 2. Phage DNA circularizes 3. New phage DNA and proteins are synthesized 4. Cell lyses releasing phages 5. Phage DNA is inserted into the bacterial chromosome 6. Prophage replicated at each normal cell division 7. Prophage may leave chromosome

6. Matching. Match the correct antimicrobial to the correct mechanism (A-F) for each of the antimicrobials that target prokaryotic ribosomes to inhibit protein synthesis. (2 pts)

A. Aminoglycosides B. Tetracycline C. Chloramphenicol D. Lincosamides E. Antisense Nucleic Acids F. Oxazolidinones

4. Fill in the blanks. Enter the correct mechanism for each of the animal virus host cell entry methods depicted in the figures (A-C).

A. Membrane Fusion B. Direct Penetration C. Endocytosis

4. Matching. The morphological arrangements of cocci and bacilli refer to the various patterns in which these cells can be found due to their division planes and whether they remain attached after division. Match each letter (A-I) to the corresponding morphological arrangement. (2 pts)

A. Palisade B. Single bacillus C. Diplobacilli D. Streptobacilli E. Staphylococci F. Tetrads G. Sarcinae H. Diplococci I. Streptococci

2. Which of the following is primarily responsible for the shape of a virion? A. the capsid B. the type of nucleic acid C. the specific host the virus targets D. the number of genes in the viral genome

Answer: A. the capsid Rationale: The capsid is the protein shell that encloses the viral genome, and its structure determines the overall shape of the virus. The capsid is made up of repeating protein subunits called capsomeres that come together to form the final shape of the virus. The type of nucleic acid, specific host, and number of genes in the viral genome can affect the replication and infectivity of the virus, but they do not determine the shape of the virion.

11. Which of the following is NOT a method for culturing viruses? A. Cell culture B. Embryonated eggs C. Animal inoculation D. Plant inoculation E. Immunoblotting

Answer: E. Immunoblotting Rationale: Immunoblotting is not a method for culturing viruses. It is a technique used to detect specific proteins, such as viral proteins, in a sample.

3. In a study of viral pathogenesis, researchers find that certain RNA viruses mutate at a higher rate than DNA viruses. Which of the following characteristics of RNA virus replication most likely contributes to this increased mutation rate? A. RNA viruses utilize the host's DNA polymerase, which has proofreading capabilities. B. RNA viruses replicate their genomes in the nucleus where mutations are less likely to occur. C. RNA viruses often replicate using RNA-dependent RNA polymerase, which lacks proofreading activity. D. RNA viruses typically integrate their genomes into the host DNA, leading to more stable replication.

answer: C. RNA viruses often replicate using RNA-dependent RNA polymerase, which lacks proofreading activity. Rationale: RNA viruses typically rely on RNA-dependent RNA polymerase (RdRp) to replicate their genomes. This enzyme is prone to making errors during replication because it lacks the proofreading capabilities that are found in DNA polymerases. Without the ability to correct errors that occur during the replication process, mutations accumulate at a higher rate in RNA viruses compared to DNA viruses, whose DNA polymerases have the ability to identify and correct mismatched nucleotides.

7. Fill in the blank. Identify the types (shapes) of viruses depicted below based on capsid structure:

A = helical, helical virus, helical viruses B = polyhedral, polyhedral virus, polyhedral viruses C = enveloped / spherical, enveloped / spherical virus, enveloped / spherical viruses D = complex / bacteriophage, complex virus, complex viruses, bacteriophage virus(s)

15. A clandestine military research organization is designing a super secret hidden research facility in the Antarctic. The facility will specialize in experiments using Yersinia pestis, yeast cells, methicillin-resistant Staphylococcus aureus (MRSA), and the handling of prions. Which biosafety level is appropriate for each organism? A. Level 1 for yeast cells, Level 2 for MRSA, Level 3 for Y. pestis, and Level 4 for prions. B. Level 1 for yeast cells, Level 3 for MRSA and Y. pestis, and Level 4 for prions. C. Level 2 for yeast cells, Level 3 for MRSA, Level 4 for Y. pestis, and Level 1 for prions. D. Level 1 for yeast cells and MRSA, Level 2 for Y. pestis, Level 3 for prions.

-Answer: A. Level 1 for yeast cells, Level 2 for MRSA, Level 3 for Y. pestis, and Level 4 for prions. Rationale: Yeast Cells: These are generally non-pathogenic and are considered safe to handle. They are typically worked with at Biosafety Level 1 (BSL-1), suitable for agents not known to consistently cause disease in healthy adults. Methicillin-resistant Staphylococcus aureus (MRSA): MRSA is a more virulent strain of Staphylococcus aureus and is resistant to many antibiotics. It is handled at Biosafety Level 2 (BSL-2), which is designed for work involving agents that pose moderate hazards to personnel and the environment. Yersinia pestis: The bacterium responsible for the plague, Y. pestis can be lethal and is capable of aerosol transmission. It requires Biosafety Level 3 (BSL-3) containment, which is for work with indigenous or exotic agents that may cause serious or potentially lethal disease through inhalation. Prions: Prions are known for causing fatal neurodegenerative diseases and are extremely difficult to deactivate. They require the highest level of biocontainment, Biosafety Level 4 (BSL-4), used for work with dangerous and exotic agents that pose a high individual risk of aerosol-transmitted laboratory infections and life-threatening disease.

4. A lab technician uses an autoclave to sterilize a set of surgical instruments and a 70% ethanol solution to clean the lab bench. Which statement accurately contrasts these two processes? A. Both processes achieve sterilization, but the autoclave is more time consuming. B. The autoclave sterilizes the instruments by killing all microbial life, while the ethanol disinfects the bench by reducing microbial load. C. Ethanol sterilizes the bench, whereas the autoclave disinfects the instruments. D. Both processes are forms of disinfection, differing only in the chemicals used.

-Answer: B. The autoclave sterilizes the instruments by killing all microbial life, while the ethanol disinfects the bench by reducing microbial load. Rationale: Autoclaving: This process involves using high-pressure steam at temperatures much higher than the boiling point of water. An autoclave effectively sterilizes surgical instruments by killing all forms of microbial life, including resistant spores. Sterilization refers to the complete elimination of all forms of life, including bacterial spores which are among the toughest organisms to kill. 70% Ethanol Solution: Ethanol is commonly used as a disinfectant. It works by denaturing proteins and dissolving lipids, effectively killing many types of microorganisms. However, it does not necessarily achieve the same level of sterility as autoclaving, especially with regard to more resistant forms like spores. Disinfection, unlike sterilization, doesn't necessarily kill all forms of life; instead, it reduces the microbial load to levels considered safe for general purposes.

3. Gerhard Domagk was awarded the Nobel Prize in Physiology or Medicine in 1939. However, the Nazi regime in Germany at the time had a decree that forbade German nationals from accepting the Nobel Prize. What groundbreaking medical discovery is Gerhard Domagk known for? A. The invention of the first vaccine B. The discovery of the first sulfa drug, Prontosil, leading to the development of antibiotics C. The identification of the polio virus D. The development of the first X ray machine

-Answer: B. The discovery of the first sulfa drug, Prontosil, leading to the development of antibiotics Rationale: Gerhard Domagk is renowned for his discovery of Prontosil, the first sulfa drug, which was a major breakthrough in medical history and laid the foundation for the development of antibiotics. Before the discovery of Prontosil in the 1930s, there were very few effective treatments for bacterial infections, which often led to high mortality rates from what are now considered treatable diseases. Domagk's work demonstrated that Prontosil was effective against a wide range of bacterial infections, thereby introducing a new era in medical treatment where bacterial diseases could be effectively combated with medication. This discovery was so significant that it earned him the Nobel Prize in Physiology or Medicine in 1939. The other options (A, C, D) are not related to Domagk's contribution to medicine.

5. In a clinical study, a 3% hydrogen peroxide solution was used for wound cleaning, while another group used a saline solution. The bacterial count in wounds treated with hydrogen peroxide showed a 99% reduction, while saline treated wounds showed no significant change. This scenario exemplifies: A. Sterilization with hydrogen peroxide. B. Antisepsis with hydrogen peroxide and sterilization with saline. C. Antisepsis with hydrogen peroxide, reducing the microbial load in the wound. D. Disinfection of the wound with both solutions.

-Answer: C. Antisepsis with hydrogen peroxide, reducing the microbial load in the wound. Rationale: Hydrogen Peroxide as an Antiseptic: Hydrogen peroxide in a 3% solution is commonly used as an antiseptic. Antiseptics are chemical agents applied to living tissue to reduce the possibility of infection, sepsis, or putrefaction. In this scenario, hydrogen peroxide's significant reduction of bacterial count by 99% in the wounds exemplifies its role as an antiseptic. It reduces the microbial load but does not necessarily achieve complete sterilization (the total destruction of all forms of microbial life). Saline Solution: Saline solution, which is essentially sterile salt water, is often used for wound cleaning due to its isotonic nature. It helps in cleaning the wound but does not have inherent antimicrobial properties. Thus, it is not considered an antiseptic, disinfectant, or sterilizing agent, as evidenced by the lack of significant change in bacterial count in wounds treated with saline. Why Not Sterilization or Disinfection: Sterilization refers to the complete destruction of all forms of microbial life, including bacteria, viruses, spores, and fungi, which is not the case here. Disinfection, on the other hand, implies the elimination of most pathogenic microorganisms (except bacterial spores) on inanimate objects, which also does not apply to this scenario where a living human tissue (wound) is involved.

17. A study shows that high temperatures achieved through moist heat (like autoclaving) effectively kill all forms of microbial life, including spores, while certain chemical disinfectants fail to eliminate spore forming bacteria. This suggests: A. Physical methods are universally superior to chemical methods. B. Chemical methods should be used for all microbial control scenarios. C. Certain physical methods can be more effective than chemical methods, particularly against resilient forms like spores. D. Both methods are equally effective against all forms of microbial life.

-Answer: C. Certain physical methods can be more effective than chemical methods, particularly against resilient forms like spores. Rationale: The study's findings highlight the effectiveness of moist heat, such as that used in autoclaving, in killing all forms of microbial life, including the more resistant spores. Autoclaving, a physical method of sterilization, employs high temperature and pressure to achieve a level of microbial control that is comprehensive and effective against even the toughest forms of microbial life. On the other hand, the study notes that certain chemical disinfectants are unable to eliminate spore-forming bacteria. This indicates that while chemical methods are broadly effective against many types of microorganisms, they may not be as effective as some physical methods (like autoclaving) in dealing with particularly resilient forms, such as spores. The answer does not suggest that physical methods are universally superior (option A) or that chemical methods should be used in all scenarios (option B). Nor does it suggest that both methods are equally effective against all forms of microbial life (option D). Instead, it indicates that the choice between physical and chemical methods should be based on the specific requirements of the microbial control scenario, particularly when dealing with hard-to-kill organisms like spores.

18. In a food processing plant, different methods are used to control microbial growth. Machines are routinely passed through high temperature steam, while work surfaces are wiped down with a quaternary ammonium compound. What does this imply about the use of physical and chemical control methods? A. The plant exclusively uses chemical methods for microbial control. B. Physical methods are favored due to their effectiveness in eliminating all microbial forms. C. The plant employs physical methods for machines and chemical methods for surfaces. D. The plant uses outdated methods and should consider modern alternatives.

-Answer: C. The plant employs physical methods (high-temperature steam) for machines and chemical methods (quaternary ammonium compound) for surfaces. Rationale: The scenario describes the use of two different types of microbial control methods for different applications within the food processing plant. High-Temperature Steam: This is a physical method of microbial control. Using high temperatures, especially in the form of steam, is a common and effective way to sterilize equipment, particularly in settings where sterilization is crucial. The high temperature is effective at killing a wide range of microorganisms, including resistant spores. Quaternary Ammonium Compounds: These are chemical agents used for disinfecting surfaces. They are a type of chemical disinfectant known for their effectiveness against a variety of microbes and are commonly used in settings that require regular and efficient surface disinfection. The use of high-temperature steam for machines and quaternary ammonium compounds for surfaces indicates a strategic approach to microbial control, utilizing the strengths of both physical and chemical methods in appropriate contexts. This strategy helps in ensuring effective microbial control while taking into account the practical aspects of the applications, such as the nature of the surfaces being treated and the type of microorganisms expected in different areas of the plant.

11. A study shows that certain microbial groups consistently survive standard disinfection procedures. These groups are identified as bacterial endospores, mycobacteria, and non enveloped viruses. What common feature contributes to their high resistance to antimicrobial agents? A. Their rapid reproduction rate B. Their ability to form biofilms C. The presence of protective structures or layers D. Their ability to mutate rapidly

-Answer: C. The presence of protective structures or layers. Rationale: Bacterial Endospores: These are highly resistant, dormant structures formed by some bacteria as a means of survival under adverse conditions. The endospore's resistance is due to its tough outer coating, which protects the genetic material inside from extreme environmental conditions, including many disinfection procedures. Mycobacteria: This group of bacteria, which includes the causative agents of tuberculosis and leprosy, is known for its robust cell wall. The cell wall of mycobacteria contains high levels of mycolic acid, making it waxy and resistant to many chemical disinfectants. Non-Enveloped Viruses: Unlike enveloped viruses, non-enveloped viruses lack a lipid envelope and instead have a more resilient protein coat. This protein coat provides them with greater resistance to environmental stresses and many disinfectants.

15. What is the fate of the prophage during the lysogenic cycle? A. It is copied every time the host DNA replicates. B. It is degraded by the activity of host defense enzymes. C. It is packaged into viral proteins and maintained until the host is exposed to environmental stress. D. It is released from the cell by cell-lysis.

Answer: A. It is copied every time the host DNA replicates. Rationale: During the lysogenic cycle, the prophage integrates into the host chromosome and is replicated along with the host DNA during cell division.

. Solving for Maximum Per Capita Rate of Increase. A bacterial population grows following a logistic model with a carrying capacity (K) of 20,000 bacteria. If the initial population size (N) is 5000 bacteria and instantaneous rate of population growth (dN/dt) is 200 bacteria per hour, calculate the maximum per capita rate of increase (rmax). Express your answer as a decimal rounded to the nearest thousandth and do NOT include the units in your answer.

Answer: 0.053 per hour Rationale: The maximum per capita rate of increase (rmax) is calculated using the rearranged logistic growth formula. It represents the growth rate per individual in the population when the population size is 5,000 and the carrying capacity is 20,000. The rate reflects how rapidly the population grows towards its carrying capacity under the given conditions. The logistic growth equation is given as: dN/dt = rmax * N * (1 - N/K) Where: dN/dt is the instantaneous rate of population growth (given as 200 bacteria per hour). N is the current population size (given as 5,000 bacteria). K is the carrying capacity (given as 20,000 bacteria). rmax is the maximum per capita rate of increase (which we need to calculate). We can rearrange the equation to solve for rmax: rmax = (dN/dt) / (N * (1 - N/K)) Now, let's plug in the values: rmax = (200 bacteria/hour) / (5,000 bacteria * (1 - 5,000 bacteria / 20,000 bacteria)) Simplify the equation: rmax = (200 bacteria/hour) / (5,000 bacteria * 0.75)

Logistic Growth Calculation at Half Carrying Capacity. A bacterial population grows following a logistic model with a carrying capacity (K) of 10,000 bacteria. If the initial population size (N) is 5000 bacteria and the maximum per capita rate of increase (rmax) is 0.1 per hour, calculate the rate of population growth at this instant in bacteria per hour. Round your answer to the nearest whole bacterium and do NOT include the units in your answer.

Answer: 250 bacteria per hour Rationale: The logistic growth equation is given as: dN/dt = rmax * N * (1 - N/K) Where: dN/dt is the instantaneous rate of population growth (which we need to calculate). N is the current population size (given as 5,000 bacteria). K is the carrying capacity (given as 10,000 bacteria). rmax is the maximum per capita rate of increase (given as 0.1 per hour). We need to find the rate of population growth (dN/dt) at this instant. Let's plug in the values and solve for dN/dt: dN/dt = (0.1 per hour) * (5,000 bacteria) * (1 - 5,000 bacteria / 10,000 bacteria) Simplify the equation: dN/dt = 0.1 * 5,000 * (1 - 0.5) dN/dt = 0.1 * 5,000 * 0.5 Now, calculate dN/dt: dN/dt = 0.1 * 2,500 dN/dt = 250 bacteria per hour

Exponential Growth Calculation. Suppose a bacterial culture starts with 1000 bacteria and the per capita growth rate (rmax) is 0.5 per hour. Calculate the number of bacteria after 3 hours assuming exponential growth conditions. Round your answer to the nearest whole bacterium. Formula: N = N0 × e(rmax ×t) Where: N = number of bacteria at time N0 = initial number of bacteria rmax = maximum per capita rate of increase t = time in hours e = base of the natural logarithm (approximately 2.71828)

Answer: 4482 bacteria Rationale: Exponential growth is modeled using the formula: N(t) = N0 * e^(rmax * t) Where: N(t) is the population size at time t. N0 is the initial population size (given as 1,000 bacteria). rmax is the per capita growth rate (given as 0.5 per hour). t is the time (given as 3 hours). Now, plug in the values and calculate N(t): N(t) = 1,000 * e^(0.5 * 3) N(t) = 1,000 * e^(1.5) Using the value of e^(1.5), we can calculate N(t): N(t) ≈ 1,000 * 4.48169 (rounded to 5 decimal places) N(t) ≈ 4,481.69 bacteria

What characteristic is common among chemoheterotrophic organisms regarding their nutritional requirements? A. They require pre-formed organic compounds for both energy and carbon source. B. They are capable of using carbon dioxide as their sole source of carbon. C. They can produce organic compounds from simple inorganic sources using light energy. D. They can survive without any organic compounds by utilizing inorganic chemical reactions

Answer: A. They require pre-formed organic compounds for both energy and carbon source. Rationale: Chemoheterotrophs are organisms that obtain both their energy and carbon from organic compounds. This means they consume organic matter produced by other organisms, which can include carbohydrates, fats, and proteins, for their metabolic needs.

7. In a pristine ancient lake, researchers discover a bacterium with a primitive cell structure that resembles early life forms. This bacterium is likely: A. A deeply branching bacterium adapted to ancient Earth-like conditions. B. A cyanobacterium, contributing to the transformation of the atmosphere. C. A green nonsulfur bacterium, indicative of chloroplast development. D. A purple sulfur bacterium, dependent on hydrogen sulfide for photosynthesis.

Answer: A. A deeply branching bacterium adapted to ancient Earth-like conditions. Rationale: The description of a bacterium with a primitive cell structure found in an environment resembling early Earth conditions aligns with the characteristics of deeply branching bacteria. These bacteria are thought to be similar to the earliest forms of bacterial life, and they are known to inhabit environments that mimic those of early Earth, such as hydrothermal vents and hot springs, which are often found in ancient lakes. They are termed "deeply branching" because they appear near the root of phylogenetic trees, indicating their ancient divergence from other forms of life. Option B, cyanobacteria, are known for their significant role in transforming Earth's early anaerobic atmosphere into an oxygen-rich one through oxygenic photosynthesis. However, they are not typically characterized by a primitive cell structure. Option C, green nonsulfur bacteria, are typically involved in photosynthesis but do not necessarily have primitive cell structures that resemble early life forms. Option D, purple sulfur bacteria, use hydrogen sulfide during photosynthesis but are not specifically described as having a primitive cell structure or being similar to the earliest forms of bacterial life.

12. Scientists are investigating the cause of the earthy scent in the air following rainfall, known as petrichor. Which phylum of bacteria, commonly found in soil, is recognized for producing the compound that contributes to this distinctive smell? A. Actinobacteria B. Proteobacteria C. Archaea D. Spirochaetes

Answer: A. Actinobacteria Rationale: The earthy scent that we detect after rainfall, termed petrichor, is largely due to geosmin, a compound produced by soil-dwelling bacteria. Actinobacteria, particularly the genus Streptomyces, are known for their geosmin production. These bacteria play a crucial role in the decomposition of organic materials and, upon moistening of the soil after dry conditions, release geosmin into the air, contributing to the characteristic smell of petrichor. Option B, Proteobacteria, is a broad phylum of bacteria that includes many different types of bacteria, some of which are found in soil, but they are not primarily known for producing the earthy scent associated with petrichor. Option C, Archaea, consists of a group of microorganisms that are distinct from bacteria and are not known for geosmin production. Option D, Spirochaetes, are a group of bacteria characterized by their spiral shape and are not associated with the production of petrichor-related compounds.

1. If a novel species is found to have a cell wall lacking peptidoglycan but containing ether-linked membrane lipids, it is most likely: A. An archaeon B. A Gram-positive bacterium C. A Gram-negative bacterium D. A eukaryotic microbe

Answer: A. An archaeon Rationale: Archaea are known for having unique cell membranes that contain ether-linked lipids, which differ from the ester-linked lipids found in bacteria and eukaryotes. Additionally, archaeal cell walls do not contain peptidoglycan, which is a major component of bacterial cell walls, particularly in Gram-positive bacteria. Gram-negative bacteria also have peptidoglycan, albeit in a thinner layer sandwiched between their outer and inner membranes. Eukaryotic microbes, such as fungi and protists, do not typically have cell walls with peptidoglycan. Therefore, a microorganism with ether-linked lipids and no peptidoglycan is characteristic of archaea, making option A the correct answer.

1. Selective toxicity is a principle that is fundamental to pharmacology and particularly crucial in antimicrobial therapy. What is meant by selective toxicity? A. Chemotherapeutic agents should act against the pathogen and not the host. B. Chemotherapeutic agents should work on many different targets on a pathogen. C. Chemotherapeutic agents should have only one mode of action. D. Chemotherapeutic agents should work on certain types of pathogens.

Answer: A. Chemotherapeutic agents should act against the pathogen and not the host. Rationale: Selective toxicity refers to the ability of a chemotherapeutic agent to selectively target and inhibit or kill a pathogen without significantly harming the host organism. This is a critical feature of effective antimicrobial agents, as the goal is to eliminate the infectious agent while minimizing side effects and damage to the patient's own cells and tissues. In contrast, non-selective agents, such as broad-spectrum antibiotics or chemotherapy drugs, can cause significant harm to both pathogenic and healthy cells, leading to serious side effects and toxicities.

8. There are several factors to consider when choosing a route of administration for antimicrobial drugs. Which would likely be the most effective route of administration when a patient has an infection that could result in death if not treated promptly? A. Continuous intravenous B. One-time intravenous C. Intramuscular injection D. Oral E. Topical

Answer: A. Continuous intravenous Rationale: When a patient has an infection that could result in death if not treated promptly, the most effective route of administration for antimicrobial drugs is continuous intravenous (IV) infusion. See Question 1 Rationale for additional information.

8. Which of the following is corrected regarding the viral envelope? A. It is composed of cellular phospholipid membrane but both cellular and viral proteins. B. It contains only viral proteins C. It contains only host proteins D. It is composed of mycolic acid E. It results in viruses that are stronger and more robust than naked viruses.

Answer: A. It is composed of cellular phospholipid membrane but both cellular and viral proteins. Rationale: The viral envelope is a lipid bilayer membrane that surrounds some viruses, derived from the host cell's plasma membrane or other internal membrane systems during the virus's budding process. This membrane contains both viral and host cell proteins, as the virus acquires its envelope by budding from the host cell membrane, taking some of the host cell's membrane proteins with it. Option B, "It contains only viral proteins," is incorrect, as the viral envelope contains both viral and host cell proteins. Option C, "It contains only host proteins," is also incorrect, as the viral envelope contains both viral and host cell proteins. Option D, "It is composed of mycolic acid," is incorrect, as mycolic acid is not typically a component of the viral envelope. Option E, "It results in viruses that are stronger and more robust than naked viruses," is generally true inside of hosts, but enveloped viruses are less resistant to environmental stresses than naked viruses outside of the host. Enveloped viruses are typically more fragile and susceptible to environmental stressors such as temperature, pH changes, and detergents. The lipid bilayer membrane of the viral envelope can be disrupted or destroyed by these stressors, which can lead to the inactivation of the virus. In contrast, naked viruses are generally more resistant to environmental stressors due to the stability of their protein coats.

12. In a hospital setting, it's observed that disinfectants are less effective against Mycobacterium tuberculosis, Clostridium spores, and Norovirus. What is the most likely reason for this resistance pattern? A. Mycobacterium tuberculosis has a waxy cell wall, Clostridium forms hardy spores, and Norovirus lacks an envelope, making them more resistant. B. All three microbes have developed specific enzyme systems to neutralize antibiotics. C. These microbes have acquired resistance genes through horizontal gene transfer. D. They are all capable of altering their target sites for antimicrobial agents.

Answer: A. Mycobacterium tuberculosis has a waxy cell wall, Clostridium forms hardy spores, and Norovirus lacks an envelope, making them more resistant. Rationale: Mycobacterium tuberculosis: This bacterium is known for its waxy cell wall, which contains high levels of mycolic acid. This unique cell wall composition provides a barrier that makes it resistant to many disinfectants and also to dehydration and certain antibiotics. Clostridium spores: The genus Clostridium includes several species that can form spores, such as Clostridium difficile. These spores are extremely resilient and can survive harsh environmental conditions, including exposure to many disinfectants. The spore coat provides physical and chemical protection, making them difficult to eradicate. Norovirus: This is a non-enveloped virus, meaning it lacks a lipid envelope. Non-enveloped viruses generally have a more robust protein coat compared to enveloped viruses, making them more resistant to certain disinfectants and environmental conditions.

11. During the winter months, a clinic sees an increase in patients with symptoms of pneumonia. Cultures reveal a Gram-positive coccus in chains. Which genus, according to the table, is the most likely to be associated with these pneumonia cases? A. Streptococcus B. Staphylococcus C. Mycoplasmas D. Lactobacillus

Answer: A. Streptococcus Rationale: The description of the bacteria as a Gram-positive coccus in chains is characteristic of the genus Streptococcus. Streptococcus pneumoniae, in particular, is a well-known cause of pneumonia, especially during the winter months when respiratory infections are more common. The chain formation is a distinctive growth pattern for Streptococcus species, which differentiates them from Staphylococcus species, which tend to form clusters. Option B, Staphylococcus, can cause pneumonia, but it is more commonly associated with skin infections and food poisoning, and does not typically form chains. Option C, Mycoplasmas, do cause pneumonia (often referred to as "walking pneumonia") but are not characterized as Gram-positive due to their lack of a traditional cell wall, and they do not form chains. Option D, Lactobacillus, is generally considered non-pathogenic and is not a common cause of pneumonia.

. In bacterial cells, binary fission is initiated when: A. The bacterium reaches a critical size and nutrient availability is high. B. The bacterium encounters unfavorable environmental conditions. C. The bacterium has exhausted all its genetic material. D. The bacterium's cell wall begins to break down due to external damage.

Answer: A. The bacterium reaches a critical size and nutrient availability is high. Rationale: Binary fission is the primary method of reproduction in bacteria and typically begins when the cell has grown to a size where it can split into two viable daughter cells. This process usually occurs under favorable nutritional conditions, allowing the cell to duplicate its DNA and cellular components before division.

14. Which of the following steps in the nucleic acid synthesis pathway is specifically inhibited by sulfonamides? A. The conversion of PABA to folic acid. B. The conversion of folic acid to tetrahydrofolic acid C. The conversion of tetrahydrofolic acid to purine/pyrimidine nucleotides D. The conversion of purine/pyrimidine nucleotides to DNA/RNA

Answer: A. The conversion of PABA to folic acid. Rationale: Sulfonamides are a class of antimicrobial drugs that work by inhibiting bacterial folate synthesis, which is required for nucleic acid synthesis. Folate is a precursor of tetrahydrofolic acid, which is involved in the synthesis of purine and pyrimidine nucleotides, the building blocks of DNA and RNA. Sulfonamides specifically inhibit the conversion of para-aminobenzoic acid (PABA) to folic acid by binding to and inhibiting the enzyme dihydropteroate synthase, which is required for this conversion. This inhibition leads to a decrease in the production of tetrahydrofolic acid and ultimately a decrease in the production of purine and pyrimidine nucleotides, resulting in inhibition of bacterial growth.

. A medical microbiologist isolates a bacterium from a patient's anaerobic infection site. Upon culturing, it is noted that the bacterium does not grow in the presence of oxygen. However, genomic analysis shows the presence of genes coding for catalase and peroxidase. How can this finding be reconciled with the bacterium's growth profile? A. The genes for catalase and peroxidase are likely non-functional or not expressed. B. The bacterium is actually a facultative anaerobe that prefers anaerobic conditions. C. The presence of catalase and peroxidase suggests the bacterium can neutralize singlet oxygen and superoxide radicals. D. The bacterium might be using catalase and peroxidase as an alternative carbon and energy source.

Answer: A. The genes for catalase and peroxidase are likely non-functional or not expressed. Rationale: The bacterium does not grow in the presence of oxygen, which is characteristic of obligate anaerobes. Despite having genes for catalase and peroxidase, which are enzymes that break down hydrogen peroxide into water and oxygen, the bacterium may not express these genes, or the enzymes might be non-functional. This is consistent with obligate anaerobes which often lack or have reduced capacity to deal with oxidative stress due to oxygen.

A microbiology student is learning the streak plate method for isolating individual bacterial colonies from a mixed culture. After incubation, she observes that colonies are not well-isolated on the agar plate. Which of the following could be the reason for the lack of isolation? A. The loop was not sterilized between streaks of different plate sectors. B. The inoculum size was too small, leading to insufficient growth. C. The agar plate was incubated at the wrong temperature. D. The student did not allow the loop to cool after sterilizing it, killing the bacteria.

Answer: A. The loop was not sterilized between streaks of different plate sectors. Rationale: The streak plate method aims to separate individual bacteria to form discrete colonies by gradually diluting the bacteria across the surface of an agar plate. If the loop is not sterilized between streaks, bacteria from a more concentrated sector can be carried over to the next sector, leading to a lack of isolation. The inoculum size being too small (B) typically results in fewer colonies, not poorly isolated ones. Incubation at the wrong temperature (C) could affect overall growth but would not specifically impact colony isolation. Not allowing the loop to cool (D) would indeed kill the bacteria, but this would result in no growth at all, rather than a lack of isolation. Hence, not sterilizing the loop between streaks is the most likely reason for failing to isolate individual colonies.

10. A variety of problems make it challenging to study viruses in the lab and may limit our ability to fully understand the mechanisms of viral infection that are necessary to develop effective treatments and vaccines. Which of the following is a problem with growing viruses in the laboratory? A. Viruses require a living host to replicate B. Viruses are too small to study C. Viruses are difficult to detect D. Viruses are resistant to all known disinfectants E. Most viruses can infect a wide range of hosts, posing a risk to researchers.

Answer: A. Viruses require a living host to replicate Rationale: One of the major challenges of studying viruses in the laboratory is that viruses require a living host cell to replicate. This means that viruses cannot be grown in a standard culture medium like bacteria or other cells. Instead, researchers must use specialized techniques to grow viruses in living cells or in organisms that can be infected by the virus. In addition to the challenge of growing viruses, there are other obstacles that can make studying viruses difficult, such as their small size, their ability to mutate rapidly, and the fact that some viruses can infect a wide range of hosts, but many others are more limited in their host range..

2. Broad-spectrum antibiotics target many types of bacteria. The use of broad-spectrum antibiotics to treat an infection might seem like a good idea at first, but there are some drawbacks to their use. Which of the following is NOT a concern when using broad-spectrum antibiotics? Broad spectrum antibiotics... A. may increase microbial antagonism leading to more competition between microbiota. B. can destroy normal microbiota. C. are more likely to lead to antibiotic resistant bacteria. D. may allow for secondary or superinfections to develop.

Answer: A. may increase microbial antagonism leading to more competition between microbiota. Rationale: While broad-spectrum antibiotics can have many negative effects, increasing microbial antagonism leading to more competition between microbiota is not one of them. The other options listed are all concerns when using broad-spectrum antibiotics. They can destroy normal microbiota, leading to opportunistic infections and allowing for secondary or superinfections to develop. Additionally, broad-spectrum antibiotics are more likely to lead to antibiotic-resistant bacteria due to their wider range of activity.

2. The first true antibiotic was __________, which was discovered by __________. A. penicillin; Alexander Fleming B. streptomyces; Selman Waksman C. sulfanilamide; Gerhart Domagk D. arsenic compounds; Paul Enrich

Answer: A. penicillin; Alexander Fleming Rationale: penicillin is a type of antibiotic that is effective against a wide range of bacterial infections. It was discovered by the Scottish biologist and pharmacologist Alexander Fleming in 1928. Fleming noticed that a type of mold called Penicillium notatum had contaminated one of his petri dishes and was producing a substance that killed the bacteria he had been growing. He went on to isolate and purify the substance, which he named penicillin. This discovery was a major breakthrough in the fight against bacterial infections and paved the way for the development of other antibiotics.

2. The ____________ is a defining characteristic of bacteria which is not found in eukaryotes and is therefore a good choice for chemotherapeutic agents. A. peptidoglycan cell wall B. carbohydrate glycocalyx C. protein cell wall D. phospholipid cytoplasmic membrane E. protein cell wall

Answer: A. peptidoglycan cell wall Rationale: peptidoglycan is a unique component of the cell wall of bacteria that provides structural support and protection from the environment. It is composed of a complex network of polysaccharides and peptides that is not present in eukaryotic cells. This makes the peptidoglycan cell wall a good target for chemotherapeutic agents that can selectively target and destroy bacterial cells without harming eukaryotic cells.

3. Many drugs that inhibit the synthesis of the cell wall act by A. preventing the cross-linkage of NAM subunits. B. disrupting the formation of the mycolic acid layer of the cell wall. C. preventing the formation of β-lactamases. D. blocking the secretion of cell wall molecules from the cytoplasm.

Answer: A. preventing the cross-linkage of NAM subunits. Rationale: The cell wall of bacteria is composed of peptidoglycan, a complex network of polysaccharides and peptides that provides structural support and protection from the environment. Peptidoglycan synthesis involves the sequential addition of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) subunits, which are cross-linked by peptide bridges. Many drugs that inhibit the synthesis of the cell wall act by interfering with this process, often by preventing the cross-linkage of NAM subunits. These drugs include penicillins, cephalosporins, carbapenems, and monobactams, which all target different aspects of peptidoglycan synthesis but ultimately result in the disruption of cell wall formation and bacterial cell death.

13. A _____ phage is one that displays a lysogenic life cycle in contrast to a _____ phage that does not have the ability to display lysogeny. A. temperate, virulent B. virulent, temperate C. wild-type, mutated D. mutated, prophage E. prophage, temperate

Answer: A. temperate, virulent Rationale: N/A

6. Dengue (DENG-gey) fever is a viral disease that occurs in tropical and subtropical areas of the world. Outside of blood transfusions, the only way a human is likely to contract dengue virus is through the bite of an infected Aedes aegypti mosquito. Humans are the only known vertebrate hosts of the virus. Why are viruses typically so specific for the hosts they infect? The specificity of the virus is due to... A. the affinity of viral surface proteins for complementary surface proteins on host cells. B. the nutritional requirements of the virus. C. the specific body temperature of humans. D. the aerobic requirements of the virus. E. the affinity of virus for human nucleic acids.

Answer: A. the affinity of viral surface proteins for complementary surface proteins on host cells. Rationale: The specificity of a virus for the hosts they infect is primarily due to the affinity of viral surface proteins for complementary surface proteins on host cells. These surface proteins, such as receptors and co-receptors, are essential for the virus to attach to and enter host cells. Different viruses have specific surface proteins that allow them to bind to particular receptors on the surface of host cells. The virus-host specificity is therefore determined by the compatibility between the viral surface proteins and the host cell receptors. Option B, "the nutritional requirements of the virus," is not a significant factor in virus-host specificity. Option C, "the specific body temperature of humans," is not a significant factor in virus-host specificity either. Option D, "the aerobic requirements of the virus," is also not a significant factor in virus-host specificity. Option E, "the affinity of virus for human nucleic acids," is not an accurate explanation for the specificity of viruses for their hosts, as the surface proteins, not nucleic acids, determine virus-host specificity.

17. After initially failing to reisolate bacteriophage from a bacterial culture using centrifugation and filtration, the same researcher from above decides to wait to ensure the phages have completed their lytic cycle. Considering the average burst time for bacteriophages under optimal conditions, how long should the researcher wait before attempting to reisolate the bacteriophage? A. 5 to 10 minutes B. 20 to 40 minutes C. 1 to 2 hours D. 4 to 6 hours

Answer: B. 20 to 40 minutes Rationale: The average burst time for many bacteriophages, such as the T4 bacteriophage infecting E. coli, is typically around 20 to 40 minutes under optimal conditions. This time frame includes the attachment of the phage to the bacterium, the injection of its genetic material, replication inside the host, assembly of new phage particles, and finally the lysis (bursting) of the host cell, releasing the new phages. Waiting for this duration ensures that the lytic cycle is likely to be complete, and new phages are present in the extracellular environment, allowing for successful reisolation using centrifugation and filtration. Options A, C, and D do not align with the typical burst time for bacteriophages under standard conditions.

5. In a minimum bactericidal concentration (MBC) test, Petri plates are inoculated with samples taken from clear MIC tubes. After incubation, any growth of bacterial colonies on a plate indicates that the original concentration of antimicrobial drug was bacteriostatic. The picture below depicts the results of an MBC test. Based on these data, the MIC for this drug is _____ and the MBC is _____. A. 8 g/ml; 8 g/ml B. 8 g/ml; 16 g/ml C. 8 g/ml; 25 g/ml D. 16 g/ml; 8 g/ml E. 25 g/ml; 8 g/ml

Answer: B. 8 g/ml; 16 g/ml Rationale: In the MBC test, samples taken from clear MIC tubes are plated onto nutrient agar plates and incubated. The MBC is the lowest concentration of an antimicrobial agent that completely kills the bacteria being tested, as evidenced by the absence of any visible growth on the agar plates. Looking at the options provided, the correct answer is B. There is no visible growth of bacteria on the plate with the highest concentration of antimicrobial agent (16 μg/mL), indicating that this concentration is bactericidal, and thus the MBC. The MIC can be determined by identifying the lowest concentration of antimicrobial agent that inhibits visible growth of the bacteria on the in the tube but not the agar plate. In this case, the MIC is 8 μg/mL, which is the concentration of the antimicrobial agent in the tube with no visible growth before incubation on drug-free agar.

6. A newly discovered archaeon can thrive in both high-temperature environments and saline conditions. It has a cell membrane consisting of ether-linked lipids but does not utilize or produce methane. This archaeon is likely to be classified as: A. A thermophilic methanogen, given its tolerance to heat. B. A halophilic thermophile, capable of surviving in both hot and saline conditions. C. An acidophilic halophile, due to its adaptive nature. D. A psychrophilic methanogen, misidentified in its temperature range.

Answer: B. A halophilic thermophile, capable of surviving in both hot and saline conditions. Rationale: The key aspects of the archaeon described in the scenario are its ability to thrive in high-temperature environments and saline conditions, along with the presence of ether-linked lipids in its cell membrane, which is a characteristic feature of archaea. The absence of methane utilization or production rules out its classification as a methanogen, which is known for methane production as a part of their metabolic process. Given its dual tolerance for high temperatures and high salinity, the most suitable classification would be a halophilic thermophile. Halophiles are adapted to high salinity environments, and thermophiles are organisms that thrive at relatively high temperatures. The combination of these traits in one archaeon suggests it can survive and function in environments that are both saline and hot.

7. "Drug concentration" is derived by collecting a blood sample from a patient at any time after drug administration and measuring the amount of a drug in a given volume of blood plasma. Match the correct drug concentrations with their most likely routes of administration using the graph. A. A: IM, B: IV, C: Oral B. A: IV, B: IM, C: Oral C. A: Oral, B: IM, C: IV D. A: Oral, B: IV, C: IM E. A: IV, B: Oral, C: IM

Answer: B. A: IV, B: IM, C: Oral Rationale: Drugs administered via IM injection are absorbed slowly from the injection site and enter the bloodstream over time. This results in a slower onset of action compared to IV administration but faster than oral administration. IM administration also produces lower peak drug concentrations compared to IV administration. Drugs administered via IV injection are delivered directly into the bloodstream, resulting in rapid onset of action and high drug concentrations. IV administration allows for precise control over drug dosage and can quickly deliver drugs to tissues throughout the body. Drugs administered orally are absorbed through the gastrointestinal (GI) tract and enter the bloodstream through the liver. This process can result in delayed onset of action and lower peak drug concentrations compared to IV or IM administration.

A microbiologist notes that a certain species of bacteria only grows well under high pressures, such as those found at the bottom of deep oceans. This bacterium can be categorized as a: A. Barotolerant organism B. Barophilic organism C. Osmophilic organism D. Acidophilic organism

Answer: B. Barophilic organism. Rationale: Barophilic organisms are those that thrive at high pressures, such as those found in deep-sea environments. They are adapted to these conditions and often cannot survive at the lower pressures found at the surface. Barotolerant organisms, on the other hand, can tolerate high pressures but do not necessarily require them for optimal growth. Osmophilic organisms are adapted to high solute concentrations, and acidophilic organisms thrive in acidic environments, which are not related to the pressure conditions described in the question.

8. A new antibiotic, Drug X, is found to be highly effective against a range of bacteria. Laboratory analysis shows that it impedes the synthesis of peptidoglycan. Which cellular component does Drug X primarily target? A. Cytoplasmic membranes B. Cell walls C. Nucleic acids D. Proteins

Answer: B. Cell walls Rationale: Peptidoglycan Synthesis: Peptidoglycan is a crucial component of the cell walls of bacteria, particularly in Gram-positive bacteria. It provides structural integrity and shape to the bacterial cell. An antibiotic that impedes the synthesis of peptidoglycan would primarily target the cell wall. By inhibiting the formation of this vital component, the antibiotic compromises the structural integrity of the bacterial cell wall, leading to the lysis and death of the bacteria.

15. Nearly all quinolone antibiotics in use are fluoroquinolones, which contain a fluorine atom in their chemical structure and are effective against both Gram-negative and Gram-positive bacteria. One example is ciprofloxacin, which is one of the most widely used antibiotics worldwide. Quinolones and fluoroquinolones act against what bacterial target? A. Cell walls B. DNA gyrase C. Metabolic pathways unique to bacteria D. Bacterial ribosomes E. Cell membranes

Answer: B. DNA gyrase Rationale: Quinolones and fluoroquinolones act against bacterial DNA gyrase, which is an essential enzyme involved in bacterial DNA replication and transcription. By binding to and inhibiting DNA gyrase, these antibiotics prevent the bacteria from effectively synthesizing and replicating their DNA, ultimately leading to inhibition of bacterial growth and cell death.

5. A microbiology student is examining electron micrographs of three distinct viruses. Virus A appears spherical with a lipid bilayer envelope, Virus B is helical with no envelope, and Virus C is polyhedral in shape with a protein coat. What differences in viral characteristics can be identified from this scenario and data? A. Differences in viral genome structure B. Differences in viral size and shape C. Variation in host cell range D. Variation in viral replication cycles

Answer: B. Differences in viral size and shape Rationale: The scenario and data provide information about the size and shape of the three distinct viruses, which indicates differences in viral characteristics. Virus A is described as spherical with a lipid bilayer envelope, Virus B is helical with no envelope, and Virus C is polyhedral in shape with a protein coat. These descriptions highlight variations in the size and shape of the viruses, making option B the correct choice. The scenario does not provide information about viral genome structure, host cell range, or viral replication cycles, so options A, C, and D are not directly addressed in the provided data.

8. Researchers isolate a nonmotile bacterium from a stratified lake where sunlight can penetrate but oxygen is absent below the thermocline. This bacterium contains bacteriochlorophyll, which allows it to utilize light for energy via a process that does not generate oxygen, and it relies on sulfur compounds instead of water as electron donors in its photosynthetic process. Which type of phototrophic bacteria does it likely represent? A. Cyanobacteria B. Green sulfur bacteria C. Green nonsulfur bacteria D. Purple sulfur bacteria E. Purple nonsulfur bacteria

Answer: B. Green sulfur bacteria Rationale: The scenario describes a nonmotile bacterium that uses bacteriochlorophyll for photosynthesis without generating oxygen, and relies on sulfur compounds instead of water as electron donors. These are distinctive characteristics of green sulfur bacteria. They are strict anaerobes that perform anoxygenic photosynthesis, and they are typically nonmotile, residing in the lower light zones of stratified bodies of water such as lakes. Option A, Cyanobacteria, are known for oxygenic photosynthesis, which is a process that generates oxygen, unlike the bacterium described. Option C, Green nonsulfur bacteria, can perform photosynthesis but they are not strictly anoxygenic and often not sulfur-dependent. Option D, Purple sulfur bacteria, while they do use sulfur in anoxygenic photosynthesis, are typically motile, which does not align with the nonmotile nature of the bacterium described. Option E, Purple nonsulfur bacteria, are versatile and can grow photoheterotrophically when organic compounds are present; they are not solely dependent on sulfur compounds and are generally motile.

9. Hepatitis C virus (HCV) is a small, enveloped, single-stranded RNA virus that primarily affects the liver. HCV does not integrate its genome directly into the host cell's DNA. A physician observes that a subset of patients with chronic Hepatitis C virus (HCV) infection develop liver cancer over time. What mechanism is most likely responsible for the oncogenic potential of HCV? A. HCV induces a strong cytotoxic T cell response that kills potentially cancerous cells. B. HCV leads to persistent infection and chronic liver damage, culminating in cirrhosis and increased risk of cancer. C. HCV directly inserts viral oncogenes into the host cell DNA, causing immediate transformation to a cancerous state. D. HCV infection causes the liver to stop regenerating, making it susceptible to cancer.

Answer: B. HCV leads to persistent infection and chronic liver damage, culminating in cirrhosis and increased risk of cancer. Rationale: Hepatitis C virus (HCV) infection can become chronic, leading to long-term liver inflammation. Chronic inflammation can cause continuous liver cell damage and regeneration. Over time, the regenerative processes can lead to errors in DNA replication, which may accumulate mutations. Additionally, the chronic inflammation itself can contribute to an environment that is conducive to cancer development. This chronic damage can progress to cirrhosis, which is a known risk factor for the development of hepatocellular carcinoma, a common type of liver cancer. Option A is incorrect because a strong cytotoxic T cell response is part of the body's defense mechanism against virally infected cells and would typically act to prevent the formation of cancer by destroying infected cells, not contribute to it. Option C is incorrect because HCV is an RNA virus that does not integrate its genome directly into the host cell's DNA like retroviruses do. Therefore, it does not insert viral oncogenes that could immediately transform the host cell. Option D is incorrect because HCV infection does not stop liver regeneration; rather, the virus can cause damage that leads to increased cell turnover and regeneration, which can potentially lead to cancer over time due to the reasons explained above.

3. An environmental sample from a salty lake reveals the presence of organisms that are both spherical and rod-shaped. Further analysis of these organisms reveals the presence of a pigment known as bacteriorhodopsin. These organisms are likely using what kind of energy source for survival? A. Organic carbon, as they are likely heterotrophs B. Light, as they are likely photoheterotrophs C. Methane, as they are likely methanotrophs D. Inorganic molecules, as they are likely chemolithotrophs

Answer: B. Light, as they are likely photoheterotrophs Rationale: The presence of bacteriorhodopsin, a pigment used by certain archaea known as halobacteria, is indicative of their ability to use light as an energy source. Bacteriorhodopsin allows these microorganisms to perform a type of photosynthesis that involves the absorption of light and the translocation of protons across the cellular membrane, which leads to the synthesis of ATP. This process is characteristic of photoheterotrophs—organisms that use light for energy but rely on organic compounds as a carbon source, not carbon dioxide like photoautotrophs.

9. In a controlled study, Bacteria Y was treated with Antimicrobial Agent Z. The results showed leakage of cellular contents and subsequent cell death. This indicates that Agent Z's mode of action is most likely targeting: A. The bacteria's ribosomes, disrupting protein synthesis. B. The synthesis of nucleic acids. C. The integrity of the cytoplasmic membrane. D. The cell wall synthesis process.

Answer: C. The integrity of the cytoplasmic membrane. Rationale: Leakage of Cellular Contents: The observation that Antimicrobial Agent Z causes leakage of cellular contents and subsequent cell death suggests that it is compromising the integrity of the bacterial cell's cytoplasmic membrane. The cytoplasmic membrane is crucial for maintaining the cellular environment and integrity. Damage to this membrane leads to the loss of essential cellular components and the inability to control the passage of substances into and out of the cell, eventually resulting in cell death.

13. Viruses are not considered to be alive because they do not possess all the characteristics of life that define living organisms. While they share some similarities with living things, they lack many of the key traits that define life. Which of the following characteristics do viruses share with all living organisms? A. Ability to reproduce independently B. Presence of genetic material C. Capability to perform metabolic activities D. Capability to maintain homeostasis E. Ability to infect cells

Answer: B. Presence of genetic material Rationale: Viruses contain either DNA or RNA as their genetic material, which they use to replicate and produce new virus particles. However, viruses cannot reproduce independently, perform metabolic activities, or maintain homeostasis. They also do not respond to stimuli, grow or develop, and do not have a cellular structure. Therefore, viruses are not considered living organisms. In addition, while viruses do have the ability to infect cells, not all living organisms are capable of infecting other cells or organisms.

8. The oncogene theory has been supported by numerous studies and is a key concept in the field of cancer biology. Understanding the mechanisms by which proto-oncogenes are activated and the ways in which oncogenes contribute to the development and progression of cancer is essential for the development of effective cancer therapies. Which of the following statements is true about proto-oncogenes? A. Proto-oncogenes are mutated forms of tumor suppressor genes. B. Proto-oncogenes are normal genes involved in cell growth and division. C. Proto-oncogenes are only found in cancer cells. D. Proto-oncogenes are inactive in normal cells and only become activated in response to viral infections.

Answer: B. Proto-oncogenes are normal genes involved in cell growth and division. Rationale: Proto-oncogenes are normal cellular genes that play an important role in regulating cell growth and division. They become oncogenes when they are mutated or abnormally expressed, leading to uncontrolled cell growth and division. Proto-oncogenes are present in normal cells and can become activated in response to a variety of signals, such as growth factors, hormones, and cellular stress. Mutations in proto-oncogenes can occur spontaneously or as a result of exposure to carcinogens such as radiation and chemicals. In addition, some viruses can insert their DNA into proto-oncogenes, disrupting their normal regulation and leading to their abnormal expression and activation.

A lab technician is preparing to culture a sample from a highly contaminated environmental source. The goal is to isolate and grow a specific bacterium known to be present in the sample. Which type of general culture media would be the best choice for this purpose? A. Transport media, to maintain the viability of all bacteria during transport to the lab. B. Selective media, to suppress the growth of competing microorganisms. C. Enriched media, to encourage the growth of all bacteria in the sample. D. Differential media, to visually distinguish the specific bacterium from others.

Answer: B. Selective media, to suppress the growth of competing microorganisms. Rationale: Selective media contain substances that favor the growth of particular bacteria while inhibiting the growth of others. This is particularly useful when culturing a sample from a contaminated source where the goal is to isolate a specific bacterium. Transport media are designed to preserve specimens during transport, not to suppress the growth of contaminants. Enriched media would not be appropriate in this scenario because they encourage the growth of all bacteria, not just the target bacterium. Differential media allow for the differentiation of bacteria based on biochemical reactions but do not necessarily suppress the growth of contaminants, which is the primary requirement for this task. Therefore, selective media are the best choice for isolating a specific bacterium from a highly contaminated source

The cell walls of Mycobacterium tuberculosis are different from those of other bacteria. This bacterium uses mycolic acid in its cell wall and stains acid-fast. Which of the antimicrobial drugs listed below is the best choice for treatment of M. tuberculosis infection? A. Penicillin B. Streptomycin C. Erythromycin D. Tetracycline E. Sulfonamides

Answer: B. Streptomycin Rationale: Mycobacterium tuberculosis has a unique cell wall composition, with the presence of mycolic acid making it difficult for certain antimicrobial drugs to penetrate the cell wall and reach the bacterium. Streptomycin, however, is effective against M. tuberculosis and is often used as part of the first-line therapy for tuberculosis. Penicillin, erythromycin, tetracycline, and sulfonamides are not effective against M. tuberculosis.

16. A researcher infects a bacterial culture with a small number of bacteriophage. After five minutes, the researcher attempts to reisolate the bacteriophage from the culture using centrifugation and filtration but is unable to locate any bacteriophage. What is the most likely reason for this? A. The bacteriophage were destroyed by the bacterial immune response. B. The bacteriophage have all entered the bacterial cells and are no longer in the extracellular environment. C. The centrifugation and filtration process was not sensitive enough to detect the bacteriophage. D. The bacteriophage failed to infect the bacteria and were degraded.

Answer: B. The bacteriophage have all entered the bacterial cells and are no longer in the extracellular environment. Rationale: Bacteriophages infect bacteria by attaching to their surface and injecting their genetic material. After this initial infection stage, the phages are no longer present in the extracellular environment because they are either within the bacterial cells or attached to their surface. The method of centrifugation followed by filtration is effective in separating free, unattached phages from bacterial cells. If no phages are found in the filtrate, it indicates that they are not present in the extracellular medium. This would be the case if the phages have successfully infected the bacteria and are inside the cells.

A team of researchers is investigating a bacterial strain that has been causing persistent infections in patients with implanted medical devices. The strain forms a biofilm on these devices, which is difficult to eradicate. Which aspect of biofilm biology could be targeted to disrupt this bacterial community and potentially treat the infections? A. The enhanced nutritional supply within the biofilm matrix. B. The communication pathways used in quorum sensing. C. The mucous membrane compatibility of the strain. D. The solitary behavior of microorganisms within the biofilm

Answer: B. The communication pathways used in quorum sensing. Rationale: Biofilms are complex communities of microorganisms that adhere to surfaces and are embedded in a self-produced matrix. One key aspect of biofilm formation and maintenance is quorum sensing, a system of stimulus and response correlated to population density. Quorum sensing involves the production, release, and detection of chemical signal molecules called autoinducers. This communication regulates gene expression in response to changes in cell-population density, including genes responsible for biofilm formation. Targeting the quorum sensing pathways can disrupt the coordination and regulation of genes involved in biofilm maintenance and virulence, making it a promising approach to disrupting biofilm formation and persistence.

17. Given time, heredity, and genetic variation, an organism will evolve when a selective factor is introduced. The widespread use of antibiotics is the selective factor responsible for the evolution of antibiotic resistance in bacteria. Which of the following statements is NOT true regarding the evolution of antibiotic resistance in bacteria? A. Bacteria can acquire plasmids conferring resistance through horizontal gene transfer. B. The exposure of bacteria to an antibiotic causes the bacteria to produce resistance genes. C. Resistance can emerge due to random mutations in chromosomal genes. D. Due to natural genetic variation, resistant bacteria will be present in a population, even in the absence of antibiotics.

Answer: B. The exposure of bacteria to an antibiotic causes the bacteria to produce resistance genes. Rationale: While exposure to antibiotics can select for resistant bacteria, it does not cause bacteria to produce resistance genes. Instead, bacteria can acquire resistance genes through horizontal gene transfer or random mutations in chromosomal genes. Additionally, due to natural genetic variation, some bacteria may already possess genes that confer resistance even in the absence of antibiotics.

If a thermophilic bacterium shows optimal growth at 60°C, what would likely happen to its growth rate if it were incubated at room temperature (approximately 25°C)? A. The growth rate would increase. B. The growth rate would decrease. C. The growth rate would remain unchanged. D. The bacterium would sporulate.

Answer: B. The growth rate would decrease. Rationale: Thermophilic bacteria are adapted to high temperatures and have optimal enzymatic and metabolic activity at those temperatures, such as 60°C. If the temperature is significantly reduced, such as to room temperature (approximately 25°C), the metabolic processes slow down due to the enzymes being less active at this sub-optimal temperature. This would result in a decreased growth rate. The bacterium would not necessarily sporulate, as sporulation is a response to nutrient depletion rather than temperature changes

Which of the following best describes photoautotrophs and their process of obtaining energy? A. They use organic compounds to generate energy without the production of oxygen. B. They use light as an energy source and carbon dioxide as a carbon source to produce oxygen. C. They obtain energy by oxidizing inorganic substances, such as hydrogen and sulfur. D. They require organic compounds for energy and do not produce oxygen as a by-product.

Answer: B. They use light as an energy source and carbon dioxide as a carbon source toproduce oxygen.Rationale: Photoautotrophs, which include plants, algae, and cyanobacteria, are capable ofphotosynthesis, a process in which they use light energy to convert carbon dioxide and water intoorganic compounds, such as glucose. Oxygen is released as a by-product of this process. This iswhat distinguishes them as "auto" (self) trophs that use "photo" (light) for their energy source

15. Viroids and prions are infectious agents that are simpler than viruses. What is the fundamental difference in their composition? A. Viroids are made of proteins only, while prions are made of RNA only. B. Viroids consist of a short strand of RNA without a protein coat, while prions are misfolded proteins that can induce misfolding in normal proteins. C. Viroids are composed of DNA and proteins, while prions are composed of RNA and lipids. D. Viroids and prions are both made up entirely of lipids but differ in their lipid composition.

Answer: B. Viroids consist of a short strand of RNA without a protein coat, while prions are misfolded proteins that can induce misfolding in normal proteins. Rationale: Viroids are infectious agents that consist of a small, circular piece of RNA that is capable of self-replication but does not encode for proteins and lacks a protein coat, which differentiates them from viruses. They typically infect plants and can cause various diseases. Prions, on the other hand, are infectious proteins that are characterized by their misfolded conformation. Unlike viroids and viruses, prions do not contain nucleic acids. The abnormal folding of prion proteins can induce other normally folded proteins of the same type to also adopt the misfolded prion form. This process leads to a cascade of misfolding that can disrupt cell function and is associated with a group of neurodegenerative diseases known as transmissible spongiform encephalopathies.

14. The debate over whether viruses are "alive" often centers around their lack of certain features. Which of the following is a key reason why viruses are generally not considered living organisms? A. Viruses are not capable of evolving or mutating. B. Viruses do not possess the necessary components for independent metabolic processes. C. Viruses are only composed of proteins and lack genetic material. D. Viruses cannot be seen with a microscope.

Answer: B. Viruses do not possess the necessary components for independent metabolic processes. Rationale: One of the fundamental reasons viruses are not considered living organisms is because they lack the cellular machinery required for independent metabolic processes. Metabolism includes all the biochemical reactions that occur within an organism, such as energy production, synthesis of molecules, and waste elimination. Viruses are unable to perform these functions outside of a host cell; they can only replicate by hijacking the metabolic machinery of a host.

10. The maximum daily dose of Tylenol (acetaminophen) recommended by the FDA is 3,000 mg. The estimated lethal dose of the drug is 10 grams in one day. What is the therapeutic index for Tylenol? A. ~ 300 B. ~ 3 C. ~ 0.3 D. ~ 0.003

Answer: B. ~3 Rationale: TI = TD50 / ED50 where TD50 is the dose of a drug that causes a toxic effect in 50% of the population, and ED50 is the dose of the drug that produces a therapeutic effect in 50% of the population.

13. Ergosterol, which is similar to cholesterol, regulates cell membrane permeability and fluidity. The antibiotic Amphotericin B attaches to ergosterol to form channels that damage cytoplasmic membrane integrity. Which of the following cell types would be most susceptible to this drug and which would be least susceptible? A. Bacterial, human B. Fungal, human C. Fungal, bacterial D. Bacterial, fungal

Answer: C. Fungal, bacterial Rationale: Amphotericin B is an antifungal antibiotic that targets the fungal cell membrane. Specifically, it binds to ergosterol, a fungal-specific membrane component, which leads to the formation of channels in the membrane and leakage of intracellular components, ultimately resulting in cell death. Based on this mechanism of action, fungal cells are most susceptible to Amphotericin B, as they possess ergosterol in their cell membrane. Bacterial cells, on the other hand, do not have ergosterol in their cell membrane and are therefore less susceptible to the drug. Human cells also lack ergosterol in their cell membrane, and therefore are not targeted by Amphotericin B. However, there is some potential for toxicity to human cells at higher doses of the drug due to similarity between ergosterol and human cholesterol, which is a consideration when using it therapeutically.

16. A laboratory technician is tasked with sterilizing surgical equipment and reducing microbial load on the laboratory surfaces. She uses autoclaving for the equipment and a bleach solution for the surfaces. This scenario exemplifies the use of: A. Physical methods for both equipment and surfaces. B. Chemical methods for both equipment and surfaces. C. A physical method (autoclaving) for equipment and a chemical method (bleach) for surfaces. D. A chemical method (bleach) for equipment and a physical method (autoclaving) for surfaces.

Answer: C. A physical method (autoclaving) for equipment and a chemical method (bleach) for surfaces. Rationale: Autoclaving: This is a physical method of sterilization that uses pressurized steam to achieve high temperatures, effectively killing all forms of life, including bacterial spores, viruses, and fungi. Autoclaving is a standard method for sterilizing surgical equipment and other items that can withstand high temperatures and pressure, ensuring they are free from all microbial life. Bleach Solution: Bleach (sodium hypochlorite) is a chemical disinfectant. It is widely used for disinfecting surfaces in laboratories and medical facilities due to its broad-spectrum antimicrobial activity. Bleach is effective against bacteria, viruses, and fungi. However, it does not necessarily achieve sterilization but significantly reduces the microbial load on surfaces. The scenario demonstrates the use of a physical method (autoclaving) for sterilizing equipment, which requires a higher level of microbial control, and a chemical method (bleach solution) for cleaning and disinfecting surfaces, where complete sterilization may not be necessary but a significant reduction in microbial presence is required.

10. Laboratory experiments reveal that Treatment A inhibits DNA replication, while Treatment B blocks RNA transcription in bacterial cells. These treatments affect the bacteria by: A. Interfering with cell wall synthesis and cytoplasmic membrane function, respectively. B. Disrupting protein synthesis in both cases. C. Affecting nucleic acid synthesis, with Treatment A targeting DNA and Treatment B targeting RNA. D. Altering the permeability of the cytoplasmic membrane in both cases.

Answer: C. Affecting nucleic acid synthesis, with Treatment A targeting DNA and Treatment B targeting RNA. Rationale: Treatment A - Inhibiting DNA Replication: This treatment directly affects the process of DNA replication. DNA replication is crucial for bacterial cell division and proliferation. By inhibiting this process, Treatment A prevents the bacteria from replicating their genetic material, which is essential for cell division and growth. Treatment B - Blocking RNA Transcription: This treatment interferes with the process of RNA transcription, which is the first step in protein synthesis where the DNA sequence is transcribed into messenger RNA (mRNA). RNA is vital for carrying the genetic information necessary to build proteins. Blocking RNA transcription hampers the bacteria's ability to produce proteins, which are essential for various cellular functions.

6. A microbiologist uses two agents to treat bacterial cultures: Agent A stops bacterial growth, and Agent B kills the bacteria. When she removes the agents, growth resumes in cultures treated with Agent A but not in those treated with Agent B. This illustrates: A. Agent A is bactericidal, and Agent B is bacteriostatic. B. Both agents are bactericidal, but Agent B is more effective. C. Agent A is bacteriostatic, and Agent B is bactericidal. D. Both agents are bacteriostatic, but Agent A is reversible.

Answer: C. Agent A is bacteriostatic, and Agent B is bactericidal. Rationale: Agent A: This agent stops bacterial growth but does not kill the bacteria. When Agent A is removed, the bacteria resume growth. This characteristic is indicative of a bacteriostatic agent. Bacteriostatic agents inhibit the growth or multiplication of bacteria but do not necessarily kill them. Their effect is reversible - once the agent is removed, the bacteria can potentially begin to grow again. Agent B: This agent kills the bacteria, and when it is removed, the bacterial cultures do not resume growth. This is a key feature of bactericidal agents. Bactericidal agents cause the death of bacterial cells and are not reversible - the bacteria do not recover even after the agent is removed.

An experiment is designed to assess the growth of a bacterium at different hydrogen ion concentrations. The bacterium exhibits optimal growth at a pH of 9. This bacterium can be classified as: A. Acidophile B. Neutrophile C. Alkaliphile D. Halophile

Answer: C. Alkaliphile. Rationale: Alkaliphiles are organisms that thrive at relatively high pH levels, usually above 9. Since the bacterium in question shows optimal growth at a pH of 9, it can be classified as an alkaliphile. Acidophiles prefer acidic environments (pH below 7), neutrophiles grow best in neutral pH conditions (around pH 7), and halophiles are organisms that thrive in high salt concentrations, which is unrelated to pH levels

3. A researcher is examining three different virus particles under an electron microscope. She observes that all three viruses have a protein coat surrounding their genetic material. Virus A has a single-stranded RNA genome, Virus B has a double-stranded DNA genome, and Virus C has a single-stranded DNA genome. Which common characteristic of viruses can be identified from this scenario and data? A. All viruses have a lipid bilayer envelope. B. All viruses contain double-stranded RNA genomes. C. All viruses possess a protein coat. D. All viruses replicate inside host cell nuclei.

Answer: C. All viruses possess a protein coat. Rationale: The common characteristic of viruses identified in this scenario and data is that all viruses have a protein coat, also known as a capsid, that surrounds their genetic material. This protein coat plays a crucial role in protecting the viral genome and facilitating its entry into host cells during infection. The information provided in the scenario and data does not mention the presence of a lipid bilayer envelope, double-stranded RNA genomes, or the location of viral replication within host cell nuclei, making options A, B, and D incorrect.

12. What is meant when a bacterium is said to become "resistant" to an antibiotic? A. The antibiotic kills or inhibits the bacterium B. The antibiotic is metabolized by the bacterium, providing more energy for growth C. The bacterium is neither killed nor inhibited by the antibiotic D. The antibiotic mutates in a way that benefits the bacterium

Answer: C. The bacterium is neither killed nor inhibited by the antibiotic Rationale: When a bacterium is said to become "resistant" to an antibiotic, it means that the bacterium is neither killed nor inhibited by the antibiotic. When bacteria are exposed to antibiotics, those that are susceptible to the drug are killed or have their growth inhibited, while those that are resistant are not affected and can continue to grow and reproduce. Bacteria can become resistant to antibiotics in several ways. For example, they can acquire resistance genes from other bacteria through processes like conjugation or transduction, which allow them to produce enzymes that can inactivate the antibiotic or modify its target. They can also develop mutations that prevent the antibiotic from binding to its target or that allow the bacteria to pump the drug out of the cell before it can have an effect.

1. Host cell recognition refers to the process by which a virus recognizes and attaches to a specific host cell in order to infect it. This process is critical for the virus to establish an infection and replicate itself. How do animal viruses and bacteriophages compare in terms of host cell recognition? A. Animal viruses recognize and bind to specific receptor molecules on the host cell surface, while bacteriophages do not require specific recognition and can infect a wide range of host cells. B. Bacteriophages recognize and bind to specific receptor molecules on the host cell surface, while animal viruses do not require specific recognition and can infect a wide range of host cells. C. Both animal viruses and bacteriophages recognize and bind to specific receptor molecules on the host cell surface. D. Both animal viruses and bacteriophages do not require specific recognition and can infect a wide range of ho

Answer: C. Both animal viruses and bacteriophages recognize and bind to specific receptor molecules on the host cell surface. Rationale: Both animal viruses and bacteriophages use receptor recognition to initiate the infection process, but they differ in the specific receptors that they recognize. Animal viruses typically recognize specific proteins or carbohydrates on the host cell surface that are involved in various cellular functions. These receptors are typically highly specific to particular cell types, which is why animal viruses often have a limited host range. Similarly, bacteriophages recognize and bind to specific receptors on the surface of bacterial cells, which can be proteins, carbohydrates, or lipids. However, the receptor recognition mechanisms of bacteriophages are more diverse than those of animal viruses, and some bacteriophages are highly specific for their host cells while others have a broader host range.

5. The primary difference between DNA virus and RNA virus replication is the way in which the viral genetic material is replicated. What is the main difference in nucleic acid replication between DNA viruses and RNA viruses? A. DNA viruses use reverse transcriptase for replication, while RNA viruses use RNA polymerase. B. DNA viruses use RNA polymerase for replication, while RNA viruses use DNA polymerase. C. DNA viruses replicate in the nucleus, while RNA viruses replicate in the cytoplasm. D. DNA viruses can only replicate in dividing cells, while RNA viruses can replicate in both dividing and non-dividing cells.

Answer: C. DNA viruses replicate in the nucleus, while RNA viruses replicate in the cytoplasm. Rationale: DNA viruses typically replicate their genetic material in the nucleus of the infected host cell using host cell machinery, including DNA polymerase and other enzymes. The replication of DNA viruses involves several stages, including the uncoating of the viral genome, the synthesis of complementary DNA (cDNA) strands, and the generation of progeny genomes. RNA viruses, on the other hand, typically replicate their genetic material in the cytoplasm of the host cell, where they use the host cell's ribosomes, tRNA molecules, and other enzymes to synthesize viral proteins and replicate their RNA genome. RNA viruses can replicate using either RNA-dependent RNA polymerase or reverse transcriptase, depending on the type of RNA virus.

17. To which of the methods of animal virus host cell entry is this poem most likely referring? The virus approaches the cell, A predator stalking its prey, It seeks to infiltrate and invade, And make the host cell its new domain. The virus binds to the cell surface, A lock and key connection so precise, It triggers a cascade of events, A pathway to the cell's inner sanctum. The cell engulfs the virus whole, Like a Venus flytrap closing tight, The virus now trapped inside, A captive in its new-found home. A. Direct penetration B. Membrane fusion C. Endocytosis D. Exocytosis E. Phagocytosis

Answer: C. Endocytosis Rationale: The poem describes the virus binding to the cell surface and then being engulfed by the cell, which is characteristic of endocytosis.

. In a microbiology lab, a student is tasked with selecting an appropriate culture media for growing a fastidious bacterium. Which type of general culture media should the student choose? A. Selective media, to inhibit the growth of unwanted bacteria. B. Differential media, to distinguish the desired bacterium based on biochemical characteristics. C. Enriched media, to provide the necessary nutrients and growth factors. D. Reducing media, to create an anaerobic environment.

Answer: C. Enriched media, to provide the necessary nutrients and growth factors. Rationale: Enriched media are designed to support the growth of fastidious organisms by providing the essential nutrients and growth factors that these bacteria cannot synthesize on their own. Selective media are used to suppress the growth of competing microorganisms, while differential media are used to distinguish one bacterium from another based on specific biochemical reactions. Reducing media are utilized to cultivate anaerobic bacteria by removing molecular oxygen from the medium. Therefore, for a fastidious bacterium requiring specific nutrients, enriched media would be the appropriate choice

9. In a hospital's intensive care unit, there is an increased incidence of wound infections post-surgery. The infectious agent is found to be resistant to many common antibiotics and is identified as a low G+C Gram-positive bacterium. Which genus is the most likely culprit? A. Bacillus, which can cause severe skin infections. B. Clostridia, due to the severe symptoms. C. Enterococcus, due to its presence in clinical settings. D. Actinobacteria, which are common in decomposition of organic materials.

Answer: C. Enterococcus, due to its presence in clinical settings. Rationale: Enterococci, specifically Enterococcus faecalis and Enterococcus faecium, are significant causes of hospital-acquired infections and are known for their resistance to many common antibiotics. These bacteria can cause a range of illnesses, including wound infections, particularly in the intensive care unit (ICU) setting where patients are at higher risk due to invasive procedures and compromised immune systems. Option A, Bacillus, includes some species that can infect humans, but they are not typically associated with drug-resistant infections in hospital settings. Option B, Clostridia, are more commonly associated with diseases such as tetanus and botulism and are not typically linked to hospital-acquired wound infections. Option D, Actinobacteria, is a broad phylum that includes many different genera, some of which are involved in the decomposition of organic materials. While some members of Actinobacteria (such as those in the genus Mycobacterium) can cause healthcare-associated infections. However, these are high G+C bacteria rather than low.

10. A dentist diagnoses a patient with dental caries and identifies a Gram-positive bacterium as the cause. The most likely genus responsible for this condition, given its association with low G+C content and oral health, is: A. Clostridia, which includes various pathogenic species. B. Bacillus, which can be found in various environments. C. Lactobacillus, known for its role in the normal oral microbiome and potential in caries development. D. Mycoplasmas, due to their lack of a cell wall and unique morphology.

Answer: C. Lactobacillus, known for its role in the normal oral microbiome and potential in caries development. Rationale: Among the options provided, Lactobacillus is the genus most commonly associated with dental caries. Lactobacilli are part of the normal flora of the mouth and have been implicated in the progression of caries due to their ability to ferment carbohydrates and produce lactic acid, which can lead to the demineralization of tooth enamel. Option A, Clostridia, includes species known for causing more severe and often life-threatening diseases, such as tetanus and botulism. They are not typically associated with dental caries. Option B, Bacillus, is a genus that encompasses a wide range of environmental bacteria, some of which can be pathogenic. However, they are not commonly linked to dental caries. Option D, Mycoplasmas, are unique bacteria that lack a cell wall and are primarily associated with respiratory and urogenital infections, not dental caries.

2. A patient presents with an infection characterized by a helical-shaped bacterium observed under a microscope. Which of the following bacteria from the slide's information might be the cause? A. Lactococcus lactis B. Vibrio cholerae C. Leptospira species D. Lactobacillus plantarum

Answer: C. Leptospira species Rationale: The helical or spiral shape described in the scenario is characteristic of spirochetes. Among the options given, Leptospira species are the only spirochetes, which are known for their distinctive spiral shape. Lactococcus lactis and Lactobacillus plantarum are both cocci and rod-shaped bacteria, respectively, and do not have a helical shape. Vibrio cholerae is a comma-shaped bacterium and, while somewhat curved, it does not form the tight spirals that are typical of spirochetes. Thus, the correct answer, based on the shape provided in the scenario, is Leptospira species.

5. In a wastewater treatment plant, an archaeon is identified as the primary agent in converting sewage into biogas, predominantly methane. Besides lacking peptidoglycan in its cell structure, this archaeon most likely belongs to which group? A. Halophiles, as they are commonly found in water treatment plants. B. Thermophiles, which are often involved in biogas production. C. Methanogens, due to their role in methane production. D. Acidophiles, given the acidic conditions of sewage treatment.

Answer: C. Methanogens, due to their role in methane production. Rationale: Methanogens are a distinct group of archaea known for their ability to produce methane as a metabolic byproduct, particularly in anaerobic conditions such as those found in wastewater treatment plants. They play a crucial role in the decomposition of organic matter in sewage by converting it into biogas, which is rich in methane. This process is part of the larger biogeochemical cycle of carbon. The fact that the archaeon is identified as the primary agent for methane production in the scenario strongly suggests that it belongs to the group of methanogens.

A microbiologist is cultivating a marine bacterium that requires high salt concentrations. When placed in a low-salt environment, the bacterium fails to grow. What aspect of the environment is most likely influencing microbial growth in this scenario? A. Temperature B. pH C. Osmotic pressure D. Hydrostatic pressure

Answer: C. Osmotic pressure. Rationale: Marine bacteria often require high salt concentrations for optimal growth due to their adaptation to the osmotic pressure of the ocean. When such bacteria are placed in a low-salt environment, the osmotic balance is disrupted, leading to either too much water entering the cells (causing them to burst) or not enough water, impeding cellular processes. This aspect of the environment—osmotic pressure—is the force exerted by the movement of water across a semipermeable membrane due to a difference in solute concentration. Temperature and pH could also influence growth, but the specific mention of salt concentration points directly to osmotic pressure. Hydrostatic pressure refers to the pressure exerted by a fluid at equilibrium due to the force of gravity, which is not related to salt concentration.

18. Which of the following is NOT an effective way to slow the development of antimicrobial drug resistance in bacteria? A. Ensure that patients finish their entire antimicrobial drug prescription in order to fully inhibit the pathogen. B. Take advantage of drug synergism by using a combination of more than one antimicrobial at the same time. C. Prescribe antimicrobial drugs for as many new infections as possible to ensure that pathogens are killed before they can naturally develop resistance. D. Develop new semisynthetic antimicrobial medicines to act as second and third generation drugs.

Answer: C. Prescribe antimicrobial drugs for as many new infections as possible to ensure that pathogens are killed before they can naturally develop resistance. Rationale: Overuse and misuse of antimicrobial drugs can accelerate the development of drug resistance. The correct approach is to use antimicrobial drugs only when necessary and appropriate, and to use them in the most effective way possible, such as by ensuring patients complete the entire course of treatment and using combination therapy to reduce the likelihood of resistance. The development of new drugs is also important, but it should be combined with appropriate use and stewardship of existing drugs to slow the development of resistance

16. Which method of drug resistance most commonly renders bacteria resistant to penicillin and similar drugs? A. Change in the permeability of the drug B. Removal of the drug via a pump C. Production of an enzyme that destroys the drug D. Alteration of the drug target E. Overproduction of an enzyme in a key metabolic pathway

Answer: C. Production of an enzyme that destroys the drug Rationale: Many bacteria can produce an enzyme called β-lactamase, which can hydrolyze the β-lactam ring of penicillin and related antibiotics, rendering them inactive. This is a common mechanism of resistance to penicillins and other β-lactam antibiotics. Other mechanisms of drug resistance, such as changes in the permeability of the drug or alteration of the drug target, can also contribute to resistance to penicillin and related antibiotics. However, production of β-lactamase is considered the primary mechanism for penicillin resistance in many bacteria, as it can rapidly inactivate the antibiotic and prevent it from binding to its target.

6. Which of the following best describes the pattern of virion abundance in a persistent infection? A. High levels of virions are produced constantly throughout the infection. B. Virions are produced in cycles of high and low abundance. C. Virions are produced at low levels that gradually increase over time. D. Virions are not produced at all until reactivation occurs.

Answer: C. Virions are produced at low levels that gradually increase over time. Rationale: A persistent infection is a type of viral infection that can last for a long period of time, during which the virus continues to replicate and produce virions without causing immediate death of the host cell. During a persistent infection, the virus can establish a state of equilibrium with the host cell, in which viral replication is maintained at a low level, often without causing overt cytopathic effects. Over time, the level of virus production may gradually increase, leading to a slow and steady accumulation of virions within the infected host. However, the level of virus production is typically much lower than in an acute infection, where large amounts of virions are produced rapidly and cause a rapid onset of symptoms.

7. Which category of antimicrobial drug works by changing the shape of a ribosome? A. oxazolidinones B. chloramphenicol C. aminoglycosides D. tetracyclines

Answer: C. aminoglycosides Rationale: Aminoglycosides such as gentamicin and streptomycin work by binding to the 30S ribosomal subunit and causing misreading of the mRNA, which ultimately leads to the production of non-functional proteins. This changes the shape of the ribosome and affects its ability to correctly synthesize proteins.

4. In order to determine which antibiotic might treat an infection most effectively, a plate is inoculated with a pathogen. A variety of paper disks, each containing a known amount of antibiotic, are placed on the surface of the plate. After incubation, the plates are examined for zones of inhibition. Which type of test is this? A. Etest B. minimum inhibitory concentration test C. diffusion susceptibility test D. minimum bactericidal concentration

Answer: C. diffusion susceptibility test Rationale: The test described is a diffusion susceptibility test, also known as a Kirby-Bauer test. In this test, paper disks containing known amounts of antibiotics are placed on a plate inoculated with a pathogen. The antibiotic diffuses from the disk into the agar, creating a concentration gradient. If the antibiotic is effective against the pathogen, a clear zone of inhibition will be visible around the disk where the growth of the bacteria has been inhibited. The size of the zone of inhibition is measured and compared to standard values to determine the susceptibility of the pathogen to the antibiotic.

8. Which category of antimicrobial drug essentially acts to stall a ribosome as it reads mRNA? A. chloramphenicol B. tetracyclines C. lincosamides D. antisense nucleic acids

Answer: C. lincosamides Rationale: Lincosamides such as clindamycin bind to the 50S ribosomal subunit and block the translocation step of protein synthesis. This prevents the proper movement of mRNA through the ribosome and leads to inhibition of protein synthesis. Tetracyclines bind to the 30S ribosomal subunit and block the docking site for incoming aminoacyl-tRNA in the A-site, leading to inhibition of protein synthesis. Chloramphenicol also inhibits protein synthesis, but it does so by binding to the 50S ribosomal subunit and preventing peptide bond formation. Antisense nucleic acids are a type of nucleic acid therapy that target specific genes by binding to complementary sequences of mRNA, leading to degradation or inhibition of translation of the targeted mRNA.

17. Antimicrobials known as "attachment antagonists" are particularly useful for preventing A. biofilm formation. B. nucleic acid synthesis. C. virus infection. D. bacterial protein synthesis. E. cell membrane synthesis.

Answer: C. virus infection. Rationale: Attachment antagonists are a type of antiviral drug that block the ability of a virus to attach to and enter host cells, thereby preventing infection. Examples of attachment antagonists include maraviroc, which blocks the attachment of HIV to human CD4 cells, and palivizumab, which blocks the attachment of respiratory syncytial virus (RSV) to human respiratory cells.

9. A therapeutic index is a mathematical ratio that compares the amount of a drug that provides a therapeutic effect to the amount that causes toxicity. A therapeutic window focuses on the dosage range where a drug provides a therapeutic effect without toxicity. What is the estimated therapeutic window for the drug as depicted in the graph below? A. ~ 0 - 10 mg/kg B. ~ 0 - 40 mg/kg C. ~ 0.8 - 80 mg/kg D. ~ 10 - 200 mg/kg E. ~ 30 - 200 mg/kg

Answer: C. ~ 0.8 - 80 mg/kg (I believe that 5 - 10 mg/kg is a better answer!) Rationale: The therapeutic window is the range of drug doses that provides a therapeutic effect without causing unacceptable toxicity. Some sources incorrectly define the therapeutic window as the range between effective dose (ED50) and toxic dose (TD50). A better pharmacologic definition is the range between ED95 to TD5 but this can be very narrow and difficult to dose. Since there is no overlap between the therapeutic curve and the toxic effect curve, there is no reason to not give a higher dose in which 90-100% of patients have the therapeutic effect with no toxicity.

6. The Etest is a way of determining antimicrobial sensitivity by placing a strip impregnated with antimicrobials onto an agar plate. A strain of bacterium or fungus will not grow near a concentration of antibiotic or antifungal if it is sensitive and the results can be used to determine a minimum inhibitory concentration (MIC). What is the MIC for ceftolozane/tazobactam (C/T) against the bacteria of interest based on the results below? A. ~256 g/ml B. ~24 g/ml C. ~1.0 g/ml D. ~0.25 g/ml E. ~0.016 g/ml

Answer: C. ~1.0 g/ml Rationale: The Etest is a method used to determine the minimum inhibitory concentration (MIC) of an antimicrobial agent against a microorganism. In this test, a strip impregnated with a gradient of the antimicrobial agent is placed on an agar plate inoculated with the microorganism of interest. The point at which the growth of the microorganism intersects with the strip is the MIC. Based on the results provided, the MIC for ceftolozane/tazobactam (C/T) against the bacteria of interest is approximately 1.0 µg/mL, as the growth of the microorganism intersects with the strip at that point.

1. Which of the following is NOT associated with the work of Paul Ehrlich? A. arsenic compounds B. the concept of chemotherapy C. the concept of the "magic bullet" D. sulfanilamide

Answer: D. sulfanilamide Rationale: Sulfanilamide is NOT associated with the work of Paul Ehrlich. Paul Ehrlich is known for his contributions to immunology and his pioneering work in the fields of hematology and chemotherapy. He introduced the concept of the "magic bullet," which referred to a drug that could target specific cells or pathogens without harming healthy cells. He also discovered the first effective treatment for syphilis, which involved using arsenic compounds. However, the development of sulfanilamide, which is an antibiotic, occurred after Ehrlich's time by the German biochemist Gerhard Domagk in the early 1930s.

4. The cytosol of bacteria contain ____________ which has/have components different from those of the functionally equivalent structure of eukaryotes. A. a phospholipid bilayer B. a cytoskeleton C. a nucleus D. 70S ribosomes E. 80S ribosomes

Answer: D. 70S ribosomes Rationale: The cytosol of bacteria contains 70S ribosomes, which have components different from those of the functionally equivalent structure of eukaryotes. Ribosomes are the cellular structures responsible for protein synthesis, and they are composed of two subunits: the large and the small subunit. In bacteria, the two subunits combine to form a 70S ribosome, which is functionally equivalent to the 80S ribosome found in eukaryotes. However, the components of the 70S ribosome are different from those of the 80S ribosome. This difference makes the bacterial ribosome a good target for many antibiotics that selectively inhibit bacterial protein synthesis without affecting eukaryotic cells.

15. Some bacteria are resistant to erythromycin because of a mutation in their ribosomal RNA. What type of resistance does this represent? A. Change in the permeability of the drug B. Removal of the drug via a pump C. Production of an enzyme that destroys the drug D. Alteration of the drug target E. Overproduction of an enzyme in a key metabolic pathway

Answer: D. Alteration of the drug target Rationale: Erythromycin is a macrolide antibiotic that binds to the 50S subunit of bacterial ribosomes, inhibiting protein synthesis. The mutation in ribosomal RNA alters the structure of the ribosome, preventing erythromycin from binding to its target site. This reduces the effectiveness of the drug and results in resistance to erythromycin. Resistance due to alteration of the drug target is a common mechanism of antibiotic resistance. Other examples include mutations in penicillin-binding proteins that reduce the affinity of these proteins for β-lactam antibiotics, and mutations in DNA gyrase or topoisomerase IV that reduce the binding of fluoroquinolone antibiotics.

2. Which of the following is NOT correct when comparing the replication cycles of animal viruses and bacteriophages? A. Both may involve the insertion of viral genetic material into the host cell's genome. B. Both involve the assembly of viral particles and their subsequent release from the host cell. C. Both involve the production of viral proteins using the host cell's machinery. D. Both involve the formation of a prophage in the host cell's genome.

Answer: D. Both involve the formation of a prophage in the host cell's genome. Rationale: The formation of a prophage is a step that occurs only in the replication cycle of a temperate bacteriophage, which integrates its genetic material into the bacterial chromosome to form a prophage. During the lysogenic cycle, the prophage replicates along with the host chromosome and can remain latent until environmental cues trigger its induction into the lytic cycle. Animal viruses do not have a lysogenic cycle, and therefore do not form prophages. In contrast, both animal viruses and bacteriophages may involve the insertion of viral genetic material into the host cell's genome, the production of viral proteins using the host cell's machinery, and the assembly of viral particles followed by their release from the host cell. However, the specific mechanisms involved in these steps can differ depending on the type of virus.

7. The development of cancer is a complex process that can involve multiple factors. However, understanding how viruses can contribute to cancer development can help researchers develop new treatments and preventative measures. How do viruses contribute to the development of cancer in host cells? A. By damaging the host cell's cytoskeleton B. By inducing host cell apoptosis C. By promoting the expression of tumor suppressor genes D. By promoting the expression of oncogenes E. By knocking out the expression of proto-oncogenes

Answer: D. By promoting the expression of oncogenes Rationale: Oncogenes are genes that, when mutated or expressed at high levels, can promote uncontrolled cell growth and division. Some viruses can directly or indirectly activate oncogenes in infected host cells, leading to the development of cancer. For example, the human papillomavirus (HPV) can produce proteins that inactivate tumor suppressor genes, leading to the overexpression of oncogenes and the development of cervical cancer.

12. In a teaching hospital, medical students are learning about different methods of culturing viruses. They are asked to identify the methods that can be used to culture viruses for diagnostic purposes. Which of the following methods are they likely to learn about? A. Use of selective bacterial media and incubation at 37°C. B. Inoculation of susceptible live animals and observation of disease symptoms. C. Employment of differential centrifugation to isolate viruses from patient samples. D. Cultivation in cell culture lines derived from human or animal tissues.

Answer: D. Cultivation in cell culture lines derived from human or animal tissues. Rationale: One of the standard methods for culturing viruses in a laboratory setting, particularly for diagnostic purposes, is to use cell lines derived from human or animal tissues. Viruses require living cells to replicate because they need to hijack the host cell's machinery to produce new viral particles. These cell cultures provide the necessary environment for the virus to infect and replicate. Medical students would learn how to observe the effects of viral infection in these cells, such as the appearance of cytopathic effects or the formation of plaques. Option A is incorrect because viruses cannot be cultured on bacterial media as they require living cells for replication. Option B, using live animals, is a method that has been historically important but is less common now due to ethical considerations, practicality, and the advent of cell culture techniques. It's also less specific for diagnostic purposes where quick and controlled results are needed. Option C, differential centrifugation, is a technique used to purify viruses from patient samples but is not a culture method; it is used for isolating viruses from other components in the sample prior to culture or for analytical purposes.

12. Lysozyme is important for which of the following stages of lytic replication in T4 bacteriophage? A. Attachment and Synthesis B. Attachment and Assembly C. Synthesis and Assembly D. Entry and Release

Answer: D. Entry and Release Rationale: During the entry stage, the T4 phage uses its tail fibers to recognize and attach to specific receptors on the surface of the host bacterium. The phage then injects its genetic material, consisting of double-stranded DNA, into the bacterial cytoplasm using its tail tube. Before the phage DNA can start replicating and producing new phage particles, the phage-encoded lysozyme enzyme breaks down the bacterial cell wall, allowing the phage genetic material to enter the bacterial cytoplasm more easily. This process is necessary for the successful entry of the phage genetic material into the host bacterium, and thus the initiation of the lytic replication cycle. During the release stage, T4 phages accumulate within the host cell, and a high concentration of phage particles can cause the host cell to burst or lyse. The lysozyme encoded by T4 phage can help to weaken the bacterial cell wall by breaking the bonds between the peptidoglycan strands, facilitating the release of the newly formed phage particles.

13. Which of the following statements is MOST correct regarding the development of antibiotic resistance in populations of bacteria? A. Bacteria exposed to antibiotics alter their DNA to become resistant. B. Exposure to antibiotics causes mutations which produce resistant bacteria. C. The bodies of people who take antibiotics become resistant to them. D. Exposure to antibiotics selects for the members of a bacterial population which already have a resistant phenotype.

Answer: D. Exposure to antibiotics selects for the members of a bacterial population which already have a resistant phenotype. Rationale: when bacteria are exposed to antibiotics, those that have a preexisting resistance mechanism, such as having acquired resistance genes or mutations, have a survival advantage over those that do not. The resistant bacteria can continue to grow and reproduce while susceptible bacteria are killed or inhibited by the antibiotic. Over time, the resistant bacteria become more prevalent in the population, leading to the emergence of antibiotic-resistant strains.

16. AZT (azidothymidine) is a thymidine analogue that works by selectively inhibiting human immunodeficiency virus (HIV)'s reverse transcriptase. Why are humans able to tolerate exposure to AZT while the virus cannot? A. AZT blocks viral attachment to human cell membranes which does not harm human cells. B. Humans lack thymidine and are unaffected by thymidine analogues. C. Humans lack cell walls and are thus unaffected by the action of AZT. D. Humans do not naturally possess the reverse transcriptase enzyme.

Answer: D. Humans do not naturally possess the reverse transcriptase enzyme. Rationale: Reverse transcriptase is an enzyme used by retroviruses, such as HIV, to convert their RNA genome into DNA, which can then be integrated into the host cell's genome. AZT is a thymidine analogue that is incorporated into the viral DNA by reverse transcriptase, causing chain termination and inhibiting viral replication. Since humans do not possess the reverse transcriptase enzyme, AZT is not incorporated into human DNA and does not have the same inhibitory effect on human cells. This selectivity allows AZT to specifically target and inhibit the replication of the HIV virus without causing significant harm to human cells.

10. Which of the following lists the stages of lytic replication in order from earliest to latest stages? I. Synthesis II. Assembly III. Attachment IV. Release V. Entry A. III, II, V, I, IV B. V, III, II, IV, I C. I, III, V, II, IV D. III, V, I, II, IV E. I, II, III, V, IV

Answer: D. III, V, I, II, IV Rationale: N/A

15. Which of the following correctly describes Chlamydia's unique biphasic life cycle? A. Obligate intracellular reproductive cycle throughout its life cycle B. Elementary bodies (EB) exist inside host cells C. Reticulate bodies (RB) are responsible for transmission D. It alternates between extracellular elementary bodies (EB) and intracellular reticulate bodies (RB)

Answer: D. It alternates between extracellular elementary bodies (EB) and intracellular reticulate bodies (RB) Rationale: Chlamydia's unique biphasic life cycle involves an alternation between two distinct forms: elementary bodies (EB) and reticulate bodies (RB). The elementary bodies are the extracellular, infectious form responsible for transmission, while the reticulate bodies are the intracellular, replicative form responsible for the growth and replication of the bacterium inside host cells. This cycle is characteristic of Chlamydia and is essential for its survival and propagation within the host.

A new species of bacteria has been discovered that thrives in environments with low oxygen levels. Laboratory tests reveal that it can detoxify superoxide radicals and peroxide anion but not singlet oxygen or hydroxyl radicals. Which category would these bacteria most likely fall into? A. Obligate aerobe B. Obligate anaerobe C. Facultative anaerobe D. Microaerophile

Answer: D. Microaerophile Rationale: The bacteria in question thrive in environments with low oxygen levels. This already suggests they are not obligate aerobes, which require high levels of oxygen, nor are they obligate anaerobes, which are killed by oxygen. The ability to detoxify superoxide radicals and peroxide anion indicates that these bacteria have some mechanisms to manage the presence of oxygen, which would be toxic to obligate anaerobes. However, their inability to neutralize singlet oxygen or hydroxyl radicals suggests they cannot tolerate high oxygen concentrations. This combination of characteristics fits best with microaerophiles, which require oxygen levels that are lower than atmospheric concentrations but can handle some oxygen presence with limited detoxification mechanisms. Microaerophiles typically thrive in environments where oxygen is present but at reduced levels.

11. During the lytic replication cycle, what action does a phage take to ensure that its host bacterium does NOT continue synthesizing cellular molecules? A. A phage traps the host cell DNA in an endosome. B. The host DNA is released from the cell. C. The phage integrates itself into the host cell chromosome. D. Phage enzymes degrade the bacterial DNA.

Answer: D. Phage enzymes degrade the bacterial DNA. Rationale: During the lytic replication cycle, a phage takes action to ensure that its host bacterium does not continue synthesizing cellular molecules by degrading the bacterial DNA using phage enzymes. After the phage has hijacked the bacterial host cell machinery to replicate its own genetic material and produce new phage particles, the phage will need to lyse (burst) the host cell to release the new phage particles. To accomplish this, the phage will synthesize and release enzymes that degrade the bacterial DNA, which halts cellular metabolism and ensures that the bacterial host cell focuses solely on phage replication.

5. What is the difference between prokaryotic and eukaryotic ribosomes? A. Prokaryotic ribosomes consist of a 50S large and a 40S small subunit, while eukaryotic ribosomes consist of 60S and 30S subunits. B. Prokaryotic ribosomes consist of a 60S large and a 30S small subunit, while eukaryotic ribosomes consist of a 50S large and a 40S small subunit. C. Prokaryotic ribosomes are 80S and consist of 30S and 50S subunits, while eukaryotic ribosomes are 100S and consist of 60S and 40S subunits. D. Prokaryotic ribosomes consist of a 50S large and a 30S small subunit, while eukaryotic ribosomes consist of a 60S large and a 40S small subunit.

Answer: D. Prokaryotic ribosomes consist of a 50S large and a 30S small subunit, while eukaryotic ribosomes consist of a 60S large and a 40S small subunit. Rationale: The ribosomes are the organelles responsible for protein synthesis in all cells. They are composed of two subunits, each of which contains ribosomal RNA (rRNA) and proteins. In prokaryotes, the large subunit is made up of a 50S subunit and the small subunit is made up of a 30S subunit. Together, they form the 70S ribosome. In contrast, the eukaryotic ribosome is composed of a 60S large subunit and a 40S small subunit, which combine to form an 80S ribosome. The differences in ribosomal subunit composition are exploited by antibiotics, which can selectively target bacterial ribosomes without harming eukaryotic ribosomes.

11. What benefit does the presence of an R plasmid provide to a microbe that possesses this extra piece of DNA? A. Rapid growth rate B. Rapid motility C. Removal of competitive microbes D. Resistance to a particular antibiotic E. Resistance to heat shock

Answer: D. Resistance to a particular antibiotic Rationale: The presence of an R plasmid provides resistance to a particular antibiotic to a microbe that possesses this extra piece of DNA. R plasmids are small, circular pieces of DNA that can be transferred between bacteria and carry genes that provide resistance to antibiotics, heavy metals, or other toxic substances. When a bacterium acquires an R plasmid, it gains the ability to produce enzymes or other proteins that can break down or modify antibiotics, making them ineffective. Resistance to antibiotics provided by R plasmids can be a significant problem in clinical settings as it can lead to the emergence of multidrug-resistant bacteria that are difficult to treat.

14. How is the lytic cycle different from the lysogenic cycle with respect to the infected host cell? A. The viral DNA may integrate into the host genome during the lytic stage. B. The host cell is allowed to live during the lytic stage. C. The host cell can only divide during the lytic stage. D. The host cell dies during the lytic stage.

Answer: D. The host cell dies during the lytic stage. Rationale: During the lytic cycle, the infected host cell is ultimately destroyed as the newly synthesized viral particles are released by causing lysis (bursting) of the cell. In contrast, during the lysogenic cycle, the viral DNA can integrate into the host genome and remain dormant without causing immediate harm to the host cell. The lysogenic cycle can eventually progress into the lytic cycle, leading to the death of the host cell.

13. When comparing the resistance of various microbes to antimicrobial agents, bacterial endospores, mycobacteria, and protozoan cysts are often cited as highly resistant. What is a key factor in the resistance of these microbial groups? A. Their dependence on host organisms for survival B. The presence of unique metabolic pathways C. Their ability to rapidly alter genetic material D. The protective barriers they possess, such as thick cell walls or cysts

Answer: D. The protective barriers they possess, such as thick cell walls or cysts. Rationale: Bacterial Endospores: These are extremely resistant structures formed by certain bacteria, such as Bacillus and Clostridium species, as a means of survival under adverse conditions. The endospore's resistance is mainly due to its multi-layered structure, including a tough outer coating that protects its genetic material and essential cellular components. Mycobacteria: This group, which includes Mycobacterium tuberculosis and Mycobacterium leprae, is known for its thick, waxy cell wall composed of mycolic acid and other complex lipids. This cell wall provides a strong barrier against many antimicrobial agents and environmental stresses. Protozoan Cysts: Protozoa can form cysts as a way to survive outside a host or in harsh environmental conditions. These cysts have protective layers that make them resistant to various antimicrobial treatments and environmental challenges.

11. Which of the following is NOT a beta-lactam antibiotic? A. penicillin G B. cephalothin C. methicillin D. amphotericin B

Answer: D. amphotericin B Rationale: amphotericin B is not a beta-lactam antibiotic. It is an antifungal medication that works by binding to ergosterol, a component of fungal cell membranes, and disrupting membrane integrity. Penicillin G, cephalothin, and methicillin are all beta-lactam antibiotics, which have a beta-lactam ring in their chemical structure and work by inhibiting bacterial cell wall synthesis.

10. Which category of drug is complementary to mRNA of the pathogen? A. oxazolidinones B. macrolides C. tetracyclines D. antisense nucleic acids

Answer: D. antisense nucleic acids Rationale: Antisense nucleic acids are a type of nucleic acid therapy that use synthetic nucleic acid molecules complementary to specific sequences of mRNA of the pathogen to inhibit translation or induce degradation of the mRNA. This approach allows for selective targeting of specific genes or gene products in the pathogen, and can be used as an alternative to traditional antibiotics. Oxazolidinones, macrolides, and tetracyclines are classes of antibiotics that target various steps of bacterial protein synthesis, but do not act through sequence-specific targeting of mRNA.

12. Bacitracin blocks the transport of NAG and NAM across the cytoplasmic membrane to the cell wall. Like other antimicrobials that block cell wall synthesis, this would result in weak cell walls and __________. A. inhibition of nucleic acid synthesis B. the inability to synthesize flagella C. less protein synthesis D. cell lysis due to the effect of osmotic pressure

Answer: D. cell lysis due to the effect of osmotic pressure Rationale: Bacitracin blocks the transport of NAG and NAM (building blocks of the bacterial cell wall) across the cytoplasmic membrane, which results in weak cell walls. This weak cell wall cannot withstand the normal osmotic pressure, which causes the bacterial cell to swell and lyse. Inhibition of nucleic acid synthesis, inability to synthesize flagella, and less protein synthesis are not the direct effects of bacitracin on bacterial cells.

1. Which of the following is NOT a common characteristic of viruses? Viruses... A. do not possess cytoplasmic membranes, cytosol, or organelles. B. are responsible for most of the diseases that infect humans. C. do not grow nor respond to the environment D. exhibit complex metabolic pathways E. lack the ability to reproduce independently

Answer: D. exhibit complex metabolic pathways Rationale: Viruses do not have the machinery to carry out their own metabolism and are dependent on host cells for their replication and survival. Therefore, they do not exhibit complex metabolic pathways. Options A, C, and E are true characteristics of viruses, and option B is true as many but not all human diseases are caused by viruses.

9. Chloramphenicol blocks the action of the large (50S) subunit. This essentially _____. A. prevents the attachment of tRNA to a ribosome B. changes the shape of the large subunit C. prevents a large subunit from attaching to a small subunit D. prevents the formation of peptide bonds

Answer: D. prevents the formation of peptide bonds Rationale: Chloramphenicol blocks the action of the large (50S) subunit by preventing the formation of peptide bonds. Chloramphenicol binds to the peptidyl transferase center of the ribosome, which is responsible for catalyzing the formation of peptide bonds between amino acids during protein synthesis. By blocking this activity, chloramphenicol inhibits the ability of the ribosome to produce functional proteins, which can be an effective strategy for controlling bacterial growth.

3. After the discovery and commercialization of antibiotics, Alexander Fleming developed the broth dilution technique using the turbidity of the broth for assessment of growth. This is commonly believed to be the inception of minimum inhibitory concentration tests. How many "compounds of interest" in the figure below appears to be more effective than the quality control (QC) antibiotic against the bacteria of interest? A. 1 compound B. 2 compounds C. 3 compounds D. 4 compounds E. 5 compounds F. 6 compounds G. 7 compounds

Answer: E. 5 compounds Rationale: Each well in the plate contains a different concentration of a test compound or the QC antibiotic, and the growth of bacteria in each well is being assessed by turbidity. The QC antibiotic is used as a standard of comparison to ensure that the bacteria are susceptible to inhibition, and any test compound that is more effective than the QC antibiotic at inhibiting bacterial growth would appear as a well with lower turbidity or no visible growth at a lower concentration. With this in mind, there are 5 test compounds that appear to be more effective than the QC antibiotic against the bacteria of interest, this means that there are 5 rows in the plate with no visible growth at concentrations lower than the QC antibiotic MIC.

14. In a recent Nature Medicine article (Gygli et al, 2021), the authors found that 31% of multidrug-resistant Mycobacterium tuberculosis cases could directly or indirectly be linked to prison populations. How did the multi-drug-resistant strains of M. tuberculosis most likely become so prevalent? A. Inclusion of histidine in the prisoner's diets B. A frameshift mutation C. A missense mutation D. A nonsense mutation E. Horizontal gene transfer of resistance factors located on plasmids.

Answer: E. Horizontal gene transfer of resistance factors located on plasmids. Rationale: Horizontal gene transfer is the transfer of genetic material between bacteria that are not parent and offspring. This can occur through various mechanisms, including conjugation, transformation, and transduction. One of the most common ways for antibiotic resistance genes to spread is through plasmids, which are small, circular pieces of DNA that can replicate independently of the bacterial chromosome. In the case of multidrug-resistant M. tuberculosis, it is possible that the resistance genes were acquired through horizontal gene transfer, either from other M. tuberculosis strains or from other bacteria in the prison environment.

4. Which of the following is NOT true concerning the genetic material of viruses? Virus genetic material... A. may be composed of dsDNA, ssDNA, dsRNA, or ssRNA. B. may be linear and segmented or single and circular. C. can be a component of virus classification schemes. D. may be DNA or RNA but never both. E. is generally much larger than the genome of prokaryotic cells.

Answer: E. is generally much larger than the genome of prokaryotic cells. Rationale: Option E, "is generally much larger than the genome of prokaryotic cells," is the correct answer because virus genomes can vary widely in size and are generally smaller than prokaryotic genomes. For example, some viruses have very small genomes consisting of only a few thousand nucleotides, The other options (A, B, C, and D) are true statements concerning the genetic material of viruses.

13. Matching Proteobacteria Groups to Descriptions. Match the class of Proteobacteria with the correct description. (2 pts) Column A: 1. Alphaproteobacteria 2. Betaproteobacteria 3. Gammaproteobacteria 4. Deltaproteobacteria 5. Epsilonproteobacteria 6. Zetaproteobacteria Column B: A. Largest group of Proteobacteria, includes Pseudomonas and Xanthomonas B. Nitrogen-fixing bacteria, includes Rhizobium C. Oxidize iron as an energy source and play a role in Earth's iron cycle D. Ferric iron-reducing, sulfur-reducing bacteria E. Inhabit the digestive tract of animals and are associated with symptoms or pathogens F. Ammonia oxidizing, arsenic-resistant soil bacteria

Answers: 1 - B. Nitrogen-fixing bacteria, includes Rhizobium. 2 - F. Ammonia oxidizing, arsenic-resistant soil bacteria. 3 - A. Largest group of Proteobacteria, includes Pseudomonas and Xanthomonas. 4 - D. Ferric iron-reducing, sulfur-reducing bacteria. 5 - E. Inhabit the digestive tract of animals and are associated with symptoms or pathogens. 6 - C. Oxidize iron as an energy source and play a role in Earth's iron cycle.

14. Matching Proteobacteria Groups to Diseases/Characteristics. Match the class or group of Proteobacteria (Column A) with the disease or characteristic they are known for (Column B). (2 pts) Column A: 1. Rickettsia 2. Neisseria 3. Bordetella 4. Legionella 5. Campylobacter 6. Helicobacter 7. Desulfovibrio 8. Bdellovibrio Column B: A. Causes whooping cough (pertussis) B. Known for recycling sulfur C. Associated with Legionnaires' disease D. Causes blood poisoning E. Causes Rocky Mountain spotted fever F. Causes ulcers G. Causes gonorrhea and can lead to meningitis H. Attacks and preys on other bacteria

Answers: 1 - E. Causes Rocky Mountain spotted fever. 2 - G. Causes gonorrhea and can lead to meningitis. 3 - A. Causes whooping cough (pertussis). 4 - C. Associated with Legionnaires' disease. 5 - D. Causes blood poisoning. 6 - F. Causes ulcers. 7 - B. Known for recycling sulfur. 8 - H. Attacks and preys on other bacteria.