Microbiology Quiz: Aerobes Anaerobes And Oxygen
20 questions · exam conditions
0:00
Aerobes Anaerobes And OxygenQuestion 1 of 20

A knockout mutant of a facultative anaerobe is created that lacks the gene for catalase but retains a functional gene for superoxide dismutase (SOD). The growth medium is then supplemented with ferrous iron (Fe²⁺). When this mutant is exposed to an aerobic environment, which of the following events is the most significant contributor to its subsequent cell death?

Direct damage to DNA and lipids by accumulating superoxide radicals (O₂⁻).
Inhibition of anaerobic respiratory enzymes by direct binding of molecular oxygen.
Generation of highly reactive hydroxyl radicals (•OH) via the Fenton reaction.
Depletion of cellular NADH due to excessive activity of peroxidase enzymes.
← Back to quizzes

Microbiology Quiz

Microbiology Quiz: Aerobes Anaerobes And Oxygen

Practice Aerobes Anaerobes And Oxygen in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Aerobes Anaerobes And Oxygen, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A knockout mutant of a facultative anaerobe is created that lacks the gene for catalase but retains a functional gene for superoxide dismutase (SOD). The growth medium is then supplemented with ferrous iron (Fe²⁺). When this mutant is exposed to an aerobic environment, which of the following events is the most significant contributor to its subsequent cell death?

  1. Direct damage to DNA and lipids by accumulating superoxide radicals (O₂⁻).
  2. Inhibition of anaerobic respiratory enzymes by direct binding of molecular oxygen.
  3. Generation of highly reactive hydroxyl radicals (•OH) via the Fenton reaction. (correct answer)
  4. Depletion of cellular NADH due to excessive activity of peroxidase enzymes.
Explanation: The functional SOD will convert superoxide radicals (O₂⁻) to hydrogen peroxide (H₂O₂). In the absence of catalase, H₂O₂ will accumulate. The presence of ferrous iron (Fe²⁺) in the medium will then catalyze the Fenton reaction (Fe²⁺ + H₂O₂ → Fe³⁺ + •OH + OH⁻), producing highly reactive and destructive hydroxyl radicals (•OH). These radicals cause widespread, lethal damage to cellular macromolecules.

Question 2

A research group successfully transfers a plasmid containing a constitutively expressed gene for catalase from Escherichia coli into Clostridium botulinum, an obligate anaerobe. The transformed C. botulinum is then plated on a nutrient-rich agar and incubated in an ambient air incubator (21% O₂). What is the most likely outcome?

  1. The transformed bacteria will grow robustly, forming colonies comparable to E. coli.
  2. No growth will occur because the bacteria still lack superoxide dismutase (SOD). (correct answer)
  3. The transformed bacteria will grow, but only if the medium is supplemented with reducing agents like thioglycollate.
  4. Limited, microaerophilic-like growth will occur in a thin layer just beneath the agar surface.
Explanation: Obligate anaerobes like Clostridium typically lack both superoxide dismutase (SOD) and catalase. Providing catalase only solves the problem of hydrogen peroxide (H₂O₂). The initial reactive oxygen species formed, the superoxide radical (O₂⁻), would still be generated and would be highly toxic. Without SOD to convert O₂⁻ to H₂O₂, the cells would suffer lethal damage, preventing growth.

Question 3

Some obligate anaerobes can survive brief exposure to atmospheric oxygen, even though they lack SOD and catalase. Which of the following represents a plausible non-enzymatic mechanism for this transient tolerance?

  1. Accumulation of high intracellular concentrations of manganese ions (Mn²⁺) that can scavenge superoxide radicals. (correct answer)
  2. The ability to rapidly form endospores, with the vegetative cells dying immediately upon oxygen exposure.
  3. Possession of a cell membrane that is completely impermeable to the diffusion of molecular oxygen.
  4. Rapid synthesis of a thick polysaccharide capsule that binds to and neutralizes reactive oxygen species.
Explanation: Some organisms, like Lactobacillus plantarum, have been shown to use a non-enzymatic defense. They accumulate very high intracellular concentrations of manganese (II) ions. These Mn²⁺ ions, complexed with small molecules like phosphate or lactate, can act as chemical scavengers of superoxide radicals, functionally replacing the need for the SOD enzyme. This provides a level of protection against oxidative stress.

Question 4

A research team isolates a novel bacterium from a hydrothermal vent. In the lab, it fails to grow in an incubator with ambient air (21% O₂) or in a completely anaerobic chamber (0% O₂). However, vigorous growth is observed when cultured in a specialized chamber with a 5% O₂, 10% CO₂, and 85% N₂ atmosphere. Based on these findings, what is the most likely profile of oxygen-detoxifying enzymes this bacterium possesses?

  1. High levels of both superoxide dismutase (SOD) and catalase.
  2. Complete absence of both superoxide dismutase (SOD) and catalase.
  3. Presence of superoxide dismutase (SOD) but a complete absence of catalase or peroxidase.
  4. Presence of superoxide dismutase (SOD) and low levels of either catalase or peroxidase. (correct answer)
Explanation: The growth pattern described is characteristic of a microaerophile, which requires a low concentration of oxygen for metabolism but is damaged by atmospheric levels. This implies the organism must have superoxide dismutase (SOD) to handle the superoxide radicals formed from oxygen, but its capacity to detoxify the resulting hydrogen peroxide (H₂O₂) via catalase or peroxidase is limited, making atmospheric oxygen levels toxic.

Question 5

A strain of E. coli, a facultative anaerobe, is genetically engineered to constitutively overproduce both superoxide dismutase and catalase at levels ten times higher than the wild type. What is the most likely physiological consequence when this engineered strain is grown in a hyperbaric oxygen chamber?

  1. It will die rapidly due to the overproduction of toxic oxygen gas from the hyperactive catalase.
  2. It will exhibit significantly greater resistance to oxygen toxicity compared to the wild-type strain. (correct answer)
  3. It will switch to an exclusively fermentative metabolism to avoid generating any reactive oxygen species.
  4. Its growth rate will be unchanged, as wild-type E. coli is already maximally resistant to oxygen.
Explanation: The primary cause of death in hyperbaric oxygen is overwhelming oxidative stress from reactive oxygen species (ROS). Superoxide dismutase and catalase are the key enzymes that detoxify these ROS. By overproducing these protective enzymes, the engineered strain would have a much greater capacity to neutralize the flood of superoxide radicals and hydrogen peroxide generated under hyperbaric oxygen, allowing it to survive and grow in conditions that would be lethal to the wild-type strain.

Question 6

A researcher wishes to definitively differentiate between an unknown facultative anaerobe and an unknown aerotolerant anaerobe. Which of the following experimental setups provides the clearest distinction?

  1. Inoculating a fluid thioglycollate tube and observing if growth occurs in both the oxic and anoxic zones.
  2. Plating the organism on an agar plate and incubating one half aerobically and the other half anaerobically.
  3. Comparing the colony size and cell density on plates incubated aerobically versus anaerobically. (correct answer)
  4. Performing a catalase test by adding hydrogen peroxide to a smear of the organism.
Explanation: Both organisms will grow under both conditions. The key difference is how they grow. A facultative anaerobe will grow much more robustly (larger colonies, higher density) in the presence of oxygen because it can perform highly efficient aerobic respiration. An aerotolerant anaerobe grows by fermentation regardless of oxygen presence, so its colony size and growth rate will be similar under both aerobic and anaerobic conditions. Comparing the amount of growth is the most effective differentiator.

Question 7

A deep, unstirred tube of nutrient broth is inoculated with a mixture of Pseudomonas aeruginosa (an obligate aerobe) and Clostridium sporogenes (an obligate anaerobe). After incubation, a stable community is established. Where in the tube would you expect to find the highest concentration of viable Clostridium sporogenes?

  1. Evenly distributed throughout the tube, protected from oxygen by a biofilm matrix produced by Pseudomonas.
  2. In a narrow band just below the surface, in a zone of optimal oxygen concentration for both species.
  3. Primarily at the very bottom of the tube, where oxygen has been depleted by the growth of Pseudomonas at the surface. (correct answer)
  4. Concentrated at the air-broth interface, where nutrient exchange is most rapid for both organisms.
Explanation: Pseudomonas aeruginosa is an obligate aerobe and will grow exclusively at the top of the broth where oxygen is plentiful. In the process of its aerobic respiration, it will consume the dissolved oxygen, creating a steep oxygen gradient down the tube. This process, known as oxygen scavenging, renders the deeper parts of the tube anoxic. Clostridium sporogenes, an obligate anaerobe, is killed by oxygen and can therefore only grow in the anoxic environment created at the bottom of the tube.

Question 8

Both catalase and peroxidase detoxify hydrogen peroxide (H₂O₂). However, the peroxidase reaction requires a key input that the catalase reaction does not. An organism that relies solely on peroxidase for H₂O₂ removal is metabolically constrained because the reaction...

  1. produces a toxic byproduct in addition to water.
  2. is significantly less efficient and has a lower turnover rate than catalase.
  3. only functions at a very low pH, limiting the environments where the organism can survive.
  4. consumes a cellular reductant, typically NADH, which is then unavailable for other metabolic processes. (correct answer)
Explanation: When you encounter questions about enzyme mechanisms in microbiology, focus on what each enzyme requires as inputs and produces as outputs. Both catalase and peroxidase break down hydrogen peroxide, but they use fundamentally different chemical strategies. Catalase works through a simple disproportionation reaction: 2H2O2→2H2O+O22H_2O_2 → 2H_2O + O_2. This enzyme requires only hydrogen peroxide as a substrate and generates water and oxygen without consuming any cellular energy currency or reducing equivalents. Peroxidase, however, uses a different mechanism. It reduces hydrogen peroxide to water while simultaneously oxidizing an organic substrate (often NADH or another cellular reductant): H2O2+NADH+H+→2H2O+NAD+H_2O_2 + NADH + H^+ → 2H_2O + NAD^+. This makes option D correct—the peroxidase reaction consumes valuable reducing equivalents like NADH that the cell needs for biosynthesis, energy production, and maintaining cellular redox balance. Option A is wrong because peroxidase produces only water, not toxic byproducts. Option B incorrectly suggests efficiency differences when the real constraint is substrate requirements, not reaction speed. Option C is false—peroxidase enzymes function across a wide pH range and aren't restricted to very low pH environments. The metabolic constraint occurs because organisms relying solely on peroxidase must constantly regenerate the consumed reductants through other metabolic pathways, creating an energy burden that catalase-containing organisms avoid. Remember: when comparing enzyme systems, always consider what cellular resources each reaction consumes—not just what products they make.

Question 9

An organism is classified as a capnophile, meaning it requires elevated concentrations of carbon dioxide (CO₂) for growth. This organism is also found to be a microaerophile. What is the physiological relationship between these two characteristics?

  1. The two characteristics are physiologically independent; one relates to carbon acquisition/pH balance, the other to oxygen tolerance. (correct answer)
  2. The organism uses CO₂ as an electron acceptor in a unique form of anaerobic respiration.
  3. High CO₂ concentrations are necessary to protect the organism's oxygen-sensitive respiratory enzymes.
  4. The enzyme superoxide dismutase in this organism requires a high partial pressure of CO₂ to function correctly.
Explanation: When you encounter questions about microbial growth requirements, focus on understanding that different environmental needs often serve distinct physiological functions, even when they occur in the same organism. Capnophiles require elevated CO₂ concentrations (typically 5-10%) primarily for carbon fixation pathways and to maintain optimal intracellular pH. Many capnophiles use CO₂ in biosynthetic reactions or need it to activate certain enzymes involved in metabolism. Microaerophiles, meanwhile, require oxygen concentrations lower than atmospheric levels (usually 2-10% instead of 21%) because they lack robust antioxidant systems to handle the toxic byproducts of oxygen metabolism, like superoxide radicals and hydrogen peroxide. Answer A correctly identifies that these are independent physiological requirements—one addressing carbon metabolism and pH regulation, the other managing oxygen toxicity. The organism simply has two separate environmental needs that don't directly influence each other. Answer B is incorrect because CO₂ is rarely used as a terminal electron acceptor in respiration; organisms typically use oxygen, nitrate, or sulfate for this purpose. Answer C creates a false connection—high CO₂ doesn't protect oxygen-sensitive enzymes from oxidative damage. Answer D invents a fictional requirement; superoxide dismutase (which breaks down harmful superoxide) doesn't need CO₂ to function and actually helps organisms tolerate higher oxygen levels. Remember that when studying microbial physiology, don't assume that co-occurring traits must be functionally related. Organisms often have multiple independent environmental requirements serving different cellular processes.

Question 10

An aerotolerant anaerobe that possesses superoxide dismutase and NADH peroxidase is cultured in a medium with a high concentration of free copper ions (Cu²⁺). This culture exhibits significantly reduced growth compared to a control culture. What is the most likely mechanism for this copper-induced toxicity?

  1. The copper ions directly and irreversibly denature the bacterium's fermentative enzymes.
  2. The copper ions catalyze the formation of highly damaging hydroxyl radicals from hydrogen peroxide. (correct answer)
  3. The bacteria mistake copper for iron, leading to the synthesis of non-functional, copper-containing enzymes.
  4. The copper ions act as a competitive inhibitor for the manganese cofactor at the active site of superoxide dismutase.
Explanation: This scenario describes a Fenton-like or Haber-Weiss-like reaction. The organism's SOD produces hydrogen peroxide (H₂O₂). Although it has peroxidase to detoxify H₂O₂, transition metals like copper can catalyze the conversion of H₂O₂ into extremely reactive hydroxyl radicals (•OH). The reaction is typically Cu⁺ + H₂O₂ → Cu²⁺ + •OH + OH⁻. The rate of •OH production in the presence of excess copper likely overwhelms the cell's defenses, causing significant oxidative damage and inhibiting growth.

Question 11

A microaerophilic bacterium, Campylobacter jejuni, is accidentally incubated on an agar plate in a hyperbaric oxygen chamber (98% O₂). Which statement best explains the resulting widespread cell death?

  1. The high partial pressure of oxygen directly and irreversibly inhibits the bacterium's anaerobic fermentation enzymes.
  2. The bacterium's electron transport chain becomes completely saturated by the excess oxygen, halting ATP synthesis.
  3. The rate of reactive oxygen species (ROS) formation vastly overwhelms the bacterium's limited detoxifying enzyme capacity. (correct answer)
  4. The bacterium rapidly exhausts the trace amounts of CO₂ in the chamber, which is an absolute requirement for its growth.
Explanation: Microaerophiles are adapted to low oxygen concentrations (e.g., 2-10%). They possess oxygen-detoxifying enzymes like SOD and catalase/peroxidase, but at much lower levels than obligate aerobes or facultative anaerobes. In a hyperbaric oxygen environment, the extremely high concentration of O₂ leads to a massive rate of formation of reactive oxygen species (ROS). The bacterium's limited enzymatic defenses are quickly overwhelmed, resulting in catastrophic oxidative damage to proteins, lipids, and DNA, and ultimately cell death.

Question 12

A genetic mutation renders the superoxide dismutase (SOD) of an obligate aerobe, Pseudomonas aeruginosa, non-functional. The genes for catalase and peroxidase remain intact and fully functional. What is the most likely fate of this mutant strain when plated and incubated in ambient air?

  1. It will grow normally, as its powerful catalase can directly neutralize superoxide radicals.
  2. It will grow, but much more slowly than the wild type, relying on its backup peroxidase activity.
  3. It will fail to grow because of the inability to detoxify the initial superoxide anion (O₂⁻). (correct answer)
  4. It will grow only in the crowded center of a colony where oxygen levels have been depleted.
Explanation: The detoxification of reactive oxygen species is a sequential process. Superoxide dismutase (SOD) is the crucial first step, converting the superoxide radical (O₂⁻) into hydrogen peroxide (H₂O₂). Catalase and peroxidase act on H₂O₂, the product of the SOD reaction. If SOD is non-functional, the cell cannot neutralize the initial, highly damaging superoxide radical. Neither catalase nor peroxidase can substitute for this function, so the cell will suffer fatal oxidative damage and fail to grow.

Question 13

A researcher compares Lactobacillus species (an aerotolerant anaerobe) and E. coli (a facultative anaerobe). Both are grown in separate aerobic broth cultures. The researcher then adds a 3% hydrogen peroxide solution to a sample from each culture. Vigorous bubbling is observed from the E. coli sample, while no bubbling is seen from the Lactobacillus sample, despite its successful growth in air. This difference occurs because Lactobacillus typically...

  1. does not produce hydrogen peroxide as a byproduct of its metabolism.
  2. utilizes a peroxidase enzyme that reduces H₂O₂ to water without producing oxygen gas. (correct answer)
  3. possesses a unique form of superoxide dismutase that completely bypasses H₂O₂ production.
  4. neutralizes H₂O₂ using non-enzymatic antioxidants, such as glutathione, absorbed from the medium.
Explanation: The bubbling observed is the release of O₂ gas from the breakdown of hydrogen peroxide (2H₂O₂ → 2H₂O + O₂), a reaction catalyzed by the enzyme catalase. E. coli, a facultative anaerobe, possesses catalase. Aerotolerant anaerobes like Lactobacillus often lack catalase but still need to neutralize H₂O₂. They use peroxidase enzymes, which catalyze the reduction of H₂O₂ to H₂O using a reducing agent like NADH (H₂O₂ + NADH + H⁺ → 2H₂O + NAD⁺). This reaction does not produce oxygen gas, hence no bubbling is observed.

Question 14

When cultured in a flask with nutrient broth exposed to air, a facultative anaerobe and an aerotolerant anaerobe both survive. However, after 24 hours, the culture density of the facultative anaerobe is significantly higher than that of the aerotolerant anaerobe. What is the best explanation for this difference?

  1. The aerotolerant anaerobe produces more potent toxic metabolic byproducts during fermentation.
  2. The facultative anaerobe can switch to aerobic respiration, which yields significantly more ATP per mole of glucose. (correct answer)
  3. The aerotolerant anaerobe's superoxide dismutase is less efficient at detoxifying radicals than the facultative anaerobe's.
  4. The facultative anaerobe's rapid consumption of oxygen creates a more favorable, partially anaerobic environment.
Explanation: The key difference is metabolic strategy. In the presence of oxygen, the facultative anaerobe performs aerobic respiration, generating a large amount of ATP (approx. 30-32 per glucose). The aerotolerant anaerobe can survive in oxygen but cannot use it for energy production; it relies exclusively on fermentation, which has a much lower ATP yield (approx. 2 per glucose). This vast difference in energy efficiency results in much faster growth and higher cell density for the facultative anaerobe.

Question 15

Some obligate anaerobes can survive brief exposure to atmospheric oxygen, even though they lack SOD and catalase. Which of the following represents a plausible non-enzymatic mechanism for this transient tolerance?

  1. Accumulation of high intracellular concentrations of manganese ions (Mn²⁺) that can scavenge superoxide radicals. (correct answer)
  2. The ability to rapidly form endospores, with the vegetative cells dying immediately upon oxygen exposure.
  3. Possession of a cell membrane that is completely impermeable to the diffusion of molecular oxygen.
  4. Rapid synthesis of a thick polysaccharide capsule that binds to and neutralizes reactive oxygen species.
Explanation: Some organisms, like Lactobacillus plantarum, have been shown to use a non-enzymatic defense. They accumulate very high intracellular concentrations of manganese (II) ions. These Mn²⁺ ions, complexed with small molecules like phosphate or lactate, can act as chemical scavengers of superoxide radicals, functionally replacing the need for the SOD enzyme. This provides a level of protection against oxidative stress.

Question 16

A knockout mutant of a facultative anaerobe is created that lacks the gene for catalase but retains a functional gene for superoxide dismutase (SOD). The growth medium is then supplemented with ferrous iron (Fe²⁺). When this mutant is exposed to an aerobic environment, which of the following events is the most significant contributor to its subsequent cell death?

  1. Direct damage to DNA and lipids by accumulating superoxide radicals (O₂⁻).
  2. Inhibition of anaerobic respiratory enzymes by direct binding of molecular oxygen.
  3. Generation of highly reactive hydroxyl radicals (•OH) via the Fenton reaction. (correct answer)
  4. Depletion of cellular NADH due to excessive activity of peroxidase enzymes.
Explanation: The functional SOD will convert superoxide radicals (O₂⁻) to hydrogen peroxide (H₂O₂). In the absence of catalase, H₂O₂ will accumulate. The presence of ferrous iron (Fe²⁺) in the medium will then catalyze the Fenton reaction (Fe²⁺ + H₂O₂ → Fe³⁺ + •OH + OH⁻), producing highly reactive and destructive hydroxyl radicals (•OH). These radicals cause widespread, lethal damage to cellular macromolecules.

Question 17

A researcher compares Lactobacillus species (an aerotolerant anaerobe) and E. coli (a facultative anaerobe). Both are grown in separate aerobic broth cultures. The researcher then adds a 3% hydrogen peroxide solution to a sample from each culture. Vigorous bubbling is observed from the E. coli sample, while no bubbling is seen from the Lactobacillus sample, despite its successful growth in air. This difference occurs because Lactobacillus typically...

  1. does not produce hydrogen peroxide as a byproduct of its metabolism.
  2. utilizes a peroxidase enzyme that reduces H₂O₂ to water without producing oxygen gas. (correct answer)
  3. possesses a unique form of superoxide dismutase that completely bypasses H₂O₂ production.
  4. neutralizes H₂O₂ using non-enzymatic antioxidants, such as glutathione, absorbed from the medium.
Explanation: The bubbling observed is the release of O₂ gas from the breakdown of hydrogen peroxide (2H₂O₂ → 2H₂O + O₂), a reaction catalyzed by the enzyme catalase. E. coli, a facultative anaerobe, possesses catalase. Aerotolerant anaerobes like Lactobacillus often lack catalase but still need to neutralize H₂O₂. They use peroxidase enzymes, which catalyze the reduction of H₂O₂ to H₂O using a reducing agent like NADH (H₂O₂ + NADH + H⁺ → 2H₂O + NAD⁺). This reaction does not produce oxygen gas, hence no bubbling is observed.

Question 18

A microaerophilic bacterium, Campylobacter jejuni, is accidentally incubated on an agar plate in a hyperbaric oxygen chamber (98% O₂). Which statement best explains the resulting widespread cell death?

  1. The high partial pressure of oxygen directly and irreversibly inhibits the bacterium's anaerobic fermentation enzymes.
  2. The bacterium's electron transport chain becomes completely saturated by the excess oxygen, halting ATP synthesis.
  3. The rate of reactive oxygen species (ROS) formation vastly overwhelms the bacterium's limited detoxifying enzyme capacity. (correct answer)
  4. The bacterium rapidly exhausts the trace amounts of CO₂ in the chamber, which is an absolute requirement for its growth.
Explanation: Microaerophiles are adapted to low oxygen concentrations (e.g., 2-10%). They possess oxygen-detoxifying enzymes like SOD and catalase/peroxidase, but at much lower levels than obligate aerobes or facultative anaerobes. In a hyperbaric oxygen environment, the extremely high concentration of O₂ leads to a massive rate of formation of reactive oxygen species (ROS). The bacterium's limited enzymatic defenses are quickly overwhelmed, resulting in catastrophic oxidative damage to proteins, lipids, and DNA, and ultimately cell death.

Question 19

An aerotolerant anaerobe that possesses superoxide dismutase and NADH peroxidase is cultured in a medium with a high concentration of free copper ions (Cu²⁺). This culture exhibits significantly reduced growth compared to a control culture. What is the most likely mechanism for this copper-induced toxicity?

  1. The copper ions directly and irreversibly denature the bacterium's fermentative enzymes.
  2. The copper ions catalyze the formation of highly damaging hydroxyl radicals from hydrogen peroxide. (correct answer)
  3. The bacteria mistake copper for iron, leading to the synthesis of non-functional, copper-containing enzymes.
  4. The copper ions act as a competitive inhibitor for the manganese cofactor at the active site of superoxide dismutase.
Explanation: This scenario describes a Fenton-like or Haber-Weiss-like reaction. The organism's SOD produces hydrogen peroxide (H₂O₂). Although it has peroxidase to detoxify H₂O₂, transition metals like copper can catalyze the conversion of H₂O₂ into extremely reactive hydroxyl radicals (•OH). The reaction is typically Cu⁺ + H₂O₂ → Cu²⁺ + •OH + OH⁻. The rate of •OH production in the presence of excess copper likely overwhelms the cell's defenses, causing significant oxidative damage and inhibiting growth.

Question 20

A strain of E. coli, a facultative anaerobe, is genetically engineered to constitutively overproduce both superoxide dismutase and catalase at levels ten times higher than the wild type. What is the most likely physiological consequence when this engineered strain is grown in a hyperbaric oxygen chamber?

  1. It will die rapidly due to the overproduction of toxic oxygen gas from the hyperactive catalase.
  2. It will exhibit significantly greater resistance to oxygen toxicity compared to the wild-type strain. (correct answer)
  3. It will switch to an exclusively fermentative metabolism to avoid generating any reactive oxygen species.
  4. Its growth rate will be unchanged, as wild-type E. coli is already maximally resistant to oxygen.
Explanation: The primary cause of death in hyperbaric oxygen is overwhelming oxidative stress from reactive oxygen species (ROS). Superoxide dismutase and catalase are the key enzymes that detoxify these ROS. By overproducing these protective enzymes, the engineered strain would have a much greater capacity to neutralize the flood of superoxide radicals and hydrogen peroxide generated under hyperbaric oxygen, allowing it to survive and grow in conditions that would be lethal to the wild-type strain.