Free Kingdom Prokaryotae MCQs with Answers

411. The structures that help some bacteria to attach to rocks and/or host tissues are

• A. Mesosomes
• B. Holdfast
• C. Rhizoids
• D. Fimbriae

Explanation: Fimbriae are indeed small, bristle-like structures that extend from the surface of bacterial cells and are crucial for adhesion to various surfaces, including rocks and host tissues. They allow bacteria to establish themselves in suitable environments, such as the urinary tract in the case of Neisseria gonorrhoeae. In contrast, mesosomes are involved in other cellular processes but do not primarily function in attachment. Holdfasts are specialized attachment structures, but they are not as ubiquitous as fimbriae across different bacterial species. Rhizoids, while serving a similar anchoring function in other organisms, are not characteristic of bacteria.

412. A parasite living inside body of the host is called:

• A. Ectoparasite
• B. Obligate parasite
• C. Facultative parasite
• D. Endoparasite

Explanation: [A] Ectoparasite;parasite that lives outside of its host[B] obligate parasite is the parasite that cannot complete its life cycle without exploiting a host[C] Facultative Parasite's life cycle does not completely depend on the host[D] Endoparasite e.g tapeworm that lives insides the host

413. The maximum growth rate occurs in the:

• A. Stationary phase
• B. Senescent phase
• C. Lag phase
• D. Exponential phase

Explanation: Maximum growth rate occurs in the exponential or acceleration or log phase. The point at which the exponential growth begins to slow down is known as the inflexion point. The exponential phase, also known as the exponential growth phase, refers to a stage in the growth of a population or a culture of microorganisms when the population size or biomass increases rapidly and continuously at a constant relative rate. This phase is characterized by a J-shaped growth curve on a graph where the population size or biomass increases exponentially over time.

414. The type of locomotion shown by bacteria towards carbon and energy sources is

• A. positive chemotaxis.
• B. negative chemotaxis
• C. positive magnetotaxis
• D. negative magnetotaxis

Explanation: Positive chemotaxis means moving towards a chemical stimulus i.e. O2 and energy source. So option A is correct. All other options does not match the given statement.

415. Escherichia coli is a facultative anaerobic bacterium because it can

• A. not survive in the presence of oxygen.
• B. utilise low concentration of oxygen only.
• C. survive without oxygen, but uses oxygen if it is present.
• D. neither use oxygen for its growth nor can be harmed by it.

Explanation: A facultative anaerobe, such as Escherichia coli, can grow in both aerobic (with oxygen) and anaerobic (without oxygen) environments. However, it thrives better when oxygen is available because it can generate more energy through aerobic respiration than anaerobic processes. Therefore, it does not require oxygen but will use it if present, which makes Option C correct. Option A is incorrect because Escherichia coli can indeed survive in oxygen. Option B is incorrect as facultative anaerobes are not limited to low concentrations of oxygen. Option D describes aerotolerant organisms, which makes it incorrect for this question.

416. Which of the following statements CORRECTLY describes the obligate anaerobic bacteria?

• A. They are killed by free oxygen
• B. They essentially need free oxygen
• C. They can live with or without free oxygen
• D. They obtain free oxygen from the host's body

Explanation: Obligate anaerobic bacteria are organisms that cannot survive in the presence of oxygen. Oxygen is toxic to them, and they only thrive in environments devoid of it. This distinguishes them from aerobic bacteria, which require oxygen, and facultative anaerobes, which can survive with or without oxygen. Understanding these distinctions helps clarify that the correct statement about obligate anaerobes is that they are killed by free oxygen, as described in Option A. Options B, C, and D incorrectly describe characteristics of other types of bacteria or misrepresent the nature of obligate anaerobes.

417. A microbiologist used Gram staining method in a laboratory to identify different types of infectious bacteria in four different samples (I, II, III and IV). The given table shows the results.Based on given results, gram positive bacteria are found in samples:

• A. I
• B. IV
• C. I and IV
• D. II and III

Explanation: Gram staining is a method used to differentiate bacterial species into two groups based on the chemical and physical properties of their cell walls. Gram-positive bacteria retain the crystal violet stain and appear purple under a microscope. In this question, to identify which samples contain gram-positive bacteria, you must refer to the staining results. Samples II and III exhibit the characteristics of gram-positive bacteria, while Samples I and IV do not, as they are gram-negative, which shows a red-pink colour. Therefore, the correct answer is Sample II and III.

418. A microbiologist used Gram staining method in a laboratory to identify different types of infectious bacteria in four different samples (I, II, III and IV). The given table shows the results.Based on the results of sample I and IV, he infers that the cell walls of Escherichia coli and Salmonella typhi have

• A. Thin peptidoglycan layer
• B. Thick peptidoglycan layer
• C. Outer membrane of lipoprotein
• D. Outer membrane of lipopolysaccharide

Explanation: Gram staining is a method used to classify bacteria into two major groups: Gram-positive and Gram-negative, based on the structure of their cell walls. Gram-negative bacteria, such as Escherichia coli and Salmonella typhi, have a thin peptidoglycan layer located between the cell membrane and an outer membrane composed of lipopolysaccharides, lipoproteins, and phospholipids. This structure is significant in determining the bacteria's response to Gram staining, where Gram-negative bacteria do not retain the crystal violet stain and instead take up the counterstain (safranin), appearing pink. Thus, the inference that the cell walls of E. coli and S. typhi have a thin peptidoglycan layer is correct. Other options either incorrectly describe the cell wall structure of Gram-negative bacteria or are partially correct.

419. The given diagram shows a phenomenon in bacteria.

• A. Conjugation
• B. Transduction
• C. Transformation
• D. Endospore formation

Explanation: Bacterial transformation is the process where bacteria take up foreign DNA from their environment and incorporate it into their own genome, leading to new genetic traits. So option C is correct. Other processes like conjugation, transduction and endospore formation do not relate to the figure.

420. Different methods of sterilisation are used to kill microorganisms.The method which kills the microorganisms by disrupting their DNA is

• A. Dry heat
• B. Radiation
• C. Fltration
• D. Moist heat

Explanation: The correct answer is radiation. Ionizing radiation, such as gamma rays or X-rays, is used in sterilization because it can penetrate cells and disrupt their DNA, leading to the death of the microorganisms. This method is particularly useful for sterilizing medical equipment and other materials that cannot withstand heat.Dry heat and moist heat sterilization methods kill microorganisms through thermal denaturation of proteins and other cell components, rather than by disrupting DNA.Filtration removes microorganisms from solutions by physically trapping them in a filter but does not affect the DNA or kill the microorganisms, which makes it unsuitable when complete destruction is necessary.