Cell Structure Test MCQ: Quiz!

The correct answer is "All of the Above Match" because each type of antibiotic listed inhibits a specific process in bacterial cells. Aminoglycosides inhibit protein synthesis in Gram (-) bacteria, Cephalosporins inhibit protein synthesis in a wide range of bacteria, Fluoroquinolones inhibit RNA & DNA synthesis in Gram (-) bacteria, and Macrolides inhibit protein synthesis. Therefore, all of the statements match the actions of the antibiotics listed.

Viruses are obligate intracellular parasites, meaning they can only replicate inside host cells of bacteria, plants, animals, fungi, and algae. They can contain either DNA or RNA, but not both at the same time. Viruses do not independently fulfill the characteristics of life, as they cannot reproduce or carry out metabolic processes on their own. Additionally, viruses lack enzymes for most metabolic processes. Therefore, all of the given characteristics are true for viruses.

Moderate halophiles are microorganisms that can tolerate and thrive in environments with moderate levels of salt concentration. Sea water, being a naturally occurring saline environment, contains a significant amount of salt, making it an ideal habitat for moderate halophiles. Fresh water, cold water, hot water, and bottled water typically have lower salt concentrations and therefore would not be suitable for the growth and survival of moderate halophiles.

The explanation for the correct answer is that the larger versions of the many-layered structure with dark spots observed in the cells grown in the rich media indicate an increase in protein production within the endoplasmic reticulum (ER). The rich media provides more nutrients and resources for the cells to produce proteins, leading to the enlargement of the ER and the presence of dark spots, which could be ribosomes attached to the ER. This suggests that the cells grown in the rich media are actively synthesizing more proteins compared to the cells grown in the minimal media.

Bacteria that are "acid-fast" have a waxy envelope that makes them resistant to decolorization with acid or alcohol after being stained with strong carbon fuchsin dye. This property is observed in bacteria such as Mycobacterium tuberculosis and Mycobacterium leprae.

A culture medium on which only gram-positive organisms grow is called a selective medium. Additionally, the presence of a yellow halo surrounding Staphylococcus aureus colonies indicates that the medium is also a differential medium, as it allows for the differentiation of different types of bacteria based on their ability to produce a specific pigment. Therefore, the correct answer is a and b.

Peroxisomes are mostly found in kidney and liver cells because they carry out the oxidation of toxic materials. This is supported by the fact that they have 2, which implies that they contain oxidizing agents. Therefore, the presence of peroxisomes in kidney and liver cells allows for the detoxification of harmful substances, making these organs vital for maintaining overall health and well-being.

The correct answer is the same number of cells per ml in each. In the stationary phase, the growth of cells slows down and the number of cells remains relatively constant. Since the initial number of cells inoculated in both the 100ml and 200ml cultures is the same, and they have both entered the stationary phase, the number of cells per ml in each culture would be the same.

The given correct answer is "None of The Above". This means that all the options provided in the question are matched correctly. Helical Virus is matched correctly with TMV and Influenza, Enveloped Virus is matched correctly with Pox Virus and Herpes Virus, Non Enveloped Virus is matched correctly with Adeno Virus, Idino Virus, Papo Virus, and Parvo Virus, and Enveloped RNA Virus is matched correctly with Retro Virus, Paramyxovirus, and Corona Virus.

The given information states that plating 0.1 ml of a 10-3 dilution of the microbial culture yields 50 colonies. This means that there are 50 colonies in 0.1 ml of the diluted culture. To find the viable count of the original culture, we need to account for the dilution factor. Since the dilution is 10-3, the original culture was diluted by a factor of 1000 (10-3 = 1/1000). Therefore, the viable count of the original culture is 50 x 1000 = 50,000 cells in 0.1 ml. To express this in ml-1, we multiply by 10, as there are 10 sets of 0.1 ml in 1 ml. Therefore, the viable count of the original culture is 50,000 x 10 = 500,000 cells.ml-1, which is equivalent to 5 x 105 cells.ml-1.