Nursing Elites

1. The dark, round structures observed within some bacterial cells are most likely:

• A. Ribosomes
• B. Flagella
• C. Endospores
• D. Capsules

Correct answer: C

Rationale: Endospores are the correct answer. Endospores are dark, round structures observed within some bacterial cells. They are highly resistant dormant structures formed by certain bacteria in response to harsh environmental conditions. Ribosomes are not dark, round structures within bacterial cells; they are responsible for protein synthesis. Flagella are long, whip-like appendages used for movement and are not dark, round structures within bacterial cells. Capsules are outer layers of polysaccharides that some bacteria produce for protection and adherence, not dark, round structures within bacterial cells.

2. The adaptive immune system has two main branches: humoral and cell-mediated immunity. Which type of immune cell is primarily responsible for antibody production?

• A. Helper T cells
• B. Cytotoxic T cells
• C. B cells
• D. Natural killer cells

Correct answer: C

Rationale: B cells are primarily responsible for antibody production in the adaptive immune system. When activated, B cells differentiate into plasma cells, which are specialized to produce antibodies. Helper T cells play a crucial role in activating B cells, but they do not directly produce antibodies. Cytotoxic T cells are involved in cell-mediated immunity by killing infected cells, while natural killer cells are part of the innate immune system and target virus-infected and cancerous cells. Therefore, the correct answer is C (B cells) as they are directly involved in the production of antibodies, essential for humoral immunity.

3. How do sweat glands help regulate body temperature?

• A. Producing sebum for lubrication
• B. Constricting blood vessels in the skin
• C. Releasing a watery fluid that evaporates for cooling
• D. Thickening the epidermis for insulation

Correct answer: C

Rationale: Sweat glands play a vital role in regulating body temperature by releasing a watery fluid that evaporates from the skin's surface. This evaporation process helps cool the body down by dissipating heat, which is essential for maintaining a stable internal temperature. Choice A is incorrect because sweat glands do not produce sebum but rather a watery fluid. Choice B is incorrect because sweating itself does not involve the constriction of blood vessels; instead, it promotes heat loss through evaporation. Choice D is incorrect as sweat glands do not thicken the epidermis for insulation but rather facilitate the cooling process through sweat evaporation.

4. What is the term for the maximum amount of solute that can dissolve in a solvent at a specific temperature and pressure?

• A. Molarity
• B. Solubility
• C. Concentration
• D. Saturation

Correct answer: B

Rationale: Solubility is the correct term for the maximum amount of solute that can dissolve in a solvent at a specific temperature and pressure. Molarity is a measure of concentration, not the maximum amount that can dissolve. Concentration is a general term for the amount of solute present in a given amount of solvent. Saturation is related to solubility but specifically refers to a state where no more solute can be dissolved in the solvent.

5. What is the molarity of a solution made by dissolving 4.0 grams of NaCl into enough water to make 120 mL of solution? The atomic mass of Na is 23.0 g/mol, and Cl is 35.5 g/mol.

• A. 0.34 M
• B. 0.57 M
• C. 0.034 M
• D. 0.057 M

Correct answer: B

Rationale: To find the molarity, first calculate the moles of NaCl. Moles of NaCl = 4.0 g / (23.0 g/mol + 35.5 g/mol) = 0.068 mol. Next, use the formula for molarity: Molarity = moles of solute / liters of solution. Molarity = 0.068 mol / 0.120 L = 0.57 M. Therefore, the molarity of the solution is 0.57 M. Choice A, 0.34 M, is incorrect as it does not match the calculated molarity. Choice C, 0.034 M, is incorrect as it is a decimal point off from the correct molarity. Choice D, 0.057 M, is incorrect as it does not match the calculated molarity of 0.57 M.

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