Nursing Elites

1. Balance the chemical equation: C4H10 + O2 → CO2 + H2O. What is the coefficient for oxygen?

• A. 5
• B. 6
• C. 7
• D. 8

Correct answer: B

Rationale: To balance the chemical equation, we need to ensure that the number of each type of atom is the same on both sides of the equation. In this case, there are 10 oxygen atoms on the right side (5 in CO2 and 5 in H2O). To balance this, we need to add a coefficient of 6 in front of O2 on the left side, resulting in 6 O2 molecules. This change will give us a total of 12 oxygen atoms on both sides, making the equation balanced. Choice A (5) is incorrect because it does not account for all the oxygen atoms present in the products. Choices C (7) and D (8) are incorrect as they would result in an imbalance in the number of oxygen atoms on both sides of the equation.

2. What property of a substance measures the amount of matter per unit volume?

• A. Mass
• B. Weight
• C. Density
• D. Volume

Correct answer: C

Rationale: Density is the property of a substance that measures the amount of matter per unit volume. It is calculated by dividing the mass of an object by its volume. Mass refers to the amount of matter in an object, weight is the force of gravity acting on an object, and volume is the amount of space an object occupies. Therefore, in this context, density is the correct answer as it specifically relates to the concentration of matter within a given volume.

3. Which of the following is an example of a long bone in the human body?

• A. Carpals
• B. Femur
• C. Tarsals
• D. Metacarpals

Correct answer: B

Rationale: The correct answer is B, Femur. The femur is the longest and strongest bone in the human body, located in the thigh region. Long bones are characterized by their elongated shape, such as the femur, which provides support, strength, and mobility. Choices A, C, and D are incorrect because carpal bones (A) are short bones found in the wrist, tarsal bones (C) are short bones found in the ankle, and metacarpal bones (D) are classified as long bones, but they are shorter than the femur and mainly found in the hand.

4. Which feedback loop inhibits the stimulus or the deviation from homeostasis?

• A. Negative feedback loop
• B. Positive feedback loop
• C. Inhibitory feedback loop
• D. Stimulating feedback loop

Correct answer: A

Rationale: The correct answer is A: Negative feedback loop. Negative feedback loops work to inhibit the stimulus or reduce the deviation from a set point, maintaining homeostasis by counteracting any changes from the norm. In this case, the negative feedback loop acts to minimize any deviation from the body's internal balance, ensuring stability and optimal functioning. Choice B, a positive feedback loop, amplifies the stimulus or deviation, moving systems away from homeostasis. Choice C, an inhibitory feedback loop, is not a commonly recognized term in the context of feedback mechanisms. Choice D, a stimulating feedback loop, is not a standard term and does not accurately describe a feedback loop's role in maintaining homeostasis.

5. During which stage of meiosis II are sister chromatids separated, resulting in four genetically unique daughter cells?

• A. Prophase I
• B. Prophase II
• C. Anaphase I
• D. Anaphase II

Correct answer: D

Rationale: - Prophase I occurs in meiosis I, not meiosis II. During Prophase I, homologous chromosomes pair up and exchange genetic material in a process called crossing over. - Prophase II is the stage where the nuclear envelope breaks down, and spindle fibers start to reappear, preparing the cell for division. Sister chromatids are still attached during Prophase II. - Anaphase I is the stage in meiosis I where homologous chromosomes are separated and pulled to opposite poles of the cell. - Anaphase II is the stage in meiosis II where sister chromatids are separated and pulled to opposite poles of the cell, resulting in four genetically unique daughter cells. This is the stage where the final separation of genetic material occurs, leading to the formation of haploid daughter cells.

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