Nursing Elites

1. Which is an example of a gymnosperm?

• A. Red cedar
• B. Japanese cherry
• C. Flowering dogwood
• D. American chestnut

Correct answer: A

Rationale: Red cedar is the correct answer as it is an example of a gymnosperm. Gymnosperms are plants that produce seeds not enclosed within an ovary or fruit. In the case of red cedar, it belongs to the gymnosperm group and has naked seeds that are exposed on the surface of scales or leaves. Choices B, C, and D are angiosperms, not gymnosperms. Japanese cherry, flowering dogwood, and American chestnut are all examples of angiosperms, which are flowering plants with seeds enclosed within an ovary.

2. During which phase of mitosis does the nuclear envelope disintegrate?

• A. Prophase
• B. Prophase
• C. Prometaphase
• D. Metaphase

Correct answer: C

Rationale: During the prometaphase stage of mitosis, the nuclear envelope disintegrates. This allows the condensed chromosomes to move towards the center of the cell, preparing for their alignment along the metaphase plate. The breakdown of the nuclear envelope is a crucial step in mitosis to ensure the proper segregation of genetic material into daughter cells. Choices A, B, and D are incorrect as the nuclear envelope disintegration specifically occurs during the prometaphase stage, not in prophase or metaphase.

3. Which animal has an open transport system?

• A. Grasshopper
• B. Earthworm
• C. Dolphin
• D. Chicken

Correct answer: B

Rationale: The correct answer is B: Earthworm. Earthworms have an open circulatory system, meaning their blood and interstitial fluid are not enclosed in blood vessels. Instead, the blood flows freely within the body cavity, allowing for direct exchange of nutrients and waste products with surrounding tissues. This lack of a closed transport system is a characteristic feature of earthworms. Choices A, C, and D are incorrect because grasshoppers, dolphins, and chickens have closed circulatory systems where the blood is enclosed within blood vessels, unlike earthworms.

4. A student was asked to count birds in a given location over a 24-hour period. Which count would make their data most valid?

• A. Count birds at one feeder every 6 hours.
• B. Count birds at three feeders at noon and 6:00 P.M.
• C. Count birds at one feeder at noon and 6:00 P.M.
• D. Count birds at three feeders every 6 hours.

Correct answer: C

Rationale: Counting birds at one feeder at consistent time intervals (noon and 6:00 P.M.) over a 24-hour period ensures that the data collected are evenly distributed and provide a more accurate representation of the bird population in that location throughout the day. This method allows for observations during different times of the day, capturing potential variations in bird activity and distribution. Choice A is not as effective as it lacks observations at different times of the day, potentially missing variations in bird behavior. Choice B involves multiple locations and times, which could introduce more variables and make it harder to analyze the data accurately. Choice D, counting birds at three feeders every 6 hours, may provide too frequent data points and not cover all significant time intervals for observing bird activity.

5. Which component is not found in the nucleotide of DNA?

• A. Simple sugar
• B. Nitrogen base
• C. Phosphate group
• D. Citric acid

Correct answer: D

Rationale: Citric acid is not found in the nucleotide of DNA. A DNA nucleotide is composed of a simple sugar (deoxyribose), a nitrogen base (adenine, thymine, cytosine, or guanine), and a phosphate group. Citric acid is not part of DNA nucleotides; instead, it is involved in the citric acid cycle of cellular respiration. Choices A, B, and C are essential components of DNA nucleotides, making them incorrect answers.

Similar Questions