Nursing Elites

1. Whose energy efficiency is greater?

• A. Herbivore
• B. Carnivore
• C. Omnivore
• D. Decomposer

Correct answer: D

Rationale: Decomposers have the greatest efficiency of energy among the given options. Decomposers break down organic matter, such as dead plants and animals, into simpler substances through the process of decomposition. This breakdown process results in the release of nutrients back into the ecosystem, making energy more readily available for other organisms to use. In contrast, herbivores, carnivores, and omnivores all derive their energy from the consumption of other living organisms, making their energy efficiency lower than that of decomposers. Herbivores consume plants for energy, which involves energy loss due to inefficiencies in converting plant matter into usable energy. Carnivores consume herbivores or other carnivores, leading to further energy loss through each trophic level. Omnivores consume both plant and animal matter, but their energy efficiency is still lower than decomposers because of the energy loss associated with consuming living organisms. Decomposers play a crucial role in recycling nutrients and energy in ecosystems, making them highly efficient in the utilization of energy.

2. Where is the site of cellular respiration in eukaryotic cells?

• A. Nucleus
• B. Chloroplasts
• C. Mitochondria
• D. Ribosomes

Correct answer: C

Rationale: The correct answer is C: Mitochondria. Mitochondria are often referred to as the powerhouse of the cell because they are responsible for cellular respiration in eukaryotic cells. During cellular respiration, mitochondria generate energy in the form of ATP. Choices A, B, and D are incorrect. The nucleus is the organelle that houses the cell's genetic material, chloroplasts are responsible for photosynthesis in plant cells, and ribosomes are involved in protein synthesis, not cellular respiration.

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. What is the typical result of mitosis in humans?

• A. two diploid cells
• B. two haploid cells
• C. four diploid cells
• D. four haploid cells

Correct answer: A

Rationale: The correct answer is A: two diploid cells. Mitosis is a type of cell division that results in two daughter cells, each having the same number of chromosomes as the parent cell. In humans, the typical result of mitosis is the formation of two diploid cells, not haploid or four cells. Choice B, two haploid cells, is incorrect because mitosis produces genetically identical diploid cells. Choices C and D, four diploid cells and four haploid cells, respectively, are incorrect as mitosis results in two daughter cells, not four.

5. What is the first step in the conversion of glucose to pyruvate?

• A. Glycolysis
• B. Krebs cycle
• C. Electron transport chain
• D. Aerobic respiration

Correct answer: A

Rationale: The correct answer is Glycolysis. Glycolysis is the initial step in the conversion of glucose to pyruvate. During glycolysis, glucose is broken down into pyruvate through a series of enzymatic reactions. Choice B, the Krebs cycle, occurs after glycolysis in aerobic cellular respiration. Choice C, the Electron transport chain, is the final step in aerobic respiration where the majority of ATP is produced. Choice D, Aerobic respiration, is a broader term that encompasses glycolysis, the Krebs cycle, and the electron transport chain, but it is not the specific first step in the conversion of glucose to pyruvate.

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