Nursing Elites

1. Which bacteria are spherical in shape?

• A. Clostridia
• B. Bacilli
• C. Spirilla
• D. Cocci

Correct answer: D

Rationale: The correct answer is D, Cocci. Bacteria that are spherical in shape are known as cocci. They can be found in various arrangements such as clusters, chains, or pairs. Examples of cocci bacteria include Staphylococcus aureus and Streptococcus pneumoniae. Choice A, Clostridia, consists of rod-shaped bacteria. Choice B, Bacilli, refers to rod-shaped bacteria as well. Choice C, Spirilla, are spiral-shaped bacteria, not spherical.

2. 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.

3. Which of the following shows how information is transformed to make a protein?

• A. DNA-RNA-protein
• B. gene-chromosome-protein
• C. ATP-amino acid-protein
• D. RNA-DNA-protein

Correct answer: A

Rationale: The correct sequence of events for protein synthesis is that DNA is transcribed into RNA, which is then translated into a protein. This process is known as the central dogma of molecular biology. In this sequence, DNA serves as the template for RNA synthesis (transcription), and the RNA serves as the template for protein synthesis (translation). Choice B is incorrect as genes are segments of DNA that code for proteins, and chromosomes are structures that contain DNA. Choice C is incorrect as ATP is an energy molecule and amino acids are the building blocks of proteins, but they are not directly involved in the synthesis process. Choice D is incorrect because RNA is first transcribed from DNA and then used as a template to produce proteins.

4. Huntington’s disease is carried on the dominant allele. In a situation where two heterozygous parents have the disease, what percentage of their offspring are predicted to be disease-free?

• A. 0%
• B. 25%
• C. 50%
• D. 100%

Correct answer: B

Rationale: In this scenario, both parents are heterozygous for Huntington's disease, meaning each carries one dominant allele (representing the disease) and one recessive allele (representing no disease). When they have offspring, there is a 25% chance that each child will inherit two recessive alleles, making them disease-free. The Punnett square for two heterozygous parents (Hh x Hh) yields a 25% probability of offspring being homozygous recessive (hh) and therefore disease-free. Choice A (0%) is incorrect because there is a possibility of disease-free offspring. Choice C (50%) is incorrect as it represents the likelihood of being a carrier. Choice D (100%) is incorrect as all offspring will not be disease-free in this scenario.

5. If both parents have polydactylism, what percentage of their offspring is predicted to manifest the anomaly?

• A. 25%
• B. 50%
• C. 75%
• D. 100%

Correct answer: B

Rationale: When both parents have polydactylism, they each possess at least one dominant allele for polydactylism. Through a Punnett square analysis, it can be determined that 50% of their offspring will inherit the dominant allele from both parents. This results in a 100% chance of manifesting the anomaly (2 out of 4 possibilities). Therefore, 50% of the offspring are predicted to manifest the anomaly. Choices A, C, and D are incorrect as they do not align with the principles of Mendelian genetics and Punnett square analysis.

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