genes control heredity in man and other organisms this gene is

HESI A2

HESI A2 Biology Practice Test

1. Genes control heredity in man and other organisms. This gene is ___

A. A segment of DNA
B. A bead-like structure on the chromosomes
C. A protein molecule
D. A segment of RNA

Correct answer: A

Rationale: Genes are sections of DNA that contain instructions for building and operating living organisms. These sections of DNA contain the genetic information that is passed on from one generation to the next. Genes are composed of specific sequences of nucleotides within the DNA molecule. Therefore, genes are best described as segments of DNA. Choice A is correct. Choice B is incorrect because genes are not bead-like structures on chromosomes but specific sequences of DNA. Choice C is incorrect because genes are not protein molecules but rather sequences of nucleotides. Choice D is incorrect because genes are not segments of RNA but DNA.

2. Which of the following is a tertiary consumer?

A. Owl
B. Shrew
C. Grasshopper
D. Wheat

Correct answer: A

Rationale: The correct answer is A, Owl. Tertiary consumers are organisms that feed on secondary consumers, which, in turn, feed on primary consumers. Owls are considered tertiary consumers because they primarily feed on animals such as rodents, which are secondary consumers. Shrew (choice B) is a secondary consumer, feeding on insects and worms, placing it at a lower trophic level than the owl. Grasshopper (choice C) is a primary consumer, feeding on plants. Wheat (choice D) is not a consumer in the food chain but a plant.

3. 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.

4. Select the strand of DNA that would match this segment: ACTTGCA

A. TGAACGT
B. GACCATG
C. ACTTGCA
D. None of the above

Correct answer: A

Rationale: The correct match for the DNA segment ACTTGCA is TGAACGT. In DNA, adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G). Therefore, the complementary strand to ACTTGCA should be TGAACGT, making choice A the correct answer. Choice B (GACCATG) does not follow the base pairing rules; hence, it is incorrect. Choice C (ACTTGCA) is the original segment, not its complementary strand. Choice D is incorrect as well because a matching strand does exist.

5. Which part of cellular respiration produces the greatest amount of ATP?

A. electron transport chain
B. glycolysis
C. citric acid cycle
D. fermentation

Correct answer: A

Rationale: The electron transport chain (ETC) produces the greatest amount of ATP during cellular respiration. This process occurs in the inner mitochondrial membrane and involves the transfer of electrons through a series of protein complexes, creating a proton gradient that drives the synthesis of ATP. By utilizing the energy from the electron carriers NADH and FADH2 produced in earlier stages of cellular respiration, the ETC can generate a large amount of ATP efficiently through oxidative phosphorylation. Glycolysis only produces a small amount of ATP in comparison to the ETC. The citric acid cycle generates some ATP but not as much as the ETC. Fermentation does not produce ATP through oxidative phosphorylation and yields a much smaller amount of ATP compared to the ETC.