TEAS 7 Science Practice Test

Correct Answer: A

Rationale: The correct answer is A: mRNA. mRNA (messenger RNA) contains the code for making a protein through a process called translation. mRNA is transcribed from DNA in the nucleus and carries the genetic information to the ribosomes where protein synthesis occurs. tRNA (choice B) brings amino acids to the ribosome, rRNA (choice C) is a component of the ribosome, and DNA polymerase (choice D) is involved in DNA replication, not protein synthesis.

Correct Answer: A

Rationale: The correct answer is A because this situation indicates that the results are consistent (reliable) across experiments, but the test used does not measure what it claims to measure (not valid). Reliability refers to consistency in results, while validity refers to accuracy in measuring what it intends to measure. Therefore, in this scenario, the results can be consistently replicated (reliable) even though the test does not actually measure the intended concept (not valid). Summary: A: The results are consistent (reliable) but the test is not measuring what it claims to measure (not valid). B: This choice does not fit the scenario described in the question. C: This choice indicates both reliability and validity are lacking, which does not align with the scenario. D: This choice implies that both reliability and validity are present, which contradicts the situation described.

Correct Answer: B

Rationale: In anaphase of mitosis, chromatids separate and move to opposite poles of the cell, ensuring each daughter cell receives a complete set of chromosomes. This ensures genetic stability and accurate distribution of genetic material. In metaphase, chromosomes align at the cell's equator but do not separate. Prophase is the initial stage of mitosis where chromosomes condense but do not divide, and telophase is the final stage where chromosomes decondense and nuclear envelopes reform. Therefore, anaphase is the correct phase where chromatids divide into chromosomes and migrate to opposite ends of the cell.

Correct Answer: D

Rationale: The correct answer is D because a control group is necessary in experiments to compare results with a group that does not receive the experimental treatment. In this case, if tanks A, B, or C were chosen as the control group, there would be no basis for comparison with a group that did not receive any treatment. Therefore, without a specific control group that does not undergo any experimental manipulation, it is not possible to determine the true impact of the treatment being tested. Tanks A, B, and C cannot serve as the control group in this experiment as they are all part of the experimental setup and are not kept untreated for comparison.

Correct Answer: C

Rationale: The correct answer is C. When two heterozygous mice (WwSs) are crossed, they can produce offspring with genotypes of Ww, WW, ww, Ss, SS, ss. Therefore, the most likely genotype for their offspring is WwSs. Choice A (wwss) is incorrect because it doesn't account for the possibility of capital letters in the genotypes. Choice B (WWSS) is incorrect because it represents the homozygous dominant genotype, which is less likely when crossing heterozygous mice. Choice D (WWSs) is incorrect because it assumes the offspring will have two dominant alleles for the first trait and one dominant allele for the second trait, which is less likely than having both traits in heterozygous form.