TEAS 7 Science Practice Test

Correct Answer: C

Rationale: The correct answer is C, the law of independent assortment. This law states that alleles for different traits segregate independently of each other during gamete formation. This is due to the random alignment of homologous chromosomes during meiosis. The law of segregation (A) describes how alleles of a gene separate during gamete formation. The law of dominance (B) explains that one allele can mask the expression of another in heterozygous individuals. The law of predictive traits (D) is not a recognized Mendelian law and is therefore incorrect. In summary, the law of independent assortment best describes how alleles from different genes segregate independently during gamete formation.

Correct Answer: B

Rationale: The correct answer is B: medulla oblongata. This region of the brain is responsible for controlling and regulating autonomic functions like respiration, digestion, and heart rate. It is located in the brainstem and contains vital centers for these functions. The cerebellum (A) is primarily involved in coordination and balance. The temporal lobe (C) is responsible for auditory processing and memory. The cerebral cortex (D) is involved in higher cognitive functions like thinking and decision-making. Therefore, the medulla oblongata is the correct choice as it specifically controls autonomic functions.

Correct Answer: A

Rationale: The correct answer is A. In a hypertonic solution, the concentration of solutes outside the cell is higher than inside. Water moves out of the cell via osmosis to equalize concentrations, causing the cell to shrink. This is due to the higher osmotic pressure outside the cell. Choices B and C are incorrect because in a hypertonic solution, water moves out of the cell, causing it to shrink, not swell or remain the same size. Choice D is incorrect as pH is not directly affected by osmosis.

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.