TEAS 7 practice test science

Questions 108

ATI TEAS 7

ATI TEAS 7 Test Bank

TEAS 7 practice test science Questions

Question 1 of 5

What protein complex controls the progression of mitosis through its activation and degradation?

Correct Answer: B

Rationale: A) Ribosome: Ribosomes are cellular organelles responsible for protein synthesis and are not directly involved in controlling the progression of mitosis. B) Cyclin-dependent kinase (CDK): CDKs are a family of protein kinases that regulate the cell cycle, including the progression of mitosis. CDK activity is controlled by cyclins, which bind to CDKs to activate them at specific points in the cell cycle. The activation and degradation of cyclins regulate the activity of CDKs, which in turn control the progression of mitosis. C) Centriole: Centrioles are involved in organizing the microtubules of the mitotic spindle but do not directly control the progression of mitosis. D) Microtubule: Microtubules are structural components of the cytoskeleton and are involved in various cellular processes, including mitosis, but they do not control

Question 2 of 5

What protein complex controls the progression of mitosis through its activation and degradation?

Correct Answer: B

Rationale: A) Ribosome: Ribosomes are cellular organelles responsible for protein synthesis and are not directly involved in controlling the progression of mitosis. B) Cyclin-dependent kinase (CDK): CDKs are a family of protein kinases that regulate the cell cycle, including the progression of mitosis. CDK activity is controlled by cyclins, which bind to CDKs to activate them at specific points in the cell cycle. The activation and degradation of cyclins regulate the activity of CDKs, which in turn control the progression of mitosis. C) Centriole: Centrioles are involved in organizing the microtubules of the mitotic spindle but do not directly control the progression of mitosis. D) Microtubule: Microtubules are structural components of the cytoskeleton and are involved in various cellular processes, including mitosis, but they do not control

Question 3 of 5

According to Newton's third law of motion, for every action, there is an equal and opposite _________.

Correct Answer: A

Rationale: Newton's third law of motion states that for every action, there is an equal and opposite reaction. This law emphasizes that forces always exist in pairs. When one object exerts a force on a second object (action), the second object exerts an equal force in the opposite direction back on the first object (reaction). This principle is crucial in understanding the interactions between objects and the resulting motion observed in the physical world. Choices B, C, and D are incorrect because while force is involved, the specific concept highlighted by Newton's third law is the equal and opposite reaction. Acceleration and momentum are also related to motion but are not directly tied to Newton's third law of motion, which focuses on the equality and oppositeness of forces in interactions.

Question 4 of 5

How does an increase in surface area affect the force of friction between two surfaces?

Correct Answer: A

Rationale: An increase in surface area typically increases the force of friction between two surfaces. This is because with a larger surface area in contact, there are more points of contact between the surfaces, leading to a greater frictional force resisting motion. As a result, the correct answer is that an increase in surface area increases the force of friction between two surfaces. Choice B is incorrect because a greater surface area increases the frictional force. Choice C is incorrect because an increase in surface area results in more contact points and greater friction. Choice D is incorrect as the relationship between surface area and friction is predictable - an increase in surface area generally leads to an increase in frictional force.

Question 5 of 5

Which of the following units is used to express concentration as a mass of solute per unit volume of solution?

Correct Answer: A

Rationale: Molality (m) is the unit used to express concentration as a mass of solute per unit volume of solution. It is calculated by dividing the mass of solute in grams by the mass of the solvent in kilograms. Molality is preferred over molarity when there are large temperature variations as it is temperature-independent, making it a more accurate measure of concentration. Molarity (B) is the unit used to express concentration as moles of solute per liter of solution, weight percent (C) is the mass of solute per 100 parts of the total mass of the solution, and parts per million (D) is used to express very small concentrations where 1 ppm is equivalent to 1 mg of solute per liter of solution.

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