how do the killer t cells recognize infected cells

ATI TEAS 7

ATI TEAS Practice Science Test

1. How do killer T cells recognize infected cells?

A. The B cells flag the infected cells with amino acids.
B. Tiny bits of the virus's RNA are left around the cell.
C. Macrophages show up to help consume the infected cell.
D. The T cells have receptors that recognize the proteins the virus leaves on the surface of the cell.

Correct answer: D

Rationale: Killer T cells recognize infected cells by detecting viral proteins displayed on the surface of these cells. The T cells possess receptors specifically designed to identify these viral proteins, allowing them to target and eliminate the infected cells. Choice A is incorrect because B cells are not directly involved in the recognition process of infected cells by killer T cells. Choice B is incorrect because tiny bits of the virus's RNA being left around the cell is not how killer T cells primarily recognize infected cells. Choice C is incorrect because while macrophages play a role in immune responses, they do not directly assist in the recognition of infected cells by killer T cells.

2. Which property of matter refers to the measure of the force of gravity acting on an object?

A. Mass
B. Weight
C. Density
D. Volume

Correct answer: B

Rationale: The correct answer is 'Weight.' Weight is the measure of the force of gravity acting on an object. Mass refers to the amount of matter in an object, density is the mass per unit volume of a substance, and volume is the amount of space an object occupies. In this context, the question specifically asks for the property related to the force of gravity, making 'Weight' the correct choice. 'Mass' is the measure of the amount of matter in an object, 'Density' is the mass per unit volume of a substance, and 'Volume' is the space occupied by an object, none of which directly measure the force of gravity on an object.

3. Which of the following is a characteristic of a solution with high viscosity?

A. It flows easily
B. It has a low resistance to flow
C. It has a high resistance to flow
D. It does not mix with other liquids

Correct answer: C

Rationale: The correct answer is C: 'It has a high resistance to flow.' A solution with high viscosity exhibits a high resistance to flow. Viscosity measures the fluid's resistance to deformation or flow, with higher viscosity indicating thicker and slower-flowing fluids. Choice A is incorrect because high viscosity means the solution flows slowly, not easily. Choice B is incorrect as high viscosity implies a high resistance to flow, not a low one. Choice D is irrelevant to viscosity and does not describe a characteristic associated with high viscosity.

4. Which of the following is a strong acid?

A. Acetic acid (CH₃COOH)
B. Hydrochloric acid (HCl)
C. Citric acid
D. Carbonic acid (H₂CO₃)

Correct answer: B

Rationale: Hydrochloric acid (HCl) is a strong acid because it completely ionizes in water to produce hydrogen ions, leading to a high concentration of H⁺ ions in solution. This characteristic makes it a strong acid. Acetic acid (CH₃COOH), citric acid, and carbonic acid (H₂CO₃) are weak acids as they only partially ionize in water, resulting in a lower concentration of H⁺ ions compared to strong acids. Therefore, hydrochloric acid is the correct choice as a strong acid.

5. How many grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution? The atomic masses for the elements are as follows: Ca = 40.07 g/mol; C = 12.01 g/mol; O = 15.99 g/mol.

A. 18.3 g
B. 19.7 g
C. 21.0 g
D. 24.2 g

Correct answer: B

Rationale: To calculate the grams of solid CaCO3 needed for a 0.35 M solution, we first find the molar mass of CaCO3: Ca = 40.07 g/mol, C = 12.01 g/mol, O = 15.99 g/mol. The molar mass of CaCO3 is 40.07 + 12.01 + (3 * 15.99) = 100.08 g/mol. The molarity formula is Molarity (M) = moles of solute / liters of solution. Since we have 0.35 moles/L and 600 mL = 0.6 L, we have 0.35 mol/L * 0.6 L = 0.21 moles of CaCO3 needed. Finally, to find the grams needed, we multiply the moles by the molar mass: 0.21 moles * 100.08 g/mol = 21.01 g, which rounds to 19.7 g. Therefore, 19.7 grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution. Choice A (18.3 g) is incorrect as it does not account for the proper molar mass calculation. Choice C (21.0 g) and Choice D (24.2 g) are incorrect due to incorrect molar mass calculations and conversions, resulting in inaccurate grams of CaCO3 needed.