Experience the Quality With Free ATI TEAS Exam Questions That Helps Candidates Pass

Irregular plurals that do not end in –s (children, men, women) form the plural possessive by adding ’s. Therefore, children’s temperatures (A) shows that the temperatures belong to multiple children. Option B incorrectly tacks the apostrophe after an imagined “childrens,” which is not a valid plural form. C lacks any possessive marking. D misplaces the apostrophe on “temperature’s,” implying that the temperature owns something. TEAS often tests possessives with irregular plurals: men’s room, women’s health, people’s rights. A quick check is to paraphrase with “of”: “the temperatures of the children” → children’s temperatures. Also keep agreement: a plural owner (children) + plural thing owned (temperatures). Avoid inserting apostrophes into plain plurals; apostrophes signal possession (or contractions), not plurality. Clear possessive structure is especially important in clinical documentation when attributing measurements or symptoms to the correct patient group.

A 1:4 dilution means for every 1 part stock, add 4 parts diluent (total = 5 parts). If you already have 80 mL of stock (this is the “1 part”), then the diluent volume is (4 \times 80\ \text{mL} = 320\ \text{mL}). The final volume would be (80 + 320 = 400\ \text{mL}), but the question specifically asks for diluent to add, not total volume. Thus, 320 mL (C) is correct. Choice A (160 mL) interprets the ratio as 1:2; Choice B (240 mL) mistakenly multiplies by 3; Choice D (400 mL) is the final volume, not the diluent addition. On TEAS, read dilution ratios carefully: “1:4” means 1 part stock + 4 parts diluent (5 total). A reliable method is to set up parts: stock = 1 part = 80 mL → 1 part = 80 ⇒ 4 parts = (4 \times 80). Always confirm whether the problem wants added diluent or final volume—a common test trap.

In RBC transfusion, type O lacks A and B antigens on red cells, and Rh-negative lacks the D antigen. Therefore, O-negative RBCs present no ABO or Rh antigens likely to be attacked by a recipient’s anti-A, anti-B, or anti-D antibodies, making O-negative the classic “universal donor” for RBCs. Thus any ABO/Rh recipient (A, B, AB, O; Rh+ or Rh−) can receive O-negative cells in urgent or compatible settings—answer D. Note an important distinction: this applies to red cell transfusions; plasma compatibility is reversed because plasma contains antibodies (e.g., AB plasma is the “universal donor” for plasma). Options A–C incorrectly restrict compatibility. TEAS Science often tests ABO logic:

• AB+ = universal recipient for RBCs (has A, B, and Rh antigens; no anti-A/anti-B; tolerates all).

• O− = universal donor for RBCs (no A, B, Rh antigens).

Clinical nuance: whenever possible, crossmatch is performed; O-negative is reserved for emergencies (e.g., trauma) or specific populations (e.g., Rh-negative patients of childbearing potential) to conserve supply.

Translation occurs on ribosomes, which may be free in the cytosol (producing cytosolic proteins) or bound to rough endoplasmic reticulum (producing secreted or membrane proteins). Thus B is correct. Transcription (A) occurs in the nucleus, where DNA is transcribed to pre-mRNA; it is then processed (capping, splicing, polyadenylation) and exported to the cytoplasm for translation. Glycolysis (C) takes place in the cytosol, not the mitochondrial matrix; the matrix hosts the citric acid cycle and pyruvate dehydrogenase. DNA replication (D) also occurs in the nucleus during S phase, not in the Golgi; the Golgi apparatus modifies and sorts proteins and lipids for trafficking. TEAS frequently assesses organelle–function mapping: nucleus (DNA, transcription), ribosomes (translation), rough ER (secretory pathway), smooth ER (lipids/detox), Golgi (processing/sorting), mitochondria (ATP generation via TCA/ETC), cytosol (glycolysis). Knowing these pairings prevents common distractor traps that move nuclear processes into the cytoplasm or assign metabolic pathways to incorrect organelles.

Model the draws sequentially with changing denominators (no replacement). Probability the first is defective: (\dfrac{3}{8}). After removing a defective, 2 defective remain out of 7 total: (P(\text{second defective} \mid \text{first defective}) = \dfrac{2}{7}). Multiply:

(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}).

Thus, A is correct. Option B ((\tfrac{2}{7})) ignores the first-draw probability and treats the second step as if it stood alone. C ((\tfrac{5}{28})) reflects arithmetic or counting errors. D ((\tfrac{3}{56})) divides by the wrong denominator twice (treating events as independent with a fixed 8 or erroneously halving). TEAS probability questions are simplified by a tree diagram or by combinations:

(\dfrac{\binom{3}{2}}{\binom{8}{2}} = \dfrac{3}{28}) gives the same result using unordered pairs. Always note with vs without replacement: if replaced, both factors would be (\frac{3}{8}), giving (\frac{9}{64}). Confirm your final fraction is simplified. A quick sense check: With only 3 defectives in 8, drawing two in a row should be unlikely (≈0.107), which matches (\frac{3}{28}).