Try a balanced review collection across the DAT’s academic subjects and all six PAT task types. Use explanations to connect an answer to the concept, passage evidence or spatial rule behind it.
For visual questions, use the full-size diagram link if needed. Some PAT answers are images; inspect their geometry before selecting a label.
Showing 36 of 36 questions
Answer D. Glycolysis includes substrate-level phosphorylation, which transfers phosphate directly to ADP without ATP synthase. Electron transport and oxygen diffusion do not themselves attach phosphate to ADP.
Answer B. With no opposing pressure gradient, water moves into the cell by osmosis toward the compartment with the higher concentration of nonpenetrating solutes. The outside solution is hypotonic relative to the cell, so its volume initially increases. The solutes themselves cannot cross the stated membrane.
Answer A. Lysosomes contain acid hydrolases that digest material from endocytic vesicles and worn cellular components. Normal ATP production does not establish that lysosomal digestion is functioning.
Answer B. Proteins destined for secretion enter the rough endoplasmic reticulum during translation and then pass through the Golgi for processing and sorting. Secretory vesicles subsequently fuse with the plasma membrane.
Answer A. At low temperatures, cholesterol interferes with close packing of phospholipid tails and limits solidification. Its effect depends on temperature; at higher temperatures it also restrains excessive phospholipid movement.
Answer C. Receptor-specific inhibition together with saturable uptake supports receptor-mediated endocytosis. Pinocytosis can internalize extracellular fluid, but it does not explain this selective receptor dependence. Simple diffusion and osmosis do not explain uptake of a particular protein through its surface receptor.
Answer B. The aluminum can produce 0.80 × 2/4 = 0.40 mol Al₂O₃, whereas the oxygen could produce 0.90 × 2/3 = 0.60 mol. Aluminum is limiting, so the smaller product amount is the maximum yield.
Answer B. For a 100 g sample, the mole amounts are approximately 3.33 C, 6.7 H, and 3.33 O. Dividing by the smallest gives the ratio 1:2:1, or CH₂O.
Answer C. Percent yield is actual yield divided by theoretical yield times 100%. Here, 9.00/12.0 × 100% = 75.0%; the theoretical yield belongs in the denominator.
Answer A. Dilution changes volume while preserving moles of solute. M₁V₁ = M₂V₂ gives M₂ = 0.800 × 25.0/100.0 = 0.200 M.
Answer B. Ag⁺ and Cl⁻ form AgCl in a 1:1 ratio. The 0.015 mol of chloride consumes 0.015 mol of silver ion, leaving 0.020 − 0.015 = 0.005 mol Ag⁺.
Answer B. For a fixed amount of ideal gas, V is proportional to T/P. Temperature and pressure both increase by a factor of 1.5, so their effects on volume cancel.
Answer B. A full-headed curved arrow tracks an electron pair from its source to its destination. Here the oxygen lone pair supplies both electrons of the new bond to the proton.
Answer C. A fishhook arrow represents movement of a single electron and is commonly used in radical mechanisms. A full-headed curved arrow represents movement of an electron pair.
Answer C. The product oxygen has three bonds and one lone pair. Formal charge is 6 valence electrons minus 2 nonbonding electrons minus 3 assigned bonding electrons, giving +1.
Answer B. The electronegative fluorines withdraw electron density through sigma bonds, stabilizing the negative charge of the carboxylate conjugate base. Greater conjugate-base stabilization favors deprotonation.
Answer C. The conjugate base of phenol has resonance contributors that distribute negative charge beyond oxygen. Ethoxide lacks that delocalization, making its formation by deprotonation less favorable.
Answer D. An acid–base equilibrium favors the side with the weaker acid, which has the higher pKa. Amide ion forms ammonia with pKa about 38, so it strongly favors deprotonating an alkyne with pKa about 25.
Answer C. Grazer barriers improved early survival, but storms still prevented some protected populations from persisting. The passage therefore distinguishes successful establishment from the additional conditions needed for a reproducing, self-sustaining forest.
Answer A. The key difference appeared after the rinse removed loosely associated cells. Similar counts without rinsing argue against assuming that the grooves simply prevented bacteria from reaching the surface. Similar counts before rinsing help rule out a difference in initial arrival, but the difference in retention after rinsing directly demonstrates the attachment effect.
Answer A. Temperature spikes that occur during tours but not during matched periods when tours are canceled would support a contribution from visitor activity. The same spikes on comparable closed days would instead weaken that explanation. Slower changes in rock temperature explain the stable background pattern, not why the short spikes coincide with visitors.
Answer D. The team measured carbon dioxide as an additional indicator of microbial consumption. Fragment size alone could reflect physical breakup, so the gas measurement addresses the difference between fragmentation and biological breakdown. The gas measurement does not directly measure abrasion; it supplies evidence for the biological process that fragment size alone cannot establish.
Answer A. The author notes practical benefits and opportunities while acknowledging confounding factors and missing reproductive data. That combination supports cautious interest rather than certainty or dismissal.
Answer A. The passage first describes meltwater holes, then evaluates the assumption that their microbes mainly arrive from land. Genetic comparisons lead to a mixed-source explanation and a proposal for further sampling.
Answer C. Multiply every term by 12 to clear the denominators: 3(3x − 2) − 2(x + 1) = 48. Expanding gives 9x − 6 − 2x − 2 = 48, so 7x = 56 and x = 8.
Answer B. Let a be adult tickets, leaving 180 − a student tickets. Then 12a + 7(180 − a) = 1,510. Simplifying gives 5a + 1,260 = 1,510, so a = 50. The remaining 130 student tickets account for $910, and adult tickets account for $600.
Answer B. If the path has width w, the outer dimensions are 12 + 2w and 8 + 2w. Thus (12 + 2w)(8 + 2w) = 192, giving w² + 10w − 24 = 0. Factoring gives (w + 12)(w − 2) = 0. A width must be positive, so w = 2 meters.
Answer D. Write both bases as powers of 2. The equation becomes 2^(6x − 3) = 2^(2x + 4), so 6x − 3 = 2x + 4. Therefore 4x = 7 and x = 7/4.
Answer D. Squaring gives 3x + 10 = x², or (x − 5)(x + 2) = 0. The candidates are 5 and −2, but a principal square root is nonnegative. Substitution confirms √25 = 5, whereas √4 is 2 rather than −2. Thus only x = 5 works.
Answer C. The inequality is equivalent to −5 ≤ 2x − 3 ≤ 5. Add 3 and divide by 2 to obtain −1 ≤ x ≤ 4. The integers are −1, 0, 1, 2, 3, and 4, for a total of six.
Answer E. Trace the broad face rather than the slanted extrusion edges. Its columns, from left to right in the object drawing, rise to 1, 3, 2 equal units. The correct aperture preserves those steps and proportions after rotation. The other apertures change the height or position of a step. This drill fixes the direction of passage to isolate silhouette matching.
Answer A. Looking from the right collapses the left-to-right width. At each depth, keep the tallest column across that width. Reading from front to back gives heights 1, 3, 2. Only one choice preserves that profile. The 3D reference makes this silhouette drill fully specified; it is a practice aid beyond the two views used in an exam item.
Answer C. The angle sizes are 1: 48°, 2: 42°, 3: 39°, and 4: 45°. Therefore the increasing order is 3 < 2 < 4 < 1. Mentally rotate the angles to align one ray; compare the gap to the other ray instead of the slope on the page.
Answer B. Undo the folds in reverse order, reflecting the hole across each crease. These 2 half-folds create 4 layers at the punched location, so one punch produces 4 holes. Match their positions as well as their reflection symmetry; every choice has the same number of holes. In the four-row, four-column answer lattice, numbering from the top left starting at 1, the (row, column) positions are (2, 1), (2, 4), (3, 1), (3, 4).
Answer B. Count exposed faces cube by cube, including faces at the back and inside the steps. Do not count any face touching another cube or the tabletop. For this 10-cube assembly, the tally is 1 painted face: 2 cubes; 2 painted faces: 3 cubes; 3 painted faces: 4 cubes; 5 painted faces: 1 cube. Thus exactly 2 cubes have 1 painted face.
Answer C. The opposite face pairs are ● / ◇, + / □, × / ⊙. Opposite faces cannot share an edge. For the remaining candidates, preserve the order of the three faces around their shared corner: the valid view has ● on top, + on the left and ⊙ on the right. Swapping the two side faces gives a mirror image, which rotating the cube cannot produce.
Test Ninjas · Content checked October 23, 2026
Start with one subject and complete a small block before checking the explanations. The bank includes six biology, six general chemistry, six organic chemistry, six reading comprehension, six quantitative reasoning and six perceptual ability questions. The PAT group has one item from each task type.
For science, explain the mechanism or relationship before looking at the choices. For reading, locate evidence in the supplied passage; these are short passage drills rather than full-length reading sections. For quantitative reasoning, write the setup and carry units through the calculation. PAT requires careful interpretation of the diagrams, not outside science knowledge.
Use the review filter to revisit missed or saved items. This collection reuses selected questions from our daily bank so that a useful skill is available on demand, instead of only on its scheduled day. The dedicated PAT collection provides additional visual practice.
Six questions each cover biology, general chemistry and organic chemistry. Review molecular and cellular processes, chemical quantities and equilibria, and structural or reaction reasoning. This is a foundational sample, not an exhaustive list of organic chemistry topics.
Six short original science passages support questions about meaning, evidence and inference. Read what the passage establishes rather than substituting a familiar fact about the subject. Full DAT reading preparation also requires longer passages and sustained pacing.
Practice algebra, ratios, rates, percentages, probability and applied numerical reasoning. Numeric choices are ordered without changing which response is correct. Check whether the question asks for a quantity, a ratio or a change.
One visual drill each covers keyholes, top–front–end views, angle ranking, hole punching, cube counting and pattern folding. The companion PAT page expands these to five examples per type.
For science, state the defining property that separates your answer from the nearest distractor. If the explanation introduces a concept you cannot explain, return to the underlying topic before collecting more question attempts.
A true scientific statement may still be the wrong reading answer if the passage does not establish it. Identify whether the prompt asks what is stated, implied or used as evidence.
In PAT, count edges, track folds or compare projections systematically. A shape that looks familiar can fail because of one reversed relationship or a missing hidden cube.
No. They are original Test Ninjas items from our DAT bank. ADA resources guide the exam descriptions and task categories, but the question wording and diagrams are independently produced.
No. Six organic chemistry items provide a small foundational sample. Consult the current ADA specifications, including the 2026 organic chemistry update, to plan complete topic coverage.
No. These are short passage drills intended to isolate evidence and inference skills. Add full-length practice to develop endurance and time management.
This page has one PAT item per type. The dedicated PAT bank has five per type, including these six previews. Shared questions are intentional, and saved progress is separate on each page.
No. Accuracy on an uncalibrated practice collection cannot be converted into a valid DAT scaled score or Academic Average. The progress display is for study planning.
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