DAT
DAT is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
DAT is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
10 sample questions for the DAT, with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.
Sections covered on the exam: Surveyofthenaturalsciences, Biology, Quantitative Reasoning, Structureandfunctionofsystems, Genetics, Stoichiometry, Gases, Solutions, Thermodynamicsandthermochemistry, Oxidation-Reductionreactions, Periodicproperties, Nuclearreactions, Structuralevaluation, Mechanisms, Perceptual Ability, Viewrecognition, Paperfolding, Passage-Based, Quantitativecomparison, Probabilityandstatistics, Liquidsandsolids, Laboratory.
Question 1 · Surveyofthenaturalsciences
The initial velocity of an enzyme-catalyzed reaction was measured at several substrate concentrations, with and without 10 µM of compound X. Velocities are in µmol/min. 1 33 14 2 50 25 6 75 50 50 96 89 Which conclusion is best supported by these data?
Answer: B. Compound X competes with the substrate for the active site, raising the apparent Km without changing Vmax.
Without inhibitor, half of the maximal velocity (50) is reached at 2 mM, so Km ≈ 2 mM; with compound X, 50 is reached at 6 mM, so the apparent Km has tripled. At 50 mM substrate the inhibited velocity (89) nearly matches the uninhibited one (96), showing that excess substrate overcomes the inhibitor and Vmax is unchanged, the signature of competitive inhibition. Equal binding to the free enzyme and the enzyme–substrate complex (noncompetitive) or covalent inactivation would lower Vmax, so high substrate could not restore the rate. Binding only the enzyme–substrate complex would lower the apparent Km, not raise it.
Question 2 · Thermodynamicsandthermochemistry
Using the standard enthalpies of formation below, calculate ΔH° for the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l). −103.8 −393.5 −285.8
Answer: B. −2219.9 kJ
O2 is an element in its standard state, so its ΔHf° is zero. The value −2427.5 kJ subtracts +103.8 instead of −103.8, −575.5 kJ ignores the coefficients, and +2219.9 kJ reverses products and reactants; combustion is exothermic.
Question 3 · Perceptual Ability
The three-dimensional object shown may be turned in any direction before it is passed through an opening, but once started it must pass straight through without twisting. Which opening would the object pass through exactly? (Hidden portions of the object contain no irregularities.)
Answer: A. Option A
The correct opening is the exact outline of the object viewed from the side (end-on): pushing the object straight through along that direction requires an opening matching that silhouette, including every step and notch at the same scale. Each distractor adds or removes a small section of the outline, so it matches none of the object's three silhouettes in any orientation.
Question 4 · Paperfolding
Refer to Figure 1 in Question 57. How many cubes in Figure 1 have exactly four of their sides painted?
Answer: E. 5 cubes
Figure 1 contains 19 cubes. Count painted sides cube by cube: a cube's top is painted only if nothing sits on it, each of its four vertical sides is painted if no cube touches that side, and the bottom is never painted. The tally is 2 with 1 side, 6 with 2 sides, 5 with 3 sides, 5 with 4 sides, 1 with 5 sides, so 5 cubes have exactly 4 painted sides. Remember that hidden back sides and hidden supporting cubes still count.
Question 5 · Passage-Based
Refer to Passage 2. According to the passage, Wells's demonstration in Boston was considered unsuccessful because
Answer: A. the patient groaned during the extraction
Paragraph 3 states that the demonstration "was judged a failure when the patient groaned during the extraction," probably because of an incomplete dose. Fatal arrhythmias are associated with chloroform in paragraph 5, flammability is a drawback of ether, and nothing in the passage suggests Morton interfered with Wells's demonstration.
Question 6 · Probabilityandstatistics
A data set has a mean of 40 and a standard deviation of 6. Every value in the set is multiplied by 1.5, and then 10 is subtracted from each result. What are the mean and standard deviation of the new data set?
Answer: D. Mean 50, standard deviation 9
Multiplying every value by 1.5 multiplies both the mean and the standard deviation by 1.5, giving 60 and 9. Subtracting 10 from every value shifts the mean to 50 but does not change the spread, so the standard deviation stays 9. "Mean 60, standard deviation 9" omits the subtraction, "Mean 50, standard deviation 6" wrongly treats scaling as leaving spread unchanged, and 13.5 multiplies the standard deviation by 1.52, the factor that applies to variance.
Question 7 · Stoichiometry
Complete combustion of a 0.3574 g sample of a compound containing only C, H, and O produces 0.7700 g of CO2 (44.01 g/mol) and 0.3150 g of H2O (18.02 g/mol). What is the empirical formula of the compound?
Answer: C. C5H10O2
Moles of C = 0.7700 ÷ 44.01 = 0.01750 mol (0.2102 g), and moles of H = 2 × (0.3150 ÷ 18.02) = 0.03496 mol (0.0352 g). Oxygen is found by difference: 0.3574 − 0.2102 − 0.0352 = 0.1120 g, or 0.00700 mol. Dividing by 0.00700 gives C 2.50 : H 5.00 : O 1.00, and doubling gives C5H10O2. CH2 ignores the oxygen in the sample, C2H5O rounds 2.5 down instead of doubling, and C5H10O halves the oxygen.
Question 8 · Mechanisms
Which of the following conversions would NOT occur as written under ordinary laboratory conditions?
Answer: C. CH3COOCH3 + NaCl → CH3COCl + NaOCH3
Nucleophilic acyl substitution proceeds "downhill," from a more reactive derivative to a less reactive one, and the leaving group must be a weaker base than the incoming nucleophile. Converting an ester into an acid chloride is uphill: chloride is a weak nucleophile, and it would have to expel methoxide, a much stronger base. Acid chloride to ester and anhydride to amide are both downhill conversions, and saponification of an ester by hydroxide is driven to completion because the carboxylate product is resonance-stabilized and unreactive.
Question 9 · Quantitative Reasoning
Refer to Figure 3 in Question 157. How many cubes in Figure 3 have exactly three of their sides painted?
Answer: B. 2 cubes
Figure 3 contains 14 cubes. Count painted sides cube by cube: a cube's top is painted only if nothing sits on it, each of its four vertical sides is painted if no cube touches that side, and the bottom is never painted. The tally is 1 with 1 side, 6 with 2 sides, 2 with 3 sides, 4 with 4 sides, 1 with 5 sides, so 2 cubes have exactly 3 painted sides. Remember that hidden back sides and hidden supporting cubes still count.
Question 10 · Passage-Based
Refer to Passage 4. According to the passage, transduction differs from transformation in that transduction
Answer: C. uses a bacteriophage to carry bacterial DNA from one cell to another
Paragraph 3 defines transduction as the process in which a bacteriophage packages bacterial DNA and delivers it to the next cell it infects. A direct cell-to-cell connection describes conjugation, and uptake of free DNA describes transformation itself. The passage places no same-species restriction on transduction.
TTP Prep has 194 DAT practice questions with explanations, timed and mock-exam modes, 307 flashcards, interactive lessons, a printable cheat sheet and a readiness score. Start free, no card needed.
More free questions: browse every exam.