OAT Practice Questions (2026) — Free Sample Questions with Answers

16 free OAT questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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Biology

Question 1

The pedigree shows a rare trait. Assuming the mode of inheritance most consistent with the pedigree, what is the probability that individual II-2 is a heterozygous carrier?

Pedigree of a rare trait
III12123AffectedUnaffected
Squares are males and circles are females; filled symbols are affected.
Show answer & explanation

Answer: D

  1. Unaffected parents with an affected daughter → autosomal recessive; I-1 and I-2 are both Aa.
  2. Offspring: 1/4 AA, 1/2 Aa, 1/4 aa.
  3. II-2 is unaffected, so remove aa: AA : Aa = 1/4 : 1/2.
  4. P(Aa | unaffected) = (1/2)/(3/4) = 2/3.

1/4: Gives the chance that a child of two carriers is affected, not that an unaffected child is a carrier.

1/3: Inverts the ratio: of the three unaffected genotypes, only one is homozygous normal.

1/2: Unconditioned carrier probability for any child of two carriers; it ignores that II-2 is known to be unaffected.

2/3: Correct. An affected daughter of unaffected parents shows autosomal recessive inheritance (X-linked recessive would need an affected father), so both parents are Aa. An unaffected child is AA, Aa or aA, so P(carrier) = 2/3 (Genetics: human genetics).

3/4: This is the probability that a child of two carriers is unaffected, not that II-2 is a carrier.

An unaffected sibling of an affected individual (with carrier parents) has a 2/3 chance of being a carrier.

Question 2

A rod cell in a dark-adapted retina absorbs a photon. Which sequence of events follows in the rod?

Show answer & explanation

Answer: D

cGMP rises → Na⁺ channels open → the rod depolarizes → glutamate release increases: Treats the rod like a typical excitatory receptor; photoreceptors respond to light by hyperpolarizing.

cGMP falls → Na⁺ channels close → the rod hyperpolarizes → glutamate release increases: The first three steps are right, but hyperpolarization reduces, not increases, transmitter release.

cGMP falls → K⁺ channels close → the rod depolarizes → glutamate release increases: The light-sensitive channels are cGMP-gated cation (Na⁺) channels, and their closing hyperpolarizes the cell.

cGMP falls → Na⁺ channels close → the rod hyperpolarizes → glutamate release decreases: Correct. Activated rhodopsin turns on transducin and phosphodiesterase; cGMP falls, cGMP-gated Na⁺ channels close, the dark current stops, the rod hyperpolarizes and releases less glutamate (Structure and Function of Systems: nervous/sensory).

all-trans-retinal becomes 11-cis → cGMP falls → the rod hyperpolarizes → glutamate release decreases: Reverses the isomerization: light converts 11-cis-retinal to all-trans-retinal.

Photoreceptors are depolarized in the dark (dark current) and hyperpolarize in light, which reduces their glutamate release.

Question 3

Pancreatic acinar cells are given a 3-minute pulse of a radiolabeled amino acid, then chased with unlabeled amino acid, with or without drug Q. The table shows where the label is found. Which step does drug Q most likely block?

Percentage of cellular radiolabel in each compartment
Chase time (min)ConditionRough ERGolgiSecretory vesicles
5No drug85105
20No drug256015
60No drug52075
5Drug Q8794
20Drug Q85105
60Drug Q84106
Total incorporation of label was the same with and without drug Q.
Show answer & explanation

Answer: E

Binding of the signal recognition particle to the signal peptide: If cotranslational targeting failed, labeled protein would not enter the rough ER at all; here 85–87% is in the ER.

Translation of the secretory protein on ribosomes: Total incorporation was unchanged, so protein synthesis is not inhibited.

Packaging of protein from the trans-Golgi into secretory vesicles: A Golgi exit block would make label pile up in the Golgi; under drug Q the Golgi never fills.

Fusion of secretory vesicles with the plasma membrane: An exocytosis block would trap label in secretory vesicles, the last compartment, not in the ER.

Budding of transport vesicles from the ER toward the Golgi: Correct. With drug Q the label enters the rough ER normally but stays there, so the block is in ER-to-Golgi vesicle traffic (Cell and Molecular Biology: experimental cell biology; endomembrane system).

In a pulse-chase experiment, label accumulates in the last compartment it can reach; the blocked step lies just after that compartment.

General Chemistry

Question 4

A galvanic cell is built from Ag⁺/Ag (E° = +0.80 V) and Ni²⁺/Ni (E° = −0.25 V) half-cells, with the cell reaction 2Ag⁺ + Ni → 2Ag + Ni²⁺. Using F = 96,500 C/mol, what is ΔG° for the reaction as written?

Show answer & explanation

Answer: C

  1. Cathode: Ag⁺ + e⁻ → Ag (+0.80 V); anode: Ni → Ni²⁺ + 2e⁻ (−0.25 V).
  2. E°cell = 0.80 − (−0.25) = +1.05 V (not multiplied by 2).
  3. n = 2 mol e⁻ for 2Ag⁺ + Ni.
  4. ΔG° = −nFE° = −2 × 96,500 × 1.05 = −202,650 J ≈ −203 kJ.

−405 kJ: Multiplies the Ag⁺/Ag potential by 2 as well as using n = 2: E° does not scale with coefficients (E° = 1.05 V, not 2.10 V).

−357 kJ: Uses E° = 2(0.80) + 0.25 = 1.85 V by doubling the Ag potential; E°cell = 0.80 − (−0.25) = 1.05 V.

−203 kJ: Correct. E°cell = 0.80 − (−0.25) = +1.05 V, and 2 electrons transfer (Ni → Ni²⁺ + 2e⁻). ΔG° = −nFE° = −(2)(96,500)(1.05) J ≈ −2.03 × 10⁵ J = −203 kJ; negative, as a galvanic cell requires (Oxidation-Reduction Reactions: electrochemical calculations).

−101 kJ: Uses n = 1 (one electron per Ag⁺) instead of the 2 electrons transferred in the reaction as written.

+203 kJ: Takes E° as anode − cathode (−1.05 V), giving a positive ΔG°; a galvanic cell is spontaneous.

E°cell = E°cathode − E°anode (never scaled by coefficients); ΔG° = −nFE° uses the electrons transferred in the equation as written.

Question 5

Fluorine is more electronegative than oxygen, yet water (bp 100 °C) boils at a much higher temperature than hydrogen fluoride (bp 20 °C). Which statement best explains this?

Show answer & explanation

Answer: E

The O–H bond is more polar than the H–F bond: Reverses electronegativity: H–F is the more polar bond.

H₂O has the larger molar mass and stronger London forces: HF (20 g/mol) is slightly heavier than H₂O (18 g/mol).

HF molecules are nonpolar and cannot form hydrogen bonds: HF is very polar and hydrogen-bonds strongly; it just forms fewer bonds per molecule.

Water has the larger dipole, which alone sets the boiling point: The dipole moments of H₂O and HF are almost the same; the number of hydrogen bonds is what differs.

Water forms about twice as many hydrogen bonds per molecule as HF: Correct. HF has one H to donate, so its hydrogen bonds form chains; water has two H atoms and two lone pairs, giving an extended network with about twice as many hydrogen bonds per molecule (Liquids and Solids: intermolecular forces).

Boiling point depends on both the strength and the number of intermolecular attractions per molecule.

Organic Chemistry

Question 6

An unknown compound gives the data in the table. Which compound is it?

Spectroscopic data for the unknown
MeasurementResult
Molecular formulaC₄H₈O₂
IRStrong band at 1740 cm⁻¹; no broad band at 2500–3300 cm⁻¹ or near 3400 cm⁻¹
¹H NMRδ 4.12 (2H, quartet); δ 2.05 (3H, singlet); δ 1.26 (3H, triplet)
Show answer & explanation

Answer: B

Methyl propanoate: Also an ester, but its CH₃O singlet would be near 3.7 and its CH₂ quartet near 2.3, next to the carbonyl.

Ethyl acetate: Correct. 1740 cm⁻¹ with no O–H band indicates an ester. The quartet at 4.1 is an O–CH₂ next to CH₃ (triplet at 1.3), and the singlet at 2.0 is CH₃ on the carbonyl: CH₃CO₂CH₂CH₃ (Chemical & Physical Properties: spectroscopy).

Butanoic acid: A carboxylic acid would show a broad O–H band at 2500–3300 cm⁻¹ and a signal near 11–12 ppm.

Propyl formate: A formate ester has a one-proton singlet near 8.0 ppm and a three-carbon chain (triplet, sextet, triplet).

3-Hydroxybutan-2-one: Has an O–H band near 3400 cm⁻¹ and a ketone C=O near 1715 cm⁻¹.

In ¹H NMR, an O–CH₂ appears near 4 ppm and a CH₂ next to C=O near 2.3 ppm; this places the ester oxygen.

Question 7

Methoxide ion reacts with acetyl chloride. Which species is formed directly by the electron movement shown by the two curved arrows?

Curved arrows for one mechanistic step
CH₃O−COH₃CCl
Full-headed curved arrows show the movement of electron pairs; lone pairs are not drawn.
Show answer & explanation

Answer: E

Methyl acetate (CH₃CO₂CH₃) and chloride ion: This is the overall product after the intermediate collapses, but expelling Cl⁻ needs a further arrow not drawn here.

Acetate ion and chloromethane: This would require attack at the methyl group of a different molecule; no arrow points to a methyl carbon.

An acylium ion, CH₃C≡O⁺, and methoxide: Forming an acylium ion needs an arrow from the C–Cl bond to Cl; the arrows instead add a nucleophile.

An enolate, CH₂=C(O⁻)Cl, and methanol: Would need an arrow from methoxide to an α-hydrogen; the arrow goes to the carbonyl carbon.

The tetrahedral alkoxide CH₃C(O⁻)(Cl)OCH₃: Correct. The first arrow forms a new O–C bond from the methoxide lone pair; the second moves the C=O π electrons onto oxygen. Nothing breaks the C–Cl bond, so the result is the tetrahedral intermediate (Mechanisms: curved arrows, addition).

Product-from-arrows items: make and break only the bonds the arrows show. Do not jump ahead to the overall product.

Reading Comprehension

Question 8

According to the passage, a glass photochromic lens clears when the light is removed because

Lenses That Respond to Light

1Photochromic spectacle lenses darken in bright sunlight and clear again indoors, sparing the wearer from switching between ordinary glasses and sunglasses. The word photochromic describes any material whose color changes reversibly when it absorbs light. Although the effect seems simple, it depends on a balance between two competing processes, and understanding that balance explains both the strengths and the familiar frustrations of these lenses.

2The first commercially successful photochromic lenses, introduced in the 1960s, were made of glass containing a great many microscopic crystals of silver halide, the same light-sensitive compound used in photographic film. When ultraviolet light strikes a crystal, it frees electrons that convert some silver ions into tiny clusters of metallic silver. These clusters absorb visible light, and the lens darkens. In photographic film the change is made permanent by chemical processing. In a lens, however, the halogen atoms released by the light remain trapped in the surrounding glass, close to where they were formed. When the light is removed, they recombine with the silver, the clusters disappear, and the lens clears. Small amounts of copper added to the glass help this reverse reaction proceed.

3Most lenses sold today are plastic, and plastic cannot hold silver halide crystals in the same way. Instead, plastic photochromic lenses contain organic dye molecules dissolved in the lens material or in a thin layer near its surface. In their resting form, these molecules absorb only ultraviolet light and are therefore colorless. Absorbing an ultraviolet photon causes a ring of atoms within the molecule to break open, and the molecule rearranges into a flatter, more extended shape that absorbs visible light. The opened form is less stable than the closed one, and with time it snaps back to its original shape, restoring the clear state.

4In both kinds of lens, the darkening reaction is driven by light, while the reverse reaction, at least in the most common designs, is driven mainly by heat. At any moment, molecules or silver clusters are being switched on by ultraviolet light and switched off by thermal motion. Under steady conditions the two processes reach a balance, and the darkness of the lens reflects that balance rather than the brightness of the light alone.

5This explains why temperature matters so much. On a cold, bright winter day, the reverse reaction is slow, so the lens becomes very dark; on a hot summer beach, with equally strong sunlight, the faster reverse reaction keeps the lens noticeably lighter. The same principle governs how quickly lenses clear. A wearer who walks from bright snow into a warm building may wait several minutes for cold lenses to fade, whereas lenses that are already warm clear more quickly. Manufacturers try to choose dyes that darken strongly yet fade quickly, but the two goals pull in opposite directions: a molecule that returns readily to its clear form also tends to darken less in the sun.

6Another common complaint concerns driving. Most automobile windshields block nearly all of the ultraviolet light from the sun while transmitting most visible light. Because conventional photochromic lenses are activated by ultraviolet light, they darken only slightly inside a car, even on a bright day. Some newer designs use dyes that also respond to the shortest wavelengths of visible light, allowing partial darkening behind a windshield.

7Finally, photochromic dyes do not last indefinitely. Each switching cycle carries a small chance that a molecule will undergo an unwanted side reaction and permanently lose its ability to switch, a process called fatigue. Over years of use, plastic lenses may darken less fully or retain a slight residual tint. The silver halide system in glass lenses, by contrast, is regenerated with each cycle and is highly resistant to fatigue, one reason glass photochromic lenses were long valued despite being heavier than plastic ones.

Show answer & explanation

Answer: E

copper atoms absorb the visible light that the silver clusters had been absorbing: Paragraph 2 says copper helps the reverse reaction proceed, not that it absorbs light.

chemical processing removes the halogen from the glass: Chemical processing is what makes the change permanent in photographic film, not in a lens.

a ring of atoms in each dye molecule closes again: That is the mechanism of plastic lenses (paragraph 3).

the glass blocks ultraviolet light once it has darkened: Not stated; clearing is attributed to recombination, not to blocked ultraviolet light.

halogen atoms trapped near the silver recombine with it: Correct. Paragraph 2: the halogen atoms "remain trapped in the surrounding glass" and "recombine with the silver" when the light is removed.

When a passage contrasts two systems (film vs. lens, glass vs. plastic), check that the answer belongs to the system asked about.

Question 9

Which of the following best expresses the main point of the passage?

Strength in Numbers

1Most laboratory studies of bacteria have used cells swimming freely in liquid broth, a condition described as planktonic. In nature, however, many bacteria spend much of their existence attached to surfaces, living in dense communities called biofilms. A biofilm forms when free cells settle on a surface, multiply, and surround themselves with a slimy matrix of their own making, composed largely of polysaccharides, proteins, and DNA released from the cells. The film on unbrushed teeth, the slippery coating on river stones, and the deposits inside neglected drains are all biofilms.

2Biofilms are of more than academic interest because they readily form on medical devices. Urinary catheters, artificial heart valves, joint replacements, and contact lenses all provide surfaces on which bacteria can settle, and infections associated with these devices are notoriously difficult to cure. Bacteria living in a biofilm can survive antibiotic concentrations many times higher, in some cases a thousand times higher, than those that kill the same bacteria in planktonic form. Often the only reliable remedy is to remove the device entirely.

3Remarkably, this protection usually does not depend on genetic resistance. When bacteria are scraped from a biofilm and grown again as free cells, they are typically about as susceptible to antibiotics as the original population. Microbiologists therefore distinguish resistance, an inherited ability to grow in the presence of a drug, from tolerance, the ability to survive exposure to a drug for a time without being able to grow in it. Biofilm protection is largely a matter of tolerance, and several mechanisms contribute to it.

4The first is the matrix itself. For some antibiotics, especially those that carry an electric charge, the matrix acts as a barrier, binding or slowing the drug so that cells deep in the film are exposed to a lower concentration, or are exposed later, than cells at the surface. Yet this cannot be the whole story, since many antibiotics penetrate biofilms readily and still fail to kill the cells within.

5A second mechanism arises from the structure of the community. Cells near the surface of a biofilm have ample access to oxygen and nutrients, but those in the interior receive only what diffuses past their neighbors. As a result, much of a mature biofilm consists of cells growing slowly or not at all. Many antibiotics attack processes that are most active in growing cells, such as building the cell wall or making new proteins, so slow-growing cells are hard to kill.

6The third mechanism involves persisters, a small fraction of cells that enter a dormant state in which the targets of most drugs are largely inactive. Persisters occur in planktonic cultures too, but in a biofilm they are shielded by the matrix from the cells and antibodies of the immune system, which would otherwise clear them, so they can outlast a course of treatment. When the antibiotic is withdrawn, persisters can resume growing and rebuild the population, which helps explain why device infections often return after apparently successful therapy. Because persisters are not genetically different from their neighbors, their descendants are once again mostly susceptible to the drug.

7These findings have practical consequences. The standard laboratory tests that guide the choice of an antibiotic measure the lowest concentration of the drug that prevents the growth of planktonic cells. Such tests can suggest that an infection should respond to a drug that in fact fails against the biofilm in the patient's body. Researchers are therefore developing tests that grow bacteria as biofilms before exposing them to drugs, as well as treatments that break up the matrix, interfere with the chemical signals bacteria use to coordinate biofilm formation, or rouse persisters so that conventional antibiotics can act on them.

8For now, prevention remains the most effective strategy. In eye care, for example, contact lens storage cases are a common site of biofilm growth, and contaminated cases have been linked to serious corneal infections. Practitioners advise rubbing and rinsing lenses, emptying and air-drying cases between uses, and replacing cases regularly, measures intended to remove young biofilms before they can mature.

Show answer & explanation

Answer: B

Bacteria in biofilms survive antibiotics mainly by acquiring resistance genes from neighbors.: Reversal: paragraph 3 says the protection "usually does not depend on genetic resistance."

Bacteria in biofilms tolerate antibiotics through several mechanisms that do not require genetic resistance.: Correct. Paragraph 3 states the thesis and paragraphs 4–6 give the three mechanisms; paragraphs 7–8 draw consequences.

The biofilm matrix prevents antibiotics from reaching the cells inside the film.: One mechanism only, and overstated: paragraph 4 says many antibiotics penetrate readily.

Medical devices should be avoided because bacteria colonize them readily.: Overstatement; the passage recommends prevention measures, not avoiding devices.

Standard antibiotic tests are more accurate for biofilm infections than once thought.: Reversal of paragraph 7, which says the tests can mislead.

The main point usually appears where the passage turns from description to argument (here, paragraph 3).

Question 10

Which of the following best states the main idea of the passage?

How the Eye Learns Its Length

1Few organs change as much after birth as the eye. A newborn's eyeball is only about 70 percent of its adult length, and because it is so short, the image formed by its cornea and lens typically falls slightly behind the retina. Most infants are therefore mildly farsighted, or hyperopic. Over the first several years of life the eye lengthens, while the cornea and the crystalline lens become flatter and optically weaker. These changes are not independent of one another. Across a large group of children, refractive errors that are widely scattered at birth narrow toward a common value near zero, a state in which distant objects are focused sharply on the retina without effort. Vision scientists call this coordinated process emmetropization, and for much of the twentieth century it was regarded as a largely inherited program that simply unfolded on schedule.

2That view began to change when researchers noticed what happens to eyes that are denied a clear image early in life. Children born with a dense cataract, or with a drooping eyelid that covers the pupil, often develop eyes that grow too long, so that the image falls in front of the retina and distant objects appear blurred. This condition, nearsightedness or myopia, could also be produced deliberately in laboratory animals. When a translucent diffuser was placed over one eye of a young chick, that eye elongated rapidly and became measurably myopic within days, while the uncovered eye developed normally. The finding implied that eye growth was not merely programmed; it was steered by the quality of the visual image.

3Experiments with lenses showed how precise this steering could be. A negative, or diverging, lens moves the focused image backward, behind the retina. Chicks fitted with such lenses responded by accelerating the growth of the eye until the retina had moved back to meet the image. A positive, or converging, lens has the opposite effect, pulling the image forward, and chicks wearing positive lenses slowed their eye growth until the mismatch was again eliminated. Because the eye corrected errors of either sign, it could not simply be responding to the overall amount of blur, which is similar for images in front of and behind the retina. Somehow the developing eye detects the direction of defocus and adjusts its growth accordingly. Similar compensation has since been demonstrated in tree shrews, marmosets, and rhesus monkeys, although mammals respond more slowly than birds.

4Where in the visual system is this decision made? The obvious candidate was the brain, which interprets blur in conscious vision and controls the focusing muscles of the eye. Two lines of evidence pointed elsewhere. First, when a diffuser covered only the upper or lower half of the visual field, only the corresponding half of the eye grew abnormally, producing a lopsided globe. A signal routed through the brain would be expected to affect the whole eye. Second, chicks whose optic nerves had been cut, severing communication between eye and brain, still compensated for lenses in roughly the right direction. The analysis of the image, and the command to grow, appear to originate largely within the retina itself.

5The retina, however, does not lengthen the eye directly. It sends chemical signals outward through the layers behind it. The first tissue to respond is the choroid, a spongy, blood-rich layer between the retina and the tough outer coat of the eye, the sclera. In chicks wearing positive lenses, the choroid can thicken substantially within hours, physically pushing the retina forward toward the image; with negative lenses it thins. These choroidal shifts are a quick but partial correction. The lasting change comes more slowly, as cells in the sclera alter the production and breakdown of the structural molecules that make up the wall of the eye, allowing the wall to expand or restraining its growth over a period of weeks.

6Among the candidate messengers, dopamine has received particular attention. Certain retinal neurons release dopamine in response to light, and its release follows a daily rhythm, rising during the day. In chicks, the dopamine content of the retina falls in eyes that are deprived of clear vision, and drugs that activate dopamine receptors can suppress the excessive growth that a diffuser would otherwise cause. These observations do not prove that dopamine is the only stop signal for eye growth, but they made it a natural link between the visual environment and the machinery of growth.

7The question has become pressing because myopia is no longer uncommon. In parts of East Asia, surveys of young adults finishing secondary school have found that as many as eight or nine in ten are myopic, compared with a small minority a few generations earlier. Prevalence has also climbed in Europe and North America, though less steeply. Change this rapid, occurring within the span of a few generations, cannot be explained by shifts in the gene pool. Genes clearly influence individual susceptibility, since children of myopic parents are at higher risk, but the speed of the increase points to changes in how children live. Two environmental factors stand out in study after study: the intensity of education, with its long hours of reading and close work, and the amount of time children spend outdoors.

8Of the two, time outdoors has proved the more surprising. Children who spend more time outside are less likely to become myopic, and the association remains even when researchers account for how much reading and close work the children do. Nor is it explained by exercise, since time spent playing sports indoors shows little protective effect. School-based trials, in which classes were given extra time outdoors each day, found fewer new cases of myopia among the participating children than among children in comparison schools. The leading explanation is light intensity. Even in shade, daylight commonly provides ten or more times the illumination found in a typical classroom, and in direct sun the difference can exceed a hundredfold. In animal studies, chicks reared under very bright light develop less deprivation myopia than chicks kept under ordinary laboratory lighting, and this protection is weakened when a drug that blocks dopamine receptors is given, consistent with the idea that bright light acts through retinal dopamine.

9Light is not the only candidate. Outdoors, most objects are far away, so images across the whole visual field lie near a common focus. Indoors, nearby walls, furniture, and books place parts of the scene at very different distances, and some researchers argue that the resulting pattern of defocus across the retina, especially in its periphery, encourages the eye to grow. The two explanations are not mutually exclusive, and both may contribute. What does seem clear is that outdoor time works best before myopia begins: it reduces the number of children who become myopic but appears to do comparatively little to slow the condition once it is established.

10For children who are already myopic, other approaches aim to slow progression. Eye drops containing a low concentration of atropine have reduced the rate of progression in clinical trials. Atropine blocks muscarinic acetylcholine receptors and, at high doses, paralyzes the muscle that focuses the eye for near vision, so it was once assumed to work by relieving the strain of close work. That explanation has lost favor, partly because atropine also inhibits experimental myopia in chicks, whose focusing muscle is controlled through a different type of acetylcholine receptor that atropine does not block. Its site of action is now thought to lie in the retina or in the tissues behind it. Specially designed spectacle lenses and contact lenses take yet another approach, keeping central vision sharp while deliberately placing images in front of the peripheral retina, in effect delivering the same "slow down" signal that a positive lens gave the chicks.

11These efforts matter because myopia is more than an inconvenience solved by glasses. Glasses and contact lenses correct the optical error but do not change the shape of the eye, and an eye that has grown excessively long has a stretched, thinned retina and sclera. Highly myopic adults face substantially higher lifetime risks of retinal detachment, glaucoma, and degeneration of the central retina, conditions that can permanently impair sight. Viewed in this light, myopia is less a simple refractive error than a disorder of growth, one whose course is shaped in childhood by how much light, and what kind of visual world, the developing eye receives.

Show answer & explanation

Answer: D

Myopia is mainly an inherited condition, and its apparent rise reflects better diagnosis.: Reverses paragraph 7, which says the rise is too fast to be genetic and points to changes in how children live.

Animal studies with lenses have shown that the brain controls the growth of the eye.: Reverses paragraph 4, which concludes that the growth command originates largely within the retina.

Low-dose atropine and special lenses have made childhood myopia a fully preventable disorder.: Overstatement: paragraph 10 says these treatments slow progression; nothing says myopia is fully preventable.

Visual feedback steers eye growth, and changes in children's visual world help explain rising myopia.: Correct. The passage moves from emmetropization as image-guided growth (paragraphs 1–6) to environmental causes of the myopia increase (7–9) and growth-based treatments (10–11).

Children should spend more time outdoors because sunlight raises dopamine levels throughout the body.: Too narrow and distorted: the passage discusses retinal dopamine only, and outdoor time is one topic among several.

A main-idea answer must cover the whole passage, not one paragraph, and must not overstate the author's claims.

Physics

Question 11

In the circuit shown, a 12 V battery with negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel combination of 6.0 Ω and 3.0 Ω resistors. What current flows through the battery?

Series-parallel circuit
+12 VR₁ = 2.0 Ω6.0 Ω3.0 Ω
R₁ is in series with the 6.0 Ω and 3.0 Ω resistors, which are in parallel.
Show answer & explanation

Answer: D

  1. R_parallel = (6.0 × 3.0)/(6.0 + 3.0) = 18/9 = 2.0 Ω.
  2. R_total = 2.0 + 2.0 = 4.0 Ω.
  3. I = V/R = 12/4.0 = 3.0 A.
  4. Check: the parallel pair is smaller than its smaller resistor (2.0 < 3.0 Ω) ✓.

1.1 A: Adds all three resistors in series (11 Ω).

1.5 A: Leaves out the 3.0 Ω branch (12 V / 8.0 Ω).

2.4 A: Leaves out the 6.0 Ω branch (12 V / 5.0 Ω).

3.0 A: Correct. Parallel: (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω; total 2.0 + 2.0 = 4.0 Ω; I = 12/4.0 = 3.0 A.

6.0 A: Uses only the parallel combination (12 V / 2.0 Ω) and ignores R₁.

Reduce parallel groups first (product over sum for two resistors), then add series resistances and apply Ohm's law.

Question 12

A 0.015 kg bullet moving horizontally at 400 m/s embeds itself in a 1.985 kg wooden block hanging at rest from a light string. To what maximum height above its starting point does the block rise? (g = 10 m/s²)

Show answer & explanation

Answer: C

  1. Collision (momentum conserved): (0.015)(400) = (0.015 + 1.985)v → 6.0 = 2.0v → v = 3.0 m/s.
  2. Swing (mechanical energy conserved): ½(2.0)(3.0)² = (2.0)(10)h → 9.0 = 20h.
  3. h = 0.45 m.
  4. Check: KE lost in the collision is 1200 − 9 J, as expected for a perfectly inelastic collision.

0.15 m: Forgets to square the speed: uses v/(2g) instead of v²/(2g).

0.30 m: Sets gh equal to v (mixing momentum with energy) instead of ½v².

0.45 m: Correct. Momentum: 0.015 × 400 = 2.0 v → v = 3.0 m/s; then ½v² = gh → h = 9.0/20 = 0.45 m.

0.90 m: Drops the ½ in kinetic energy: h = v²/g.

60 m: Assumes kinetic energy is conserved in the collision (½ × 0.015 × 400² = 1200 J); the embedding collision is perfectly inelastic.

Use momentum for the collision and energy only for the swing afterward; kinetic energy is not conserved when objects stick together.

Question 13

A 2.0 kg block slides down a 37° incline, as shown. The coefficient of kinetic friction is 0.25. What is the magnitude of the block's acceleration? (g = 10 m/s², sin 37° = 0.60, cos 37° = 0.80)

Block on an incline
mgNf37°μk = 0.25
The block slides down the slope; friction f acts up the slope.
Show answer & explanation

Answer: C

  1. Along the slope: mg sin θ = 2.0 × 10 × 0.60 = 12 N (down the slope).
  2. N = mg cos θ = 2.0 × 10 × 0.80 = 16 N; f = μk N = 0.25 × 16 = 4.0 N (up the slope).
  3. F_net = 12 − 4.0 = 8.0 N; a = 8.0/2.0 = 4.0 m/s².
  4. Check: less than the frictionless value g sin θ = 6.0 m/s².

2.0 m/s²: This is only the friction deceleration μk g cos θ; it omits the gravity component down the slope.

3.5 m/s²: Uses friction = μk·mg (5 N) instead of μk·mg cos θ (4 N).

4.0 m/s²: Correct. a = g sin θ − μk g cos θ = 10(0.60) − 0.25(10)(0.80) = 6.0 − 2.0 = 4.0 m/s².

6.0 m/s²: Ignores friction: this is g sin θ alone.

8.0 m/s²: Uses cos θ for the component of gravity along the slope and ignores friction.

On an incline, gravity's component along the slope uses sin θ and the normal force uses cos θ; friction is μk times the normal force.

Quantitative Reasoning

Question 14

Tolliver Outfitters raised the price of a jacket by 20% and later reduced the new price by 25%. The jacket now sells for $72. What was its original price?

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Answer: B

  1. Final = p × 1.20 × 0.75 = 0.90p.
  2. 0.90p = 72 → p = 80.
  3. Check: 80 × 1.20 = 96; 96 × 0.75 = 72 ✓.

$75.79: Treats the changes as a net 5% decrease (72/0.95); successive percents multiply.

$80.00: Correct. p × 1.20 × 0.75 = 0.90p = 72 → p = $80.00.

$86.40: Applies the 20% increase to $72 instead of undoing the changes.

$90.00: Undoes the 25% cut by adding 25% to $72 (72 × 1.25).

$96.00: Undoes only the 25% decrease (72/0.75) and ignores the earlier increase.

Successive percent changes multiply (1.20 × 0.75 = 0.90); undo them by dividing by the combined factor.

Question 15

Use the table. Which clinic had the greatest percent increase in eye examinations from 2024 to 2025?

Eye examinations performed at four clinics
Clinic20242025
Norland1,2501,480
Quarry Hill9801,176
Ridgeview1,6001,520
Westbrook720846
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Answer: B

  1. Norland: 230/1,250 = 18.4%.
  2. Quarry Hill: 196/980 = 20.0%.
  3. Ridgeview: −80/1,600 = −5.0%.
  4. Westbrook: 126/720 = 17.5%.
  5. Greatest: Quarry Hill.

Norland: Largest increase in number (+230), but 230/1,250 = 18.4%, less than Quarry Hill's.

Quarry Hill: Correct. 196/980 = 20.0%, the largest percent increase (Norland 18.4%, Westbrook 17.5%, Ridgeview −5.0%).

Ridgeview: Largest totals, but its examinations fell (−80, −5.0%).

Westbrook: 126/720 = 17.5%, below Quarry Hill's 20.0%.

Quarry Hill and Westbrook equally: The percentages differ (20.0% vs 17.5%); rounding both to "about 20%" hides the difference.

Percent change = change ÷ original value; the biggest absolute change need not be the biggest percent change.

Question 16

At Mercer Eye Clinic, 40% of patients are under 30 years old. Of the patients under 30, 25% wear contact lenses; of the patients 30 or older, 10% wear contact lenses. If a randomly chosen patient wears contact lenses, what is the probability that the patient is under 30?

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Answer: E

  1. P(under 30 and contacts) = 0.40 × 0.25 = 0.10.
  2. P(30+ and contacts) = 0.60 × 0.10 = 0.06.
  3. P(contacts) = 0.10 + 0.06 = 0.16.
  4. P(under 30 | contacts) = 0.10/0.16 = 0.625.

0.10: This is P(under 30 and contacts) = 0.40 × 0.25, not the conditional probability.

0.16: This is P(contacts) overall, the denominator of the answer.

0.25: This is P(contacts | under 30), the reverse condition.

0.40: This is the unconditional P(under 30); it ignores the information that the patient wears contacts.

0.625: Correct. P(under 30 | contacts) = 0.10/(0.10 + 0.06) = 0.10/0.16 = 0.625.

P(A | B) = P(A and B)/P(B); find P(B) by adding the joint probabilities from every group.

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