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PASSAGE (Questions 26–32 — same passage as Q26): The human immune system is divided into two broad arms: innate immunity and adaptive immunity. Innate immunity provides the first, rapid line of defense. It is nonspecific—it responds the same way to any perceived threat—and does not improve with repeated exposure. Key components include physical barriers (skin, mucous membranes), phagocytic cells (neutrophils, macrophages, dendritic cells), natural killer (NK) cells, and the complement system. Pathogen recognition is mediated by pattern recognition receptors (PRRs), notably Toll-like receptors (TLRs), which detect conserved microbial features called pathogen-associated molecular patterns (PAMPs). Upon TLR activation, cells produce pro-inflammatory cytokines (e.g., IL-1beta, TNF-alpha, IL-6) that recruit additional immune cells and induce fever. Adaptive immunity is slower to mount (days to weeks) but is highly specific and develops immunological memory. It is mediated by lymphocytes: B cells and T cells. B cells, when activated, differentiate into plasma cells that secrete antibodies. Antibodies bind antigens with high specificity via their variable regions, facilitating neutralization, opsonization, and complement activation. T cells are subdivided into helper T cells (CD4+) and cytotoxic T cells (CD8+). CD4+ T cells secrete cytokines that coordinate both humoral (antibody-mediated) and cell-mediated immune responses, while CD8+ T cells directly kill infected or transformed cells expressing foreign peptides on MHC class I molecules. Immunological memory—the basis of vaccination—arises from the production of long-lived memory B and T cells after initial antigen exposure. Upon re-exposure to the same antigen, memory cells mount a faster, stronger, and more prolonged secondary immune response compared to the primary response. Vaccines exploit this principle by exposing the immune system to an antigen (or its mimic) without causing disease, generating protective memory. The concept of herd immunity relies on a sufficient proportion of a population being immune—either through vaccination or prior infection—to interrupt pathogen transmission and protect susceptible individuals. Question 27: According to the passage, which cells directly kill infected cells displaying foreign peptides on MHC class I molecules?
A
B
C
D
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2026 Statistics
Key Facts: OAT Reading Comprehension Exam
50 questions across 3 passages
OAT Reading Comprehension question count
2026 OAT Candidate Guide (ADA)
60 minutes
Time allowed for OAT Reading Comprehension
2026 OAT Candidate Guide (ADA)
1,100–1,500 words per passage
OAT Reading Comprehension passage length
OAT User's Guide (ADA/ASCO, 2025)
200–400 scale; median 300
OAT Reading Comprehension score range
OAT Candidate Guide (ADA)
$520 full OAT exam fee
OAT examination cost (2025–2026)
Shorelight / ASCO official sources
No prior topic knowledge required
OAT Reading Comprehension passage prerequisite
2026 OAT Candidate Guide (ADA)
The OAT Reading Comprehension Test contains 3 scientific passages of approximately 1,100–1,500 words each, followed by a total of 50 questions to be completed in 60 minutes (per the 2026 OAT Candidate Guide). The section is scored on a 200–400 scale with a national median of 300. The test is administered by the ADA on behalf of ASCO and is required for admission to all U.S. optometry schools and the University of Waterloo, Canada. Prior knowledge of the passage topic is explicitly not a prerequisite according to official guidelines.
Sample OAT Reading Comprehension Practice Questions
Try these sample questions to test your OAT Reading Comprehension exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1PASSAGE (Questions 1–6): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 1: According to the passage, what direct effect does phototransduction have on the photoreceptor cell's membrane potential?
A.It depolarizes the cell by opening additional sodium channels
B.It hyperpolarizes the cell by causing sodium channels to close
C.It hyperpolarizes the cell by opening potassium channels
D.It depolarizes the cell by increasing cGMP concentration
Explanation: The passage states that the reduction in cGMP causes sodium channel closure, which hyperpolarizes the cell. This is the direct membrane effect described in the phototransduction cascade.
2PASSAGE (Questions 1–6 — same passage as Q1): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 2: What is the primary purpose of the foveal pit as described in the passage?
A.To concentrate rod photoreceptors for enhanced dim-light sensitivity
B.To route the optic nerve fibers away from the line of sight
C.To reduce light scattering by removing overlying retinal neurons
D.To house all three cone subtypes in a single compact region
Explanation: The passage directly states that the foveal pit is a thinning caused by the absence of overlying retinal neurons, and this minimizes light scattering while maximizing resolution. The question asks for the 'primary purpose' as stated in the passage.
3PASSAGE (Questions 1–6 — same passage as Q1): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 3: Based on the passage, a person with a severe vitamin A deficiency would MOST likely experience difficulty with which of the following?
A.Color discrimination under bright lighting conditions
B.Distinguishing fine details at the center of the visual field
C.Vision in low-light (scotopic) conditions
D.Processing visual signals in the occipital lobe
Explanation: The passage states that retinal, the chromophore in rhodopsin (the rod photopigment), is a derivative of vitamin A. Rods are responsible for scotopic (low-light) vision. Without sufficient vitamin A, rhodopsin cannot function normally, impairing dim-light vision first.
4PASSAGE (Questions 1–6 — same passage as Q1): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 4: Which of the following best describes the main idea of the passage?
A.The fovea is the most important structure in the human eye because it provides the highest visual acuity.
B.The passage explains the anatomy of retinal photoreceptors, the biochemical mechanism of phototransduction, and the key structural features of the retina.
C.Vitamin A is essential for photoreception, and deficiency leads to blindness.
D.Rods are more numerous than cones and therefore play a more important role in human vision.
Explanation: The passage covers three interrelated topics: photoreceptor anatomy (rods and cones), the phototransduction cascade (rhodopsin, cGMP, hyperpolarization), and key retinal structures (fovea, optic disc, visual pathway). Option B accurately captures this broad scope.
5PASSAGE (Questions 1–6 — same passage as Q1): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 5: As used in the passage, the term 'chromophore' most nearly means:
A.A protein that forms the structural backbone of a photopigment
B.A light-absorbing molecule responsible for initiating the visual response
C.An enzyme that catalyzes the breakdown of cyclic GMP
D.A G-protein intermediate in the phototransduction cascade
Explanation: The passage introduces retinal as 'a light-sensitive chromophore,' stating it absorbs photons and isomerizes to initiate the signaling cascade. From context, 'chromophore' means a light-absorbing molecule that starts the visual response.
6PASSAGE (Questions 1–6 — same passage as Q1): The human retina contains two primary types of photoreceptor cells: rods and cones. Rods, numbering approximately 120 million per retina, are exquisitely sensitive to dim light and are responsible for scotopic (low-light) vision. They are densely packed in the peripheral retina but notably absent from the fovea centralis. Cones, of which there are roughly 6–7 million per retina, are concentrated in the fovea and are responsible for photopic (bright-light) vision and color discrimination. Three subtypes of cones exist—S-cones (short wavelength, sensitive to blue light), M-cones (medium wavelength, sensitive to green light), and L-cones (long wavelength, sensitive to red light)—enabling trichromatic color vision in humans. Phototransduction begins when light strikes a photopigment molecule. In rods, the photopigment is rhodopsin, composed of the protein opsin bound to a light-sensitive chromophore called retinal (a derivative of vitamin A). Absorption of a photon isomerizes retinal from the 11-cis conformation to the all-trans conformation, initiating a G-protein signaling cascade. This activates transducin, which in turn activates phosphodiesterase (PDE). PDE hydrolyzes cyclic GMP (cGMP), lowering its intracellular concentration. Because cGMP normally holds sodium channels open, the reduction in cGMP causes channel closure, hyperpolarizing the cell. This hyperpolarization reduces glutamate release from the photoreceptor's synaptic terminal, signaling the onset of light to downstream bipolar and ganglion cells. The fovea, roughly 1.5 mm in diameter, is the region of highest visual acuity. Its center, the foveola, contains only cones with no overlying retinal neurons—a thinning called the foveal pit—which minimizes light scattering and maximizes resolution. The optic disc, where the optic nerve exits the eye, contains no photoreceptors and corresponds to the physiological blind spot. The neural signals from retinal ganglion cells travel along the optic nerve to the lateral geniculate nucleus (LGN) of the thalamus, then to the primary visual cortex (V1) in the occipital lobe for further processing. Question 6: The author organizes the passage primarily by:
A.Presenting a problem with current understanding and then proposing a solution
B.Moving from the cellular level of photoreceptors outward to the anatomical and neural pathway levels
C.Comparing and contrasting the properties of rods and cones throughout each paragraph
D.Listing the major diseases that affect each retinal structure in order of clinical importance
Explanation: The passage begins at the cellular level (rod vs. cone subtypes), then zooms into the molecular mechanism of phototransduction, and finally describes the anatomical structures (fovea, optic disc) and the visual neural pathway (LGN → V1). This is a progressive, level-by-level organization.
7PASSAGE (Questions 7–13): Glucose metabolism is central to cellular energy production. In glycolysis, a molecule of glucose (a six-carbon sugar) is split into two molecules of pyruvate (three carbons each) in the cytosol, yielding a net gain of 2 ATP and 2 NADH per glucose. Glycolysis proceeds through ten enzyme-catalyzed steps. The first is glucose phosphorylation by hexokinase, consuming one ATP to produce glucose-6-phosphate (G6P). G6P cannot cross the plasma membrane, effectively trapping glucose inside the cell. A second ATP is consumed later in the process when fructose-6-phosphate is converted to fructose-1,6-bisphosphate by phosphofructokinase-1 (PFK-1), the primary regulatory step. PFK-1 is allosterically inhibited by high concentrations of ATP and citrate (signals of energy sufficiency) and activated by AMP and fructose-2,6-bisphosphate. Under aerobic conditions, pyruvate enters the mitochondrial matrix, where it is oxidatively decarboxylated by the pyruvate dehydrogenase complex (PDC) to form acetyl-CoA, releasing one CO2 and one NADH per pyruvate. Acetyl-CoA then enters the citric acid cycle (Krebs cycle), where it is completely oxidized over eight steps, generating per acetyl-CoA: 3 NADH, 1 FADH2, and 1 GTP (equivalent to 1 ATP). The high-energy electrons carried by NADH and FADH2 are donated to the electron transport chain (ETC) on the inner mitochondrial membrane. As electrons flow through Complexes I–IV, protons are pumped from the matrix to the intermembrane space, creating a proton-motive force. ATP synthase (Complex V) harnesses this force to synthesize ATP from ADP and inorganic phosphate (Pi)—a process called oxidative phosphorylation. Under anaerobic conditions—when oxygen is absent or insufficient—cells regenerate NAD+ from NADH via fermentation. In animals and many bacteria, pyruvate is reduced to lactate by lactate dehydrogenase (LDH), reoxidizing NADH to NAD+. This allows glycolysis to continue producing ATP even without oxygen, albeit at a much lower yield (2 ATP per glucose compared to approximately 30–32 ATP under full aerobic conditions). In yeast, fermentation produces ethanol and CO2 instead of lactate. The switch between aerobic and anaerobic metabolism is regulated by oxygen availability and the cellular energy charge. Question 7: According to the passage, what is the primary function of PFK-1 in glycolysis?
A.It phosphorylates glucose to form glucose-6-phosphate, trapping glucose in the cell
B.It catalyzes the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate at the main regulatory step
C.It donates electrons to the electron transport chain to initiate ATP synthesis
D.It converts pyruvate to acetyl-CoA in the mitochondrial matrix
Explanation: The passage states that PFK-1 converts fructose-6-phosphate to fructose-1,6-bisphosphate and identifies this as the 'primary regulatory step' of glycolysis. This is a detail question answered directly in the second sentence of the first paragraph.
8PASSAGE (Questions 7–13 — same passage as Q7): Glucose metabolism is central to cellular energy production. In glycolysis, a molecule of glucose (a six-carbon sugar) is split into two molecules of pyruvate (three carbons each) in the cytosol, yielding a net gain of 2 ATP and 2 NADH per glucose. Glycolysis proceeds through ten enzyme-catalyzed steps. The first is glucose phosphorylation by hexokinase, consuming one ATP to produce glucose-6-phosphate (G6P). G6P cannot cross the plasma membrane, effectively trapping glucose inside the cell. A second ATP is consumed later in the process when fructose-6-phosphate is converted to fructose-1,6-bisphosphate by phosphofructokinase-1 (PFK-1), the primary regulatory step. PFK-1 is allosterically inhibited by high concentrations of ATP and citrate (signals of energy sufficiency) and activated by AMP and fructose-2,6-bisphosphate. Under aerobic conditions, pyruvate enters the mitochondrial matrix, where it is oxidatively decarboxylated by the pyruvate dehydrogenase complex (PDC) to form acetyl-CoA, releasing one CO2 and one NADH per pyruvate. Acetyl-CoA then enters the citric acid cycle (Krebs cycle), where it is completely oxidized over eight steps, generating per acetyl-CoA: 3 NADH, 1 FADH2, and 1 GTP (equivalent to 1 ATP). The high-energy electrons carried by NADH and FADH2 are donated to the electron transport chain (ETC) on the inner mitochondrial membrane. As electrons flow through Complexes I–IV, protons are pumped from the matrix to the intermembrane space, creating a proton-motive force. ATP synthase (Complex V) harnesses this force to synthesize ATP from ADP and inorganic phosphate (Pi)—a process called oxidative phosphorylation. Under anaerobic conditions—when oxygen is absent or insufficient—cells regenerate NAD+ from NADH via fermentation. In animals and many bacteria, pyruvate is reduced to lactate by lactate dehydrogenase (LDH), reoxidizing NADH to NAD+. This allows glycolysis to continue producing ATP even without oxygen, albeit at a much lower yield (2 ATP per glucose compared to approximately 30–32 ATP under full aerobic conditions). In yeast, fermentation produces ethanol and CO2 instead of lactate. The switch between aerobic and anaerobic metabolism is regulated by oxygen availability and the cellular energy charge. Question 8: Which of the following conditions, according to the passage, would INHIBIT PFK-1 activity?
A.Rising AMP concentration combined with falling ATP levels
B.High intracellular ATP and elevated citrate
C.Low oxygen availability and high fructose-2,6-bisphosphate
D.Increased ADP concentration and depleted NAD+
Explanation: The passage explicitly states that PFK-1 is allosterically inhibited by high concentrations of ATP and citrate—both signals of energy sufficiency. The other conditions described in the wrong options either activate PFK-1 or are not mentioned in the passage.
9PASSAGE (Questions 7–13 — same passage as Q7): Glucose metabolism is central to cellular energy production. In glycolysis, a molecule of glucose (a six-carbon sugar) is split into two molecules of pyruvate (three carbons each) in the cytosol, yielding a net gain of 2 ATP and 2 NADH per glucose. Glycolysis proceeds through ten enzyme-catalyzed steps. The first is glucose phosphorylation by hexokinase, consuming one ATP to produce glucose-6-phosphate (G6P). G6P cannot cross the plasma membrane, effectively trapping glucose inside the cell. A second ATP is consumed later in the process when fructose-6-phosphate is converted to fructose-1,6-bisphosphate by phosphofructokinase-1 (PFK-1), the primary regulatory step. PFK-1 is allosterically inhibited by high concentrations of ATP and citrate (signals of energy sufficiency) and activated by AMP and fructose-2,6-bisphosphate. Under aerobic conditions, pyruvate enters the mitochondrial matrix, where it is oxidatively decarboxylated by the pyruvate dehydrogenase complex (PDC) to form acetyl-CoA, releasing one CO2 and one NADH per pyruvate. Acetyl-CoA then enters the citric acid cycle (Krebs cycle), where it is completely oxidized over eight steps, generating per acetyl-CoA: 3 NADH, 1 FADH2, and 1 GTP (equivalent to 1 ATP). The high-energy electrons carried by NADH and FADH2 are donated to the electron transport chain (ETC) on the inner mitochondrial membrane. As electrons flow through Complexes I–IV, protons are pumped from the matrix to the intermembrane space, creating a proton-motive force. ATP synthase (Complex V) harnesses this force to synthesize ATP from ADP and inorganic phosphate (Pi)—a process called oxidative phosphorylation. Under anaerobic conditions—when oxygen is absent or insufficient—cells regenerate NAD+ from NADH via fermentation. In animals and many bacteria, pyruvate is reduced to lactate by lactate dehydrogenase (LDH), reoxidizing NADH to NAD+. This allows glycolysis to continue producing ATP even without oxygen, albeit at a much lower yield (2 ATP per glucose compared to approximately 30–32 ATP under full aerobic conditions). In yeast, fermentation produces ethanol and CO2 instead of lactate. The switch between aerobic and anaerobic metabolism is regulated by oxygen availability and the cellular energy charge. Question 9: The passage implies that the regeneration of NAD+ during anaerobic fermentation is essential because:
A.NAD+ is required to pump protons across the inner mitochondrial membrane
B.NAD+ is required as an electron acceptor to sustain continued glycolysis
C.NAD+ is consumed by the pyruvate dehydrogenase complex and must be replenished
D.NAD+ directly powers ATP synthase by binding to Complex V
Explanation: Glycolysis produces NADH from NAD+; if NAD+ is not regenerated, there would be no NAD+ available and glycolysis would halt. The passage states that fermentation reoxidizes NADH to NAD+, 'allowing glycolysis to continue producing ATP even without oxygen.' NAD+ functions as the electron acceptor in glycolysis.
10PASSAGE (Questions 7–13 — same passage as Q7): Glucose metabolism is central to cellular energy production. In glycolysis, a molecule of glucose (a six-carbon sugar) is split into two molecules of pyruvate (three carbons each) in the cytosol, yielding a net gain of 2 ATP and 2 NADH per glucose. Glycolysis proceeds through ten enzyme-catalyzed steps. The first is glucose phosphorylation by hexokinase, consuming one ATP to produce glucose-6-phosphate (G6P). G6P cannot cross the plasma membrane, effectively trapping glucose inside the cell. A second ATP is consumed later in the process when fructose-6-phosphate is converted to fructose-1,6-bisphosphate by phosphofructokinase-1 (PFK-1), the primary regulatory step. PFK-1 is allosterically inhibited by high concentrations of ATP and citrate (signals of energy sufficiency) and activated by AMP and fructose-2,6-bisphosphate. Under aerobic conditions, pyruvate enters the mitochondrial matrix, where it is oxidatively decarboxylated by the pyruvate dehydrogenase complex (PDC) to form acetyl-CoA, releasing one CO2 and one NADH per pyruvate. Acetyl-CoA then enters the citric acid cycle (Krebs cycle), where it is completely oxidized over eight steps, generating per acetyl-CoA: 3 NADH, 1 FADH2, and 1 GTP (equivalent to 1 ATP). The high-energy electrons carried by NADH and FADH2 are donated to the electron transport chain (ETC) on the inner mitochondrial membrane. As electrons flow through Complexes I–IV, protons are pumped from the matrix to the intermembrane space, creating a proton-motive force. ATP synthase (Complex V) harnesses this force to synthesize ATP from ADP and inorganic phosphate (Pi)—a process called oxidative phosphorylation. Under anaerobic conditions—when oxygen is absent or insufficient—cells regenerate NAD+ from NADH via fermentation. In animals and many bacteria, pyruvate is reduced to lactate by lactate dehydrogenase (LDH), reoxidizing NADH to NAD+. This allows glycolysis to continue producing ATP even without oxygen, albeit at a much lower yield (2 ATP per glucose compared to approximately 30–32 ATP under full aerobic conditions). In yeast, fermentation produces ethanol and CO2 instead of lactate. The switch between aerobic and anaerobic metabolism is regulated by oxygen availability and the cellular energy charge. Question 10: How many total NADH molecules are produced per glucose molecule through glycolysis AND the pyruvate dehydrogenase step combined (considering both pyruvates from one glucose)?
A.2 NADH
B.4 NADH
C.6 NADH
D.8 NADH
Explanation: Glycolysis yields 2 NADH per glucose. The PDC converts each pyruvate to acetyl-CoA releasing 1 NADH per pyruvate; since one glucose yields 2 pyruvates, the PDC step contributes 2 NADH. Total: 2 (glycolysis) + 2 (PDC) = 4 NADH. This requires applying two pieces of passage information together.
About the OAT Reading Comprehension Exam
The OAT Reading Comprehension section tests the ability to read, comprehend, and analytically evaluate dense scientific passages. It consists of 50 questions distributed across 3 passages (approximately 1,100–1,500 words each) covering various scientific topics. No prior knowledge of the topic is required—all answers are derivable from the passage text.
Questions
50 scored questions
Time Limit
60 minutes
Passing Score
200–400 scale; median 300; competitive schools prefer 320–340+
Exam Fee
$520 for the full OAT (partial fee waivers available) (American Dental Association (ADA) on behalf of ASCO)
OAT Reading Comprehension Exam Content Outline
~20%
Locating Specific Details
Direct fact retrieval from passage text including numbers, definitions, and explicitly stated information
~20%
Inference and Logical Reasoning
Drawing conclusions implied but not directly stated; connecting two or more ideas from the passage
~15%
Main Idea and Author's Purpose
Identifying the primary argument or purpose of a passage or specific paragraph
~15%
Applying Information
Using passage-provided formulas, definitions, or mechanisms to answer scenario-based questions
~10%
Tone and Attitude
Determining the author's perspective, tone, or evaluative stance toward the subject
~10%
Vocabulary in Context
Determining the meaning of scientific or technical terms from surrounding context
~10%
Passage Structure and Organization
Identifying the organizational logic, paragraph function, and structural devices used by the author
How to Pass the OAT Reading Comprehension Exam
What You Need to Know
- Passing score: 200–400 scale; median 300; competitive schools prefer 320–340+
- Exam length: 50 questions
- Time limit: 60 minutes
- Exam fee: $520 for the full OAT (partial fee waivers available)
Keys to Passing
- Complete 500+ practice questions
- Score 80%+ consistently before scheduling
- Focus on highest-weighted sections
- Use our AI tutor for tough concepts
OAT Reading Comprehension Study Tips from Top Performers
1Preview the questions associated with a passage BEFORE reading it so you know what information to look for as you read.
2Annotate key terms, definitions, and numerical facts as you encounter them in the passage—these are frequent detail question targets.
3For tone and attitude questions, look for evaluative language (e.g., 'however,' 'despite,' 'although') that signals the author's stance or concessions.
4For inference questions, your answer must follow logically from the passage; avoid answers that require outside knowledge or go beyond what the text supports.
5Practice with dense scientific texts outside of OAT prep materials—journals, science news articles, and review papers will build reading speed and comfort with technical vocabulary.
Frequently Asked Questions
How many passages are in the OAT Reading Comprehension section?
There are 3 reading passages, each approximately 1,100–1,500 words long. The 50 questions are distributed across the three passages, with approximately 15–20 questions per passage.
Do I need prior science knowledge to answer OAT Reading Comprehension questions?
No. The official OAT Candidate Guide explicitly states that prior understanding of the science topics is not a prerequisite. All answers can be derived directly from careful reading of the passage.
How long do I have for the Reading Comprehension section?
You have 60 minutes to complete all 50 questions across the 3 passages. This averages to about 1.2 minutes per question, so managing time between passages is important.
What topics appear in OAT Reading Comprehension passages?
Passages cover various scientific topics including biology, chemistry, health sciences, ecology, and vision science. Passages are dense and academic in nature; the ability to extract information efficiently from challenging text is the key skill tested.
How is the OAT Reading Comprehension section scored?
The Reading Comprehension section is scored on a scale of 200–400, with a national median of approximately 300. It contributes to both the section score and the Academic Average (AA) score reported to optometry schools.
What is the best strategy for OAT Reading Comprehension?
Active reading strategies are essential: preview the questions before reading the passage so you know what to look for, annotate key terms and topic sentences, and return to the passage to verify each answer. Avoid relying on outside knowledge—stay strictly within the text.