TEAS 7 Science Practice Test

25 practice questions

Choose an answer, then check it under Show answer.

1. Question 1 of 25Easy

   Which phase change occurs when a solid becomes a liquid?

   1. AA. Melting
   2. BB. Boiling
   3. CC. Freezing
   4. DD. Condensation
   Show answer to question 1

   Answer: A. Melting

   Melting is the change from a solid to a liquid. During melting, a substance absorbs energy that allows its particles to move more freely while remaining in the liquid state.

   Why the other options are wrong

   • BBoiling changes a liquid into a gas, not a solid into a liquid.
   • CFreezing changes a liquid into a solid, the reverse of melting.
   • DCondensation changes a gas into a liquid, not a solid into a liquid.

2. Question 2 of 25Medium

   How does the integumentary system respond to an elevated core body temperature to facilitate cooling?

   1. AA. Vasodilation of dermal blood vessels and increased eccrine sweat production
   2. BB. Vasoconstriction of dermal blood vessels and contraction of arrector pili muscles
   3. CC. Vasoconstriction of dermal blood vessels and increased apocrine sweat production
   4. DD. Vasodilation of dermal blood vessels and increased sebum production
   Show answer to question 2

   Answer: A. Vasodilation of dermal blood vessels and increased eccrine sweat production

   When body temperature rises, dermal blood vessels undergo vasodilation to increase blood flow near the surface, dissipating heat via radiation, while eccrine glands increase sweat output for evaporative cooling. Vasoconstriction of dermal vessels and arrector pili contraction occur in response to cold temperatures to conserve heat.

   Why the other options are wrong

   • BThis confuses cold responses with cooling responses: vasoconstriction reduces heat loss, and arrector pili contraction causes goosebumps rather than effective cooling.
   • CThis confuses heat conservation with heat loss and apocrine glands with eccrine glands. Vasoconstriction reduces heat loss, and eccrine glands, not apocrine glands, provide the sweat responsible for thermoregulatory cooling.
   • DThis mistakes sebum for sweat. Although vasodilation promotes heat loss, sebum lubricates skin and hair rather than providing evaporative cooling.

3. Question 3 of 25Easy

   Which physiological response occurs as a direct result of sympathetic nervous system stimulation?

   1. AA. Increased motility in the gastrointestinal tract
   2. BB. Decreased heart rate and contractility
   3. CC. Constriction of the pupils
   4. DD. Dilation of the bronchioles
   Show answer to question 3

   Answer: D. Dilation of the bronchioles

   Activation of the sympathetic nervous system triggers the fight-or-flight response, which causes dilation of the bronchioles to maximize airflow for oxygen exchange. In contrast, constriction of the pupils and increased gastrointestinal motility are mediated by the parasympathetic nervous system, which promotes rest-and-digest activities.

   Why the other options are wrong

   • AThis confuses parasympathetic activity with sympathetic activity, which generally reduces gastrointestinal motility rather than increasing it.
   • BThis reverses the sympathetic response: sympathetic stimulation increases heart rate and contractility to support greater blood flow during stress.
   • CThis confuses parasympathetic pupil constriction with sympathetic pupil dilation, which allows more light to enter the eyes.

4. Question 4 of 25Medium

   Why do physiological positive feedback loops require an outside event or limiting factor to terminate?

   1. AA. They prevent the endocrine system from maintaining cellular sensitivity to ligands.
   2. BB. They irreversibly alter the original set point of the regulated variable.
   3. CC. They drive the variable progressively further away from the initial resting state.
   4. DD. They bypass cellular effectors and act directly on sensory afferent pathways.
   Show answer to question 4

   Answer: C. They drive the variable progressively further away from the initial resting state.

   Positive feedback systems are self-amplifying cycles in which the effector response enhances the original stimulus, pushing the variable farther from its baseline until the stimulus is removed or a limiting factor interrupts the cascade. They do not bypass effectors; effectors are essential in producing the amplified response. These loops do not permanently destroy or reset physiological set points, but operate transiently until a specific objective is achieved.

   Why the other options are wrong

   • AThis confuses positive feedback with receptor desensitization. Reduced cellular responsiveness to ligands does not explain why an amplifying feedback cycle needs a stopping event or limiting factor.
   • BThis confuses movement away from a resting state with a permanent change in the set point. Positive feedback amplifies a stimulus without necessarily changing the regulated variable’s set point.
   • DThis confuses sensory detection with the response that amplifies a stimulus. Positive feedback depends on effectors to reinforce the original stimulus, not on bypassing them.

5. Question 5 of 25Hard

   A person has persistently slow, shallow breathing that causes arterial carbon dioxide to rise. After several days, which kidney response would help compensate for the resulting acid-base disturbance?

   1. AA. Increase net hydrogen ion excretion and decrease bicarbonate reabsorption.
   2. BB. Decrease net hydrogen ion excretion and increase bicarbonate reabsorption.
   3. CC. Decrease net hydrogen ion excretion and decrease bicarbonate reabsorption.
   4. DD. Increase net hydrogen ion excretion and increase bicarbonate reabsorption.
   Show answer to question 5

   Answer: D. Increase net hydrogen ion excretion and increase bicarbonate reabsorption.

   Inadequate ventilation retains carbon dioxide, increasing carbonic acid formation and lowering blood pH. The kidneys compensate by increasing net hydrogen ion excretion and conserving bicarbonate, which buffers acid. This response moves blood pH toward normal without correcting the underlying respiratory problem.

   Why the other options are wrong

   • ADecreasing bicarbonate reabsorption removes a blood buffer, opposing compensation for the increased acid load.
   • BDecreasing hydrogen ion excretion retains acid, opposing compensation even though increased bicarbonate reabsorption would help.
   • CDecreasing hydrogen ion excretion retains acid, while decreasing bicarbonate reabsorption loses buffer; both changes worsen the respiratory acidosis.

6. Question 6 of 25Medium

   Which organelle relies on the enzyme catalase to neutralize toxic hydrogen peroxide produced during the beta-oxidation of very long-chain fatty acids?

   1. AA. Lysosome
   2. BB. Peroxisome
   3. CC. Proteasome
   4. DD. Spliceosome
   Show answer to question 6

   Answer: B. Peroxisome

   Peroxisomes contain oxidases that break down fatty acids, generating toxic hydrogen peroxide as a byproduct, which catalase then safely converts into water and oxygen. Lysosomes also contain degradation enzymes, but they utilize acid hydrolases rather than catalase and oxidative enzymes.

   Why the other options are wrong

   • AConfusing lysosomal digestion with peroxisomal oxidation is the error: lysosomes use acid hydrolases to digest cellular material, not catalase to neutralize hydrogen peroxide from very long-chain fatty acid breakdown.
   • CConfusing protein degradation with fatty acid oxidation is the error: proteasomes break down proteins, not very long-chain fatty acids, and do not use catalase to neutralize hydrogen peroxide.
   • DConfusing RNA processing with fatty acid oxidation is the error: spliceosomes remove introns from precursor messenger RNA, rather than breaking down fatty acids or neutralizing hydrogen peroxide with catalase.

7. Question 7 of 25Easy

   A clinical trial is designed to evaluate the efficacy of a novel antihypertensive drug in reducing blood pressure. Which experimental feature best minimizes both participant expectancy bias and researcher observation bias?

   1. AA. Testing the novel medication across multiple unrelated age groups simultaneously
   2. BB. Administering a double-blind protocol using an identical inert placebo
   3. CC. Allowing participants to select between the active drug and standard care
   4. DD. Informing the investigators of treatment assignments while blinding participants
   Show answer to question 7

   Answer: B. Administering a double-blind protocol using an identical inert placebo

   A double-blind, placebo-controlled design keeps participants and researchers assessing outcomes unaware of treatment assignments, helping reduce participant expectancy bias and researcher observation bias. An identical inert placebo helps conceal assignments and account for placebo related responses when comparing groups, but does not eliminate placebo effects. Testing multiple age groups changes the study population but does not conceal assignments or directly address either bias. Informing investigators while blinding participants leaves researcher observation bias possible. Allowing participants to choose their treatment introduces selection bias.

   Why the other options are wrong

   • AConfusing age diversity with bias control overlooks that testing multiple age groups does not conceal treatment assignments or prevent participant expectations and researcher observations from influencing results.
   • CConfusing participant choice with bias control introduces selection bias and leaves participants aware of their treatment, so expectancy bias remains and researcher observation bias is not controlled.
   • DConfusing single blinding with double blinding overlooks that investigators who know treatment assignments may interpret or record outcomes differently, leaving researcher observation bias uncontrolled.

8. Question 8 of 25Hard

   Consider the reaction: 2H₂ + O₂ → 2H₂O. If 4.0 g of H₂ and 40.0 g of O₂ are reacted, what is the limiting reactant? (Atomic masses: H = 1, O = 16)

   1. AA. Both reactants are limiting
   2. BB. O₂
   3. CC. H₂
   4. DD. Neither reactant is limiting
   Show answer to question 8

   Answer: C. H₂

   Moles of H₂ = 4.0 g / 2.0 g/mol = 2.0 mol. Moles of O₂ = 40.0 g / 32.0 g/mol = 1.25 mol. The reaction requires 2 mol H₂ for every 1 mol O₂. With 2.0 mol H₂ and 1.25 mol O₂, H₂ is completely consumed (2.0 mol H₂ needs 1.0 mol O₂, but 1.25 mol O₂ is available), so H₂ is the limiting reactant.

   Why the other options are wrong

   • AOnly H₂ is limiting in this reaction; O₂ remains in excess.
   • BO₂ is in excess; 1.25 mol O₂ would require 2.5 mol H₂ to be fully used, but only 2.0 mol H₂ is present.
   • DThe stoichiometric ratio is 2:1, but here the ratio is 2.0:1.25 = 1.6:1, not exactly 2:1, so one reactant is limiting.

9. Question 9 of 25Easy

   Which of the following components of blood is normally prevented from passing across the glomerular filtration barrier into the filtrate?

   1. AA. Sodium ions
   2. BB. Urea waste
   3. CC. Plasma proteins
   4. DD. Glucose molecules
   Show answer to question 9

   Answer: C. Plasma proteins

   The glomerular filtration barrier normally retains most plasma proteins, including albumin, primarily because of its size selectivity and structural integrity. Small amounts of protein can enter the filtrate and are normally reabsorbed by the renal tubules. Small solutes such as glucose, urea, and sodium ions are freely filtered into the capsular space.

   Why the other options are wrong

   • ASodium ions pass freely through the glomerular filtration barrier. Choosing them confuses later regulation of sodium reabsorption with exclusion during filtration.
   • BUrea is small enough to pass freely into the filtrate. Choosing it mistakes a waste product removed by the kidneys for a substance blocked by the filtration barrier.
   • DGlucose passes freely into the filtrate and is normally reabsorbed in the proximal tubule. Choosing it confuses absence from normal urine with exclusion during glomerular filtration.

10. Question 10 of 25Easy

    Which of the following instruments is most suitable for measuring the mass of a small chemical sample in a laboratory setting?

    1. AA. Graduated cylinder
    2. BB. Barometer
    3. CC. Triple beam balance
    4. DD. Spring scale
    Show answer to question 10

    Answer: C. Triple beam balance

    A triple beam balance is used to measure the mass of an object by comparing it to known masses. A spring scale measures weight (force), a graduated cylinder measures volume, and a barometer measures atmospheric pressure.

    Why the other options are wrong

    • AA graduated cylinder measures liquid volume, not mass. Choosing it confuses the space a sample occupies with the amount of matter it contains.
    • BA barometer measures atmospheric pressure, not sample mass. Choosing it confuses the pressure exerted by the atmosphere with the amount of matter in a sample.
    • DA spring scale measures weight, a force caused by gravity, rather than mass directly. Choosing it confuses an object's gravitational force with its mass.

11. Question 11 of 25Medium

    During the light-dependent reactions of photosynthesis, what is the direct source of the oxygen gas released into the atmosphere?

    1. AA. The oxidation of ribulose bisphosphate
    2. BB. The breakdown of stored glucose
    3. CC. The reduction of carbon dioxide
    4. DD. The splitting of water molecules
    Show answer to question 11

    Answer: D. The splitting of water molecules

    Photolysis splits water molecules inside the thylakoid lumen, generating electrons to replace those excited in photosystem II, releasing hydrogen ions and oxygen gas. Carbon dioxide supplies carbon and oxygen for the synthesis of carbohydrates in the Calvin cycle, not atmospheric oxygen.

    Why the other options are wrong

    • AThis confuses reactions involving ribulose bisphosphate with oxygen production. Ribulose bisphosphate accepts carbon dioxide during carbon fixation, and its reaction with oxygen during photorespiration consumes oxygen rather than releasing it.
    • BThis confuses cellular respiration with photosynthesis. Glucose breakdown releases stored chemical energy, not oxygen gas; aerobic respiration consumes oxygen.
    • CThis mistakes carbon dioxide for the source of released oxygen gas. Carbon dioxide is reduced to form carbohydrates in the Calvin cycle, while the released oxygen comes from water.

12. Question 12 of 25Hard

    A researcher studies how humidity affects seedling growth. One growth chamber is set to 40% humidity and another to 80%. Each chamber receives 12 randomly selected, similar seedlings, and other planned conditions are matched. Each seedling is measured weekly for 4 weeks. The researcher treats the resulting 48 readings at each humidity as 48 independent treatment replicates. Which redesign best addresses the main limitation in attributing growth differences to humidity?

    1. AA. Randomly assign several chambers to each humidity level and compare chamber-level averages of plant growth.
    2. BB. Randomly assign more seedlings to each original chamber and compare seedling-level averages of plant growth.
    3. CC. Measure each seedling more frequently in each original chamber and compare individual readings as additional treatment replicates.
    4. DD. Randomly assign seedlings to shelves within each original chamber and compare shelf-level averages of plant growth.
    Show answer to question 12

    Answer: A. Randomly assign several chambers to each humidity level and compare chamber-level averages of plant growth.

    The 12 × 4 = 48 readings per humidity are repeated observations from plants sharing one chamber, not 48 independent assignments of humidity. With one chamber at each humidity, humidity is confounded with chamber identity. Randomly assigning multiple chambers to each humidity provides independent treatment replication and helps distinguish a humidity effect from chamber-specific effects.

    Why the other options are wrong

    • BAdds observations within the existing treatment units but leaves each humidity assigned to just one chamber.
    • CMistakes repeated measurements for independent treatment replication; more readings do not separate humidity from chamber identity.
    • DReduces possible shelf-position bias but does not independently replicate humidity because shelves remain within the same chamber.

13. Question 13 of 25Medium

    How does oogenesis differ fundamentally from spermatogenesis in terms of functional gamete production from a single primary germ cell?

    1. AA. Oogenesis completes all meiotic divisions continuously before the individual reaches puberty.
    2. BB. Oogenesis produces one viable ovum and nonviable polar bodies due to unequal cytokinesis.
    3. CC. Oogenesis produces four equally sized, functional gametes through rapid mitotic divisions.
    4. DD. Oogenesis relies on symmetric cytokinesis to divide cytoplasm evenly among daughter cells.
    Show answer to question 13

    Answer: B. Oogenesis produces one viable ovum and nonviable polar bodies due to unequal cytokinesis.

    Oogenesis involves unequal division of the cytoplasm during meiosis, yielding only one functional ovum and up to three nonviable polar bodies that degenerate. Spermatogenesis involves symmetric cytokinesis, producing four functional, motile spermatozoa from each primary spermatocyte. Oogenesis is also arrested at specific stages rather than progressing continuously before puberty.

    Why the other options are wrong

    • AThis overlooks meiotic arrest: primary oocytes pause in prophase I before birth, and meiosis II is completed only if fertilization occurs, not before puberty.
    • CThis confuses the four functional gametes produced by spermatogenesis with oogenesis and mistakes meiosis for mitosis. Oogenesis produces one functional ovum through meiosis.
    • DThis mistakes unequal cytokinesis for equal division: oogenesis preserves most cytoplasm in one functional ovum while polar bodies receive very little.

14. Question 14 of 25Medium

    Which statement correctly describes the relationship between atomic radius and the number of energy levels in a neutral atom?

    1. AA. Atomic radius increases as the number of energy levels increases.
    2. BB. Atomic radius decreases as the number of energy levels increases.
    3. CC. Atomic radius remains constant regardless of energy levels.
    4. DD. Atomic radius is determined solely by the number of neutrons.
    Show answer to question 14

    Answer: A. Atomic radius increases as the number of energy levels increases.

    As the number of energy levels (shells) increases, the electrons are located further from the nucleus, thereby increasing the atomic radius. The number of neutrons has a negligible effect on the size of the electron cloud.

15. Question 15 of 25Easy

    An atom of phosphorus has an atomic number of 15 and a mass number of 31. How many neutrons are located in the nucleus of this atom?

    1. AA. 15
    2. BB. 46
    3. CC. 16
    4. DD. 31
    Show answer to question 15

    Answer: C. 16

    The number of neutrons is determined by subtracting the atomic number (number of protons) from the mass number (protons plus neutrons), giving 31 minus 15, which equals 16 neutrons. Selecting 15 incorrectly identifies the number of protons or electrons rather than neutrons.

    Why the other options are wrong

    • AThis confuses the atomic number, 15, with the neutron count. The atomic number gives the number of protons, not neutrons.
    • BAdding the mass number and atomic number, 31 + 15 = 46, incorrectly counts the protons twice instead of subtracting them to find the neutron count.
    • DThis confuses the mass number, 31, with the neutron count. The mass number includes both protons and neutrons, so the 15 protons must be subtracted.

16. Question 16 of 25Hard

    A woman is a carrier of an X-linked recessive condition, and her partner is an unaffected man. Assume each child is equally likely to be male or female and that births are independent. What is the probability that their next two children will be one affected son and one unaffected carrier daughter, in either birth order?

    1. AA. 1 in 16
    2. BB. 1 in 8
    3. CC. 1 in 4
    4. DD. 1 in 2
    Show answer to question 16

    Answer: B. 1 in 8

    Each son has a 1 ÷ 2 chance of inheriting the mother's recessive allele, and each daughter has a 1 ÷ 2 chance of being an unaffected carrier. Accounting for sex, P(affected son) = (1 ÷ 2) × (1 ÷ 2) = 1 ÷ 4, and P(carrier daughter) = (1 ÷ 2) × (1 ÷ 2) = 1 ÷ 4. Either child can be the son, giving two mutually exclusive birth orders. Thus, P = 2 × (1 ÷ 4) × (1 ÷ 4) = 1 ÷ 8.

    Why the other options are wrong

    • AMultiplying (1 ÷ 4) × (1 ÷ 4) counts only one birth order, omitting the equally likely reverse order.
    • CThis result counts every daughter as a carrier, rather than only the half who inherit the mother's recessive allele.
    • DUsing 2 × (1 ÷ 2) × (1 ÷ 2) accounts for allele transmission but omits the probabilities of having a son and a daughter.

17. Question 17 of 25Medium

    A researcher designs an experiment to test the effect of temperature on the activity of an enzyme. She sets up three test tubes with identical substrate and enzyme concentrations, then incubates them at 20°C, 37°C, and 60°C. After 10 minutes, she measures the product concentration. She observes the highest product concentration at 37°C and very low product at 60°C. To improve the conclusion that temperature affects enzyme activity, which additional control should she include?

    1. AA. A test tube with boiled enzyme and substrate at 37°C
    2. BB. A test tube with enzyme but no substrate at 37°C only
    3. CC. A test tube with twice the substrate concentration at 37°C
    4. DD. A test tube with substrate but no enzyme at each temperature
    Show answer to question 17

    Answer: D. A test tube with substrate but no enzyme at each temperature

    Including substrate without enzyme at each temperature measures any product formed without the enzyme under that temperature condition. Comparing each experimental tube with its matched control helps distinguish enzyme-dependent product formation from background product formation.

    Why the other options are wrong

    • ABoiled enzyme is denatured but still tests for non-enzymatic reaction at one temperature only; it lacks controls at other temperatures.
    • BAn enzyme-only tube can help detect background signal from the enzyme preparation, depending on the assay. However, it cannot reveal product formation from substrate without enzyme and provides a control at only one temperature.
    • CChanging substrate concentration alters the reaction rate and does not serve as a control for temperature effects.

18. Question 18 of 25Easy

    An aquatic biologist measures dissolved oxygen concentration in microcosms containing Elodea plants subjected to different light levels, while keeping temperature, water volume, carbon dioxide concentration, and duration of light exposure constant. In this study, which factor constitutes the dependent variable?

    1. AA. The duration of light exposure
    2. BB. The carbon dioxide concentration
    3. CC. The dissolved oxygen concentration
    4. DD. The intensity of the light source
    Show answer to question 18

    Answer: C. The dissolved oxygen concentration

    The dependent variable is the measured outcome, dissolved oxygen concentration. Light intensity is the independent variable being manipulated, while temperature, water volume, carbon dioxide concentration, and exposure duration are controlled variables. Photosynthesis produces oxygen, while cellular respiration consumes oxygen, so both processes affect dissolved oxygen concentration.

    Why the other options are wrong

    • AExposure duration is a controlled variable because it is held constant. The measured response is dissolved oxygen concentration.
    • BChoosing carbon dioxide concentration confuses a controlled variable with the dependent variable. Carbon dioxide is held constant rather than measured as the response.
    • DChoosing light intensity confuses the independent variable with the dependent variable. Light intensity is deliberately changed to test its effect on dissolved oxygen concentration.

19. Question 19 of 25Hard

    An enzyme converts a colorless substrate into a blue product. A researcher uses color intensity to estimate product formation and finds less blue color when compound X is added. A vehicle control contains the same solvent volume without X. A blank containing substrate and X, but no enzyme, remains colorless. Which additional comparison best distinguishes reduced product formation from X changing the product's color after it forms?

    1. AA. Compare substrate mixed with heat-inactivated enzyme, with and without X, while matching the solvent volume in both mixtures.
    2. BB. Compare equal amounts of preformed blue product, with and without X, while omitting enzyme and matching final volumes.
    3. CC. Compare substrate mixed with twice the original amount of active enzyme, with and without X, while matching final volumes.
    4. DD. Compare active enzyme mixed with X, with and without substrate, while matching the solvent volume in both mixtures.
    Show answer to question 19

    Answer: B. Compare equal amounts of preformed blue product, with and without X, while omitting enzyme and matching final volumes.

    The preformed-product control tests whether X changes the measured color independently of enzyme activity. If adding X lowers the signal from preformed product, reduced color in the original experiment does not by itself establish reduced product formation. The colorless blank rules out color produced by substrate and X alone, but it does not test their effect on an existing blue product.

    Why the other options are wrong

    • ATests background color without active enzyme but does not expose a known amount of blue product to X.
    • CChanges the capacity for product formation without isolating X's effect on the color of product already present.
    • DTests whether substrate is needed for the observed color but does not distinguish reduced product formation from altered product color.

20. Question 20 of 25Medium

    An adult receives antidiuretic hormone (ADH). The kidneys filter 180 L of water per day both before and after treatment, while urine output falls from 9.0 L/day to 3.6 L/day. Which pair correctly describes the percentage of filtered water reabsorbed and the mechanism?

    1. AA. Water reabsorption decreases from 5% to 2%, and collecting duct water permeability increases.
    2. BB. Water reabsorption increases from 95% to 98%, and collecting duct water permeability increases.
    3. CC. Water reabsorption increases from 95% to 98%, and collecting duct water permeability decreases.
    4. DD. Water reabsorption decreases from 5% to 2%, and collecting duct water permeability decreases.
    Show answer to question 20

    Answer: B. Water reabsorption increases from 95% to 98%, and collecting duct water permeability increases.

    The percentage reabsorbed equals (filtered volume − urine volume) ÷ filtered volume × 100%. Before treatment, (180 − 9.0) ÷ 180 × 100% = 95%; afterward, (180 − 3.6) ÷ 180 × 100% = 98%. ADH, normally released from the posterior pituitary, increases collecting duct water permeability, allowing more filtered water to return to the blood.

    Why the other options are wrong

    • AThe values 5% and 2% represent the fractions excreted as urine, not the fractions reabsorbed.
    • CADH increases collecting duct water permeability; decreasing permeability would reduce water reabsorption, contrary to the observed response.
    • DThe percentages describe urine excretion rather than reabsorption, and ADH increases rather than decreases collecting duct water permeability.

21. Question 21 of 25Medium

    A man with type AB blood marries a woman with type O blood. What is the probability that their first child will have type O blood?

    1. AA. 0%
    2. BB. 25%
    3. CC. 100%
    4. DD. 50%
    Show answer to question 21

    Answer: A. 0%

    A person with type AB blood can only pass on either an A or a B allele, whereas a person with type O blood can only pass on an O allele. Therefore, the offspring can only be type A or type B, making it impossible for a child to have type O blood.

    Why the other options are wrong

    • BCalculating 1/2 × 1/2 = 25% assumes both parents carry one O allele, but the father carries none and the mother carries two.
    • CThe mother passes O in 2/2 = 100% of conceptions, but equating that with type O confuses inheriting one O allele with inheriting the two required for type O.
    • DCalculating 1/2 × 1 = 50% wrongly assigns the AB father a 1/2 chance of passing O, but he has no O allele to pass.

22. Question 22 of 25Medium

    Which layer of the heart wall consists predominantly of cardiac muscle tissue responsible for pumping blood?

    1. AA. Myocardium
    2. BB. Pericardium
    3. CC. Endocardium
    4. DD. Epicardium
    Show answer to question 22

    Answer: A. Myocardium

    The myocardium is the thick, middle layer of the heart wall composed of contractile cardiac muscle cells. The endocardium is the thin, smooth endothelial lining of the heart chambers and valves.

23. Question 23 of 25Hard

    A student dissolves 11.7 g of NaCl in water and adjusts the total solution volume to 250.0 mL. The student then transfers 30.0 mL of this stock solution to another container and adds water until the total volume is 200.0 mL. What is the final NaCl molarity? Use a molar mass of 58.5 g/mol for NaCl.

    1. AA. 0.104 M
    2. BB. 0.150 M
    3. CC. 0.800 M
    4. DD. 0.120 M
    Show answer to question 23

    Answer: D. 0.120 M

    The stock contains 11.7 g ÷ (58.5 g/mol) = 0.200 mol, so its concentration is 0.200 mol ÷ 0.2500 L = 0.800 M. The transferred portion contains 0.800 mol/L × 0.0300 L = 0.0240 mol NaCl. Dilution preserves this amount of solute, giving a final concentration of 0.0240 mol ÷ 0.2000 L = 0.120 M.

    Why the other options are wrong

    • ATreats 200.0 mL as the volume of water added rather than the final solution volume, giving 0.0240 mol ÷ 0.2300 L = 0.104 M.
    • BUses 0.2000 L instead of 0.2500 L to calculate the stock concentration, then dilutes that incorrect concentration: (0.200 ÷ 0.2000) × (30.0 ÷ 200.0) = 0.150 M.
    • CReports the stock concentration, 0.200 mol ÷ 0.2500 L = 0.800 M, without accounting for the subsequent dilution.

24. Question 24 of 25Medium

    A solution has a hydrogen ion concentration of 1 × 10⁻⁴ M. If the hydrogen ion concentration increases by a factor of 100, what is the new pH of the solution?

    1. AA. pH 4
    2. BB. pH 2
    3. CC. pH 6
    4. DD. pH 8
    Show answer to question 24

    Answer: B. pH 2

    The original pH is 4. Multiplying [H⁺] by 100 gives (1 × 10⁻⁴ M) × 10² = 1 × 10⁻² M, so the new pH is 2.

    Why the other options are wrong

    • ApH 4 is the original value, before the concentration increased.
    • CpH 6 moves the wrong way; it corresponds to a 100-fold decrease in hydrogen ions.
    • DpH 8 would mean the concentration fell to 1 × 10⁻⁸ M, a 10,000-fold decrease.

25. Question 25 of 25Hard

    A DNA template segment is 3′-TAC AAA ACC TGG ATT-5′. A substitution changes the third template triplet from ACC to ACT. Translation starts at AUG; the stop codons are UAA, UAG, and UGA. What effect does this substitution have on the translated chain?

    1. AA. One amino acid is replaced, leaving a chain of four amino acids.
    2. BB. No amino acid is changed, leaving a chain of four amino acids.
    3. CC. A premature stop codon forms, reducing the chain to three amino acids.
    4. DD. A premature stop codon forms, reducing the chain to two amino acids.
    Show answer to question 25

    Answer: D. A premature stop codon forms, reducing the chain to two amino acids.

    Complementary transcription gives original mRNA 5′-AUG UUU UGG ACC UAA-3′ and altered mRNA 5′-AUG UUU UGA ACC UAA-3′. UGA terminates translation without adding an amino acid, so only AUG and UUU are translated in the altered sequence. The chain therefore decreases from four to two amino acids: 4 − 2 = 2 fewer amino acids.

    Why the other options are wrong

    • AThe substitution changes UGG to the stop codon UGA, not to a codon specifying a replacement amino acid.
    • BA third-base substitution is not necessarily silent: complementary transcription changes UGG to the stop codon UGA.
    • CThis count incorrectly includes the stop codon as an amino acid; only two amino acids are added before termination.

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