MCAT Biological and Biochemical Foundations of Living Systems Quiz: 1c Evolution Natural Selection

What this quiz covers

This quiz focuses on 1c Evolution Natural Selection, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Biological and Biochemical Foundations of Living Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Allele r was eliminated because it reduced fitness, and selection consistently removed it each generation.
2. Allele r was eliminated because small populations experience random allele-frequency changes that can fix or lose neutral alleles. (correct answer)
3. Allele r was eliminated because individuals lacking r actively suppressed r in their offspring through nonrandom inheritance.
4. Allele r was eliminated because migration introduced a dominant allele that converted r into another allele during reproduction.

Explanation: This question tests understanding of genetic drift leading to random allele loss in small populations. Genetic drift causes random fluctuations in allele frequencies that can lead to fixation or loss of alleles regardless of their fitness effects, with stronger effects in smaller populations. The neutral allele r shows no consistent directional change but fluctuates randomly until eventually lost by chance, a common outcome in small populations with limited genetic sampling each generation. The correct answer B accurately describes this stochastic process in small populations. Answer A incorrectly invokes selection on a neutral allele that has no fitness effect. To recognize drift-driven allele loss, look for: (1) small population size, (2) neutral or near-neutral alleles, and (3) random fluctuations leading to eventual fixation or loss.

Question 2

1. Genetic drift favored thicker shells because the predator selectively removed alleles at random.
2. Stabilizing selection favored intermediate shells because both thin and thick shells were crushed by fish.
3. Directional selection favored thicker shells because individuals with thicker shells survived predation and reproduced more. (correct answer)
4. Shells became thicker because individual snails thickened their shells during life and transmitted that change to offspring.

Explanation: This question tests understanding of directional selection on quantitative traits. Directional selection shifts the population mean toward one extreme by favoring variants with higher fitness. The introduced fish predator selectively preys on thin-shelled snails, increasing the mean shell thickness over generations. Thicker shells are favored because they enhance survival against predation, leading to directional selection as in choice C. Choice D fails by invoking Lamarckian inheritance, where acquired traits are passed on, which is not genetically supported. To detect directional selection, monitor if trait means shift consistently with selective pressures. This applies to traits like antibiotic resistance or pesticide tolerance.

Question 3

1. Allele A stays rare because it reduces survival on the island, even though the effect is too small to measure.
2. Allele A stays rare because the initial colonists were not a representative sample of the mainland gene pool. (correct answer)
3. Allele A increases rapidly because all individuals mutate toward allele A under island conditions.
4. Allele A returns to 0.50 because populations tend to restore original allele frequencies over time.

Explanation: This question tests understanding of genetic drift in small populations. Genetic drift is the random fluctuation of allele frequencies due to chance events, especially pronounced in small populations. Here, the small founding group of lizards on the island has a lower frequency of allele A than the mainland, and it persists without detected fitness differences. Allele A stays rare because the colonists were not representative, exemplifying the founder effect in genetic drift, as in choice B. Choice A is incorrect because it assumes undetected selection, but no evidence supports this over drift. To assess drift, evaluate if population size is small and changes lack fitness correlations. This helps differentiate drift from selection in isolated populations.

Question 4

1. Resistance rises and falls because allele frequencies fluctuate randomly each month regardless of environment.
2. Resistance rises during antibiotic exposure because resistant cells leave more descendants under that condition. (correct answer)
3. Resistance decreases after removal because antibiotic exposure permanently eliminates the resistance allele from the population.
4. Resistance rises during exposure because bacteria evolve the resistance gene purposefully when threatened.

Explanation: This question tests understanding of natural selection in microbial populations. Natural selection favors alleles that increase fitness in specific environments, leading to shifts in their frequencies. In this bacterial population, the resistance allele rises during antibiotic exposure due to higher survival of resistant cells but falls afterward due to its growth cost. Resistance changes because resistant cells leave more descendants under selection, reflecting natural selection as in choice B. Choice D fails as it implies directed evolution, which contradicts random mutation and selection. To confirm selection, observe if allele shifts align with environmental pressures on fitness. This applies to antibiotic resistance evolution in clinical settings.

Question 5

1. Allele R decreased because it reduced storm survival, making selection the primary cause of the change.
2. Allele R decreased because the storm randomly changed which individuals contributed genes to the next generation. (correct answer)
3. Allele R decreased because storms increase mutation rates specifically at microsatellite loci.
4. Allele R decreased because population growth after the storm drives allele frequencies back toward 0.50.

Explanation: This question tests understanding of genetic drift via bottlenecks. Genetic drift causes allele frequency changes by random sampling, amplified in population bottlenecks. The storm drastically reduces the bird population, randomly altering allele R frequency among survivors, with no fitness differences post-event. Allele R decreases due to this random sampling error, consistent with genetic drift as in choice B. Choice A is wrong because it attributes the change to selection, but no genotype-fitness link is evident. To identify drift, check for small effective population size and lack of selective pressures. This distinguishes bottlenecks from adaptive evolution in recovering populations.

Question 6

1. Salt tolerance cannot evolve because environmental change affects all individuals equally.
2. Lower salt tolerance becomes more common because stressful environments favor less specialized phenotypes.
3. Salt tolerance increases because individual larvae acclimate and permanently pass that acclimation to offspring.
4. Higher larval survival in salty water becomes more common because tolerant genotypes contribute more offspring. (correct answer)

Explanation: This question tests understanding of adaptation to environmental change. Natural selection favors heritable traits enhancing fitness in altered conditions. Increased salinity selects for tolerant mosquito larvae, shifting the population distribution. Higher larval survival in salty water becomes common via tolerant genotypes' reproduction, as in choice D. Choice C is wrong, suggesting Lamarckian inheritance. To assess, correlate trait shifts with survival benefits. This applies to salinity or pollution adaptations.

Question 7

1. Allele b decreased because black coloration increased fitness in the canyon, driving directional selection.
2. Allele b decreased because random variation in reproductive success had a large effect in the small population. (correct answer)
3. Allele b decreased because all brown beetles changed to black during development and transmitted that change genetically.
4. Allele b decreased because heterozygotes always have lower fitness than both homozygotes.

Explanation: This question tests understanding of genetic drift in isolated populations. Genetic drift causes unpredictable allele shifts in small populations without selection. In the small beetle population, allele b decreases without fitness differences between morphs. Allele b decreased due to random reproductive variation, consistent with drift as in choice B. Choice A fails by assuming undetected selection, but evidence points to drift. To differentiate, check population size and fitness neutrality. This is key for neutral variation in small groups.

Question 8

1. Allele B rose because broad-beaked birds had higher reproductive success when hard seeds dominated. (correct answer)
2. Allele B rose because birds needed broad beaks and therefore developed them, passing the change to offspring.
3. Allele B rose because a random founder event occurred each generation, independent of seed type.
4. Allele B rose because the environment directly increased the mutation rate specifically from N to B.

Explanation: This question tests understanding of natural selection on morphological traits. Natural selection increases alleles for traits improving resource acquisition and reproduction. Broad beaks aid hard seed handling, boosting fledgling production and allele B frequency. Allele B rose as broad-beaked birds reproduced more, as in choice A. Choice B fails by invoking need-based development, not selection. To verify, link trait to fitness outcomes. This mirrors Darwin's finch beak evolution.

Question 9

1. Cells with E reduce intracellular antibiotic concentration and leave more viable offspring than cells with e. (correct answer)
2. Cells exposed to antibiotic convert e into E during the exposure, increasing E without reproduction.
3. E becomes common because small population size amplifies random fluctuations in allele frequency.
4. E becomes common because antibiotic exposure increases mutation rate equally in all cells, guaranteeing adaptation.

Explanation: This question tests understanding of natural selection through differential survival in response to environmental challenges. Natural selection requires heritable variation that affects fitness in a given environment. The efflux pump variants show clear fitness differences: variant E pumps out antibiotics more effectively, allowing those bacterial cells to survive and reproduce in the presence of the drug. The shift from 50% to 90% E in just 24 hours demonstrates strong selection, as E-bearing cells leave more offspring. Answer A correctly identifies the mechanism - reduced intracellular antibiotic concentration leads to higher survival and reproduction. Answer B incorrectly suggests Lamarckian conversion of alleles, while C invokes drift which cannot explain such rapid, directional change. The control condition (no change in antibiotic-free media) confirms this is environment-specific selection. To identify selection in microbes, look for: (1) rapid frequency changes, (2) environment-specific fitness differences, and (3) the trait must be heritable, not induced.

Question 10

1. Increased gene flow across the highway restores a single shared flowering-time distribution.
2. Reinforcement strengthens prezygotic isolation as selection disfavors low-fertility hybrids, reducing interbreeding. (correct answer)
3. A neutral allele randomly fixes in both populations, making them genetically identical at most loci.
4. Individuals alter their flowering time plastically each season, eliminating heritable differences between populations.

Explanation: This question tests understanding of speciation through reproductive isolation mechanisms. Speciation occurs when populations evolve reproductive barriers that prevent gene flow, even if they come back into contact. The highway creates initial geographic isolation (allopatry), allowing the populations to diverge in flowering time due to different pollinator communities. The key is that hybrids have reduced fertility - a postzygotic reproductive barrier. Answer B correctly identifies reinforcement, where selection against low-fitness hybrids strengthens prezygotic barriers (like flowering time differences), completing reproductive isolation. Answer A would prevent speciation by restoring gene flow, while C describes neutral evolution unrelated to reproductive barriers. To recognize speciation scenarios, look for: (1) initial isolation allowing divergence, (2) evolution of reproductive barriers, and (3) selection maintaining or strengthening those barriers when populations meet again.

Question 11

1. Allele frequencies change because hypoxia directly induces t to mutate into T in surviving juveniles.
2. Allele T increases because individuals with TT and Tt leave more offspring after differential juvenile survival. (correct answer)
3. Allele T increases because random chance has a larger effect when population size is large.
4. Allele T increases because all genotypes have equal fitness and mating is random in the lake.

Explanation: This question tests understanding of natural selection through differential survival based on genotype. Natural selection acts when genetic variants affect fitness in ways that change allele frequencies across generations. The hypoxic events create strong selection pressure where tt juveniles die more frequently, while TT and Tt individuals survive better due to their oxygen tolerance. This differential survival means T-bearing individuals contribute more offspring, increasing T frequency from 0.20 to 0.55. Answer B correctly identifies this mechanism of differential juvenile survival leading to increased reproductive contribution. The deviation from Hardy-Weinberg equilibrium specifically during hypoxic events confirms ongoing selection. Answer A incorrectly suggests direct mutation induction, while D contradicts the evidence of unequal fitness. To identify selection, look for: (1) consistent fitness differences among genotypes, (2) directional change in allele frequencies, and (3) deviations from random mating expectations.

Question 12

1. Allele frequencies shifted because predators created different selective pressures in each environment, changing relative reproductive success of morphs. (correct answer)
2. Allele frequencies shifted because the fish intentionally changed color to match the background and passed that change to offspring.
3. Allele frequencies shifted because random genetic drift is stronger in large enclosures, overriding environmental differences.
4. Allele frequencies shifted because each enclosure began with different allele frequencies, so selection cannot be implicated without identical starting conditions.

Explanation: This question tests understanding of natural selection through predator-mediated differential survival. Natural selection occurs when environmental factors cause consistent differences in reproductive success among genetic variants. In each lake, bird predation created different selective pressures: green fish were more vulnerable in murky water, while silver fish were more vulnerable among plants. When fish were transplanted, local predation patterns caused allele frequencies to shift toward the locally camouflaged morph over 5 generations. Answer B incorrectly invokes intentional color change and inheritance of acquired traits, C misunderstands drift in large populations, and D wrongly suggests selection requires identical starting conditions. To identify predator-mediated selection, look for: (1) environment-specific survival advantages, (2) consistent directional changes matching local conditions, and (3) heritable traits affecting predation risk.

Question 13

1. The evolution of behavioral isolation via divergence in mating calls that reduces gene flow between the two populations. (correct answer)
2. A temporary increase in food availability that raises population sizes on both sides, increasing gene flow across the lava field.
3. A single season in which a storm randomly changes allele frequencies at neutral loci, after which mating patterns return to normal.
4. The appearance of an allele that improves larval growth equally in both populations, preventing divergence by natural selection.

Explanation: This question tests understanding of speciation through the evolution of reproductive isolation. Speciation occurs when populations accumulate differences that prevent interbreeding, with behavioral isolation being one important mechanism. The populations already show divergence in mating calls and female preferences, creating prezygotic isolation where individuals from different populations rarely mate despite being physically capable of producing viable offspring. This behavioral isolation reduces gene flow and allows further divergence, potentially leading to complete reproductive isolation. Answer B would increase gene flow (opposing speciation), C describes temporary drift without lasting effect, and D describes parallel evolution that would maintain similarity. To identify speciation mechanisms, look for: (1) reduced gene flow between populations, (2) evolution of mating barriers, and (3) divergence in traits affecting reproduction.

Question 14

1. R increased because pesticide exposure caused insects to mutate into the resistant genotype during their lifetime.
2. R increased because resistant individuals had higher survival and left more offspring when pesticide was present. (correct answer)
3. R increased because allele frequencies change only by random drift in large populations.
4. R increased because costly traits are always favored when environments are stable.

Explanation: This question tests understanding of natural selection for resistance. Natural selection amplifies beneficial alleles under pressure, despite costs elsewhere. Pesticide application favors resistant insects, increasing allele R. R increased as resistant individuals survived and reproduced more, as in choice B. Choice A fails by suggesting induced mutations in adults. To confirm, observe frequency rises with application. This parallels insecticide resistance evolution.

Question 15

1. D increased in the polluted area because darker moths were less likely to be eaten and therefore reproduced more. (correct answer)
2. D increased in the polluted area because soot caused moths to darken during life and pass that change to offspring.
3. D increased in the polluted area because random drift always produces the same allele-frequency change in neighboring regions.
4. D remained rare in the unpolluted area because selection cannot differ across environments within the same species.

Explanation: This question tests understanding of spatially varying selection. Natural selection drives local adaptations when environments differ, as in industrial melanism. Soot darkens bark, favoring dark moths via reduced predation, increasing D in polluted areas. D increased because darker moths reproduced more, as in choice A. Choice B is wrong, invoking Lamarckian change. To check, compare frequencies across environments. This classic example illustrates selection gradients.

Question 16

1. Reduced gene flow and divergent selection on flower tube length lead to reproductive isolation between the two sides. (correct answer)
2. Speciation is unlikely because both sides experience pollination, so selection pressures are effectively identical.
3. Speciation occurs because plants intentionally lengthen or shorten tubes during life and pass the new length genetically.
4. Cross-pollination becomes rare because allele frequencies must become equal on both sides as a population grows.

Explanation: This question tests understanding of speciation mechanisms, particularly allopatric speciation driven by natural selection. Speciation occurs when populations become reproductively isolated, often due to geographic barriers reducing gene flow and allowing divergent evolution under different selection pressures. In this scenario, the highway acts as a barrier splitting the plant population, with different pollinators on each side imposing distinct selection on flower tube length. Choice A is correct because reduced gene flow combined with divergent selection leads to adaptations that cause reproductive isolation, as plants on each side evolve tube lengths suited to their local pollinators, making cross-pollination rare. Choice C is incorrect because it describes Lamarckian inheritance, where acquired traits are passed on, which does not align with modern evolutionary theory based on genetic variation and natural selection. To check similar problems, identify if a barrier reduces gene flow and if environments differ enough to drive divergent selection toward isolation. Remember, speciation requires heritable changes accumulating over generations, not individual adaptations within a lifetime.

Question 17

1. The increase in D is most consistent with random sampling error because allele frequencies can fluctuate each generation.
2. The ash caused snails to mutate from L to D during their lifetimes, increasing D in the next generation.
3. Differential survival of dark-shelled snails increased their reproductive contribution, raising the frequency of D. (correct answer)
4. Predation reduced population size, so D increased because fewer individuals always increases adaptive alleles.

Explanation: This question tests understanding of natural selection as differential survival and reproduction based on heritable traits. Natural selection occurs when individuals with certain heritable traits leave more offspring than others due to environmental pressures. In this scenario, the volcanic ash creates a new selective environment where dark shells provide camouflage from predatory birds, while light shells become more visible and vulnerable. The increase in allele D frequency from 0.40 to 0.78 results from dark-shelled snails (DD and DL genotypes) surviving predation better and contributing more offspring to subsequent generations. Answer C correctly identifies this as differential survival leading to increased reproductive contribution. Answer B incorrectly invokes Lamarckian inheritance where traits acquired during life are passed on, which doesn't occur in genetic systems. To identify natural selection, look for: (1) heritable variation, (2) differential fitness, and (3) change in allele frequencies matching the fitness differences.

Question 18

1. Temporal isolation increases as breeding times diverge, reducing gene flow between shallow- and deep-spawning groups. (correct answer)
2. Both groups converge on the same breeding time, increasing interbreeding and homogenizing allele frequencies.
3. A random storm changes allele frequencies equally in both groups, ensuring they remain a single species.
4. Individuals in warm water develop earlier breeding behavior during life and pass it directly to offspring without genetic change.

Explanation: This question tests understanding of speciation via reproductive isolation. Speciation occurs when populations diverge genetically due to barriers to gene flow, such as temporal isolation. The temperature shift causes divergent breeding times between shallow- and deep-water fish, reducing hybridization. Temporal isolation increases as breeding diverges, limiting gene flow and promoting speciation as in choice A. Choice D fails by describing acquired traits without genetic basis, not leading to isolation. To check for speciation potential, examine if reproductive barriers reduce interbreeding. This applies to allopatric or sympatric speciation in varying environments.

Question 19

1. Allele L decreases because traits that conserve water are disadvantageous when water is scarce.
2. Increased stomatal density becomes more common because drought triggers plants to mutate from L to H.
3. No change in allele frequencies is expected because selection cannot act on traits controlled by a single locus.
4. Increased water-use efficiency becomes more common because L carriers leave more offspring under drought conditions. (correct answer)

Explanation: This question tests understanding of adaptive advantages under natural selection. Natural selection promotes alleles that confer survival or reproductive benefits in challenging environments. During the drought, low stomatal density (allele L) reduces water loss, leading to higher seed set and allele frequency increase. Increased water-use efficiency becomes common as L carriers reproduce more, providing an adaptive advantage as in choice D. Choice B is incorrect because it suggests drought induces mutations, but selection acts on existing variation. To evaluate adaptation, assess if trait frequencies change with environmental fitness benefits. This principle extends to drought resistance in crops or wild plants.

Question 20

1. Allele a declines because herbivores randomly remove plants regardless of phenotype, shifting allele frequencies by chance.
2. Allele a declines because individuals with the pale phenotype contribute fewer alleles to the next generation. (correct answer)
3. Allele a declines because green plants induce mutations in neighboring pale plants, converting a to A.
4. Allele a declines because populations evolve toward traits that are aesthetically favored by herbivores.

Explanation: This question tests understanding of natural selection on recessive traits. Natural selection reduces frequencies of deleterious alleles if they lower fitness. The pale leaf phenotype (recessive a) increases visibility to herbivores, reducing seed production and allele frequency. Allele a declines because pale plants contribute fewer alleles, reflecting selection as in choice B. Choice C fails by suggesting induced mutations, not supported by evidence. To identify selection, link phenotype fitness to allele changes. This extends to camouflage or defense traits in plants.