Evolutionary Mechanisms and Natural Selection (1C)
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MCAT Biological and Biochemical Foundations of Living Systems › Evolutionary Mechanisms and Natural Selection (1C)
A rare allele r in a small isolated snail population has no measurable effect on survival or reproduction. Over 20 generations, allele r fluctuates in frequency and is eventually lost, despite no consistent environmental change. The population size remains low throughout due to limited habitat area. Based on the scenario, which outcome is most consistent with genetic drift?
Allele r was eliminated because individuals lacking r actively suppressed r in their offspring through nonrandom inheritance.
Allele r was eliminated because it reduced fitness, and selection consistently removed it each generation.
Allele r was eliminated because small populations experience random allele-frequency changes that can fix or lose neutral alleles.
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.
In a population of freshwater snails, shell thickness varies continuously. A new fish predator is introduced that can crush only thin shells. After several generations, the mean shell thickness increases and variance decreases. Which statement best reflects the process of natural selection described?
Genetic drift favored thicker shells because the predator selectively removed alleles at random.
Stabilizing selection favored intermediate shells because both thin and thick shells were crushed by fish.
Directional selection favored thicker shells because individuals with thicker shells survived predation and reproduced more.
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.
A small island is colonized by 12 lizards from a mainland population where allele A has frequency 0.50. On the island, allele A frequency in the founding cohort is 0.17. No fitness differences between genotypes are detected in mark–recapture studies over the next two generations, yet allele A remains near 0.20. Based on the scenario, which outcome is most consistent with genetic drift?
Allele A returns to 0.50 because populations tend to restore original allele frequencies over time.
Allele A stays rare because the initial colonists were not a representative sample of the mainland gene pool.
Allele A increases rapidly because all individuals mutate toward allele A under island conditions.
Allele A stays rare because it reduces survival on the island, even though the effect is too small to measure.
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.
A bacterial population contains a plasmid-borne gene that confers resistance to antibiotic X but imposes a small growth cost when X is absent. In a chemostat, antibiotic X is introduced continuously for 30 days, then removed for 30 days. The resistance allele increases during exposure and decreases after removal. Which statement best reflects the process of natural selection described?
Resistance rises during antibiotic exposure because resistant cells leave more descendants under that condition.
Resistance rises and falls because allele frequencies fluctuate randomly each month regardless of environment.
Resistance rises during exposure because bacteria evolve the resistance gene purposefully when threatened.
Resistance decreases after removal because antibiotic exposure permanently eliminates the resistance allele from the population.
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.
A storm reduces a coastal bird population from 5,000 to 40 survivors in a single season. Before the storm, allele R at a neutral microsatellite locus had frequency 0.60. Among survivors, allele R frequency is 0.15. Over the next year, the population rebounds to 1,200 with allele R near 0.18 and no evidence of genotype-dependent survival. Based on the scenario, which outcome is most consistent with genetic drift?
Allele R decreased because storms increase mutation rates specifically at microsatellite loci.
Allele R decreased because the storm randomly changed which individuals contributed genes to the next generation.
Allele R decreased because it reduced storm survival, making selection the primary cause of the change.
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.
In a coastal marsh, mosquitoes vary in salt tolerance. A seawall failure increases salinity, and larvae with higher salt tolerance have higher survival to adulthood. After several generations, the distribution shifts toward higher tolerance. What adaptive advantage is most likely to occur?
Higher larval survival in salty water becomes more common because tolerant genotypes contribute more offspring.
Salt tolerance cannot evolve because environmental change affects all individuals equally.
Lower salt tolerance becomes more common because stressful environments favor less specialized phenotypes.
Salt tolerance increases because individual larvae acclimate and permanently pass that acclimation to offspring.
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.
A population of beetles has two color morphs controlled by alleles B (black) and b (brown). In a small isolated canyon population of 30 beetles, allele b frequency changes from 0.40 to 0.05 over 5 generations. No differences in survival or mating success between morphs are detected. Based on the scenario, which outcome is most consistent with genetic drift?
Allele b decreased because random variation in reproductive success had a large effect in the small population.
Allele b decreased because heterozygotes always have lower fitness than both homozygotes.
Allele b decreased because all brown beetles changed to black during development and transmitted that change genetically.
Allele b decreased because black coloration increased fitness in the canyon, driving directional selection.
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.
A bird species has two beak shapes determined largely by a single locus: narrow (N) and broad (B). During a period when only large, hard seeds are available, broad-beaked birds have higher feeding efficiency and produce more fledglings. Allele B rises from 0.25 to 0.55. Which statement best reflects the process of natural selection described?
Allele B rose because a random founder event occurred each generation, independent of seed type.
Allele B rose because birds needed broad beaks and therefore developed them, passing the change to offspring.
Allele B rose because the environment directly increased the mutation rate specifically from N to B.
Allele B rose because broad-beaked birds had higher reproductive success when hard seeds dominated.
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.
A bacterial population contains two heritable variants at a gene affecting an efflux pump: variant E (high efflux) and variant e (low efflux). When a low dose of antibiotic is introduced, cultures started with equal frequencies of E and e show that, after 24 hours, E composes 90% of the population. In antibiotic-free media, E and e remain near 50/50. What adaptive advantage is most likely to occur in the antibiotic environment?
E becomes common because small population size amplifies random fluctuations in allele frequency.
E becomes common because antibiotic exposure increases mutation rate equally in all cells, guaranteeing adaptation.
Cells exposed to antibiotic convert e into E during the exposure, increasing E without reproduction.
Cells with E reduce intracellular antibiotic concentration and leave more viable offspring than cells with e.
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.
A flowering plant species occupies a continuous habitat. A new highway creates a physical barrier that prevents pollen and seed dispersal between north and south sides. Over 200 generations, the two sides experience different pollinator communities, and flowering time becomes heritably shifted earlier in the north and later in the south. Crosses between north and south plants produce viable seeds, but F1 hybrids have greatly reduced fertility due to mismatched flowering time with local pollinators. Which event would most likely lead to speciation?
Individuals alter their flowering time plastically each season, eliminating heritable differences between populations.
Increased gene flow across the highway restores a single shared flowering-time distribution.
Reinforcement strengthens prezygotic isolation as selection disfavors low-fertility hybrids, reducing interbreeding.
A neutral allele randomly fixes in both populations, making them genetically identical at most loci.
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.