College Biology Quiz: Variations In Populations
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Variations In PopulationsQuestion 1 of 17

A plant population exhibits continuous variation in leaf length controlled by four genes (A, B, C, D), each with two alleles. The minimum leaf length (aabbccdd) is 8 cm and maximum length (AABBCCDD) is 24 cm. An individual with genotype AaBbCcDd would most likely have a leaf length closest to:

12 cm, because heterozygotes show reduced expression compared to homozygotes
14 cm, because each contributing allele adds an equal amount to the base length
16 cm, because the individual has exactly half of the maximum contributing alleles
18 cm, because heterozygote effects are amplified in quantitative traits
20 cm, because dominant alleles mask the effects of recessive alleles completely
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College Biology Quiz

College Biology Quiz: Variations In Populations

Practice Variations In Populations in College Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Variations In Populations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Biology.

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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.

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Question 1

A plant population exhibits continuous variation in leaf length controlled by four genes (A, B, C, D), each with two alleles. The minimum leaf length (aabbccdd) is 8 cm and maximum length (AABBCCDD) is 24 cm. An individual with genotype AaBbCcDd would most likely have a leaf length closest to:

  1. 12 cm, because heterozygotes show reduced expression compared to homozygotes
  2. 14 cm, because each contributing allele adds an equal amount to the base length
  3. 16 cm, because the individual has exactly half of the maximum contributing alleles (correct answer)
  4. 18 cm, because heterozygote effects are amplified in quantitative traits
  5. 20 cm, because dominant alleles mask the effects of recessive alleles completely
Explanation: This question tests quantitative genetics, where multiple genes contribute additively to a continuous trait. When you encounter problems involving multiple genes controlling one trait, think about how each contributing allele adds incrementally to the phenotype. In this additive model, each capital letter allele contributes equally to leaf length. The range is 24 - 8 = 16 cm across 8 possible contributing alleles (two per gene). Therefore, each contributing allele adds 16 ÷ 8 = 2 cm to the base length of 8 cm. The genotype AaBbCcDd has exactly 4 contributing alleles (one from each gene pair). Starting from the base length: 8 cm + (4 × 2 cm) = 16 cm. This represents exactly half the maximum contributing alleles (4 out of 8), placing this individual at the midpoint between minimum and maximum lengths. Choice A incorrectly suggests heterozygotes have reduced expression - this confuses dominance relationships with additive effects. In quantitative traits, heterozygotes contribute normally. Choice B gives 14 cm, which would result from miscalculating the contribution per allele as 1.5 cm instead of 2 cm. Choice D claims heterozygote amplification at 18 cm, but there's no biological basis for this - additive models assume equal, independent contributions. For quantitative genetics problems, always identify the total range, count the maximum possible contributing alleles, calculate the contribution per allele, then count contributing alleles in your specific genotype. The math should place heterozygotes at intermediate positions between the extremes.

Question 2

Researchers studying a bird population find that beak depth follows a normal distribution with mean = 12 mm and standard deviation = 2 mm. After a drought reduces available food to only large, hard seeds, birds with beak depths less than 14 mm have reduced survival. What evolutionary change is most likely in the next generation?

  1. Mean beak depth will decrease because small-beaked birds are more efficient feeders
  2. Mean beak depth will remain unchanged because the trait is not heritable
  3. Mean beak depth will increase because selection favors individuals with deeper beaks (correct answer)
  4. Beak depth variance will increase because extreme phenotypes have advantages
  5. Beak depth distribution will become bimodal with two distinct size classes
Explanation: This question tests natural selection and how environmental pressures drive evolutionary change in populations. When you encounter scenarios describing trait distributions and selective pressures, think about which individuals will have higher reproductive success and how this affects the population mean. The drought creates directional selection against birds with beaks less than 14 mm deep, since they can't effectively crack the remaining large, hard seeds. In a normal distribution with mean = 12 mm and standard deviation = 2 mm, a 14 mm beak depth is one standard deviation above the mean. This means roughly 84% of the current population has beaks smaller than 14 mm and faces reduced survival. The birds with deeper beaks (≥14 mm) will have higher survival rates and greater reproductive success, passing their genes for deeper beaks to offspring. This shifts the population mean toward deeper beaks in the next generation. Looking at the wrong answers: A) contradicts the scenario since small-beaked birds actually have reduced survival, not increased efficiency. B) assumes the trait isn't heritable, but beak depth in birds is typically a heritable trait influenced by genetics. D) suggests variance increases, but directional selection actually tends to reduce variance by eliminating one extreme of the distribution. Remember that directional selection occurs when one extreme phenotype has a survival advantage. Always identify which trait values are favored by the selective pressure, then predict how this will shift the population mean in that direction.

Question 3

In a population genetics experiment, two isolated populations of the same species are maintained for 50 generations. Population X has 1000 individuals while Population Y has 50 individuals. Both start with identical allele frequencies (p = q = 0.5) at a neutral locus. Which outcome is most probable after 50 generations?

  1. Both populations will maintain p = q = 0.5 due to the stability of neutral alleles
  2. Population X will show greater change in allele frequencies due to stronger selection
  3. Population Y will more likely reach fixation (p = 1.0 or p = 0.0) for one allele (correct answer)
  4. Both populations will evolve identical allele frequencies through parallel evolution
  5. Population X will lose more alleles due to increased mutation pressure
Explanation: When you encounter population genetics problems involving different population sizes and neutral alleles, you're dealing with genetic drift—the random fluctuation of allele frequencies due to chance events during reproduction. Genetic drift's strength is inversely related to population size. In smaller populations, random sampling during gamete formation has a much stronger effect on allele frequencies across generations. Think of it like flipping coins: if you flip 10 coins, getting 7 heads (70%) is reasonably likely, but getting 700 heads out of 1000 flips is extremely unlikely. Population Y, with only 50 individuals, experiences strong genetic drift. Over 50 generations, this random "sampling error" compounds, making it highly probable that one allele will eventually reach fixation (frequency = 1.0) while the other disappears entirely. Population X, with 1000 individuals, experiences much weaker drift, so its allele frequencies will fluctuate but remain closer to the starting values. Option A is incorrect because neutral alleles aren't inherently stable—they're subject to drift regardless of selection. Option B wrongly suggests selection is involved; the question specifically mentions neutral alleles, and drift (not selection) is the driving force here. Option D is wrong because genetic drift is random—each population will drift in unpredictable directions, making identical outcomes extremely unlikely. Remember: when comparing populations of different sizes with neutral alleles, always predict that smaller populations will show more extreme changes in allele frequencies, including higher probability of fixation, due to stronger genetic drift.

Question 4

A population of moths exhibits continuous variation in wing length, with measurements ranging from 12 mm to 28 mm. After a severe storm, only moths with wing lengths between 18-22 mm survive to reproduce. If wing length has a heritability of 0.6 and the mean wing length of survivors is 20 mm while the original population mean was 17 mm, what is the expected mean wing length in the next generation?

  1. 17.0 mm, because environmental selection pressure does not affect heritable trait expression
  2. 18.8 mm, because the response equals heritability times the selection differential (correct answer)
  3. 20.0 mm, because all offspring will inherit the same average wing length as survivors
  4. 22.4 mm, because directional selection amplifies the trait beyond the parental mean value
  5. 25.0 mm, because extreme selection events lead to maximum possible trait expression
Explanation: When you encounter questions about evolutionary response to selection, you're dealing with quantitative genetics—specifically how traits change across generations when both genetics and environment play a role. The key formula here is the breeder's equation: R=h2×SR = h^2 \times S, where R is the response to selection (change in trait mean), h² is heritability (0.6), and S is the selection differential (difference between selected parents and original population). First, calculate the selection differential: survivors averaged 20 mm while the original population averaged 17 mm, so S = 20 - 17 = 3 mm. Then apply the formula: R = 0.6 × 3 = 1.8 mm. The expected mean for the next generation is the original mean plus the response: 17 + 1.8 = 18.8 mm. This confirms answer B. Answer A incorrectly assumes environmental selection has no effect on heritable traits—but that's exactly how evolution works when traits have genetic components. Answer C makes the mistake of assuming offspring will perfectly match their parents' average, ignoring that heritability is less than 1.0, meaning some variation comes from non-heritable factors. Answer D suggests selection amplifies traits beyond what the math predicts, but the response is always proportional to heritability—it can't exceed what the genetic component allows. Remember: heritability acts like a "discount factor" on selection. When h² = 1.0, offspring fully inherit parental traits; when h² = 0.6, they inherit only 60% of the difference between selected parents and the original population.

Question 5

A researcher studies genetic variation in a lizard population by examining allele frequencies at a single locus over three generations. The data shows: Generation 1: A₁ = 0.7, A₂ = 0.3; Generation 2: A₁ = 0.65, A₂ = 0.35; Generation 3: A₁ = 0.58, A₂ = 0.42. Which factor most likely explains this pattern of change?

  1. Directional selection favoring the A₂ allele, causing systematic frequency increases over time (correct answer)
  2. Genetic drift in a small population, producing random fluctuations in allele frequencies
  3. Gene flow from neighboring populations with higher A₁ allele frequencies
  4. Mutation pressure converting A₁ alleles to A₂ alleles at high rates
  5. Balancing selection maintaining both alleles at intermediate frequencies indefinitely
Explanation: When analyzing changes in allele frequencies across generations, you need to identify which evolutionary force best explains the observed pattern. Look for consistent directional changes versus random fluctuations, and consider the magnitude and consistency of the shifts. The data shows a systematic, consistent decrease in A₁ frequency (0.7 → 0.65 → 0.58) with a corresponding increase in A₂ frequency (0.3 → 0.35 → 0.42). This represents a steady, directional change of approximately 0.05-0.07 per generation, indicating strong selective pressure favoring A₂. The consistency and magnitude suggest directional selection is driving these changes, making A correct. B is wrong because genetic drift produces random, unpredictable changes in allele frequencies. You'd expect to see fluctuations going both directions—sometimes A₁ increasing, sometimes decreasing—not this consistent unidirectional pattern over three generations. C is incorrect because gene flow from populations with higher A₁ frequencies would increase A₁ in this population, not decrease it as observed. The data shows A₁ declining consistently, which contradicts this explanation. D fails because mutation rates high enough to cause 0.05-0.07 frequency changes per generation are unrealistic. Typical mutation rates are around 10⁻⁶ to 10⁻⁹ per base pair per generation—far too low to produce these dramatic shifts. Study tip: When interpreting allele frequency data, consistent directional changes over multiple generations strongly indicate selection, while random back-and-forth fluctuations suggest drift. The pattern's consistency is your key diagnostic tool.

Question 6

A population of snails exhibits shell banding patterns controlled by multiple genes. Researchers observe that in coastal areas, 40% of snails have thick bands, 35% have thin bands, and 25% have no bands. In forest populations of the same species, the frequencies are 15%, 25%, and 60% respectively. What type of variation pattern best explains these differences?

  1. Discrete variation maintained by single gene effects across all populations consistently
  2. Continuous variation modified by environmental gradients affecting gene expression patterns
  3. Polymorphic variation maintained by different selection pressures in distinct habitat types (correct answer)
  4. Neutral variation showing random drift effects independent of environmental factors
  5. Developmental variation caused by nutritional differences during shell growth periods
Explanation: When you encounter questions about trait frequencies varying dramatically between different environments, you're looking at how natural selection shapes populations in response to local conditions. The key evidence here is the striking difference in banding patterns between habitats: coastal snails are predominantly banded (75% combined thick and thin bands), while forest snails are mostly unbanded (60%). This dramatic shift suggests each environment favors different phenotypes, which is the hallmark of polymorphic variation maintained by habitat-specific selection pressures. Answer C correctly identifies this pattern. Different environments likely impose different selective advantages - perhaps banded shells provide better camouflage against coastal rocks and seaweed, while unbanded shells blend better with forest substrates. The multiple genes controlling banding create a polymorphism (multiple distinct forms) that natural selection maintains differently in each habitat. Answer A is wrong because discrete variation from single genes wouldn't show such environment-specific frequency differences - the ratios would remain consistent across populations. Answer B incorrectly describes continuous variation, but shell banding represents discrete categories (thick, thin, none), not a continuous spectrum. Answer D suggests neutral variation with random drift, but random processes wouldn't produce such consistent, environment-specific patterns - you'd expect more random frequency differences between populations. For college biology exams, remember that when trait frequencies shift dramatically between environments in predictable ways, think "local adaptation through natural selection" rather than random genetic drift or simple Mendelian inheritance patterns.

Question 7

In a fish population, scale color is controlled by two alleles (R and r) where RR individuals are red, Rr individuals are orange, and rr individuals are yellow. A breeding program starts with frequencies of R = 0.6 and r = 0.4. After random mating for one generation, what percentage of the population will display the orange phenotype?

  1. 24%, representing the frequency of heterozygous individuals in the population
  2. 36%, representing the combined frequency of red homozygotes and heterozygotes
  3. 48%, representing twice the frequency of heterozygous genotypes in offspring (correct answer)
  4. 52%, representing the majority phenotype class after random mating occurs
  5. 60%, representing the frequency of the dominant allele in the population
Explanation: When you encounter questions about allele frequencies and phenotype prediction, you're working with Hardy-Weinberg equilibrium principles. The key is recognizing that allele frequencies allow you to predict genotype frequencies after random mating. Starting with allele frequencies R = 0.6 and r = 0.4, you can calculate the expected genotype frequencies using the Hardy-Weinberg equation: p2+2pq+q2=1p^2 + 2pq + q^2 = 1, where p = frequency of R and q = frequency of r. The orange phenotype corresponds to the heterozygous genotype (Rr), which has frequency 2pq=2(0.6)(0.4)=0.48=48%2pq = 2(0.6)(0.4) = 0.48 = 48\%. This confirms answer C is correct. Answer A (24%) represents half the heterozygote frequency, suggesting a calculation error where someone forgot to multiply by 2 in the 2pq2pq term. Answer B (36%) incorrectly calculates p2=(0.6)2=0.36p^2 = (0.6)^2 = 0.36, which actually represents the red homozygote (RR) frequency alone, not red plus orange. Answer D (52%) might result from adding the dominant allele frequency (60%) plus something else incorrectly, but doesn't correspond to any meaningful Hardy-Weinberg calculation. For Hardy-Weinberg problems, always remember the pattern: homozygote frequencies are the square of individual allele frequencies (p2p^2 and q2q^2), while heterozygote frequency is always 2pq2pq. The "2" in the heterozygote calculation is crucial because there are two ways to form a heterozygote (R from mom, r from dad OR r from mom, R from dad).

Question 8

A plant breeder crosses two varieties that differ in height: one averages 100 cm (genotype aabbcc) and another averages 130 cm (genotype AABBCC). The F₁ offspring average 115 cm tall. When F₁ individuals are self-fertilized, which statement best describes the expected variation in F₂ offspring?

  1. All F₂ plants will be 115 cm tall, maintaining the intermediate parental height exactly
  2. F₂ heights will range from 100-130 cm, with most individuals near the extreme values
  3. F₂ heights will range from 100-130 cm, with most individuals near 115 cm intermediate height (correct answer)
  4. F₂ heights will exceed the original range, showing transgressive segregation beyond parental limits
  5. F₂ heights will show three distinct classes: 100 cm, 115 cm, and 130 cm only
Explanation: This question tests quantitative genetics, specifically polygenic inheritance where multiple genes contribute additively to a single trait. When you see crosses involving multiple gene pairs (like aabbcc × AABBCC) affecting one measurable characteristic, think about how alleles combine to produce continuous variation. The F₁ offspring (AaBbCc) average 115 cm, exactly intermediate between the parents (100 cm and 130 cm). This tells us each dominant allele contributes equally to height. Since there are 6 contributing alleles total in AABBCC, each adds 5 cm above the baseline of 100 cm (130-100=30 cm ÷ 6 alleles = 5 cm each). The F₁ has 3 contributing alleles, giving 100 + (3×5) = 115 cm. In the F₂ generation from AaBbCc × AaBbCc, offspring can inherit anywhere from 0 to 6 contributing alleles. However, the distribution follows a bell curve pattern. Most F₂ individuals will have 2, 3, or 4 contributing alleles (heights of 110, 115, and 120 cm), with fewer having the extreme combinations of 0-1 or 5-6 alleles. This creates the characteristic normal distribution with most individuals near the middle value of 115 cm. Answer A is wrong because genetic segregation creates variation, not uniformity. Answer B incorrectly suggests most individuals cluster at extremes rather than the middle. Answer D describes transgressive segregation, which occurs when parental strains have different combinations of contributing alleles—not applicable here since parents represent the true extremes. Remember: In polygenic inheritance, F₂ distributions are bell-shaped with peaks at intermediate values, not at the extremes.

Question 9

Researchers studying coat color in mice discover that dark coat color is controlled by multiple genes with additive effects. In a cross between very light mice (mean color score = 2) and very dark mice (mean color score = 18), the F₁ offspring have a mean color score of 10. What can be concluded about the genetic architecture of this trait?

  1. Coat color shows complete dominance, with dark alleles being recessive to light alleles
  2. Coat color demonstrates codominance, with both allele types expressed equally in heterozygotes
  3. Coat color exhibits additive gene action, with no dominance effects between contributing alleles (correct answer)
  4. Coat color involves epistatic interactions, with some genes masking the effects of others
  5. Coat color shows environmental determination, with genetic factors having minimal influence
Explanation: When you encounter questions about quantitative traits controlled by multiple genes, look for clues about the genetic architecture in the phenotypic patterns of crosses. The key insight here is examining what the F₁ generation tells us about gene action. The F₁ offspring have a mean color score of 10, which is exactly intermediate between the parental means (2 and 18). This perfect intermediate phenotype is the hallmark of additive gene action. In additive inheritance, each contributing allele adds a small, equal effect to the phenotype, and heterozygotes fall precisely halfway between the homozygous parents. Since 2+182=10\frac{2 + 18}{2} = 10, the data perfectly matches this expectation. Choice A is incorrect because complete dominance would produce F₁ offspring that resemble one parent (either very light or very dark), not an intermediate phenotype. Choice B misapplies codominance, which typically refers to situations where both alleles are simultaneously expressed (like AB blood type), not quantitative traits with intermediate values. Choice D describes epistasis, where genes interact by masking each other's effects, but epistatic interactions would likely produce more complex ratios and wouldn't necessarily yield this clean intermediate phenotype. Remember this pattern: when F₁ offspring from extreme parents show a mean phenotype exactly halfway between the parental means, you're seeing additive gene action. This is a key signature of polygenic inheritance where multiple genes contribute small, cumulative effects to produce continuous variation in traits like height, weight, or coat color.

Question 10

In a study of flower color variation, researchers find that red flowers (RR) have fitness = 1.0, pink flowers (Rr) have fitness = 0.9, and white flowers (rr) have fitness = 0.8. Starting with allele frequencies R = 0.5 and r = 0.5, what type of selection is occurring?

  1. Directional selection favoring the R allele, because red homozygotes have highest fitness (correct answer)
  2. Balancing selection maintaining both alleles, because heterozygotes have intermediate fitness
  3. Disruptive selection favoring extremes, because homozygotes outperform heterozygotes consistently
  4. Stabilizing selection favoring intermediates, because pink flowers have moderate fitness values
  5. Neutral selection with no fitness effects, because all genotypes survive to reproduce
Explanation: When analyzing selection types, you need to examine how fitness values relate to genotypes and determine which trait is being favored. The key is identifying whether selection favors one extreme, both extremes, or the intermediate phenotype. Looking at the fitness values: RR (red) = 1.0, Rr (pink) = 0.9, and rr (white) = 0.8, there's a clear pattern where fitness decreases as you move from red to white flowers. The RR genotype has the highest fitness, followed by Rr, then rr. This creates a fitness gradient that consistently favors the R allele over the r allele. When one allele consistently confers higher fitness across genotypic combinations, this indicates directional selection. Answer A correctly identifies this as directional selection favoring the R allele, since red homozygotes have the highest fitness and there's a clear fitness advantage associated with having more R alleles. Answer B incorrectly suggests balancing selection, which would require some mechanism maintaining both alleles long-term, such as heterozygote advantage or frequency-dependent selection. Here, the R allele is consistently favored. Answer C misidentifies this as disruptive selection, which would favor both homozygotes (RR and rr) over the heterozygote (Rr). Instead, we see the opposite pattern. Answer D incorrectly calls this stabilizing selection, which would favor the intermediate phenotype (pink flowers) with the highest fitness. However, pink flowers have intermediate, not maximum, fitness. Study tip: For selection questions, rank the fitness values and ask: "Which phenotype wins?" One extreme = directional; both extremes = disruptive; middle = stabilizing.

Question 11

A conservation biologist studying an endangered butterfly species finds that wing pattern variation is controlled by a single gene with three alleles (A₁, A₂, A₃) showing the following fitness values in the current environment: A₁A₁ = 0.8, A₁A₂ = 1.0, A₁A₃ = 0.9, A₂A₂ = 0.7, A₂A₃ = 0.95, A₃A₃ = 0.6. Which prediction about long-term allele frequency changes is most accurate?

  1. A₁ will reach fixation because it appears in the highest fitness genotype
  2. A₂ will be eliminated because it has the lowest homozygote fitness
  3. A₃ will increase in frequency because it shows the highest average fitness
  4. A₁ and A₂ will be maintained by balancing selection due to heterozygote advantage (correct answer)
  5. All three alleles will be lost due to inbreeding depression in small populations
Explanation: When analyzing natural selection with multiple alleles, you need to examine how each allele performs across all possible genotype combinations, not just focus on individual genotypes in isolation. Let's trace what happens to each allele. The key insight is recognizing heterozygote advantage: notice that A₁A₂ has the highest fitness (1.0), while both homozygotes involving these alleles have lower fitness (A₁A₁ = 0.8, A₂A₂ = 0.7). This creates balancing selection—natural selection actually maintains both A₁ and A₂ in the population because they perform best together, even though they're less fit when paired with themselves. Answer D correctly identifies this balancing selection. When heterozygotes have higher fitness than either homozygote, both alleles persist long-term because eliminating either one would reduce the frequency of the superior heterozygote combination. Answer A incorrectly assumes that appearing in the fittest genotype guarantees fixation, but A₁ alone (as A₁A₁) has relatively low fitness. Answer B makes the opposite error—while A₂A₂ has low fitness, A₂ is actually favored because of its performance in the A₁A₂ combination. Answer C misunderstands how to calculate "average fitness"—you can't simply average across genotypes without considering how allele frequencies affect which combinations actually occur in the population. For multiple-allele selection problems, always look for heterozygote advantage patterns first. When the best genotype is a heterozygote, expect balancing selection to maintain genetic diversity rather than driving any allele to fixation or elimination.

Question 12

A population of beetles shows the following genotype frequencies at a locus affecting wing coloration: AA = 0.49, Aa = 0.42, aa = 0.09. If this population undergoes random mating for one generation with no other evolutionary forces acting, what will be the frequency of the A allele in the offspring generation?

  1. 0.49, equal to the frequency of AA homozygotes in the parent generation
  2. 0.65, calculated from the weighted average of parental genotype contributions
  3. 0.70, calculated from adding AA frequency plus half the Aa frequency (correct answer)
  4. 0.91, representing the combined frequency of all individuals carrying the A allele
  5. 0.42, equal to the frequency of heterozygotes in the parent generation
Explanation: When you encounter questions about allele frequencies and random mating, you're working with Hardy-Weinberg principles. The key insight is that allele frequencies remain constant across generations when only random mating occurs, regardless of the starting genotype frequencies. To find the frequency of the A allele, you need to account for all the ways this allele appears in the population. The A allele is present in two forms: as both alleles in AA homozygotes, and as one allele in Aa heterozygotes. Since AA individuals contribute two A alleles and Aa individuals contribute one A allele per two total alleles, the calculation is: frequency of A = (frequency of AA) + ½(frequency of Aa) = 0.49 + ½(0.42) = 0.49 + 0.21 = 0.70. Answer A incorrectly assumes the A allele frequency equals the AA genotype frequency, ignoring the A alleles carried by heterozygotes. Answer B uses an incorrect "weighted average" approach that doesn't properly account for how alleles are distributed among genotypes. Answer D represents a common trap—it adds AA and Aa frequencies together (0.49 + 0.42 = 0.91), but this counts individuals carrying A alleles, not the actual allele frequency. The correct answer is C: 0.70. Remember this formula for any Hardy-Weinberg problem: allele frequency = homozygote frequency + ½(heterozygote frequency). This relationship holds because heterozygotes carry one copy of each allele, while homozygotes carry two copies of the same allele.

Question 13

In a laboratory population of fruit flies, researchers maintain constant population size but vary the number of breeding individuals each generation. Population A uses 50 breeding pairs, while Population B uses 5 breeding pairs. After 20 generations, which outcome regarding genetic variation is most likely?

  1. Both populations will maintain identical levels of genetic variation due to controlled breeding
  2. Population A will retain more genetic variation due to reduced impact of genetic drift (correct answer)
  3. Population B will show greater genetic variation due to more intense selection pressure
  4. Population A will lose variation faster due to increased competition among individuals
  5. Both populations will gain variation at equal rates through spontaneous mutations
Explanation: When you encounter questions about population genetics and breeding group sizes, focus on the relationship between effective population size and genetic drift. Genetic drift—the random change in allele frequencies—has a much stronger impact in smaller populations than larger ones. Population A, with 50 breeding pairs (100 individuals), represents a relatively large effective population size. In larger populations, random sampling effects during reproduction have less influence on overall allele frequencies, so genetic variation is better preserved across generations. Population B, with only 5 breeding pairs (10 individuals), experiences much stronger genetic drift. With so few breeding individuals, random events can dramatically shift allele frequencies or even eliminate alleles entirely from the gene pool. Looking at the wrong answers: Choice A incorrectly assumes that controlled breeding conditions eliminate the effects of population size—genetic drift still operates regardless of laboratory control. Choice C makes the opposite error, suggesting smaller populations gain variation. However, smaller populations lose variation faster, and "selection pressure" isn't necessarily more intense just because the population is smaller. Choice D reverses the relationship entirely, incorrectly claiming larger populations lose variation faster due to competition, when competition doesn't directly cause loss of genetic variation the way drift does. For college biology exams, remember this key principle: smaller effective population sizes always lead to faster loss of genetic variation due to stronger genetic drift effects. When you see population genetics questions comparing different breeding group sizes, the larger group will typically maintain more genetic diversity over time.

Question 14

In a population genetics study, researchers find that a recessive lethal allele (q) has a frequency of 0.02 in newborns but 0.015 in adults. If the population has random mating and the lethal allele causes death before reproduction, what is the expected allele frequency in the next generation of newborns?

  1. 0.010, because selection reduces the allele frequency by half each generation
  2. 0.015, because the adult frequency represents the new equilibrium value (correct answer)
  3. 0.018, because selection removes homozygous recessive individuals before reproduction
  4. 0.020, because newborn frequencies remain constant despite adult mortality
  5. 0.025, because reduced population size increases the relative frequency
Explanation: When you encounter population genetics problems involving lethal alleles, focus on understanding that selection changes allele frequencies between generations, and the key is identifying which frequency represents the breeding population. In this scenario, the recessive lethal allele kills homozygous recessive individuals (qq) before they can reproduce. The frequency drops from 0.02 in newborns to 0.015 in adults because selection eliminates the qq individuals. Crucially, only the surviving adults (with frequency 0.015) actually reproduce to create the next generation. Since these adults mate randomly, and no further evolutionary forces are mentioned, the allele frequency in their offspring will match the parental frequency of 0.015. Looking at the wrong answers: Choice A incorrectly assumes selection reduces frequency by a fixed proportion each generation, but selection against recessive alleles doesn't follow this simple pattern. Choice C attempts a calculation but misunderstands that we need the breeding population frequency, not a mathematical reduction from the newborn frequency. Choice D fails to recognize that selection pressure fundamentally changes allele frequencies between generations - the newborn frequency can't remain constant when individuals are dying before reproduction. The adult frequency of 0.015 represents the actual breeding population that produces the next generation, making B correct. Study tip: In lethal allele problems, always identify which generation is doing the breeding. The allele frequency in the reproductive population determines the next generation's frequency, not the frequency before selection occurs.

Question 15

In a plant population, flower color varies continuously from light pink (genotype aabbcc) to deep red (genotype AABBCC), with each capital letter contributing equally to red pigmentation. If two plants with genotypes AaBbCc are crossed, what fraction of their offspring will have exactly the same phenotype as the parents?

  1. 1/64, representing only the exact parental genotype combination in offspring
  2. 6/64, representing all offspring with exactly three dominant alleles like parents
  3. 20/64, representing the most common phenotypic class in the distribution (correct answer)
  4. 32/64, representing offspring with intermediate pigmentation levels overall
  5. 48/64, representing all offspring that inherit at least three dominant alleles
Explanation: This question tests quantitative genetics, specifically polygenic inheritance where multiple genes contribute additively to a single trait. When you see continuous variation described with multiple gene pairs, think about how dominant alleles combine to create the phenotype. Since each capital letter contributes equally to red pigmentation, the phenotype depends only on the total number of dominant alleles. The parents (AaBbCc) each have exactly 3 dominant alleles out of 6 total, placing them in the middle of the phenotypic range between light pink (0 dominant alleles) and deep red (6 dominant alleles). To find offspring with the same phenotype as the parents, you need to calculate how many offspring will have exactly 3 dominant alleles, regardless of which specific alleles they are. This follows a binomial distribution where each offspring has a 50% chance of inheriting a dominant allele at each locus. Using the binomial coefficient: (63)×(0.5)6=20×164=2064\binom{6}{3} \times (0.5)^6 = 20 \times \frac{1}{64} = \frac{20}{64} Answer A (1/64) incorrectly focuses only on the exact parental genotype AaBbCc, ignoring that other genotype combinations like AAbbcc or aaBBCC produce the same phenotype. Answer B (6/64) appears to count something incorrectly, perhaps confusing the number of ways to arrange 3 dominant alleles. Answer D (32/64) represents half the offspring but doesn't correspond to any meaningful phenotypic class. Remember: in polygenic inheritance, focus on the total number of contributing alleles, not the specific genotype combinations. Multiple genotypes can produce identical phenotypes.

Question 16

Refer to the data below. A butterfly population shows variation in wing spot number, with the following distribution in parents and their offspring after one generation of natural selection:

  1. Directional selection favoring fewer wing spots, as shown by the shift toward lower values
  2. Stabilizing selection favoring intermediate spot numbers, as shown by reduced variance in offspring (correct answer)
  3. Disruptive selection favoring extreme phenotypes, as shown by increased frequency at both ends
  4. No selection occurring, because the mean spot number remains approximately unchanged
  5. Sexual selection favoring specific spot patterns, as shown by the non-random distribution
Explanation: The offspring distribution shows a higher peak at intermediate values (4-6 spots) and reduced frequencies at extremes compared to parents, indicating stabilizing selection. The mean changes little, but variance decreases. Choice A is wrong because the mean doesn't shift significantly. Choice C would show increased frequencies at extremes. Choice D ignores the change in distribution shape. Choice E has no evidence for sexual selection from this data.

Question 17

Based on the graph shown, which conclusion about the relationship between genetic diversity and population fitness is most strongly supported?

  1. Genetic diversity has no measurable effect on population fitness across different environments
  2. Higher genetic diversity consistently improves fitness, but the benefit decreases at very high diversity levels
  3. Moderate genetic diversity optimizes fitness, while both low and high diversity reduce performance
  4. Genetic diversity effects depend on environmental conditions, showing benefits only in variable environments
Explanation: D