Biology Flashcards: Analyze Population Data For Evolution

Study Analyze Population Data For Evolution in Biology with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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Analyze Population Data For Evolution

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What is genotype frequency in a population?

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ANSWER

Proportion of individuals with a specific genotype. Calculated by dividing individuals with that genotype by total individuals.

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Flashcard 1: What is genotype frequency in a population?

Answer: Proportion of individuals with a specific genotype. Calculated by dividing individuals with that genotype by total individuals.

Flashcard 2: What is frequency-dependent selection?

Answer: Fitness depends on how common or rare a phenotype is. Selection strength varies with phenotype abundance in population.

Flashcard 3: What does it mean if allele frequency changes across generations in a dataset?

Answer: Evolution is occurring in that population. Frequency changes indicate evolutionary forces are acting.

Flashcard 4: What does it mean if allele frequency changes across generations in a dataset?

Answer: Evolution is occurring in that population. Frequency changes indicate evolutionary forces are acting.

Flashcard 5: Calculate pp when AA=36AA=36, Aa=48Aa=48, aa=16aa=16, and N=100N=100.

Answer: p=0.60p=0.60. p=2(36)+482(100)=120200=0.60p = \frac{2(36) + 48}{2(100)} = \frac{120}{200} = 0.60

Flashcard 6: Identify the Hardy–Weinberg condition: what must be true about population size?

Answer: Population is very large (minimizes genetic drift). Large populations reduce random sampling effects.

Flashcard 7: Identify the formula for allele frequency pp from genotype counts AAAA, AaAa, aaaa in NN individuals.

Answer: p=2AA+Aa2Np=\frac{2AA+Aa}{2N}. Counts each homozygote twice and heterozygote once for allele AA.

Flashcard 8: What is the Hardy–Weinberg equilibrium definition?

Answer: Allele frequencies remain constant when no evolution occurs. Predicts frequencies when evolutionary forces are absent.

Flashcard 9: Choose the best evolutionary mechanism if allele frequencies change randomly after a population crash.

Answer: Genetic drift (bottleneck effect). Population crashes cause random allele loss through sampling effects.

Flashcard 10: What is heterozygote advantage?

Answer: Heterozygotes have highest fitness, maintaining both alleles. Balancing selection maintains genetic diversity.

Flashcard 11: Choose the best evolutionary mechanism if a new island population has unusual allele frequencies from few founders.

Answer: Genetic drift (founder effect). Few founding individuals create non-representative gene pools.

Flashcard 12: What is the operational definition of evolution in population genetics?

Answer: Change in allele frequencies in a population over time. This is the quantitative definition used in population genetics studies.

Flashcard 13: Identify the Hardy–Weinberg condition: what must be true about population size?

Answer: Population is very large (minimizes genetic drift). Large populations reduce random sampling effects.

Flashcard 14: Which Hardy–Weinberg genotype frequency corresponds to homozygous recessive?

Answer: q2q^2. Frequency of aaaa genotype in Hardy-Weinberg equilibrium.

Flashcard 15: What is a selective pressure in the context of population data?

Answer: Environmental factor that causes differential survival or reproduction. Environmental factors create fitness differences between phenotypes.

Flashcard 16: Choose the best evolutionary mechanism if allele frequencies become more similar between two populations over time.

Answer: Gene flow. Migration homogenizes allele frequencies between populations over time.

Flashcard 17: Identify the formula for allele frequency pp from genotype counts AAAA, AaAa, aaaa in NN individuals.

Answer: p=2AA+Aa2Np=\frac{2AA+Aa}{2N}. Counts each homozygote twice and heterozygote once for allele AA.

Flashcard 18: Calculate expected aaaa frequency when q=0.20q=0.20 under Hardy–Weinberg equilibrium.

Answer: q2=0.04q^2=0.04. q2=(0.20)2=0.04q^2 = (0.20)^2 = 0.04

Flashcard 19: Choose the best evolutionary mechanism if allele frequencies change randomly after a population crash.

Answer: Genetic drift (bottleneck effect). Population crashes cause random allele loss through sampling effects.

Flashcard 20: What is mutation as an evolutionary mechanism?

Answer: Source of new alleles via DNA sequence changes. Introduces novel alleles but usually at low rates.

Flashcard 21: What is the Hardy–Weinberg equilibrium definition?

Answer: Allele frequencies remain constant when no evolution occurs. Predicts frequencies when evolutionary forces are absent.

Flashcard 22: Identify the strongest evidence for natural selection in population data across generations.

Answer: Consistent allele frequency change correlated with a selective pressure. Correlation between selection pressure and frequency change indicates causation.

Flashcard 23: Calculate expected carrier (heterozygote) frequency when q=0.10q=0.10 and p=0.90p=0.90.

Answer: 2pq=0.182pq=0.18. 2pq=2(0.90)(0.10)=0.182pq = 2(0.90)(0.10) = 0.18

Flashcard 24: Choose the best selection pattern if the mean phenotype increases steadily across generations.

Answer: Directional selection. Consistent shift toward one extreme indicates directional selection.

Flashcard 25: Identify the allele frequency change if pp shifts from 0.500.50 to 0.620.62 in one generation.

Answer: Δp=0.12\Delta p=0.12. Δp=0.620.50=0.12\Delta p = 0.62 - 0.50 = 0.12

Flashcard 26: What is nonrandom mating and how does it affect genotype frequencies?

Answer: Mate choice/inbreeding changes genotype frequencies, not allele frequencies directly. Affects Hardy-Weinberg genotype predictions without changing allele frequencies.

Flashcard 27: What conclusion is supported if observed genotype frequencies differ from Hardy–Weinberg expectations?

Answer: At least one Hardy–Weinberg condition is violated (evolution likely). Deviations indicate evolutionary forces are acting on the population.

Flashcard 28: Which populations experience stronger genetic drift: small or large?

Answer: Small populations. Smaller populations have greater sampling variance.

Flashcard 29: Identify the Hardy–Weinberg condition: what must be true about mating?

Answer: Mating is random. Prevents preferential mating patterns that alter genotype frequencies.

Flashcard 30: Calculate pp if q=0.30q=0.30 for a two-allele locus under Hardy–Weinberg notation.

Answer: p=0.70p=0.70. Since p+q=1p + q = 1, then p=10.30=0.70p = 1 - 0.30 = 0.70

Flashcard 31: Calculate qq when AA=36AA=36, Aa=48Aa=48, aa=16aa=16, and N=100N=100.

Answer: q=0.40q=0.40. q=2(16)+482(100)=80200=0.40q = \frac{2(16) + 48}{2(100)} = \frac{80}{200} = 0.40

Flashcard 32: Which Hardy–Weinberg genotype frequency corresponds to homozygous dominant?

Answer: p2p^2. Frequency of AAAA genotype in Hardy-Weinberg equilibrium.

Flashcard 33: Which Hardy–Weinberg genotype frequency corresponds to heterozygotes?

Answer: 2pq2pq. Frequency of AaAa genotype; factor of 2 accounts for both combinations.

Flashcard 34: Choose the best selection pattern if the mean phenotype increases steadily across generations.

Answer: Directional selection. Consistent shift toward one extreme indicates directional selection.

Flashcard 35: Identify the strongest evidence for natural selection in population data across generations.

Answer: Consistent allele frequency change correlated with a selective pressure. Correlation between selection pressure and frequency change indicates causation.

Flashcard 36: What conclusion is supported if observed genotype frequencies differ from Hardy–Weinberg expectations?

Answer: At least one Hardy–Weinberg condition is violated (evolution likely). Deviations indicate evolutionary forces are acting on the population.

Flashcard 37: Identify the Hardy–Weinberg condition: what must be true about mutation?

Answer: No mutations occur. Mutations would change allele frequencies over time.

Flashcard 38: Identify the Hardy–Weinberg condition: what must be true about migration (gene flow)?

Answer: No migration; no gene flow into or out of the population. Migration introduces or removes alleles from the population.

Flashcard 39: Choose the best selection pattern if extremes decline and intermediate phenotypes become most common.

Answer: Stabilizing selection. Intermediate optimization indicates selection against extremes.

Flashcard 40: Which inference is supported if a beneficial allele rises quickly in frequency after an environmental change?

Answer: Positive selection increased the allele's frequency. Rapid increase after environmental change indicates adaptive advantage.

Flashcard 41: What is an allele frequency in a population?

Answer: Proportion of all gene copies that are a specific allele. Calculated by dividing copies of that allele by total gene copies.

Flashcard 42: Identify the Hardy–Weinberg condition: what must be true about migration (gene flow)?

Answer: No migration; no gene flow into or out of the population. Migration introduces or removes alleles from the population.

Flashcard 43: Calculate pp if q=0.30q=0.30 for a two-allele locus under Hardy–Weinberg notation.

Answer: p=0.70p=0.70. Since p+q=1p + q = 1, then p=10.30=0.70p = 1 - 0.30 = 0.70

Flashcard 44: Calculate expected carrier (heterozygote) frequency when q=0.10q=0.10 and p=0.90p=0.90.

Answer: 2pq=0.182pq=0.18. 2pq=2(0.90)(0.10)=0.182pq = 2(0.90)(0.10) = 0.18

Flashcard 45: What conclusion is supported if observed genotype frequencies match Hardy–Weinberg expectations?

Answer: No evidence of evolution at that locus in that population. Meeting Hardy-Weinberg expectations indicates no evolutionary change.

Flashcard 46: Identify the Hardy–Weinberg condition: what must be true about natural selection?

Answer: No natural selection; equal fitness among genotypes. All genotypes must have equal reproductive success.

Flashcard 47: What is natural selection in terms of population data?

Answer: Nonrandom allele frequency change due to differential survival/reproduction. Fitness differences cause predictable frequency changes.

Flashcard 48: Choose the best selection pattern if extremes decline and intermediate phenotypes become most common.

Answer: Stabilizing selection. Intermediate optimization indicates selection against extremes.

Flashcard 49: What is heterozygote advantage?

Answer: Heterozygotes have highest fitness, maintaining both alleles. Balancing selection maintains genetic diversity.

Flashcard 50: Calculate expected AAAA frequency when p=0.70p=0.70 under Hardy–Weinberg equilibrium.

Answer: p2=0.49p^2=0.49. p2=(0.70)2=0.49p^2 = (0.70)^2 = 0.49

Flashcard 51: What is the key population-level evidence that a trait is heritable?

Answer: Trait variation correlates between parents and offspring. Parent-offspring correlation indicates genetic basis for the trait.

Flashcard 52: Calculate qq if the recessive phenotype frequency is 0.090.09 and the trait is recessive.

Answer: q=0.30q=0.30. Since q2=0.09q^2 = 0.09, then q=0.09=0.30q = \sqrt{0.09} = 0.30

Flashcard 53: What is the key population-level evidence that a trait is heritable?

Answer: Trait variation correlates between parents and offspring. Parent-offspring correlation indicates genetic basis for the trait.

Flashcard 54: State the Hardy–Weinberg genotype frequency equation for two alleles.

Answer: p2+2pq+q2=1p^2+2pq+q^2=1. Expands (p+q)2(p+q)^2 to show all genotype frequencies.

Flashcard 55: Which Hardy–Weinberg genotype frequency corresponds to homozygous recessive?

Answer: q2q^2. Frequency of aaaa genotype in Hardy-Weinberg equilibrium.

Flashcard 56: What is genetic drift?

Answer: Random change in allele frequencies due to chance. Sampling error causes unpredictable frequency changes.

Flashcard 57: Identify the Hardy–Weinberg condition: what must be true about natural selection?

Answer: No natural selection; equal fitness among genotypes. All genotypes must have equal reproductive success.

Flashcard 58: Which inference is supported if a phenotype is rare and has higher fitness specifically because it is rare?

Answer: Negative frequency-dependent selection. Rare phenotypes gain advantage specifically from their rarity.

Flashcard 59: Which inference is supported if a harmful allele persists because heterozygotes have higher fitness?

Answer: Balancing selection via heterozygote advantage. Heterozygote advantage maintains harmful alleles in populations.

Flashcard 60: State the Hardy–Weinberg allele frequency equation for two alleles.

Answer: p+q=1p+q=1. The two allele frequencies must sum to 1.

Flashcard 61: Choose the best selection pattern if both extremes increase while intermediates decrease in frequency.

Answer: Disruptive selection. Both extremes increasing indicates selection against intermediate forms.

Flashcard 62: What is the gene pool of a population?

Answer: All alleles present in the population. The collective genetic variation available for evolution.

Flashcard 63: Identify the Hardy–Weinberg condition: what must be true about mating?

Answer: Mating is random. Prevents preferential mating patterns that alter genotype frequencies.

Flashcard 64: Calculate expected heterozygote frequency when p=0.60p=0.60 and q=0.40q=0.40.

Answer: 2pq=0.482pq=0.48. 2pq=2(0.60)(0.40)=0.482pq = 2(0.60)(0.40) = 0.48

Flashcard 65: Which inference is supported if a harmful allele persists because heterozygotes have higher fitness?

Answer: Balancing selection via heterozygote advantage. Heterozygote advantage maintains harmful alleles in populations.

Flashcard 66: What is gene flow (migration) in evolutionary biology?

Answer: Movement of alleles between populations via migration and breeding. Homogenizes allele frequencies between connected populations.

Flashcard 67: Which inference is supported if a beneficial allele rises quickly in frequency after an environmental change?

Answer: Positive selection increased the allele's frequency. Rapid increase after environmental change indicates adaptive advantage.

Flashcard 68: Identify the formula for allele frequency qq from genotype counts AAAA, AaAa, aaaa in NN individuals.

Answer: q=2aa+Aa2Nq=\frac{2aa+Aa}{2N}. Counts each homozygote twice and heterozygote once for allele aa.